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Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 3 Permutations and Combination Ex 3.1 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 3 Permutations and Combination Ex 3.1

Question 1.
A teacher wants to select the class monitor in a class of 30 boys and 20 girls. In how many ways can the monitor be selected if the monitor must be a girl?
Solution:
There are 30 boys and 20 girls.
A teacher can select any boy as a class monitor from 30 boys in 30 different ways and he can select any girl as a class monitor from 20 girls in 20 different ways.
∴ by the fundamental principle of addition, the total number of ways a teacher can select a class monitor = 30 + 20 = 50
Hence, there are 50 different ways to select a class monitor.

Question 2.
A Signal is generated from 2 flags by putting one flag above the other. If 4 flags of different colours are available, how many different signals can be generated?
Solution:
A signal is generated from 2 flags and there are 4 flags of different colours available.
∴ 1st flag can be any one of the available 4 flags.
∴ It can be selected in 4 ways.
Now, 2nd flag is to be selected for which 3 flags are available for a different signal.
∴ 2nd flag can be anyone from these 3 flags.
∴ It can be selected in 3 ways.
∴ By using the fundamental principle of multiplication, total no. of ways a signal can be generated = 4 × 3 = 12
∴ 12 different signals can be generated.

Question 3.
How many two-letter words can be formed using letters from the word SPACE, when repetition of letters (i) is allowed (ii) is not allowed
Solution:
A two-letter word is to be formed out of the letters of the word SPACE.
(i) When repetition of the letters is allowed 1st letter can be selected in 5 ways.
2nd letter can be selected in 5 ways.
∴ By using the fundamental principle of multiplication, the total number of 2-letter words = 5 × 5 = 25

(ii) When repetition of the letters is not allowed 1st letter can be selected in 5 ways.
2nd letter can be selected in 4 ways.
∴ By using the fundamental principle of multiplication, the total number of 2-letter words = 5 × 4 = 20

Question 4.
How many three-digit numbers can be formed from the digits 0, 1, 3, 5, 6 if repetitions of digits (i) are allowed (ii) are not allowed?
Solution:
A three-digit number is to be formed from the digits 0, 1, 3, 5, 6.
(i) When repetition of digits is allowed,
100’s place digit should be a non-zero number.
Hence, it can be anyone from digits 1, 3, 5,6.
∴ 100’s place digit can be selected in 4 ways.
10’s and unit’s place digit can be zero and digits can be repeated.
∴ 10’s place digit can be selected in 5 ways and the unit’s place digit can be selected in 5 ways.
∴ By using the fundamental principle of multiplication, the total number of three-digit numbers = 4 × 5 × 5 = 100

(ii) When repetition of digits is not allowed,
100’s place digit should be a non-zero number.
Hence, it can be anyone from digits 1, 3, 5, 6.
∴ 100’s place digit can be selected in 4 ways.
10’s and unit’s place digit can be zero and digits can’t be repeated.
∴ 10’s place digit can be selected in 4 ways and the unit’s place digit can be selected in 3 ways.
∴ By using the fundamental principle of multiplication, the total number of two-digit numbers = 4 × 4 × 3 = 48

Question 5.
How many three-digit numbers can be formed using the digits 2, 3, 4, 5, 6 if digits can be repeated?
Solution:
A 3-digit number is to be formed from the digits 2, 3, 4, 5, 6 where digits can be repeated.
∴ Unit’s place digit can be selected in 5 ways.
10’s place digit can be selected in 5 ways.
100’s place digit can be selected in 5 ways.
∴ By using fundamental principle of multiplication, total number of 3-digit numbers = 5 × 5 × 5 = 125

Question 6.
A letter lock contains 3 rings and each ring contains 5 letters. Determine the maximum number of false trails that can be made before the lock is opened.
Solution:
A letter lock has 3 rings, each ring containing 5 different letters.
∴ A letter from each ring can be selected in 5 ways.
∴ By using the fundamental principle of multiplication, a total number of trials that can be made = 5 × 5 × 5 = 125.
Out of these 124 wrong attempts are made and in the 125th attempt, the lock gets opened.
∴ A maximum number of false trials = 124.

Question 7.
In a test, 5 questions are of the form ‘state, true or false. No student has got all answers correct. Also, the answer of every student is different. Find the number of students who appeared for the test.
Solution:
Every question can be answered in 2 ways. (True or False)
∴ By using the fundamental principle of multiplication, the total number of set of answers possible = 2 × 2 × 2 × 2 × 2 = 32.
Since One of them is the case where all questions are answered correctly,
The number of wrong answers = 32 – 1 = 31.
Since no student has answered all the questions correctly, the number of students who appeared for the test are 31.

Question 8.
How many numbers between 100 and 1000 have 4 in the unit’s place?
Solution:
Numbers between 100 and 1000 are 3-digit numbers.
A 3-digit number is to be formed from the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, where the unit place digit is 4.
Unit’s place digit is 4.
∴ it can be selected in 1 way only.
10’s place digit can be selected in 10 ways.
For a 3-digit number, 100’s place digit should be a non-zero number.
∴ 100’s place digit can be selected in 9 ways.
∴ By using the fundamental principle of multiplication,
total numbers between 100 and 1000 which have 4 in the units place = 1 × 10 × 9 = 90.

Question 9.
How many numbers between 100 and 1000 have the digit 7 exactly once?
Solution:
Numbers between 100 and 1000 are 3-digit numbers.
A 3-digit number is to be formed from the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, where exactly one of the digits is 7.
When 7 is in unit’s place:
Unit’s place digit is 7.
∴ it can be selected in 1 way only.
10’s place digit can be selected in 9 ways.
100’s place digit can be selected in 8 ways.
Total numbers which have 7 in unit’s place = 1 × 9 × 8 = 72.

When 7 is in 10’s place:
Unit’s place digit can be selected in 9 ways.
10’s place digit is 7.
∴ it can be selected in 1 way only.
100’s place digit can be selected in 8 ways.
∴ A total number of numbers which have 7 in 10’s place = 9 × 1 × 8 = 72.

When 7 is in 100’s place:
Unit’s place digit can be selected in 9 ways.
10’s place digit can be selected in 9 ways.
100’s place digit is 7.
∴ it can be selected in 1 way only.
∴ Total numbers which have 7 in 100’s place = 9 × 9 × 1 = 81.
∴ Total numbers between 100 and 1000 having digit 7 exactly once = 72 + 72 + 81 = 225

Question 10.
How many four-digit numbers will not exceed 7432 if they are formed using the digits 2, 3, 4, 7 without repetition?
Solution:
Between any set of digits, the greatest number is possible when digits are arranged in descending order.
∴ 7432 is the greatest number, formed from the digits 2, 3, 4, 7.
Since a 4-digit number is to be formed from the digits 2, 3, 4, 7, where repetition of the digit is not allowed,
1000’s place digit can be selected in 4 ways,
100’s place digit can be selected in 3 ways,
10’s place digit can be selected in 2 ways,
Unit’s place digit can be selected in 1 way.
∴ Total number of numbers not exceeding 7432 that can be formed with the digits 2, 3, 4, 7 = Total number of four-digit numbers possible from the digits 2, 3, 4, 7
= 4 × 3 × 2 × 1
= 24

Question 11.
If numbers are formed using digits 2, 3, 4, 5, 6 without repetition, how many of them will exceed 400?
Solution:
Case I: Number of three-digit numbers formed from 2, 3, 4, 5, 6, greater than 400.
100’s place can be filled by any one of the numbers 4, 5, 6.
100’s place digit can be selected in 3 ways.
Since repetition is not allowed, 10’s place can be filled by any one of the remaining four numbers.
∴ 10’s place digit can be selected in 4 ways.
Unit’s place digit can be selected in 3 ways.
∴ Total number of three-digit numbers formed = 3 × 4 × 3 = 36

Case II: Number of four-digit numbers formed from 2, 3, 4, 5, 6.
Since repetition of digits is not allowed,
1000’s place digit can be selected in 5 ways.
100’s place digit can be selected in 4 ways.
10’s place digit can be selected in 3 ways.
Unit’s place digit can be selected in 2 ways.
∴ Total number of four-digit numbers formed = 5 × 4 × 3 × 2 = 120

Case III: Number of five-digit numbers formed from 2, 3, 4, 5, 6.
Similarly, since repetition of digits is not allowed,
Total number of five digit numbers formed = 5 × 4 × 3 × 2 × 1 = 120.
∴ Total number of numbers that exceed 400 = 36 + 120 + 120 = 276

Question 12.
How many numbers formed with the digits 0, 1, 2, 5, 7, 8 will fall between 13 and 1000 if digits can be repeated?
Solution:
Case I: 2-digit numbers more than 13, less than 20, formed from the digits 0, 1, 2, 5, 7, 8.
10’s place digit is 1.
∴ it can be selected in 1 way only.
Unit’s place can be filled by any one of the numbers 5, 7, 8.
∴ Unit’s place digit can be selected in 3 ways.
∴ Total number of such numbers = 1 × 3 = 3.

Case II: 2-digit numbers more than 20 formed from 0, 1, 2, 5, 7, 8.
10’s place can be filled by any one of the numbers 2, 5, 7, 8.
∴ 10’s place digit can be selected in 4 ways.
Since repetition is allowed, the unit’s place can be filled by one of the remaining 6 digits.
∴ Unit’s place digit can be selected in 6 ways.
∴ Total number of such numbers = 4 × 6 = 24.

Case III: 3-digit numbers formed from 0, 1, 2, 5, 7, 8.
Similarly, since repetition of digits is allowed, the total number of such numbers = 5 × 6 × 6 = 180.
All cases are mutually exclusive.
∴ Total number of required numbers = 3 + 24 + 180 = 207

Question 13.
A school has three gates and four staircases from the first floor to the second floor. How many ways does a student have to go from outside the school to his classroom on the second floor?
Solution:
A student can go inside the school from outside in 3 ways and from the first floor to the second floor in 4 ways.
∴ A number of ways to choose gates = 3.
The number of ways to choose a staircase = 4.
By using the fundamental principle of multiplication,
number of ways in which a student has to go from outside the school to his classroom = 4 × 3 = 12

Question 14.
How many five-digit numbers formed using the digit 0, 1, 2, 3, 4, 5 are divisible by 5 if digits are not repeated?
Solution:
Here, repetition of digits is not allowed.
For a number to be divisible by 5,
unit’s place digit should be 0 or 5.
Case I: when unit’s place is 0
Unit’s place digit can be selected in 1 way.
10’s place digit can be selected in 5 ways.
100’s place digit can be selected in 4 ways.
1000’s place digit can be selected in 3 ways.
10000’s place digit can be selected in 2 ways.
∴ Total number of numbers = 1 × 5 × 4 × 3 × 2 = 120.

Case II: when the unit’s place is 5
Unit’s place digit can be selected in 1 way.
10000’s place should be a non-zero number.
∴ It can be selected in 4 ways.
1000’s place digit can be selected in 4 ways.
100’s place digit can be selected in 3 ways.
10’s place digit can be selected in 2 ways.
∴ Total number of numbers = 1 × 4 × 4 × 3 × 2 = 96
∴ Total number of required numbers = 120 + 96 = 216

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 6 Functions Miscellaneous Exercise 6 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 6 Functions Miscellaneous Exercise 6

(I) Select the correct answer from the given alternatives.

Question 1.
If log (5x – 9) – log (x + 3) = log 2, then x = ________
(A) 3
(B) 5
(C) 2
(D) 7
Answer:
(B) 5
Hint:
log (5x – 9) – log (x + 3) = log 2
∴ \(\frac{5 x-9}{x+3}\) = 2
∴ 3x = 9 + 6
∴ x = 5

Question 2.
If log₁₀ (log₁₀ (log₁₀ x)) = 0, then x = ________
(A) 1000
(B) 10¹⁰
(C) 10
(D) 0
Answer:
(B) 10¹⁰
Hint:
log₁₀ log₁₀ log₁₀ x = 0
∴ log₁₀ (log₁₀ (x)) = 10⁰ = 1
∴ log₁₀ x = 10¹ = 10
∴ x = 10¹⁰

Question 3.
Find x, if 2 log₂ x = 4
(A) 4, -4
(B) 4
(C) -4
(D) not defined
Answer:
(B) 4
Hint:
2 log₂ x = 4, x > 0
∴ log₂ (x²) = 4
∴ x² = 16
∴ x = ±4
∴ x = 4

Question 4.
The equation \(\log _{x^{2}} 16+\log _{2 x} 64=3\) has,
(A) one irrational solution
(B) no prime solution
(C) two real solutions
(D) one integral solution
Answer:
(A), (B), (C), (D)
Hint:
\(\log _{x^{2}} 16+\log _{2 x} 64=3\)
∴ \(\frac{\log 16}{\log x^{2}}+\frac{\log 64}{\log 2 x}=3\)
∴ 4 log 2 [log x + log 2] + (6 log 2) (2 log x) = 3 (2 log x) (log 2 + log x)
Let log 2 = a, log x = t. Then
∴ 4at + 4a² + 12at = 6at + 6t²
∴ 6t² – 10at – 4a² = 0
∴ 3t² – 5at – 2a² = 0
∴ (3t + a) (t – 2a) = 0
∴ t = \(-\frac{1}{3}\)a, 2a
∴ log x = \(\log (2)^{-\frac{1}{3}}\), log (2²)
∴ x = \(2^{-\frac{1}{3}}\), 4
∴ x = \(\frac{1}{\sqrt[3]{2}}\), 4

Question 5.
If f(x) = \(\frac{1}{1-x}\), then f(f(f(x))) is
(A) x – 1
(B) 1 – x
(C) x
(D) -x
Answer:
(C) x
Hint:

Question 6.
If f: R → R is defined by f(x) = x³, then f^-1 (8) is equal to:
(A) {2}
(B) {-2.2}
(C) {-2}
(D) (-2.2)
Answer:
(A) {2}
Hint:

Question 7.
Let the function f be defined by f(x) = \(\frac{2 x+1}{1-3 x}\) then f^-1 (x) is:
(A) \(\frac{x-1}{3 x+2}\)
(B) \(\frac{x+1}{3 x-2}\)
(C) \(\frac{2 x+1}{1-3 x}\)
(D) \(\frac{3 x+2}{x-1}\)
Answer:
(A) \(\frac{x-1}{3 x+2}\)
Hint:
f(x) = \(\frac{2 x+1}{1-3 x}\) = y, say. then
2x + 1 = y(1 – 3x)
∴ y – 1 = x(2 + 3y)
∴ x = \(\frac{y-1}{2+3 y}\) = f^-1 (y)
∴ f^-1 (x) = \(\frac{x-1}{2+3 x}\)

Question 8.
If f(x) = 2x² + bx + c and f(0) = 3 and f(2) = 1, then f(1) is equal to
(A) -2
(B) 0
(C) 1
(D) 2
Answer:
(B) 0
Hint:
f(x) = 2x² + bx + c
f(0) = 3
∴ 2(0) + b(0) + c = 3
∴ c = 3 ……..(i)
∴ f(2) = 1
∴ 2(4) + 2b + c = 1
∴ 2b + c = -7
∴ 2b + 3 = -7 …..[From (i)]
∴ b = -5
∴ f(x) = 2x² – 5x + 3
∴ f(1) = 2(1)² – 5(1) + 3 = 0

Question 9.
The domain of \(\frac{1}{[x]-x}\), where [x] is greatest integer function is
(A) R
(B) Z
(C) R – Z
(D) Q – {0}
Answer:
(C) R – Z
Hint:
f(x) = \(\frac{1}{[x]-x}=\frac{1}{-\{x\}}\)
For f to be defined, {x} ≠ 0
∴ x cannot be integer.
∴ Domain = R – Z

Question 10.
The domain and range of f(x) = 2 – |x – 5| are
(A) R+, (-∞, 1]
(B) R, (-∞, 2]
(C) R, (-∞, 2)
(D) R+, (-∞, 2]
Answer:
(B) R, (-∞, 2]
Hint:

f(x) = 2 – |x – 5|
= 2 – (5 – x), x < 5
= 2 – (x – 5), x ≥ 5
∴ f(x) = x – 3, x < 5
= 7 – x, x ≥ 5
Domain = R,
Range (from graph) = (-∞, 2]

(II) Answer the following:

Question 1.
Which of the following relations are functions? If it is a function determine its domain and range.
(i) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5) (12, 6), (14, 7)}
(ii) {(0, 0), (1, 1), (1, -1), (4, 2), (4, -2) (9, 3), (9, -3), (16, 4), (16, -4)}
(iii) {(2, 1), (3, 1), (5, 2)}
Solution:
(i) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5) (12, 6), (14, 7)}

Every element of set A has been assigned a unique element in set B
∴ Given relation is a function
Domain = {2, 4, 6, 8, 10, 12, 14}, Range = {1, 2, 3, 4, 5, 6, 7}

(ii) {(0, 0), (1, 1), (1, -1), (4, 2), (4, -2) (9, 3), (9, -3) (16, 4), (16, -4)}
∵ (1, 1), (1, -1) ∈ the relation
∴ Given relation is not a function.
As element 1 of the domain has not been assigned a unique element of co-domain.

(iii) {(2, 1), (3, 1), (5, 2)}

Every element of set A has been assigned a unique element in set B.
∴ Given relation is a function.
Domain = {2, 3, 5}, Range = {1, 2}

Question 2.
Find whether the following functions are one-one.
(i) f: R → R defined by f(x) = x² + 5
(ii) f: R – {3} → R defined by f(x) = \(\frac{5 x+7}{x-3}\) for x ∈ R – {3}
Solution:
(i) f: R → R, defined by f(x) = x² + 5
Note that f(-x) = f(x) = x² + 5
∴ f is not one-one (i.e., many-one) function.

(ii) f: R – {3} → R, defined by f(x) = \(\frac{5 x+7}{x-3}\)
Let f(x₁) = f(x₂)
∴ \(\frac{5 x_{1}+7}{x_{1}-3}=\frac{5 x_{2}+7}{x_{2}-3}\)
∴ 5x₁ x₂ – 15x₁ + 7x₂ – 21 = 5x₁ x₂ – 15x₂ + 7x₁ – 21
∴ 22(x₁ – x₂) = 0
∴ x₁ = x₂
∴ f is a one-one function.

Question 3.
Find whether the following functions are onto or not.
(i) f: Z → Z defined by f(x) = 6x – 7 for all x ∈ Z
(ii) f: R → R defined by f(x) = x² + 3 for all x ∈ R
Solution:
(i) f(x) = 6x – 7 = y (say)
(x, y ∈ Z)
∴ x = \(\frac{7+y}{6}\)
Since every integer y does not give integer x, f is not onto.

(ii) f(x) = x² + 3 = y (say)
(x, y ∈ R)
Clearly y ≥ 3 …..[x² ≥ 0]
∴ All the real numbers less than 3 from codomain R, have not been pre-assigned any element from the domain R.
∴ f is not onto.

Question 4.
Let f: R → R be a function defined by f(x) = 5x³ – 8 for all x ∈ R. Show that f is one-one and onto. Hence find f^-1.
Solution:

Question 5.
A function f: R → R defined by f(x) = \(\frac{3 x}{5}\) + 2, x ∈ R. Show that f is one-one and onto. Hence find f^-1.
Solution:

Question 6.
A function f is defined as f(x) = 4x + 5, for -4 ≤ x < 0. Find the values of f(-1), f(-2), f(0), if they exist.
Solution:
f(x) = 4x + 5, -4 ≤ x < 0
f(-1) = 4(-1) + 5 = -4 + 5 = 1
f(-2) = 4(-2) + 5 = -8 + 5 = -3
x = 0 ∉ domain of f
∴ f(0) does not exist.

Question 7.
A function f is defined as f(x) = 5 – x for 0 ≤ x ≤ 4. Find the values of x such that
(i) f(x) = 3
(ii) f(x) = 5
Solution:
(i) f(x) = 3
∴ 5 – x = 3
∴ x = 5 – 3 = 2

(ii) f(x) = 5
∴ 5 – x = 5
∴ x = 0

Question 8.
If f(x) = 3x⁴ – 5x² + 7, find f(x – 1).
Solution:
f(x) = 3x⁴ – 5x² + 7
∴ f(x – 1) = 3(x – 1)⁴ – 5(x – 1)² + 7
= 3(x⁴ – ⁴C₁ x³ + ⁴C₂ x² – ⁴C₃ x + ⁴C₄) – 5(x² – 2x + 1) + 7
= 3(x⁴ – 4x³ + 6x² – 4x + 1) – 5(x² – 2x + 1) + 7
= 3x⁴ – 12x³ + 18x² – 12x + 3 – 5x² + 10x – 5 + 7
= 3x⁴ – 12x³ + 13x² – 2x + 5

Question 9.
If f(x) = 3x + a and f(1) = 7, find a and f(4).
Solution:
f(x) = 3x + a, f(1) = 7
∴ 3(1) + a = 7
∴ a = 7 – 3 = 4
∴ f(x) = 3x + 4
∴ f(4) = 3(4) + 4 = 12 + 4 = 16

Question 10.
If f(x) = ax² + bx + 2 and f(1) = 3, f(4) = 42, find a and b.
Solution:
f(x) = ax² + bx + 2
f(1) = 3
∴ a(1)² + b(1) + 2 = 3
∴ a + b = 1 ….(i)
f(4) = 42
∴ a(4)² + b(4) + 2 = 42
∴ 16a + 4b = 40
Dividing by 4, we get
4a + b = 10 …..(ii)
Solving (i) and (ii), we get
a = 3, b = -2

Question 11.
Find composite of f and g:
(i) f = {(1, 3), (2, 4), (3, 5), (4, 6)}
g = {(3, 6), (4, 8), (5, 10), (6, 12)}
(ii) f = {(1, 1), (2, 4), (3, 4), (4, 3)}
g = {(1, 1), (3, 27), (4, 64)}
Solution:
(i) f = {(1, 3), (2, 4), (3, 5), (4, 6)}
g = {(3, 6), (4, 8), (5, 10), (6, 12)}
∴ f(1) = 3, g(3) = 6
f(2) = 4, g(4) = 8
f(3) = 5, g(5)=10
f(4) = 6, g(6) = 12
(gof) (x) = g (f(x))
(gof)(1) = g(f(1)) = g(3) = 6
(gof)(2) – g(f(2)) = g(4) = 8
(gof)(3) = g(f(3)) = g(5) = 10
(gof)(4) = g(f(4)) = g(6) = 12
∴ gof = {(1, 6), (2, 8), (3, 10), (4, 12)}

(ii) f = {(1, 1), (2, 4), (3, 4), (4, 3)}
g = {(1, 1), (3, 27), (4, 64)}
f(1) = 1, g(1) = 1
f(2) = 4, g(3) = 27
f(3) = 4, g(4) = 64
f(4) = 3
(gof) (x) = g(f(x))
(gof) (1) = g(f(1)) = g(1) = 1
(gof) (2) = g(f(2)) = g(4) = 64
(gof) (3) = g(f(3)) = g(4) = 64
(gof) (4) = g(f(4)) = g(3) = 27
∴ gof = {(1, 1), (2, 64), (3, 64), (4, 27)}

Question 12.
Find fog and gof:
(i) f(x) = x² + 5, g(x) = x – 8
(ii) f(x) = 3x – 2, g(x) = x²
(iii) f(x) = 256x⁴, g(x) = √x
Solution:
(i) f(x) = x² + 5, g(x) = x – 8
(fog) (x) = f(g(x))
= f(x – 8)
= (x – 8)² + 5
= x² – 16x + 64 + 5
= x² – 16x + 69
(gof) (x) = g(f(x))
= g(x² + 5)
= x² + 5 – 8
= x – 3

(ii) f(x) = 3x – 2, g(x) = x²
(fog) (x) = f(g(x)) = f(x²) = 3x² – 2
(gof) (x) = g(f(x))
= g(3x – 2)
= (3x – 2)²
= 9x² – 12x + 4

(iii) f(x) = 256x⁴, g(x) = √x
(fog) (x) = f(g(x)) = f (√x) = 256 (√x)⁴ = 256x²
(gof) (x) = g(f(x)) = g(256x⁴) = \(\sqrt{256 x^{4}}\) = 16x²

Question 13.
If f(x) = \(\frac{2 x-1}{5 x-2}, x \neq \frac{2}{5}\), show that (fof) (x) = x.
Solution:

Question 14.
If f(x) = \(\frac{x+3}{4 x-5}\), g(x) = \(\frac{3+5 x}{4 x-1}\), then show that (fog) (x) = x.
Solution:
f(x) = \(\frac{x+3}{4 x-5}\), g(x) = \(\frac{3+5 x}{4 x-1}\)
(fog)(x) = f(g(x))
= f(\(\frac{3+5 x}{4 x-1}\))

Question 15.
Let f: R – {2} → R be defined by f(x) = \(\frac{x^{2}-4}{x-2}\) and g: R → R be defined by g(x) = x + 2. Examine whether f = g or not.
Solution:
f(x) = \(\frac{x^{2}-4}{x-2}\), x ≠ 2
∴ f(x) = x + 2, x ≠ 2 and g(x) = x + 2,
The domain of f = R – {2}
The domain of g = R
Here, f and g have different domains.
∴ f ≠ g

Question 16.
Let f: R → R be given by f(x) = x + 5 for all x ∈ R. Draw its graph.
Solution:
f(x) = x + 5

Question 17.
Let f: R → R be given by f(x) = x³ + 1 for all x ∈ R. Draw its graph.
Solution:
Let y = f(x) = x³ + 1

Question 18.
For any base show that log(1 + 2 + 3) = log 1 + log 2 + log 3
Solution:
L.H.S. = log(1 + 2 + 3) = log 6
R.H.S. = log 1 + log 2 + log 3
= 0 + log (2 × 3)
= log 6
∴ L.H.S. = R.H.S.

Question 19.
Find x, if x = \(3^{3 \log _{3} 2}\).
Solution:
x = \(3^{3 \log _{3} 2}\)
= \(3^{\log 3\left(2^{3}\right)}\)
= 2³ ….[\(a^{\log _{a} b}\) = b]
= 8

Question 20.
Show that, log|\(\sqrt{x^{2}+1}\) + x| + log|\(\sqrt{x^{2}+1}\) – x| = 0.
Solution:
L.H.S. = log|\(\sqrt{x^{2}+1}\) + x| + log|\(\sqrt{x^{2}+1}\) – x|
= \(\log \left|\left(\sqrt{x^{2}+1}+x\right)\left(\sqrt{x^{2}+1}-x\right)\right|\)
= log|x² + 1 – x²|
= log 1
= 0
= R.H.S.

Question 21.
Show that \(\log \frac{\mathrm{a}^{2}}{\mathrm{bc}}+\log \frac{\mathrm{b}^{2}}{\mathrm{ca}}+\log \frac{\mathrm{c}^{2}}{\mathrm{ab}}=0\).
Solution:

Question 22.
Simplify log (log x⁴) – log(log x).
Solution:
log (log x⁴) – log (log x)
= log (4 log x) – log (log x) …..[log mⁿ = n log m]
= log 4 + log (log x) – log (log x) …..[log (mn) = log m + log n]
= log 4

Question 23.
Simplify \(\log _{10} \frac{28}{45}-\log _{10} \frac{35}{324}+\log _{10} \frac{325}{432}-\log _{10} \frac{13}{15}\)
Solution:

Question 24.
If log (\(\frac{a+b}{2}\)) = \(\frac{1}{2}\) (log a + log b), then show that a = b.
Solution:
log (\(\frac{a+b}{2}\)) = \(\frac{1}{2}\) (log a + log b)
∴ 2 log (\(\frac{a+b}{2}\)) = log a + log b
∴ log \(\left(\frac{a+b}{2}\right)^{2}\) = log ab
∴ \(\frac{(a+b)^{2}}{4}\) = ab
∴ a² + 2ab + b² = 4ab
∴ a² + 2ab – 4ab + b² = 0
∴ a² – 2ab + b² = 0
∴ (a – b)² = 0
∴ a – b = 0
∴ a = b

Question 25.
If b² = ac. Prove that, log a + log c = 2 log b.
Solution:
b² = ac
Taking log on both sides, we get
log b² = log ac
∴ 2 log b = log a + log c
∴ log a + log c = 2 log b

Question 26.
Solve for x, log_x (8x – 3) – log_x 4 = 2.
Solution:
log_x (8x – 3) – log_x 4 = 2
∴ \(\log _{x}\left(\frac{8 x-3}{4}\right)\) = 2
∴ x² = \(\frac{8 x-3}{4}\)
∴ 4x² = 8x – 3
∴ 4x² – 8x + 3 = 0
∴ 4x² – 2x – 6x + 3 = 0
∴ 2x(2x – 1) – 3(2x – 1) = 0
∴ (2x – 1)(2x – 3) = 0
∴ 2x – 1 = 0 or 2x – 3 = 0
∴ x = \(\frac{1}{2}\) or x = \(\frac{3}{2}\)

Question 27.
If a² + b² = 7ab, show that \(\log \left(\frac{a+b}{3}\right)=\frac{1}{2} \log a+\frac{1}{2} \log b\)
Solution:
a² + b² = 7ab
a² + 2ab + b² = 7ab + 2ab
(a + b)² = 9ab
\(\frac{(a+b)^{2}}{9}\) = ab
\(\left(\frac{a+b}{3}\right)^{2}\) = ab
Taking log on both sides, we get
log \(\left(\frac{a+b}{3}\right)^{2}\) = log (ab)
2 log \(\left(\frac{a+b}{3}\right)\) = log a + log b
Dividing throughout by 2, we get
\(\log \left(\frac{a+b}{3}\right)=\frac{1}{2} \log a+\frac{1}{2} \log b\)

Question 28.
If \(\log \left(\frac{x-y}{5}\right)=\frac{1}{2} \log x+\frac{1}{2} \log y\), show that x² + y² = 27xy.
Solution:

Question 29.
If log₃ [log₂ (log₃ x)] = 1, show that x = 6561.
Solution:
log₃ [log₂ (log₃ x)] = 1
∴ log₂ (log₃ x) = 3¹
∴ log₃ x = 2³
∴ log₃ x = 8
∴ x = 3⁸
∴ x = 6561

Question 30.
If f(x) = log(1 – x), 0 ≤ x < 1, show that f(\(\frac{1}{1+x}\)) = f(1 – x) – f(-x).
Solution:

Question 31.
Without using log tables, prove that \(\frac{2}{5}\) < log₁₀ 3 < \(\frac{1}{2}\).
Solution:
We have to prove that, \(\frac{2}{5}\) < log₁₀ 3 < \(\frac{1}{2}\)
i.e., to prove that \(\frac{2}{5}\) < log₁₀ 3 and log₁₀ 3 < \(\frac{1}{2}\)
i.e., to prove that 2 < 5 log₁₀ 3 and 2 log₁₀ 3 < 1
i.e., to prove that 2 log₁₀ 10 < 5 log₁₀ 3 and 2 log₁₀ 3 < log₁₀ 10 ……[∵ log_a a = 1]
i.e., to prove that log₁₀ 10² < log₁₀ 3⁵ and log₁₀ 3² < log₁₀ 10
i.e., to prove that 10² < 3⁵ and 3² < 10
i.e., to prove that 100 < 243 and 9 < 10 which is true
∴ \(\frac{2}{5}\) < log₁₀ 3 < \(\frac{1}{2}\)

Question 32.
Show that \(7 \log \left(\frac{15}{16}\right)+6 \log \left(\frac{8}{3}\right)+5 \log \left(\frac{2}{5}\right)+\log \left(\frac{32}{25}\right)\) = log 3
Solution:

Question 33.
Solve : \(\sqrt{\log _{2} x^{4}}+4 \log _{4} \sqrt{\frac{2}{x}}=2\)
Solution:

Question 34.
Find the value of \(\frac{3+\log _{10} 343}{2+\frac{1}{2} \log _{10}\left(\frac{49}{4}\right)+\frac{1}{2} \log _{10}\left(\frac{1}{25}\right)}\)
Solution:

Question 35.
If \(\frac{\log a}{x+y-2 z}=\frac{\log b}{y+z-2 x}=\frac{\log c}{z+x-2 y}\), show that abc = 1.
Solution:
Let \(\frac{\log a}{x+y-2 z}=\frac{\log b}{y+z-2 x}=\frac{\log c}{z+x-2 y}\) = k
∴ log a = k(x + y – 2z), log b = k(y + z – 2x), log c = k(z + x – 2y)
log a + log b + log c = k(x + y – 2z) + k(y + z – 2x) + k(z + x – 2y)
= k(x + y – 2z + y + z – 2x + z + x – 2y)
= k(0)
= 0
∴ log (abc) = log 1 …….[∵ log 1 = 0]
∴ abc = 1

Question 36.
Show that, log_y x³ . log_z y⁴ . log_x z⁵ = 60.
Solution:

Question 37.
If \(\frac{\log _{2} \mathrm{a}}{4}=\frac{\log _{2} \mathrm{~b}}{6}=\frac{\log _{2} \mathrm{c}}{3 \mathrm{k}}\) and a³b²c = 1, find the value of k.
Solution:

Question 38.
If a² = b³ = c⁴ = d⁵, show that loga bcd = \(\frac{47}{30}\).
Solution:

Question 39.
Solve the following for x, where |x| is modulus function, [x] is the greatest integer function, {x} is a fractional part function.
(i) 1 < |x – 1| < 4
(ii) |x² – x – 6| = x + 2
(iii) |x² – 9| + |x² – 4| = 5
(iv) -2 < [x] ≤ 7
(v) 2[2x – 5] – 1 = 7
(vi) [x]2 – 5 [x] + 6 = 0
(vii) [x – 2] + [x + 2] + {x} = 0
(viii) \(\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]=\frac{5 x}{6}\)
Solution:
(i) 1 < |x – 1| < 4
∴ -4 < x – 1 < -1 or 1 < x – 1 < 4
∴ -3 < x < 0 or 2 < x < 5
∴ Solution set = (-3, 0) ∪ (2, 5)

(ii) |x² – x – 6| = x + 2 …..(i)
R.H.S. must be non-negative
∴ x ≥ -2 …..(ii)
|(x – 3) (x + 2)| = x + 2
∴ (x + 2) |x – 3| = x + 2 as x + 2 ≥ 0
∴ |x – 3| = 1 if x ≠ -2
∴ x – 3 = ±1
∴ x = 4 or 2
∴ x = -2 also satisfies the equation
∴ Solution set = {-2, 2, 4}

(iii) |x² – 9| + |x² – 4| = 5
∴ |(x – 3) (x + 3)| + |(x – 2) ( x + 2)| = 5 ………(i)
Case I: x < -3
Also, x < -2, x < 2, x < 3
∴ (x – 3) (x + 3) > 0 and (x – 2) (x + 2) > 0
Equation (i) reduces to
x² – 9 + x² – 4 = 5
∴ 2x² = 18
∴ x = -3 or 3 (both rejected as x < -3)

Case II: -3 ≤ x < -2
As x < -2, x < 3
∴ (x – 3) (x + 3) < 0, (x – 2) (x + 2) > 0
Equation (i) reduces to
-(x² – 9) + x² – 4 = 5
∴ 5 = 5 (true)
-3 ≤ x < -2 is a solution ….(ii)

Case III: -2 ≤ x < 2 As x > -3, x < 3
∴ (x – 3) (x + 3) < 0,
(x – 2) (x + 2) < 0
Equation (i) reduces to
9 – x² + 4 – x2 = 5
∴ 2x² = 13 – 5
∴ x² = 4
∴ x = -2 is a solution …..(iii)

Case IV: 2 ≤ x < 3 As x > -3, x > -2
∴ (x – 3) (x + 3) < 0, (x – 2) (x + 2) > 0
Equation (i) reduces to
9 – x² + x² – 4 = 5
∴ 5 = 5 (true)
∴ 2 ≤ x < 3 is a solution ……(iv)

Case V: 3 ≤ x As x > -3, x > -2, x > 2
∴ (x + 3) (x – 3) > 0,
(x – 2) (x + 2) > 0
Equation (i) reduces to
x² – 9 + x² – 4 = 5
∴ 2x² = 18
∴ x² = 9
∴ x = 3 …..(v)
(x = -3 rejected as x ≥ 3)
From (ii), (iii), (iv), (v), we get
∴ Solution set = [-3, -2] ∪ [2, 3]

(iv) -2 < [x] ≤ 7
∴ -2 < x < 8
∴ Solution set = (-2, 8)

(v) 2[2x – 5] – 1 = 7
∴ [2x – 5] = \(\frac{7+1}{2}\) = 4
∴ [2x] – 5 = 4
∴ [2x] = 9
∴ 9 ≤ 2x < 10
∴ \(\frac{9}{2}\) ≤ x < 5
∴ Solution set = [\(\frac{9}{2}\), 5)

(vi) [x]² – 5[x] + 6 = 0
∴ ([x] – 3)([x] – 2) = 0
∴ [x] = 3 or 2
If [x] = 2, then 2 ≤ x < 3
If [x] = 3, then 3 ≤ x < 4
∴ Solution set = [2, 4)

(vii) [x – 2] + [x + 2] + {x} = 0
∴ [x] – 2 + [x] + 2 + {x} = 0
∴ [x] + x = 0 …..[{x} + [x] = x]
∴ x = 0

(viii) \(\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]=\frac{5 x}{6}\)
L.H.S. = an integer
R.H.S. = an integer
∴ x = 6k, where k is an integer

Question 40.
Find the domain of the following functions.
(i) f(x) = \(\frac{x^{2}+4 x+4}{x^{2}+x-6}\)
(ii) f(x) = \(\sqrt{x-3}+\frac{1}{\log (5-x)}\)
(iii) f(x) = \(\sqrt{1-\sqrt{1-\sqrt{1-x^{2}}}}\)
(iv) f(x) = x!
(v) f(x) = \({ }^{5-x} P_{x-1}\)
(vi) f(x) = \(\sqrt{x-x^{2}}+\sqrt{5-x}\)
(vii) f(x) = \(\sqrt{\log \left(x^{2}-6 x+6\right)}\)
Solution:
(i) f(x) = \(\frac{x^{2}+4 x+4}{x^{2}+x-6}=\frac{x^{2}+4 x+4}{(x+3)(x-2)}\)
For f to be defined, x ≠ -3, 2
∴ Domain of f = (-∞, -3) ∪ (-3, 2) ∪ (2, ∞)

(ii) f(x) = \(\sqrt{x-3}+\frac{1}{\log (5-x)}\)
For f to be defined,
x – 3 ≥ 0, 5 – x > 0 and 5 – x ≠ 1
x ≥ 3, x < 5 and x ≠ 4
∴ Domain of f = [3, 4) ∪ (4, 5)

(iii) f(x) = \(\sqrt{1-\sqrt{1-\sqrt{1-x^{2}}}}\)

Equation (i) gives solution set = [-1, 1]
∴ Domain of f = [-1, 1]

(iv) f(x) = x!
∴ Domain of f = set of whole numbers (W)

(v) f(x) = \({ }^{5-x} P_{x-1}\)
5 – x > 0, x – 1 ≥ 0, x – 1 ≤ 5 – x
∴ x < 5, x ≥ 1 and 2x ≤ 6
∴ x ≤ 3
∴ Domain of f = {1, 2, 3}

(vi) f(x) = \(\sqrt{x-x^{2}}+\sqrt{5-x}\)
x – x² ≥ 0
∴ x² – x ≤ 0
∴ x(x – 1) ≤ 0
∴ 0 ≤ x ≤ 1 …..(i)
5 – x ≥ 0
∴ x ≤ 5 …..(ii)
Intersection of intervals given in (i) and (ii) gives
Solution set = [0, 1]
∴ Domain of f = [0, 1]

(vii) f(x) = \(\sqrt{\log \left(x^{2}-6 x+6\right)}\)
For f to be defined,
log (x² – 6x + 6) ≥ 0
∴ x² – 6x + 6 ≥ 1
∴ x² – 6x + 5 ≥ 0
∴ (x – 5)(x – 1) ≥ 0
∴ x ≤ 1 or x ≥ 5 …..(i)
[∵ The solution of (x – a) (x – b) ≥ 0 is x ≤ a or x ≥ b, for a < b]
and x² – 6x + 6 > 0
∴ (x – 3)² > -6 + 9
∴ (x – 3)² > 3
∴ x < 3 – √3 0r x > 3 + √3 ……..(ii)
From (i) and (ii), we get
x ≤ 1 or x ≥ 5
Solution set = (-∞, 1] ∪ [5, ∞)
∴ Domain of f = (-∞, 1] ∪ [5, ∞)

Question 41.
(i) f(x) = |x – 5|
(ii) f(x) = \(\frac{x}{9+x^{2}}\)
(iii) f(x) = \(\frac{1}{1+\sqrt{x}}\)
(iv) f(x) = [x] – x
(v) f(x) = 1 + 2x + 4x
Solution:
(i) f(x) = |x – 5|

∴ Range of f = [0, ∞)

(ii) f(x) = \(\frac{x}{9+x^{2}}\) = y (say)
∴ x²y – x + 9y = 0
For real x, Discriminant > 0
∴ 1 – 4(y)(9y) ≥ 0
∴ y² ≤ \(\frac{1}{36}\)
∴ \(\frac{-1}{6}\) ≤ y ≤ \(\frac{1}{6}\)
∴ Range of f = [\(\frac{-1}{6}\), \(\frac{1}{6}\)]

(iii) f(x) = \(\frac{1}{1+\sqrt{x}}\) = y, (say)
∴ √x y + y = 1
∴ √x = \(\frac{1-y}{y}\) ≥ 0
∴ \(\frac{y-1}{y}\) ≤ 0
∴ o < y ≤ 1
∴ Range of f = (0, 1]

(iv) f(x) = [x] – x = -{x}
∴ Range of f = (-1, 0] …..[0 ≤ {x} < 1]

(v) f(x) = 1 + 2x + 4x
Since, 2x > 0, 4x > 0
∴ f(x) > 1
∴ Range of f = (1, ∞)

Question 42.
Find (fog) (x) and (gof) (x)
(i) f(x) = e^x, g(x) = log x
(ii) f(x) = \(\frac{x}{x+1}\), g(x) = \(\frac{x}{1-x}\)
Solution:
(i) f(x) = e^x, g(x) = log x
(fog) (x) = f(g(x))
= f(log x)
= e^{log x}
= x
(gof) (x) = g(f(x))
= g(e^x)
= log (e^x)
= x log e
= x …..[∵ log e = 1]

(ii) f(x) = \(\frac{x}{x+1}\), g(x) = \(\frac{x}{1-x}\)

Question 43.
Find f(x), if
(i) g(x) = x² + x – 2 and (gof) (x) = 4x² – 10x + 4
(ii) g(x) = 1 + √x and f [g(x)] = 3 + 2√x + x.
Solution:
(i) g(x) = x² + x – 2
(gof) (x) = 4x² – 10x + 4
= (2x – 3)² + (2x – 3) – 2
= g(2x – 3)
= g(f(x))
∴ f(x) = 2x – 3
(gof) (x) = 4x² – 10x + 4
= (-2x + 2)² + (-2x + 2) – 2
= g(-2x + 2)
= g(f(x))
∴ f(x) = -2x + 2

(ii) g(x) = 1 + √x
f(g(x)) = 3 + 2√x + x
= x + 2√x + 1 + 2
= (√x + 1)² + 2
f(√x + 1) = (√x + 1)² + 2
∴ f(x) = x² + 2

Question 44.
Find (fof) (x) if
(i) f(x) = \(\frac{x}{\sqrt{1+x^{2}}}\)
(ii) f(x) = \(\frac{2 x+1}{3 x-2}\)
Solution:

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 3 Permutations and Combination Ex 3.4 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 3 Permutations and Combination Ex 3.4

Question 1.
Find the number of permutations of letters in each of the following words.
(i) DIVYA
(ii) SHANTARAM
(iii) REPRESENT
(iv) COMBINE
(v) BAL BHARATI
Solution:
(i) There are 5 distinct letters in the word DIVYA.
∴ Number of permutations of the letters of the word DIVYA = 5! = 120

(ii) There are 9 letters in the word SHANTARAM in which ‘A’ is repeated 3 times.
∴ Number of permutations of the letters of the word SHANTARAM = \(\frac{9 !}{3 !}\)
= 9 × 8 × 7 × 6 × 5 × 4
= 60480

(iii) There are 9 letters in the word REPRESENT in which ‘E’ is repeated 3 times and ‘R’ is repeated 2 times.
∴ Number of permutations of the letters of the word REPRESENT = \(\frac{9 !}{3 ! 2 !}\)
= \(\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4}{2}\)
= 30240

(iv) There are 7 distinct letters in the word COMBINE.
∴ Number of permutations of the letters of the word COMBINE = 7! = 5040

(v) There are 10 letters in the word BALBHARATI in which ‘B’ is repeated 2 times and ‘A’ is repeated 3 times.
∴ Number of permutations of the letters of the word BALBHARATI = \(\frac{10 !}{2 ! 3 !}\)
= \(\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 3 \times 2}\)
= 302400

Question 2.
You have 2 identical books on English, 3 identical books on Hindi, and 4 identical books on Mathematics. Find the number of distinct ways of arranging them on a shelf.
Solution:
There are total 9 books to be arranged on a shelf.
Out of these 9 books, 2 books on English, 3 books on Hindi and 4 books on mathematics are identical.
∴ Total number of arrangements possible = \(\frac{9 !}{2 ! 3 ! 4 !}\)
= \(\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 !}{2 \times 3 \times 2 \times 4 !}\)
= 9 × 4 × 7 × 5
= 1260

Question 3.
A coin is tossed 8 times. In how many ways can we obtain (a) 4 heads and 4 tails? (b) at least 6 heads?
Solution:
A coin is tossed 8 times. All heads are identical and all tails are identical.
(a) 4 heads and 4 tails are to be obtained.
∴ Number of ways it can be obtained = \(\frac{8 !}{4 ! 4 !}\)
= \(\frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2}\)
= 70

(b) At least 6 heads are to be obtained.
∴ Outcome can be (6 heads and 2 tails) or (7 heads and 1 tail) or (8 heads)
∴ Number of ways it can be obtained = \(\frac{8 !}{6 ! 2 !}+\frac{8 !}{7 ! 1 !}+\frac{8 !}{8 !}\)
= \(\frac{8 \times 7}{2}\) + 8 + 1
= 28 + 8 + 1
= 37

Question 4.
A bag has 5 red, 4 blue, and 4 green marbles. If all are drawn one by one and their colours are recorded, how many different arrangements can be found?
Solution:
There is a total of 13 marbles in a bag.
Out of these 5 are Red, 4 Blue, and 4 are Green marbles.
All balls of the same colour are taken to be identical.
∴ Required number of arrangements = \(\frac{13 !}{5 ! 4 ! 4 !}\)

Question 5.
Find the number of ways of arranging letters of the word MATHEMATICAL. How many of these arrangements have all vowels together?
Solution:
There are 12 letters in the word MATHEMATICAL in which ‘M’ is repeated 2 times, ‘A’ repeated 3 times and ‘T’ repeated 2 times.
∴ Required number of arrangements = \(\frac{12 !}{2 ! 3 ! 2 !}\)
When all the vowels,
i.e., ‘A’, ‘A’, ‘A’, ‘E’, ‘I’ are to be kept together.
Let us consider them as one unit.
Number of arrangements of these vowels among themselves = \(\frac{5 !}{3 !}\) ways.
This unit is to be arranged with 7 other letters in which ‘M’ and ‘T’ repeated 2 times each.
∴ Number of such arrangements = \(\frac{8 !}{2 ! 2 !}\)
∴ Required number of arrangements = \(\frac{8 ! \times 5 !}{2 ! 2 ! 3 !}\)

Question 6.
Find the number of different arrangements of letters in the word MAHARASHTRA. How many of these arrangements have (a) letters R and H never together? (b) all vowels together?
Solution:
There are 11 letters in the word MAHARASHTRA in which ‘A’ is repeated 4 times, ‘H’ repeated 2 times, and ‘R’ repeated 2 times.
∴ Total number of words can be formed = \(\frac{11 !}{4 ! 2 ! 2 !}\)

(a) When letters R and H are never together.
Other than 2R, 2H there are 4A, 1S, 1T, 1M.
These letters can be arranged in \(\frac{7 !}{4 !}\) ways = 210.
These seven letters create 8 gaps in which 2R, 2H are to be arranged.
Number of ways to do = \(\frac{{ }^{8} \mathrm{P}_{4}}{2 ! 2 !}\) = 420
Required number of arrangements = 210 × 420 = 88200.

(b) When all vowels are together.
There are 4 vowels in the word MAHARASHTRA, i.e., A, A, A, A.
Let us consider these 4 vowels as one unit, which can be arranged among themselves in \(\frac{4 !}{4 !}\) = 1 way.
This unit is to be arranged with 7 other letters in which ‘H’ is repeated 2 times, ‘R’ is repeated 2 times.
∴ Total number of arrangements = \(\frac{8 !}{2 ! 2 !}\)

Question 7.
How many different words are formed if the letter R is used thrice and letters S and T are used twice each?
Solution:
To find the number of different words when ‘R’ is taken thrice, ‘S’ is taken twice and ‘T’ is taken twice.
∴ Total number of letters available = 7, of which ‘S’ and ‘T’ repeat 2 times each, ‘R’ repeats 3 times.
∴ Required number of words = \(\frac{7 !}{2 ! 2 ! 3 !}\)
= \(\frac{7 \times 6 \times 5 \times 4 \times 3 !}{2 \times 1 \times 2 \times 1 \times 3 !}\)
= 7 × 6 × 5
= 210

Question 8.
Find the number of arrangements of letters in the word MUMBAI so that the letter B is always next to A.
Solution:
There are 6 letters in the word MUMBAI.
These letters are to be arranged in such a way that ‘B’ is always next to ‘A’.
Let us consider AB as one unit.
This unit with the other 4 letters in which ‘M’ repeats twice is to be arranged.
∴ Required number of arrangements = \(\frac{5 !}{2 !}\)
= \(\frac{5 \times 4 \times 3 \times 2 !}{2 !}\)
= 60

Question 9.
Find the number of arrangements of letters in the word CONSTITUTION that begin and end with N.
Solution:
There are 12 letters in the word CONSTITUTION, in which ‘O’, ‘N’, ‘I’ repeat two times each, ‘T’ repeats 3 times.
When the arrangement starts and ends with ‘N’,
other 10 letters can be arranged between two N,
in which ‘O’ and ‘I’ repeat twice each and ‘T’ repeats 3 times.
∴ Required number of arrangements = \(\frac{10 !}{2 ! 2 ! 3 !}\)

Question 10.
Find the number of different ways of arranging letters in the word ARRANGE. How many of these arrangements do not have the two R’s and two A’s together?
Solution:
There are 7 letters in the word ARRANGE in which ‘A’ and ‘R’ repeat 2 times each.
∴ Number of ways to arrange the letters of word ARRANGE = \(\frac{7 !}{2 ! 2 !}\) = 1260
Consider the words in which 2A are together and 2R are together.
Let us consider 2A as one unit and 2R as one unit.
These two units with remaining 3 letters can be arranged in = \(\frac{5 !}{2 ! 2 !}\) = 30 ways.
Number of arrangements in which neither 2A together nor 2R are together = 1260 – 30 = 1230

Question 11.
How many distinct 5 digit numbers can be formed using the digits 3, 2, 3, 2, 4, 5.
Solution:
5 digit numbers are to be formed from 2, 3, 2, 3, 4, 5.
Case I: Numbers formed from 2, 2, 3, 4, 5 OR 2, 3, 3, 4, 5
Number of such numbers = \(\frac{5 !}{2 !}+\frac{5 !}{2 !}\) = 5! = 120
Case II: Numbers are formed from 2, 2, 3, 3 and any one of 4 or 5
Number of such numbers = \(\frac{5 !}{2 ! 2 !}+\frac{5 !}{2 ! 2 !}\) = 60
Required number of numbers = 120 + 60 = 180

Question 12.
Find the number of distinct numbers formed using the digits 3, 4, 5, 6, 7, 8, 9, so that odd positions are occupied by odd digits.
Solution:
A number is to be formed with digits 3, 4, 5, 6, 7, 8, 9 such that odd digits always occupy the odd places.
There are 4 odd digits, i.e. 3, 5, 7, 9.
∴ They can be arranged at 4 odd places among themselves in 4! = 24 ways.
There are 3 even digits, i.e. 4, 6, 8.
∴ They can be arranged at 3 even places among themselves in 3! = 6 ways.
∴ Required number of numbers formed = 24 × 6 = 144

Question 13.
How many different 6-digit numbers can be formed using digits in the number 659942? How many of them are divisible by 4?
Solution:
A 6-digit number is to be formed using digits of 659942, in which 9 repeats twice.
∴ Required number of numbers formed = \(\frac{6 !}{2 !}\)
= \(\frac{6 \times 5 \times 4 \times 3 \times 2 !}{2 !}\)
= 360
A 6-digit number is to be formed using the same digits that are divisible by 4.
For a number to be divisible by 4, the last two digits should be divisible by 4,
i.e. 24, 52, 56, 64, 92 or 96.
Case I: When the last two digits are 24, 52, 56 or 64.
As the digit 9 repeats twice in the remaining four numbers, the number of arrangements = \(\frac{4 !}{2 !}\) = 12
∴ 6-digit numbers that are divisible by 4 so formed are 12 + 12 + 12 + 12 = 48.
Case II: When the last two digits are 92 or 96.
As each of the remaining four numbers are distinct, the number of arrangements = 4! = 24
∴ 6-digit numbers that are divisible by 4 so formed are 24 + 24 = 48.
∴ Required number of numbers framed = 48 + 48 = 96

Question 14.
Find the number of distinct words formed from letters in the word INDIAN. How many of them have the two N’s together?
Solution:
There are 6 letters in the word INDIAN in which I and N are repeated twice.
Number of different words that can be formed using the letters of the word INDIAN = \(\frac{6 !}{2 ! 2 !}\)
= \(\frac{6 \times 5 \times 4 \times 3 \times 2 !}{2 \times 2 !}\)
= 180
When two N’s are together.
Let us consider the two N’s as one unit.
They can be arranged with 4 other letters in \(\frac{5 !}{2 !}\)
= \(\frac{5 \times 4 \times 3 \times 2 !}{2 !}\)
= 60 ways.
∴ 2 N can be arranged in \(\frac{2 !}{2 !}\) = 1 way.
∴ Required number of words = 60 × 1 = 60

Question 15.
Find the number of different ways of arranging letters in the word PLATOON if (a) the two O’s are never together. (b) consonants and vowels occupy alternate positions.
Solution:
There are 7 letters in the words PLATOON in which ‘O’ repeat 2 times.
(a) When the two O’s are never together.
Let us arrange the other 5 letters first, which can be done in 5! = 120 ways.
The letters P, L, A, T, N create 6 gaps, in which O’s are arranged.
Two O’s can take their places in ⁶P₂ ways.
But ‘O’ repeats 2 times.
∴ Two O’s can be arranged in \(\frac{{ }^{6} \mathrm{P}_{2}}{2 !}\)
= \(\frac{\frac{6 !}{(6-2) !}}{2 !}\)
= \(\frac{6 \times 5 \times 4 !}{4 ! \times 2 \times 1}\)
= 3 × 5
= 15 ways
∴ Required number of arrangements = 120 × 15 = 1800

(b) When consonants and vowels occupy alternate positions.
There are 4 consonants and 3 vowels in the word PLATOON.
∴ At odd places, consonants occur and at even places, vowels occur.
4 consonants can be arranged among themselves in 4! ways.
3 vowels in which O occurs twice and A occurs once.
∴ They can be arranged in \(\frac{3 !}{2 !}\) ways.
Now, vowels and consonants should occupy alternate positions.
∴ Required number of arrangements = 4! × \(\frac{3 !}{2 !}\)
= 4 × 3 × 2 × \(\frac{3 \times 2 !}{2 !}\)
= 72

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 7 Limits Miscellaneous Exercise 7 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7

I. Select the correct answer from the given alternatives.

Question 1.
\(\lim _{x \rightarrow 2}\left(\frac{x^{4}-16}{x^{2}-5 x+6}\right)=\)
(A) 23
(B) 32
(C) -32
(D) -16
Answer:
(C) -32
Hint:

Question 2.
\(\lim _{x \rightarrow-2}\left(\frac{x^{7}+128}{x^{3}+8}\right)=\)
(A) \(\frac{56}{3}\)
(B) \(\frac{112}{3}\)
(C) \(\frac{121}{3}\)
(D) \(\frac{28}{3}\)
Answer:
(B) \(\frac{112}{3}\)
Hint:

Question 3.
\(\lim _{x \rightarrow 3}\left(\frac{1}{x^{2}-11 x+24}+\frac{1}{x^{2}-x-6}\right)=\)
(A) \(-\frac{2}{25}\)
(B) \(\frac{2}{25}\)
(C) \(\frac{7}{25}\)
(D) \(-\frac{7}{25}\)
Answer:
(A) \(-\frac{2}{25}\)
Hint:

Question 4.
\(\lim _{x \rightarrow 5}\left(\frac{\sqrt{x+4}-3}{\sqrt{3 x-11-2}}\right)=\)
(A) \(\frac{-2}{9}\)
(B) \(\frac{2}{7}\)
(C) \(\frac{5}{9}\)
(D) \(\frac{2}{9}\)
Answer:
(D) \(\frac{2}{9}\)
Hint:

Question 5.
\(\lim _{x \rightarrow \frac{\pi}{3}}\left(\frac{\tan ^{2} x-3}{\sec ^{3} x-8}\right)=\)
(A) 1
(B) \(\frac{1}{2}\)
(C) \(\frac{1}{3}\)
(D) \(\frac{1}{4}\)
Answer:
(C) \(\frac{1}{3}\)
Hint:

Question 6.
\(\lim _{x \rightarrow 0}\left(\frac{5 \sin x-x \cos x}{2 \tan x-3 x^{2}}\right)=\)
(A) 0
(B) 1
(C) 2
(D) 3
Answer:
(C) 2
Hint:

Question 7.
\(\lim _{x \rightarrow \frac{\pi}{2}}\left[\frac{3 \cos x+\cos 3 x}{(2 x-\pi)^{3}}\right]=\)
(A) \(\frac{3}{2}\)
(B) \(\frac{1}{2}\)
(C) \(-\frac{1}{2}\)
(D) \(\frac{1}{4}\)
Answer:
(C) \(-\frac{1}{2}\)
Hint:

Question 8.
\(\lim _{x \rightarrow 0}\left(\frac{15^{x}-3^{x}-5^{x}+1}{\sin ^{2} x}\right)=\)
(A) log 15
(B) log 3 + log 5
(C) log 3 . log 5
(D) 3 log 5
Answer:
(C) log 3 . log 5
Hint:

Question 9.
\(\lim _{x \rightarrow 0}\left(\frac{3+5 x}{3-4 x}\right)^{\frac{1}{x}}=\)
(A) e³
(B) e⁶
(C) e⁹
(D) e^-3
Answer:
(A) e³
Hint:

Question 10.
\(\lim _{x \rightarrow 0}\left[\frac{\log (5+x)-\log (5-x)}{\sin x}\right]=\)
(A) \(\frac{3}{2}\)
(B) \(-\frac{5}{2}\)
(C) \(-\frac{1}{2}\)
(D) \(\frac{2}{5}\)
Answer:
(D) \(\frac{2}{5}\)
Hint:

Question 11.
\(\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{3^{\cos x}-1}{\frac{\pi}{2}-x}\right)=\)
(A) 1
(B) log 3
(C) \(3^{\frac{\pi}{2}}\)
(D) 3 log 3
Answer:
(B) log 3
Hint:

Question 12.
\(\lim _{x \rightarrow 0}\left[\frac{x \cdot \log (1+3 x)}{\left(e^{3 x}-1\right)^{2}}\right]=\)
(A) \(\frac{1}{\mathrm{e}^{9}}\)
(B) \(\frac{1}{\mathrm{e}^{3}}\)
(C) \(\frac{1}{9}\)
(D) \(\frac{1}{3}\)
Answer:
(D) \(\frac{1}{3}\)
Hint:

Question 13.
\(\lim _{x \rightarrow 0}\left[\frac{\left(3^{\sin x}-1\right)^{3}}{\left(3^{x}-1\right) \cdot \tan x \cdot \log (1+x)}\right]=\)
(A) 3 log 3
(B) 2 log 3
(C) (log 3)²
(D) (log 3)³
Answer:
(C) (log 3)²
Hint:

Question 14.
\(\lim _{x \rightarrow 3}\left[\frac{5^{x-3}-4^{x-3}}{\sin (x-3)}\right]=\)
(A) log 5 – 4
(B) log \(\frac{5}{4}\)
(C) \(\frac{\log 5}{\log 4}\)
(D) \(\frac{\log 5}{4}\)
Answer:
(B) log \(\frac{5}{4}\)
Hint:

Question 15.
\(\lim _{x \rightarrow \infty}\left[\frac{(2 x+3)^{7}(x-5)^{3}}{(2 x-5)^{10}}\right]=\)
(A) \(\frac{3}{8}\)
(B) \(\frac{1}{8}\)
(C) \(\frac{1}{6}\)
(D) \(\frac{1}{4}\)
Answer:
(B) \(\frac{1}{8}\)
Hint:

(II) Evaluate the following.

Question 1.
\(\lim _{x \rightarrow 0}\left[\frac{(1-x)^{5}-1}{(1-x)^{3}-1}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow 0}[x]\) ([*] is a greatest integer function.)
Solution:

Question 3.
If f(r) = πr² then find \(\lim _{h \rightarrow 0}\left[\frac{f(r+h)-f(r)}{h}\right]\)
Solution:

Question 4.
\(\lim _{x \rightarrow 0}\left[\frac{x}{|x|+x^{2}}\right]\)
Solution:

Question 5.
Find the limit of the function, if it exists, at x = 1
\(f(x)=\left\{\begin{array}{lll}
7-4 x & \text { for } & x<1 \\
x^{2}+2 & \text { for } & x \geq 1
\end{array}\right.\)
Solution:

Question 6.
Given that 7x ≤ f(x) ≤ 3x² – 6 for all x. Determine the value of \(\lim _{x \rightarrow 3} f(x)\)
Solution:

Question 7.
\(\lim _{x \rightarrow 0}\left[\frac{\sec x^{2}-1}{x^{4}}\right]\)
Solution:

Question 8.
\(\lim _{x \rightarrow 0}\left[\frac{e^{x}+e^{-x}-2}{x \cdot \tan x}\right]\)
Solution:

Question 9.
\(\lim _{x \rightarrow 0}\left[\frac{x\left(6^{x}-3^{x}\right)}{\cos (6 x)-\cos (4 x)}\right]\)
Solution:

Question 10.
\(\lim _{x \rightarrow 0}\left[\frac{a^{3 x}-a^{2 x}-a^{x}+1}{x \cdot \tan x}\right]\)
Solution:

Question 11.
\(\lim _{x \rightarrow a}\left[\frac{\sin x-\sin a}{x-a}\right]\)
Solution:

Question 12.
\(\lim _{x \rightarrow 2}\left[\frac{\log x-\log 2}{x-2}\right]\)
Solution:

Question 13.
\(\lim _{x \rightarrow 1}\left[\frac{a b^{x}-a^{x} b}{x^{2}-1}\right]\)
Solution:

Question 14.
\(\lim _{x \rightarrow 0}\left[\frac{\left(5^{x}-1\right)^{2}}{\left(2^{x}-1\right) \log (1+x)}\right]\)
Solution:

Question 15.
\(\lim _{x \rightarrow \infty}\left[\frac{(2 x+1)^{2}(7 x-3)^{3}}{(5 x+2)^{5}}\right]\)
Solution:

Question 16.
\(\lim _{x \rightarrow a}\left[\frac{x \cos a-a \cos x}{x-a}\right]\)
Solution:

Question 17.
\(\lim _{x \rightarrow \frac{\pi}{4}}\left[\frac{(\sin x-\cos x)^{2}}{\sqrt{2}-\sin x-\cos x}\right]\)
Solution:

Question 18.
\(\lim _{x \rightarrow 1}\left[\frac{2^{2 x-2}-2^{x}+1}{\sin ^{2}(x-1)}\right]\)
Solution:

Question 19.
\(\lim _{x \rightarrow 1}\left[\frac{4^{x-1}-2^{x}+1}{(x-1)^{2}}\right]\)
Solution:

Question 20.
\(\lim _{x \rightarrow 1}\left[\frac{\sqrt{x}-1}{\log x}\right]\)
Solution:

Question 21.
\(\lim _{x \rightarrow 0}\left(\frac{\sqrt{1-\cos x}}{x}\right)\)
Solution:

Question 22.
\(\lim _{x \rightarrow 1}\left(\frac{x+3 x^{2}+5 x^{3}+\cdots \cdots \cdots \cdots \cdots+(2 n-1) x^{n}-n^{2}}{x-1}\right)\)
Solution:

Question 23.
\(\lim _{x \rightarrow 0} \frac{1}{x^{12}}\left[1-\cos \left(\frac{x^{2}}{2}\right)-\cos \left(\frac{x^{4}}{4}\right)+\cos \left(\frac{x^{2}}{2}\right) \cdot \cos \left(\frac{x^{4}}{4}\right)\right]\)
Solution:

Question 24.
\(\lim _{x \rightarrow \infty}\left(\frac{8 x^{2}+5 x+3}{2 x^{2}-7 x-5}\right)^{\frac{4 x+3}{8 x-1}}\)
Solution:

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 2 Sequences and Series Miscellaneous Exercise 2 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2

(I) Select the correct answer from the given alternative:

Question 1.
The common ratio for the G.P. 0.12, 0.24, 0.48, is
(A) 0.12
(B) 0.2
(C) 0.02
(D) 2
Answer:
(D) 2

Question 2.
The tenth term of the geometric sequence is \(\frac{1}{4}, \frac{-1}{2}, 1,-2, \ldots\) is
(A) 1024
(B) \(\frac{1}{1024}\)
(C) -128
(D) \(\frac{-1}{128}\)
Answer:
(C) -128
Hint:

Question 3.
If for a G.P. \(\frac{t_{6}}{t_{3}}=\frac{1458}{54}\) then r = ?
(A) 3
(B) 2
(C) 1
(D) -1
Answer:
(A) 3
Hint:

Question 4.
Which term of the geometric progression 1, 2, 4, 8, ….. is 2048.
(A) 10th
(B) 11th
(C) 12th
(D) 13th
Answer:
(C) 12th
Hint:
Here, a = 1, r = 2
nth term of geometric progression = ar^n-1
∴ ar^n-1 = 2048
2^n-1 = 2¹¹
n – 1 = 11
∴ n = 12

Question 5.
If the common ratio of the G.P. is 5, the 5th term is 1875, the first term is
(A) 3
(B) 5
(C) 15
(D) -5
Answer:
(A) 3

Question 6.
The sum of 3 terms of a G.P. is \(\frac{21}{4}\) and their product is 1, then the common ratio is
(A) 1
(B) 2
(C) 4
(D) 8
Answer:
(C) 4
Hint:
Let three terms be \(\frac{a}{r}\), a, ar
According to the given conditions,
\(\frac{a}{r}\) + a + ar = \(\frac{21}{4}\) …..(i)
and \(\frac{a}{r}\) × a × ar = 1,
i.e., a³ = 1
∴ a = 1
∴ from equation (i), we get
\(\frac{1}{r}\) + 1 + r = \(\frac{21}{4}\)
By solving this, we get r = 4.

Question 7.
Sum to infinity of a G.P. 5, \(-\frac{5}{2}, \frac{5}{4},-\frac{5}{8}, \frac{5}{16}, \ldots\) is
(A) 5
(B) \(-\frac{1}{2}\)
(C) \(\frac{10}{3}\)
(D) \(\frac{3}{10}\)
Answer:
(C) \(\frac{10}{3}\)
Hint:
Here, a = 5, r = \(\frac{-1}{2}\), |r| < 1
∴ Sum to the infinity = \(\frac{a}{1-r}=\frac{5}{1+\frac{1}{2}}=\frac{10}{3}\)

Question 8.
The tenth term of H.P. \(\frac{2}{9}, \frac{1}{7}, \frac{2}{19}, \frac{1}{12}, \ldots\) is
(A) \(\frac{1}{27}\)
(B) \(\frac{9}{2}\)
(C) \(\frac{5}{2}\)
(D) 27
Answer:
(A) \(\frac{1}{27}\)
Hint:

Question 9.
Which of the following is not true, where A, G, H are the AM, GM, HM of a and b respectively, (a, b > 0)
(A) A = \(\frac{a+b}{2}\)
(B) G = \(\sqrt{a b}\)
(C) H = \(\frac{2 a b}{a+b}\)
(D) A = GH
Answer:
(D) A = GH

Question 10.
The G.M. of two numbers exceeds their H.M. by \(\frac{6}{5}\), the A.M. exceeds G.M. by \(\frac{3}{2}\) the two numbers are
(A) 6, \(\frac{15}{2}\)
(B) 15, 25
(C) 3, 12
(D) \(\frac{6}{5}\), \(\frac{3}{2}\)
Answer:
(C) 3, 12
Hint:

(II) Answer the following:

Question 1.
In a G.P., the fourth term is 48 and the eighth term is 768. Find the tenth term.
Solution:

Question 2.
Find the sum of the first 5 terms of the G.P. whose first term is 1 and the common ratio is \(\frac{2}{3}\).
Solution:

Question 3.
For a G.P. a = \(\frac{4}{3}\) and t₇ = \(\frac{243}{1024}\), find the value of r.
Solution:

Question 4.
For a sequence, if \(t_{n}=\frac{5^{n-2}}{7^{n-3}}\), verify whether the sequence is a G.P. If it is a G.P., find its first term and the common ratio.
Solution:
The sequence (t_n) is a G.P.
if \(\frac{\mathrm{t}_{\mathrm{n}+1}}{\mathrm{t}_{\mathrm{n}}}\) = constant for all n ∈ N.

Question 5.
Find three numbers in G.P. such that their sum is 35 and their product is 1000.
Solution:
Let the three numbers in G.P. be \(\frac{a}{r}\), a, ar.
According to the given conditions,
\(\frac{a}{r}\) + a + ar = 35
a(\(\frac{1}{r}\) + 1 + r) = 35 …..(i)
Also, (\(\frac{a}{r}\))(a)(ar) = 1000
a³ = 1000
∴ a = 10
Substituting the value of a in (i), we get

Hence, the three numbers in G.P. are 20, 10, 5, or 5, 10, 20.

Question 6.
Find five numbers in G.P. such that their product is 243 and the sum of the second and fourth numbers is 10.
Solution:
Let the five numbers in G.P. be \(\frac{a}{r^{2}}, \frac{a}{r}, a, a r, a r^{2}\).
According to the given condition,

Question 7.
For a sequence, S_n = 4(7ⁿ – 1), verify that the sequence is a G.P.
Solution:

∴ The given sequence is a G.P.

Question 8.
Find 2 + 22 + 222 + 2222 + … upto n terms.
Solution:
S_n = 2 + 22 + 222 +… upto n terms
= 2(1 + 11 + 111 + ….. upto n terms)
= \(\frac{2}{9}\) (9 + 99 + 999 + … upto n terms)
= \(\frac{2}{9}\) [(10 – 1) + (100 – 1) + (1000 – 1) + …… upto n terms]
= \(\frac{2}{9}\) [(10 + 100 + 1000 + … upto n terms) – (1 + 1 + 1 + ….. n times)]
Since 10, 100, 1000, ….. n terms are in G.P. with
a = 10, r = \(\frac{100}{10}\) = 10,

Question 9.
Find the nth term of the sequence 0.6, 0.66, 0.666, 0.6666,…
Solution:
0.6, 0.66, 0.666, 0.6666, …
∴ t₁ = 0.6
t₂ = 0.66 = 0.6 + 0.06
t₃ = 0.666 = 0.6 + 0.06 + 0.006
Hence, in general
t_n = 0.6 + 0.06 + 0.006 + …..upto n terms.
The terms are in G.P. with
a = 0.6, r = \(\frac{0.06}{0.6}\) = 0.1
∴ t_n = the sum of first n terms of the G.P.

Question 10.
Find \(\sum_{r=1}^{n}\left(5 r^{2}+4 r-3\right)\)
Solution:

Question 11.
Find \(\sum_{r=1}^{n} r(r-3)(r-2)\)
Solution:

Question 12.
Find \(\sum_{r=1}^{n} \frac{1^{2}+2^{2}+3^{2}+\ldots+r^{2}}{2 r+1}\)
Solution:
We know that

Question 13.
Find \(\sum_{r=1}^{n} \frac{1^{3}+2^{3}+3^{3}+\ldots r^{3}}{(r+1)^{2}}\)
Solution:

Question 14.
Find 2 × 6 + 4 × 9 + 6 × 12 + ….. upto n terms.
Solution:
2, 4, 6, ….. are in A.P.
∴ rth term = 2 + (r – 1) 2 = 2r
6, 9, 12, ….. are in A.P.
∴ rth term = 6 + (r – 1)(3) = (3r + 3)
∴ 2 × 6 + 4 × 9 + 6 × 12 + ….. to n terms

= n(n + 1) [2n + 1 + 3]
= 2n(n + 1)(n + 2)

Question 15.
Find 2 × 5 × 8 + 4 × 7 × 10 + 6 × 9 × 12 + …… upto n terms.
Solution:
2, 4, 6,… are in A.P.
∴ rth term = 2 + (r – 1) 2 = 2r
5, 7, 9, … are in A.P.
∴ rth term = 5 + (r – 1) (2) = (2r + 3)
8, 10, 12, … are in A.P.
∴ rth term = 8 + (r – 1) (2) = (2r + 6)
2 × 5 × 8 + 4 × 7 × 10 + 6 × 9 × 12 + ….. to n terms

= 2n (n + 1) [n(n + 1) + 3(2n + 1) + 9]
= 2n (n + 1)(n² + 7n + 12)
= 2n (n + 1) (n + 3) (n + 4)

Question 16.
Find \(\frac{1^{2}}{1}+\frac{1^{2}+2^{2}}{2}+\frac{1^{2}+2^{2}+3^{2}}{3}+\ldots\) upto n terms.
Solution:

Question 17.
Find 12² + 13² + 14² + 15² + ….. 20²
Solution:

Question 18.
If \(\frac{1+2+3+4+5+\ldots \text { upto } \mathrm{n} \text { terms }}{1 \times 2+2 \times 3+3 \times 4+4 \times 5+\ldots \text { upto } \mathrm{n} \text { terms }}=\frac{3}{22}\), Find the value of n.
Solution:

Question 19.
Find (50² – 49²) + (48² – 47²) + (46² – 45²) +… + (2² – 1²).
Solution:

Question 20.
If \(\frac{1 \times 3+2 \times 5+3 \times 7+\ldots \text { upto } \mathrm{n} \text { terms }}{1^{3}+2^{3}+3^{3}+\ldots \text { upto } \mathrm{n} \text { terms }}=\frac{5}{9}\), find the value of n.
Solution:

Question 21.
For a G.P. if t₂ = 7, t₄ = 1575, find a.
Solution:

Question 22.
If for a G.P. t₃ = \(\frac{1}{3}\), t₆ = \(\frac{1}{81}\) find r.
Solution:

Question 23.
Find \(\sum_{r=1}^{n}\left(\frac{2}{3}\right)^{r}\).
Solution:

Question 24.
Find k so that k – 1, k, k + 2 are consecutive terms of a G.P.
Solution:
Since k – 1, k, k + 2 are consecutive terms of a G.P.,
\(\frac{k}{k-1}=\frac{k+2}{k}\)
k² = k² + k – 2
k – 2 = 0
∴ k = 2

Question 25.
If for a G.P. first term is (27)² and the seventh term is (8)², find S₈.
Solution:

Question 26.
If pth, qth and rth terms of a G.P. are x, y, z respectively. Find the value of \(x^{q-r} \cdot y^{r-p} \cdot z^{p-q}\).
Solution:
Let a be the first term and R be the common ratio of the G.P.

Question 27.
Which 2 terms are inserted between 5 and 40 so that the resulting sequence is G.P.
Solution:
Let the required numbers be G₁ and G₂.

∴ For the resulting sequence to be in G.P. we need to insert numbers 10 and 20.

Question 28.
If p, q, r are in G.P. and \(\mathrm{p}^{1 / \mathrm{x}}=\mathrm{q}^{1 / \mathrm{y}}=\mathrm{r}^{1 / \mathrm{z}}\), verify whether x, y, z are in A.P. or G.P. or neither.
Solution:

Question 29.
If a, b, c are in G.P. and ax² + 2bx + c = 0 and px² + 2qx + r = 0 have common roots, then verify that pb² – 2qba + ra² = 0.
Solution:
a, b, c are in G.P.
∴ b² = ac
ax² + 2bx + c = 0 becomes

Question 30.
If p, q, r, s are in G.P., show that (p² + q² + r²)(q² + r² + s²) = (pq + qr + rs)².
Solution:
p, q, r, s are in G.P.

Question 31.
If p, q, r, s are in G.P., show that (pⁿ + qⁿ), (qⁿ + rⁿ), (rⁿ + sⁿ) are also in G.P.
Solution:
p, q, r, s are in G.P.
Let the common ratio be R
∴ let p = \(\frac{a}{R^{3}}\), q = \(\frac{a}{R}\), r = aR and s = aR³
To show that (pⁿ + qⁿ), (qⁿ + rⁿ), (rⁿ + sⁿ) are in G.P,
i.e., we have to show

Question 32.
Find the coefficient x⁶ in the expression of e^2x using series expansion.
Solution:

Question 33.
Find the sum of infinite terms of \(1+\frac{4}{5}+\frac{7}{25}+\frac{10}{125}+\frac{13}{625}+\ldots\)
Solution:

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 6 Circle Ex 6.3 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 6 Circle Ex 6.3

Question 1.
Write the parametric equations of the circles:
(i) x² + y² = 9
(ii) x² + y² + 2x – 4y – 4 = 0
(iii) (x – 3)² + (y + 4)² = 25
Solution:
(i) Given equation of the circle is
x² + y² = 9
⇒ x² + y² = 3²
Comparing this equation with x² + y² = r², we get r = 3
The parametric equations of the circle in terms of θ are
x = r cos θ and y = r sin θ
⇒ x = 3 cos θ and y = 3 sin θ

(ii) Given equation of the circle is
x² + y² + 2x – 4y – 4 = 0
⇒ x² + 2x + y² – 4y – 4 = 0
⇒ x² + 2x + 1 – 1 + y² – 4y + 4 – 4 – 4 = 0
⇒ (x² + 2x + 1 ) + (y² – 4y + 4) – 9 = 0
⇒ (x + 1)² + (y – 2)² = 9
⇒ (x + 1)² + (y – 2)² = 3²
Comparing this equation with (x – h)² + (y – k)² = r², we get
h = -1, k = 2 and r = 3
The parametric equations of the circle in terms of θ are
x = h + r cos θ and y = k + r sin θ
⇒ x = -1 + 3 cos θ and y = 2 + 3 sin θ

(iii) Given equation of the circle is
(x – 3)² + (y + 4)² = 25
⇒ (x – 3)² + (y + 4)² = 5²
Comparing this equation with (x – h)² + (y – k)² = r², we get
h = 3, k = -4 and r = 5
The parametric equations of the circle in terms of θ are
x = h + r cos θ and y = k + r sin θ
⇒ x = 3 + 5 cos θ and y = -4 + 5 sin θ

Question 2.
Find the parametric representation of the circle 3x² + 3y² – 4x + 6y – 4 = 0.
Solution:
Given equation of the circle is 3x² + 3y² – 4x + 6y – 4 = 0
Dividing throughout by 3, we get

Comparing this equation with (x – h)² + (y – k)² = r², we get
h = \(\frac{2}{3}\), k = -1 and r = \(\frac{5}{3}\)
The parametric representation of the circle in terms of θ are
x = h + r cos θ and y = k + r sin θ
⇒ x = \(\frac{2}{3}\) + \(\frac{5}{3}\) cos θ and y = -1 + \(\frac{5}{3}\) sin θ

Question 3.
Find the equation of a tangent to the circle x² + y² – 3x + 2y = 0 at the origin.
Solution:
Given equation of the circle is x² + y² – 3x + 2y = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = -3, 2f = 2, c = 0
⇒ g = \(-\frac{3}{2}\), f = 1, c = 0
The equation of a tangent to the circle
x² + y² + 2gx + 2fy + c = 0 at (x₁, y₁) is xx₁ +yy₁ + g(x + x₁) + f(y + y₁) + c = 0
The equation of the tangent at (0, 0) is
x(0) + y(0) + (\(-\frac{3}{2}\)) (x + 0) + 1(y + 0) + 0 = 0
⇒ \(-\frac{3}{2}\)x + y = 0
⇒ 3x – 2y = 0

Question 4.
Show that the line 7x – 3y – 1 = 0 touches the circle x² + y² + 5x – 7y + 4 = 0 at point (1, 2).
Solution:
Given equation of the circle is x² + y² + 5x – 7y + 4 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = 5, 2f = -7, c = 4
⇒ g = \(\frac{5}{2}\), f = \(\frac{-7}{2}\), c = 4
The equation of a tangent to the circle x² + y² + 2gx + 2fy + c = 0 at (x₁, y₁) is
xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
The equation of the tangent at (1, 2) is

7x – 3y – 1 = 0, which is same as the given line.
The line 7x – 3y – 1 = 0 touches the given circle at (1, 2).

Question 5.
Find the equation of tangent to the circle x² + y² – 4x + 3y + 2 = 0 at the point (4, -2).
Solution:
Given equation of the circle is x² + y² – 4x + 3y + 2 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = -4, 2f = 3, c = 2
g = -2, f = \(\frac{3}{2}\), c = 2
The equation of a tangent to the circle x² + y² + 2gx + 2fy + c = 0 at (x₁, y₁) is
xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
The equation of the tangent at (4, -2) is
x(4) + y(-2) – 2(x + 4) + \(\frac{3}{2}\)(y – 2) + 2 = 0
⇒ 4x – 2y – 2x – 8 + \(\frac{3}{2}\) y – 3 + 2 = 0
⇒ 2x – \(\frac{1}{2}\)y – 9 = 0
⇒ 4x – y – 18 = 0

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 8 Measures of Dispersion Ex 8.2 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.2

Question 1.
Find variance and S.D. for the following set of numbers.
7, 11, 2, 4, 9, 6, 3, 7, 11, 2, 5, 8, 3, 6, 8, 8, 2, 6
Solution:
Given data:
7, 11, 2, 4, 9, 6, 3, 7, 11, 2, 5, 8, 3, 6, 8, 8, 2, 6
The tabulated form of the above data is

Calculation of variance and S.D.

Question 2.
Find variance and S.D. for the following set of numbers.
65, 77, 81, 98, 100, 80, 129
Solution:

Question 3.
Compute variance and standard deviation for the following data:

Solution:

Question 4.
Compute the variance and S.D.

Solution:
Let u = \(\frac{x-\mathrm{A}}{\mathrm{h}}=\frac{x-34}{1}\)
Calculation of variance of u:

Question 5.
Following data gives ages of 100 students in a college. Calculate variance and S.D.

Solution:
Let u = \(\frac{x-\mathrm{A}}{\mathrm{h}}=\frac{x-19}{1}\)

Question 6.
Find mean, variance and S.D. of the following data.

Solution:


Alternate Method:
Let u = \(\frac{x-\mathrm{A}}{\mathrm{h}}=\frac{x-55}{10}\)
Calculation of variance of u:

Question 7.
Find the variance and S.D. of the following frequency distribution which gives the distribution of 200 plants according to their height.

Solution:
Since data is not continuous, we have to make it continuous.
Let u = \(\frac{x-\mathrm{A}}{\mathrm{h}}=\frac{x-31}{5}\)
Calculation of variance of u:

Question 8.
The mean of 5 observations is 4.8 and the variance is 6.56. If three of the five observations are 1, 3, and 8, find the other two observations.
Solution:
\(\bar{x}\) = 4.8, Var (X) = 6.56, n = 5, x₁ = 1, x₂ = 3, x₃ = 8 ……(given)
Let the remaining two observations be x₄ and x₅.


From (i), we get
x₅ = 5 or x₅ = 7
∴ The two numbers are 5 and 7.

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 9 Probability Ex 9.1 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 9 Probability Ex 9.1

Question 1.
There are four pens: Red, Green, Blue, and Purple in a desk drawer of which two pens are selected at random one after the other with replacement. State the sample space and the following events.
(a) A : Select at least one red pen.
(b) B : Two pens of the same colour are not selected.
Solution:
The drawer contains 4 pens out of which one is red (R), one is green (G), one is blue (B) and the other one is purple (P).
From this drawer, two pens are selected one after the other with replacement.
∴ The sample space S is given by
S = {RR, RG, RB, RP, GR, GG, GB, GP, BR, BG, BB, BP, PR, PG, PB, PP}
(a) A : Select at least one red pen.
At least one means one or more than one.
∴ A = {RR, RG, RB, RP, GR, BR, PR}

(b) B : Two pens of the same colour are not selected.
B = {RG, RB, RP, GR, GB, GP, BR, BG, BP, PR, PG, PB}

Question 2.
A coin and a die are tossed simultaneously. Enumerate the sample space and the following events.
(a) A : Getting a tail and an odd number.
(b) B : Getting a prime number.
(c) C : Getting head and a perfect square.
Solution:
When a coin and a die are tossed simultaneously, the sample space S is given by
S = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}
(a) A : Getting a tail and an odd number.
∴ A = {(T, 1), (T, 3), (T, 5)}

(b) B : Getting a prime number.
∴ B = {(H, 2), (H, 3), (H, 5), (T, 2), (T, 3), (T, 5)}

(c) C : Getting a head and a perfect square.
∴ C = {(H, 1), (H, 4)}

Question 3.
Find n(S) for each of the following random experiments.
(a) From an urn containing 5 gold and 3 silver coins, 3 coins are drawn at random.
(b) 5 letters are to be placed into 5 envelopes such that no envelope is empty.
(c) 6 books of different subjects are arranged on a shelf.
(d) 3 tickets are drawn from a box containing 20 lottery tickets.
Solution:
(a) There are 5 gold and 3 silver coins, i.e., 8 coins.
3 coins can be drawn from these 8 coins in \({ }^{8} \mathrm{C}_{3}\) ways.
∴ n(s) = \({ }^{8} \mathrm{C}_{3}=\frac{8 !}{5 ! 3 !}=\frac{8 \times 7 \times 6 \times 5 !}{5 ! \times 3 \times 2 \times 1}=56\)

(b) 5 letters have to be placed in 5 envelopes in such a way that no envelope is empty.
∴ The first letter can be placed into 5 envelopes in 5 different ways, the second letter in 4 ways.
Similarly, the third, fourth and fifth letters can be placed in 3 ways, 2 ways and 1 way, respectively.
∴ Total number of ways = 5!
= 5 × 4 × 3 × 2 × 1
= 120
∴ n(S) = 120

(c) 6 books can be arranged on a shelf in \({ }^{6} \mathrm{P}_{6}\) = 6! ways.
∴ n(S) = 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720

(d) 3 tickets are drawn at random from 20 tickets.
∴ 3 tickets can be selected in \({ }^{20} \mathrm{C}_{3}\) ways.
∴ n(S) = \({ }^{20} \mathrm{C}_{3}=\frac{20 !}{17 ! 3 !}=\frac{20 \times 19 \times 18 \times 17 !}{17 ! \times 3 \times 2 \times 1}\) = 1140

Question 4.
Two fair dice are thrown. State the sample space and write the favourable outcomes for the following events.
(a) A : Sum of numbers on two dice is divisible by 3 or 4.
(b) B : The sum of numbers on two dice is 7.
(c) C : Odd number on the first die.
(d) D : Even number on the first die.
(e) Check whether events A and B are mutually exclusive and exhaustive.
(f) Check whether events C and D are mutually exclusive and exhaustive.
Solution:
When two dice are thrown, the sample space is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),
(6, 1), (6, 2), (6, 3), (6,4), (6, 5), (6, 6)}
∴ n(S) = 36
(a) A: Sum of the numbers on two dice is divisible by 3 or 4.
∴ A = {(1, 2), (1, 3), (1, 5), (2, 1), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (3, 6), (4, 2), (4, 4), (4, 5), (5, 1), (5, 3), (5, 4), (6, 2), (6, 3), (6, 6)}

(b) B: Sum of the numbers on two dice is 7.
∴ B = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}

(c) C: Odd number on the first die.
∴ C = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}

(d) D: Even number on the first die.
∴ D = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

(e) A and B are mutually exclusive events as A ∩ B = Φ.
A ∪ B = {(1, 2), (1, 3), (1, 5), (1, 6), (2, 1), (2, 2), (2, 4), (2, 5), (2, 6), (3, 1), (3, 3), (3, 4), (3, 5), (3, 6), (4, 2), (4, 3), (4, 4), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (6, 1), (6, 2), (6, 3), (6, 6)} ≠ S
∴ A and B are not exhaustive events as A ∪ B ≠ S.

(f) C and D are mutually exclusive events as C ∩ D = Φ.
C ∪ D = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} = S
∴ C and D are exhaustive events.

Question 5.
A bag contains four cards marked as 5, 6, 7, and 8. Find the sample space if two cards are drawn at random
(a) with replacement.
(b) without replacement.
Solution:
The bag contains 4 cards marked 5, 6, 7, and 8. Two cards are to be drawn from this bag.
(a) If the two cards are drawn with replacement, then the sample space is
S = {(5, 5), (5, 6), (5, 7), (5, 8), (6, 5), (6, 6), (6, 7), (6, 8), (7, 5), (7, 6), (7, 7), (7, 8), (8, 5), (8, 6), (8, 7), (8, 8)}

(b) If the two cards are drawn without replacement, then the sample space is
S = {(5, 6), (5, 7), (5, 8), (6, 5), (6, 7), (6, 8), (7, 5), (7, 6), (7, 8), (8, 5), (8, 6), (8, 7)}

Question 6.
A fair die is thrown two times. Find the probability that
(a) the sum of the numbers on them is 5.
(b) the sum of the numbers on them is at least 8.
(c) the first throw gives a multiple of 2 and the second throw gives a multiple of 3.
(d) product of numbers on them is 12.
Solution:
When two dice are thrown, the sample space is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6,4), (6, 5), (6, 6)}
∴ n(S) = 36
(a) Let event A: Sum of the numbers on uppermost face is 5.
∴ A = {(1, 4), (2, 3), (3, 2), (4, 1)}
∴ n(A) = 4
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{4}{36}=\frac{1}{9}\)

(b) Let event B: Sum of the numbers on uppermost face is at least 8 (i.e., 8 or more than 8)
∴ B = {(2, 6), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6), (5, 3), (5, 4), (5, 5), (5, 6), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
∴ n(B) = 15
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{15}{36}=\frac{5}{12}\)

(c) Let event C: First throw gives a multiple of 2 and second throw gives a multiple of 3.
∴ C = {(2, 3), (2, 6), (4, 3), (4, 6), (6, 3), (6, 6)}
∴ n(C) = 6
∴ P(C) = \(\frac{n(C)}{n(S)}=\frac{6}{36}=\frac{1}{6}\)

(d) Let event D: The product of the numbers on uppermost face is 12.
∴ D = {(2, 6), (3, 4), (4, 3), (6, 2)}
∴ n(D) = 4
∴ P(D) = \(\frac{n(D)}{n(S)}=\frac{4}{36}=\frac{1}{9}\)

Question 7.
Two cards are drawn from a pack of 52 cards. Find the probability that
(a) one is a face card and the other is an ace card.
(b) one is a club and the other is a diamond.
(c) both are from the same suit.
(d) both are red cards.
(e) one is a heart card and the other is a non-heart card.
Solution:
Two cards can be drawn from a pack of 52 cards in \({ }^{52} \mathrm{C}_{2}\) ways.
∴ n(S) = \({ }^{52} \mathrm{C}_{2}\)

(a) Let event A: Out of the two cards drawn, one is a face card and the other is an ace card.
There are 12 face cards and 4 ace cards in a pack of 52 cards.
∴ One face card can be drawn from 12 face cards in \({ }^{12} \mathrm{C}_{1}\) ways and one ace card can be drawn from 4 ace cards in \({ }^{4} \mathrm{C}_{1}\) ways.
∴ n(A) = \({ }^{12} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{1}\)
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{12} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{2}}=\frac{12 \times 4}{\frac{52 \times 51}{2 \times 1}}=\frac{8}{221}\)

(b) Let event B: Out of the two cards drawn, one is club and the other is a diamond card.
There are 13 club cards and 13 diamond cards.
∴ One club card can be drawn from 13 club cards in \({ }^{13} \mathrm{C}_{1}\) ways and one diamond card can be drawn from 13 diamond cards in \({ }^{13} \mathrm{C}_{1}\) ways.
∴ n(B) = \({ }^{13} \mathrm{C}_{1} \times{ }^{13} \mathrm{C}_{1}\)
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{{ }^{13} C_{1} \times{ }^{13} C_{1}}{{ }^{52} C_{2}}=\frac{13 \times 13}{\frac{52 \times 51}{2 \times 1}}=\frac{13}{102}\)

(c) Let event C: Both the cards drawn are of the same suit.
A pack of 52 cards consists of 4 suits each containing 13 cards.
∴ 2 cards can be drawn from the same suit in \({ }^{13} \mathrm{C}_{2}\) ways.
∴ n(C) = \({ }^{13} \mathrm{C}_{2} \times 4\)
∴ P(C) = \(\frac{\mathrm{n}(\mathrm{C})}{\mathrm{n}(\mathrm{S})}=\frac{4 \times{ }^{13} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}=\frac{4 \times 13 \times 12}{52 \times 51}=\frac{4}{17}\)

(d) Let event D: Both the cards drawn are red.
There are 26 red cards in the pack of 52 cards.
∴ 2 cards can be drawn from them in \({ }^{26} \mathrm{C}_{2}\) ways.
∴ n(D) = \({ }^{26} \mathrm{C}_{2}\)
∴ P(D) = \(\frac{n(D)}{n(S)}=\frac{{ }^{26} C_{2}}{{ }^{52} C_{2}}=\frac{26 \times 25}{52 \times 51}=\frac{25}{102}\)

(e) Let event E: Out of the two cards drawn, one is heart and other is non-heart.
There are 13 heart cards in a pack of 52 cards, i.e., 39 cards are non-heart.
∴ One heart card can be drawn from 13 hdart cards in \({ }^{13} \mathrm{C}_{1}\) ways and one non-heart card can be drawn from 39 cards in \({ }^{39} \mathrm{C}_{1}\) ways.
∴ n(E) = \({ }^{13} \mathrm{C}_{1} \times{ }^{39} \mathrm{C}_{1}\)
∴ P(E) = \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{13} \mathrm{C}_{1} \times{ }^{39} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{2}}=\frac{13 \times 39}{\frac{52 \times 51}{2 \times 1}}=\frac{13}{34}\)

Question 8.
Three cards are drawn from a pack of 52 cards. Find the chance that
(a) two are queen cards and one is an ace card.
(b) at least one is a diamond card.
(c) all are from the same suit.
(d) they are a king, a queen, and a jack.
Solution:
3 cards can be drawn from a pack of 52 cards in \({ }^{52} \mathrm{C}_{3}\) ways.
∴ n(S) = \({ }^{52} \mathrm{C}_{3}\)

(a) Let event A: Out of the three cards drawn, 2 are queens and 1 is an ace card.
There are 4 queens and 4 aces in a pack of 52 cards.
∴ 2 queens can be drawn from 4 queens in \({ }^{4} \mathrm{C}_{2}\) ways and 1 ace can be drawn out of 4 aces in \({ }^{4} \mathrm{C}_{1}\) ways.

(b) Let event B: Out of the three cards drawn, at least one is a diamond.
∴ B’ is the event that all 3 cards drawn are non-diamond cards.
In a pack of 52 cards, there are 39 non-diamond cards.
∴ 3 non-diamond cards can be drawn in \({ }^{39} \mathrm{C}_{3}\) ways.

(c) Let event C: All the cards drawn are from the same suit.
A pack of 52 cards consists of 4 suits each containing 13 cards.
∴ 3 cards can be drawn from the same suit in \({ }^{13} \mathrm{C}_{3}\) ways.

(d) Let event D: The cards drawn are a king, a queen, and a jack.
There are 4 kings, 4 queens and 4 jacks in a pack of 52 cards.
∴ 1 king can be drawn from 4 kings in \({ }^{4} \mathrm{C}_{1}\) ways,
1 queen can be drawn from 4 queens in \({ }^{4} \mathrm{C}_{1}\) ways and
1 jack can be drawn from 4 jacks in \({ }^{4} \mathrm{C}_{1}\) ways.

Question 9.
From a bag containing 10 red, 4 blue, and 6 black balls, a ball is drawn at random. Find the probability of drawing
(a) a red bail.
(b) a blue or black ball.
(c) not a black ball.
Solution:
The bag contains 10 red, 4 blue, and 6 black balls,
i.e., 10 + 4 + 6 = 20 balls.
One ball can be drawn from 20 balls in \({ }^{20} \mathrm{C}_{1}\) ways.
∴ n(S) = \({ }^{20} \mathrm{C}_{1}\) = 20

(a) Let event A: Ball drawn is red.
There are total 10 red balls.
∴ 1 red ball can be drawn from 10 red balls in \({ }^{10} \mathrm{C}_{1}\) ways.
∴ n(A) = \({ }^{10} \mathrm{C}_{1}\) = 10
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{10}{20}=\frac{1}{2}\)

(b) Let event B: The ball drawn is blue or black.
There are 4 blue and 6 black balls.
∴ 1 blue ball can be drawn from 4 blue balls in \({ }^{4} \mathrm{C}_{1}\) ways
or 1 black ball can be drawn from 6 black balls in \({ }^{6} \mathrm{C}_{1}\) ways.
∴ n(B) = \({ }^{4} \mathrm{C}_{1}\) + \({ }^{6} \mathrm{C}_{1}\) = 4 + 6 = 10
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{10}{20}=\frac{1}{2}\)

(c) Let event C: Ball drawn is not black,
i.e., ball drawn is red or blue.
There are total 14 red and blue balls.
∴ 1 ball can be drawn from 14 balls in \({ }^{14} \mathrm{C}_{1}\) ways.
∴ n(C) = \({ }^{14} \mathrm{C}_{1}\) = 14
∴ P(C) = \(\frac{n(C)}{n(S)}=\frac{14}{20}=\frac{7}{10}\)

Question 10.
A box contains 75 tickets numbered 1 to 75. A ticket is drawn at random from the box. Find the probability that,
(a) number on the ticket is divisible by 6.
(b) the number on the ticket is a perfect square.
(c) the number on the ticket is prime.
(d) the number on the ticket is divisible by 3 and 5.
Solution:
The box contains 75 tickets numbered 1 to 75.
∴ 1 ticket can be drawn from the box in \({ }^{75} \mathrm{C}_{1}\) = 75 ways.
∴ n(S) = 75

(a) Let event A: Number on the ticket is divisible by 6.
∴ A = {6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72}
∴ n(A) = 12
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{12}{75}=\frac{4}{25}\)

(b) Let event B: Number on the ticket is a perfect square.
∴ B = (1, 4, 9, 16, 25, 36, 49, 64}
∴ n(B) = 8
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{8}{75}\)

(c) Let event C: Number on the ticket is a prime number.
∴C = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73}

(d) Let event D: Number on the ticket is divisible by 3 and 5,
i.e., divisible by L.C.M. of 3 and 5,
i.e., 15.
∴D = {15, 30, 45, 60, 75}
∴ n(D) = 5
∴ P(D) = \(\frac{n(D)}{n(S)}=\frac{5}{75}=\frac{1}{15}\)

Question 11.
What is the chance that a leap year, selected at random, will contain 53 Sundays?
Solution:
A leap year consists of 366 days.
It has 52 complete weeks and two more days.
These two days can be {(Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thur), (Thur, Fri), (Fri, Sat), (Sat, Sun)}.
∴ n(S) = 7
Let event E : There are 53 Sundays.
∴ E = {(Sun, Mon), (Sat, Sun)}
∴ n(E) = 2
∴ P(E) = \(\frac{n(E)}{n(S)}=\frac{2}{7}\)

Question 12.
Find the probability of getting both red balls, when from a bag containing 5 red and 4 black balls, two balls are drawn,
(i) with replacement
(ii) without replacement
Solution:
The bag contains 5 red and 4 black balls,
i.e., 5 + 4 = 9 balls.
(i) 2 balls can be drawn from 9 balls with replacement in \({ }^{9} \mathrm{C}_{1} \times{ }^{9} \mathrm{C}_{1}\) ways.
∴ n(S) = \({ }^{9} \mathrm{C}_{1} \times{ }^{9} \mathrm{C}_{1}\) = 9 × 9 = 81
Let event A: Balls drawn are red.
2 red balls can be drawn from 5 red balls with replacement in \({ }^{5} \mathrm{C}_{1} \times{ }^{5} \mathrm{C}_{1}\) ways.
∴ n(A) = \({ }^{5} \mathrm{C}_{1} \times{ }^{5} \mathrm{C}_{1}\) = 5 × 5 = 25
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{25}{81}\)

(ii) 2 balls can be drawn from 9 balls without replacement in \({ }^{9} C_{1} \times{ }^{8} C_{1}\) ways.
∴ n(S) = \({ }^{9} C_{1} \times{ }^{8} C_{1}\) = 9 × 8 = 72
2 red balls can be drawn from 5 red balls without replacement in \({ }^{5} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{1}\) ways.
∴ n(B) = \({ }^{5} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{1}\) = 5 × 4 = 20
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{20}{72}=\frac{5}{18}\)

Question 13.
A room has three sockets for lamps. From a collection of 10 bulbs of which 6 are defective. At night a person selects 3 bulbs, at random and puts them in sockets. What is the probability that
(i) room is still dark.
(ii) the room is lit.
Solution:
Total number of bulbs = 10
Number of defective bulbs = 6
∴ Number of non-defective bulbs = 4
3 bulbs can be selected out of 10 bulbs in \({ }^{10} \mathrm{C}_{3}\) ways.
∴ n(S) = \({ }^{10} \mathrm{C}_{3}\)

(i) Let event A: The room is dark.
For event A to happen the bulbs should be selected from the 6 defective bulbs. This can be done in \({ }^{6} \mathrm{C}_{3}\) ways.
∴ n(A) = \({ }^{6} \mathrm{C}_{3}\)
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{{ }^{6} C_{3}}{{ }^{10} C_{3}}=\frac{6 \times 5 \times 4}{10 \times 9 \times 8}=\frac{1}{6}\)

(ii) Let event A’: The room is lit.
∴ P(Room is lit) = 1 – P(Room is not lit)
∴ P(A’) = 1 – P(A) = 1 – \(\frac{1}{6}\) = \(\frac{5}{6}\)

Question 14.
Letters of the word MOTHER are arranged at random. Find the probability that in the arrangement
(a) vowels are always together.
(b) vowels are never together.
(c) O is at the beginning and end with T.
(d) starting with a vowel and ending with a consonant.
Solution:
There are 6 letters in the word MOTHER.
These letters can be arranged among themselves in \({ }^{6} \mathrm{P}_{6}\) = 6! ways.
∴ n(S) = 6!
(a) Let event A: Vowels are always together.
The word MOTHER consists of 2 vowels (O, E) and 4 consonants (M, T, H, R).
2 vowels can be arranged among themselves in \({ }^{2} \mathbf{P}_{2}\) = 2! ways.
Let us consider 2 vowels as one group.
This one group with 4 consonants can be arranged in \({ }^{5} \mathrm{P}_{5}\) = 5! ways.
∴ n(A) = 2! × 5!
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{2 ! \times 5 !}{6 !}=\frac{1}{3}\)

(b) Let event B: Vowels are never together.
4 consonants create 5 gaps, in which vowels are arranged.
Consider the following arrangement of consonants
_C_C_C_C_
2 vowels can be arranged in 5 gaps in \({ }^{5} \mathbf{P}_{2}\) ways.
Also 4 consonants can be arranged among themselves in \({ }^{4} \mathbf{P}_{4}\) = 4! ways.
∴ n(B) = 4! × \({ }^{5} \mathbf{P}_{2}\)
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{4 ! \times{ }^{5} P_{2}}{6 !}=\frac{4 ! \times 5 \times 4}{6 \times 5 \times 4 !}=\frac{2}{3}\)

(iii) Let event C: Word begin with O and end with T.
Thus first and last letters can be arranged in one way each and the remaining 4 letters can be arranged in \({ }^{4} \mathrm{P}_{4}\) = 4! ways
∴ n(C) = 4! × 1 × 1 = 4!
∴ P(C) = \(\frac{n(C)}{n(S)}=\frac{4 !}{6 !}=\frac{1}{30}\)

(d) Let event D: Word starts with a vowel and ends with a consonant.
There are 2 vowels and 4 consonants in the word MOTHER.
∴ The first place can be arranged in 2 different ways and the last place can be arranged in 4 different ways.
Now, the remaining 4 letters (3 consonants and 1 vowel) can be arranged in \({ }^{4} \mathrm{P}_{4}\) = 4! ways.
∴ n(D) = 2 × 4 × 4!
∴ P(D) = \(\frac{n(D)}{n(S)}=\frac{2 \times 4 \times 4 !}{6 !}=\frac{4}{15}\)

Question 15.
4 letters are to be posted in 4 post boxes. If any number of letters can be posted in any of the 4 post boxes, what is the probability that each box contains only one letter?
Solution:
There are 4 letters and 4 post boxes.
Since any number of letters can be posted in all 4 post boxes,
so each letter can be posted in different ways.
∴ n(S) = 4 × 4 × 4 × 4
Let event A: Each box contains only one letter.
∴ 1st letter can be posted in 4 different ways.
Since each box contains only one letter, 2nd letter can be posted in 3 different ways.
Similarly, 3rd and 4th letters can be posted in 2 different ways and 1 way respectively.
∴ n(A) = 4 × 3 × 2 × 1
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{4 \times 3 \times 2 \times 1}{4 \times 4 \times 4 \times 4}=\frac{3}{32}\)

Question 16.
15 professors have been invited for a round table conference by the Vice-chancellor of a university. What is the probability that two particular professors occupy the seats on either side of the Vice-chancellor during the conference?
Solution:
Since a Vice-chancellor invited 15 professors for a round table conference, there were all 16 persons in the conference.
These 16 persons can be arranged among themselves around a round table in (16 – 1)! = 15! ways.
∴ n(S) = 15!
Let event A: Two particular professors be seated on either side of the Vice-chancellor.
Those two particular persons sit on either side of a Vice chancellor in \({ }^{2} \mathrm{P}_{2}\) = 2! ways.
Thus the remaining 13 persons can be arranged in \({ }^{13} \mathrm{P}_{13}\) = 13! ways.
∴ n(A) = 13! 2!
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{13 ! \times 2 !}{15 !}=\frac{13 ! \times 2 \times 1}{15 \times 14 \times 13 !}=\frac{1}{105}\)

Question 17.
A bag contains 7 black and 4 red balls. If 3 balls are drawn at random, find the probability that
(i) all are black.
(ii) one is black and two are red.
Solution:
The bag contains 7 black and 4 red balls,
i.e., 7 + 4 = 11 balls.
∴ 3 balls can be drawn out of 11 balls in \({ }^{11} \mathrm{C}_{3}\) ways.
∴ n(S) = \({ }^{11} \mathrm{C}_{3}\)
(i) Let event A: All 3 balls drawn are black.
There are 7 black balls.
∴ 3 black balls can be drawn from 7 black balls in \({ }^{7} \mathrm{C}_{3}\) ways.
∴ n(A) = \({ }^{7} \mathrm{C}_{3}\)
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{7} \mathrm{C}_{3}}{{ }^{11} \mathrm{C}_{3}}=\frac{7 \times 6 \times 5}{11 \times 10 \times 9}=\frac{7}{33}\)

(ii) Let event B: Out of 3 balls drawn, one is black and two are red.
There are 7 black and 4 red balls.
∴ One black ball can be drawn from 7 black balls in \({ }^{7} \mathrm{C}_{1}\) ways and 2 red balls can be drawn from 4 red balls in \({ }^{4} \mathrm{C}_{2}\) ways.
∴ n(A) = \({ }^{7} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{2}\)
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{{ }^{7} C_{1} \times{ }^{4} C_{2}}{{ }^{11} C_{3}}=\frac{7 \times \frac{4 \times 3}{2}}{\frac{11 \times 10 \times 9}{3 \times 2 \times 1}}=\frac{14}{55}\)

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 8 Measures of Dispersion Ex 8.1 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 8 Measures of Dispersion Ex 8.1

Question 1.
Find a range of the following data:
19, 27, 15, 21, 33, 45, 7, 12, 20, 26
Solution:
Here, largest value (L) = 45, smallest value (S) = 7
∴ Range = L – S = 45 – 7 = 38

Question 2.
Find range of the following data:
575, 609, 335, 280, 729, 544, 852, 427, 967, 250
Solution:
Here, largest value (L) = 967, smallest value (S) = 250
∴ Range = L – S = 967 – 250 = 717

Question 3.
The following data gives a number of typing mistakes done by Radha during a week. Find the range of the data.

Solution:
Here, largest value (L) = 21, smallest value (S) = 10
∴ Range = L – S = 21 – 10 = 11

Question 4.
The following results were obtained by rolling a die 25 times. Find the range of the data.

Solution:
Here, largest value (L) = 6, smallest value (S) = 1
∴ Range = L – S = 6 – 1 = 5

Question 5.
Find range for the following data:

Solution:
Here, upper limit of the highest class (L) = 72
lower limit of the lowest class (S) = 62
∴ Range = L – S = 72 – 62 = 10

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 7 Conic Sections Ex 7.2 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.2

Question 1.
Find the
(i) lengths of the principal axes
(ii) co-ordinates of the foci
(iii) equations of directrices
(iv) length of the latus rectum
(v) distance between foci
(vi) distance between directrices of the ellipse:
(a) \(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\)
(b) 3x² + 4y² = 12
(c) 2x² + 6y² = 6
(d) 3x² + 4y² = 1
Solution:
(a) Given equation of the ellipse is \(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a² = 25 and b² = 9
a = 5 and b = 3
Since a > b,
X-axis is the major axis and Y-axis is the minor axis.

(i) Length of major axis = 2a = 2(5) = 10
Length of minor axis = 2b = 2(3) = 6
Lengths of the principal axes are 10 and 6.

(ii) We know that e = \(\frac{\sqrt{a^{2}-b^{2}}}{a}\)
= \(\frac{\sqrt{25-9}}{5}\)
= \(\frac{\sqrt{16}}{5}\)
= \(\frac{4}{5}\)
Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0),
i.e., S(5(\(\frac{4}{5}\)), 0) and S'(-5(\(\frac{4}{5}\)), 0)
i.e., S(4, 0) and S'(-4, 0)

(iii) Equations of the directrices are x = ±\(\frac{\mathrm{a}}{\mathrm{e}}\)
= \(\pm \frac{5}{\frac{4}{5}}\)
= \(\pm \frac{25}{4}\)

(iv) Length of latus rectum = \(\frac{2 b^{2}}{a}=\frac{2(3)^{2}}{5}=\frac{18}{5}\)

(v) Distance between foci = 2ae
= 2(5)(\(\frac{4}{5}\))
= 8

(vi) Distance between directrices = \(\frac{2 \mathrm{a}}{\mathrm{e}}\)
= \(\frac{2(5)}{\frac{4}{5}}\)
= \(\frac{25}{2}\)

(b) Given equation of the ellipse is 3x² + 4y² = 12
\(\frac{x^{2}}{4}+\frac{y^{2}}{3}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a² = 4 and b² = 3
a = 2 and b = √3
Since a > b,
X-axis is the major axis and Y-axis is the minor axis.

(i) Length of major axis = 2a = 2(2) = 4
Length of minor axis = 2b = 2√3
Lengths of the principal axes are 4 and 2√3.

(ii) We know that e = \(\frac{\sqrt{a^{2}-b^{2}}}{a}\)
= \(\frac{\sqrt{4-3}}{2}\)
= \(\frac{1}{2}\)
Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0),
i.e., S(2(\(\frac{1}{2}\)), 0) and S'(-2(\(\frac{1}{2}\)), 0)
i.e., S(1, 0) and S'(-1, 0)

(iii) Equations of the directrices are x = ±\(\frac{\mathrm{a}}{\mathrm{e}}\)
= \(\pm \frac{2}{\frac{1}{2}}\)
= ±4

(iv) Length of latus rectum = \(\frac{2 b^{2}}{a}=\frac{2(\sqrt{3})^{2}}{2}=3\)

(v) Distance between foci = 2ae = 2(2)(\(\frac{1}{2}\)) = 2

(vi) Distance between directrices = \(\frac{2 \mathrm{a}}{\mathrm{e}}\)
= \(\frac{2(2)}{\frac{1}{2}}\)
= 8

(c) Given equation of the ellipse is 2x² + 6y² = 6
\(\frac{x^{2}}{3}+\frac{y^{2}}{1}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a² = 3 and b² = 1
a = √3 and b = 1
Since a > b,
X-axis is the major axis and Y-axis is the minor axis.

(i) Length of major axis = 2a = 2√3
Length of minor axis = 2b = 2(1) = 2
Lengths of the principal axes are 2√3 and 2.

(ii) We know that e = \(\frac{\sqrt{a^{2}-b^{2}}}{a}\)
= \(\frac{\sqrt{3-1}}{\sqrt{3}}\)
= \(\frac{\sqrt{2}}{\sqrt{3}}\)
Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0),
i.e., S(√3(\(\frac{\sqrt{2}}{\sqrt{3}}\)), o) and S'(-√3(\(\frac{\sqrt{2}}{\sqrt{3}}\)), 0)
i.e., S(√2, 0) and S'(-√2, 0)

(iii) Equations of the directrices are x = ±\(\frac{a}{e}\),
= \(\pm \frac{\sqrt{3}}{\frac{\sqrt{2}}{\sqrt{3}}}\)
= \(\pm \frac{3}{\sqrt{2}}\)

(iv) Length of latus rectum = \(\frac{2 b^{2}}{a}=\frac{2(1)^{2}}{\sqrt{3}}=\frac{2}{\sqrt{3}}\)

(v) Distance between foci = 2ae
= \(2(\sqrt{3})\left(\frac{\sqrt{2}}{\sqrt{3}}\right)\)
= 2√2

(vi) Distance between directrices = \(\frac{2 \mathrm{a}}{\mathrm{e}}\)
= \(\frac{2 \sqrt{3}}{\frac{\sqrt{2}}{\sqrt{3}}}\)
= \(\frac{2 \times 3}{\sqrt{2}}\)
= 3√2

(d) Given equation of the ellipse is 3x² + 4y = 1.
\(\frac{x^{2}}{\frac{1}{3}}+\frac{y^{2}}{\frac{1}{4}}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a² = \(\frac{1}{3}\) and b² = \(\frac{1}{4}\)
a = \(\frac{1}{\sqrt{3}}\) and b = \(\frac{1}{2}\)
Since a > b,
X-axis is the major axis and Y-axis is the minor axis.

(i) Length of major axis = 2a = 2(\(\frac{1}{\sqrt{3}}\)) = \(\frac{2}{\sqrt{3}}\)
Length of minor axis = 2b = 2(\(\frac{1}{2}\)) = 1
Lengths of the principal axes are \(\frac{2}{\sqrt{3}}\) and 1.

(ii) We know that e = \(\frac{\sqrt{a^{2}-b^{2}}}{a}\)
e = \(\frac{\sqrt{\frac{1}{3}-\frac{1}{4}}}{\frac{1}{\sqrt{3}}}=\frac{\sqrt{\frac{1}{12}}}{\frac{1}{\sqrt{3}}}=\sqrt{\frac{3}{12}}=\sqrt{\frac{1}{4}}=\frac{1}{2}\)
Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0),
i.e., S\(\left(\frac{1}{\sqrt{3}}\left(\frac{1}{2}\right), 0\right)\) and S’\(\left(-\frac{1}{\sqrt{3}}\left(\frac{1}{2}\right), 0\right)\)
i.e., S(\(\frac{1}{2 \sqrt{3}}\), 0) and S'(-\(\frac{1}{2 \sqrt{3}}\), 0)

(iii) Equations of the directrices are x = ±\(\frac{a}{e}\),
= \(\pm \frac{\frac{1}{\sqrt{3}}}{\frac{1}{2}}\)
= \(\pm \frac{2}{\sqrt{3}}\)

(iv) Length of latus rectum = \(\frac{2 b^{2}}{a}\)
= \(\frac{2\left(\frac{1}{2}\right)^{2}}{\frac{1}{\sqrt{3}}}\)
= \(\frac{\sqrt{3}}{2}\)

(v) Distance between foci = 2ae
= \(2\left(\frac{1}{\sqrt{3}}\right)\left(\frac{1}{2}\right)\)
= \(\frac{1}{\sqrt{3}}\)

(vi) Distance between directrices = \(\frac{2 a}{e}\)
= \(\frac{2\left(\frac{1}{\sqrt{3}}\right)}{\frac{1}{2}}\)
= \(\frac{4}{\sqrt{3}}\)

Question 2.
Find the equation of the ellipse in standard form if
(i) eccentricity = \(\frac{3}{8}\) and distance between its foci = 6.
(ii) the length of the major axis is 10 and the distance between foci is 8.
(iii) distance between directrices is 18 and eccentricity is \(\frac{1}{3}\).
(iv) minor axis is 16 and eccentricity is \(\frac{1}{3}\).
(v) the distance between foci is 6 and the distance between directrices is \(\frac{50}{3}\).
(vi) the latus rectum has length 6 and foci are (±2, 0).
(vii) passing through the points (-3, 1) and (2, -2).
(viii) the distance between its directrices is 10 and which passes through (-√5, 2).
(ix) eccentricity is \(\frac{2}{3}\) and passes through (2, \(\frac{-5}{3}\)).
Solution:
(i) Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.
Given, eccentricity (e) = \(\frac{3}{8}\)
Distance between foci = 2ae
Given, distance between foci = 6
2ae = 6

The required equation of ellipse is \(\frac{x^{2}}{64}+\frac{y^{2}}{55}=1\).

(ii) Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.
Length of major axis = 2a
Given, length of major axis = 10
2a = 10
a = 5
a² = 25
Distance between foci = 2ae
Given, distance between foci = 8
2ae = 8
2(5)e = 8

The required equation of ellipse is \(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\).

(iii) Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.
Given, eccentricity (e) = \(\frac{1}{3}\)
Distance between directrices = \(\frac{2a}{e}\)
Given, distance between directrices = 18

The required equation of ellipse is \(\frac{x^{2}}{9}+\frac{y^{2}}{8}=1\)

(iv) Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.
Length of minor axis = 2b
Given, length of minor axis = 16
2b = 16
b = 8
b² = 64
Given, eccentricity (e) = \(\frac{1}{3}\)
Now, b² = a² (1 – e²)

The required equation of ellipse is \(\frac{x^{2}}{72}+\frac{y^{2}}{64}=1\).

(v) Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.
Distance between foci = 2ae
Given, distance between foci = 6
2ae = 6
ae = 3
a = \(\frac{3}{e}\) …….(i)
Distance between directrices = \(\frac{2a}{e}\)
Given, distance between directrices = \(\frac{50}{3}\)

The required equation of ellipse is \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\).

(vi) Given, the length of the latus rectum is 6, and co-ordinates of foci are (±2, 0).
The foci of the ellipse are on the X-axis.
Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.
Length of latus rectum = \(\frac{2 b^{2}}{a}\)
\(\frac{2 b^{2}}{a}\) = 6
b² = 3a …..(i)
Co-ordinates of foci are (±ae, 0)
ae = 2
a²e² = 4 …..(ii)
Now, b² = a² (1 – e²)
b² = a² – a² e²
3a = a² – 4 …..[From (i) and (ii)]
a² – 3a – 4 = 0
a² – 4a + a – 4 = 0
a(a – 4) + 1(a – 4) = 0
(a – 4) (a + 1) = 0
a – 4 = 0 or a + 1 = 0
a = 4 or a = -1
Since a = -1 is not possible,
a = 4
a² = 16
Substituting a = 4 in (i), we get
b² = 3(4) = 12
The required equation of ellipse is \(\frac{x^{2}}{16}+\frac{y^{2}}{12}=1\).

(vii) Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.
The ellipse passes through the points (-3, 1) and (2, -2).
Substituting x = -3 and y = 1 in equation of ellipse, we get

Equations (i) and (ii) become
9A + B = 1 …..(iii)
4A + 4B = 1 …..(iv)
Multiplying (iii) by 4, we get
36A + 4B = 4 …..(v)
Subtracting (iv) from (v), we get
32A = 3
A = \(\frac{3}{32}\)
Substituting A = \(\frac{3}{32}\) in (iv), we get
4(\(\frac{3}{32}\)) + 4B = 1
\(\frac{3}{8}\) + 4B = 1
4B = 1 – \(\frac{3}{8}\)
4B = \(\frac{5}{8}\)
B = \(\frac{5}{32}\)

(viii) Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.
Distance between directrices = \(\frac{2 a}{e}\)
Given, distance between directrices = 10
\(\frac{2 a}{e}\) = 10
a = 5e …..(i)
The ellipse passes through (-√5, 2).
Substituting x = -√5 and y = 2 in equation of ellipse, we get

b² = 15(\(\frac{2}{5}\))
b² = 6
The required equation of ellipse is \(\frac{x^{2}}{15}+\frac{y^{2}}{6}=1\).

(ix) Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.
Given, eccentricity (e) = \(\frac{2}{3}\)
The ellipse passes through (2, \(\frac{-5}{3}\)).
Substituting x = 2 and y = \(\frac{-5}{3}\) in equation of ellipse, we get

The required equation of ellipse is \(\frac{x^{2}}{9}+\frac{y^{2}}{5}=1\).

Question 3.
Find the eccentricity of an ellipse, if the length of its latus rectum is one-third of its minor axis.
Solution:
Let the equation of ellipse be \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), where a > b.
Length of latus rectum = \(\frac{2 b^{2}}{a}\)
Length of minor axis = 2b
According to the given condition,
Length of latus rectum = \(\frac{1}{3}\) (Minor axis)

Question 4.
Find the eccentricity of an ellipse, if the distance between its directrices is three times the distance between its foci.
Solution:
Let the required equation of ellipse be \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), where a > b.
Distance between directrices = \(\frac{2 \mathrm{a}}{\mathrm{e}}\)
Distance between foci = 2ae
According to the given condition,
distance between directrices = 3(distance between foci)
\(\frac{2 \mathrm{a}}{\mathrm{e}}\) = 3(2ae)
\(\frac{1}{\mathrm{e}}\) = 3e
\(\frac{1}{3}\) = e²
e = \(\frac{1}{\sqrt{3}}\) ……[∵ 0 < e < 1]
Eccentricity of the ellipse is \(\frac{1}{\sqrt{3}}\)

Question 5.
Show that the product of the lengths of the perpendicular segments drawn from the foci to any tangent line to the ellipse \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\) is equal to 16.
Solution:
Given equation of the ellipse is \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\).
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a² = 25, b² = 16
a = 5, b = 4

Question 6.
A tangent having slope \(\left(-\frac{1}{2}\right)\) the ellipse 3x² + 4y² = 12 intersects the X and Y axes in the points A and B respectively. If O is the origin, find the area of the triangle AOB.
Solution:
Given equation of the ellipse is 3x² + 4y² = 12.
\(\frac{x^{2}}{4}+\frac{y^{2}}{3}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get
a² = 4, b² = 3
Equations of tangents to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) having slope m are
y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\)
Here, m = \(-\frac{1}{2}\)
Equations of the tangents are
y = \(\frac{-1}{2} x \pm \sqrt{4\left(-\frac{1}{2}\right)^{2}+3}=\frac{-1}{2} x \pm 2\)
2y = -x ± 4
x + 2y ± 4 = 0
Consider the tangent x + 2y – 4 = 0
Let this tangent intersect the X-axis at A(x₁, 0) and Y-axis at B(0, y₁).
x₁ + 0 – 4 = 0 and 0 + 2y₁ – 4 = 0
x₁ = 4 and y₁ = 2
A = (4, 0) and B = (0, 2)
l(OA) = 4 and l(OB) = 2

Area of ∆AOB = \(\frac{1}{2}\) × l(OA) × l(OB)
= \(\frac{1}{2}\) × 4 × 2
= 4 sq.units

Question 7.
Show that the line x – y = 5 is a tangent to the ellipse 9x² + 16y² = 144. Find the point of contact.
Solution:
Given equation of the ellipse is 9x² + 16y² = 144
\(\frac{x^{2}}{16}+\frac{y^{2}}{9}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get
a² = 16 and b² = 9
Given equation of line is x – y = 5, i.e., y = x – 5
c² = a² m² + b²
Comparing this equation with y = mx + c, we get
m = 1 and c = -5
For the line y = mx + c to be a tangent to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=\)1, we must have
c² = a² m² + b²
c² = (-5)² = 25
a² m² + b² = 16(1)² + 9 = 16 + 9 = 25 = c²
The given line is a tangent to the given ellipse and point of contact
= \(\left(\frac{-\mathrm{a}^{2} \mathrm{~m}}{\mathrm{c}}, \frac{\mathrm{b}^{2}}{\mathrm{c}}\right)\)
= \(\left(\frac{(-16)(1)}{-5}, \frac{9}{-5}\right)\)
= \(\left(\frac{16}{5}, \frac{-9}{5}\right)\)

Question 8.
Show that the line 8y + x = 17 touches the ellipse x² + 4y² = 17. Find the point of contact.
Solution:
Given equation of the ellipse is x² + 4y² = 17.
\(\frac{x^{2}}{17}+\frac{y^{2}}{\frac{17}{4}}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a² = 17 and b² = \(\frac{17}{4}\)
Given equation of line is 8y + x = 17,
y = \(\frac{-1}{8} x+\frac{17}{8}\)
Comparing this equation with y = mx + c, we get
m = \(\frac{-1}{8}\) and c = \(\frac{17}{8}\)
For the line y = mx + c to be a tangent to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=\)1, we must have

Question 9.
Determine whether the line x + 3y√2 = 9 is a tangent to the ellipse \(\frac{x^{2}}{9}+\frac{y^{2}}{4}=1\). If so, find the co-ordinates of the point of contact.
Solution:
Given equation of the ellipse is \(\frac{x^{2}}{9}+\frac{y^{2}}{4}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a² = 9 and b² = 4
Given equation of line is x + 3y√2 = 9,
i.e., y = \(\frac{-1}{3 \sqrt{2}} x+\frac{3}{\sqrt{2}}\)
Comparing this equation with y = mx + c, we get
m = \(\frac{-1}{3 \sqrt{2}}\) and c = \(\frac{3}{\sqrt{2}}\)
For the line y = mx + c to be a tangent to the ellipse \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=\)1, we must have

Question 10.
Find k, if the line 3x + 4y + k = 0 touches 9x² + 16y² = 144.
Solution:
Given equation of the ellipse is 9x² + 16y² = 144.
\(\frac{x^{2}}{16}+\frac{y^{2}}{9}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a² = 16 and b² = 9
Given equation of line is 3x + 4y + k = 0,
i.e., y = \(-\frac{3}{4} x-\frac{k}{4}\)
Comparing this equation with y = mx + c, we get
m = \(\frac{-3}{4}\) and c = \(\frac{-\mathrm{k}}{4}\)
For the line y = mx + c to be a tangent to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=\)1, we must have
c² = a² m² + b²
\(\left(\frac{-k}{4}\right)^{2}=16\left(\frac{-3}{4}\right)^{2}+9\)
\(\frac{\mathrm{k}^{2}}{16}\) = 9 + 9
\(\frac{\mathrm{k}^{2}}{16}\) = 18
k² = 288
k = ±12√2

Question 11.
Find the equations of the tangents to the ellipse:
(i) \(\frac{x^{2}}{5}+\frac{y^{2}}{4}=1\) passing through the point (2, -2).
(ii) 4x² + 7y² = 28 from the point (3, -2).
(iii) 2x² + y² = 6 from the point (2, 1).
(iv) x² + 4y² = 9 which are parallel to the line 2x + 3y – 5 = 0.
(v) \(\frac{x^{2}}{25}+\frac{y^{2}}{4}=1\) which are parallel to the line x + y + 1 = 0.
(vi) 5x² + 9y² = 45 which are ⊥ to the line 3x + 2y + 1 = 0.
(vii) x² + 4y² = 20 which are ⊥ to the line 4x + 3y = 7.
Solution:
(i) Given equation of the ellipse is \(\frac{x^{2}}{5}+\frac{y^{2}}{4}=1\).
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a² = 5 and b² = 4
Equations of tangents to the ellipse \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\) having slope m are
y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\)
Since (2, -2) lies on both the tangents,
-2 = 2m ±\(\sqrt{5 m^{2}+4}\)
-2 – 2m = ±\(\sqrt{5 m^{2}+4}\)
Squaring both the sides, we get
4m² + 8m + 4 = 5m² + 4
m² – 8m = 0
m(m – 8) = 0
m = 0 or m = 8
These are the slopes of the required tangents.
By slope point form y – y₁ = m(x – x₁),
the equations of the tangents are
y + 2 = 0(x – 2) and y + 2 = 8(x – 2)
y + 2 = 0 and y + 2 = 8x – 16
y + 2 = 0 and 8x – y – 18 = 0

(ii) Given equation of the ellipse is 4x² + 7y² = 28.
\(\frac{x^{2}}{7}+\frac{y^{2}}{4}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a² = 7 and b² = 4
Equations of tangents to the ellipse \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\) having slope m are
y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\)
Since (3, -2) lies on both the tangents,
-2 = 3m ± \(\sqrt{7 \mathrm{~m}^{2}+4}\)
-2 – 3m = ±\(\sqrt{7 \mathrm{~m}^{2}+4}\)
Squaring both the sides, we get
9m² + 12m + 4 = 7m² + 4
2m² + 12m = 0
2m(m + 6) = 0
m = 0 or m = -6
These are the slopes of the required tangents.
By slope point form y – y₁ = m(x – x₁),
the equations of the tangents are
y + 2 = 0(x – 3) and y + 2 = -6(x – 3)
y + 2 = 0 and y + 2 = -6x + 18
y + 2 = 0 and 6x + y – 16 = 0

(iii) Given equation of the ellipse is 2x² + y² = 6.
\(\frac{x^{2}}{3}+\frac{y^{2}}{6}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get
a² = 3 and b² = 6
Equations of tangents to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) having slope m are
y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\)
Since (2, 1) lies on both the tangents,
1 = 2m ± \(\sqrt{3 m^{2}+6}\)
1 – 2m = ±\(\sqrt{3 m^{2}+6}\)
Squaring both the sides, we get
1 – 4m + 4m² = 3m² + 6
m² – 4m – 5 = 0
(m – 5)(m + 1) = 0
m = 5 or m = -1
These are the slopes of the required tangents.
By slope point form y – y₁ = m(x – x₁),
the equations of the tangents are
y – 1 = 5(x – 2) and y – 1 = -1(x – 2)
y – 1 = 5x – 10 and y – 1 = -x + 2
5x – y – 9 = 0 and x + y – 3 = 0

(iv) Given equation of the ellipse is x² + 4y² = 9.
\(\frac{x^{2}}{9}+\frac{y^{2}}{\frac{9}{4}}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get
a² = 9 and b² = \(\frac{9}{4}\)
Slope of the line 2x + 3y – 5 = 0 is \(\frac{-2}{3}\).
Since the given line is parallel to the required tangents, slope of the required tangents is
m = \(\frac{-2}{3}\)
Equations of tangents to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) having slope m are

(v) Given equation of the ellipse is \(\frac{x^{2}}{25}+\frac{y^{2}}{4}=1\).
Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get
a² = 25 and b² = 4
Slope of the given line x + y + 1 = 0 is -1.
Since the given line is parallel to the required tangents,
the slope of the required tangents is m = -1.
Equations of tangents to the ellipse

(vi) Given equation of the ellipse is 5x² + 9y² = 45.
\(\frac{x^{2}}{9}+\frac{y^{2}}{5}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get
a² = 9 and b² = 5
Slope of the given line 3x + 2y + 1 = 0 is \(\frac{-3}{2}\)
Since the given line is perpendicular to the required tangents, slope of the required tangents is
m = \(\frac{2}{3}\)
Equations of tangents to the ellipse

(vii) Given equation of the ellipse is x² + 4y² = 20.
\(\frac{x^{2}}{20}+\frac{y^{2}}{5}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a² = 20 and b² = 5
Slope of the given line 4x + 3y = 7 is \(\frac{-4}{3}\).
Since the given line is perpendicular to the required tangents,
slope of the required tangents is m = \(\frac{3}{4}\).
Equations of tangents to the ellipse \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\) having slope m are
y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\)
y = \(\frac{3}{4} x \pm \sqrt{20\left(\frac{3}{4}\right)^{2}+5}\)

Question 12.
Find the equation of the locus of a point, the tangents from which to the ellipse 3x² + 5y² = 15 are at right angles.
Solution:
Given equation of the ellipse is 3x² + 5y² = 15.
\(\frac{x^{2}}{5}+\frac{y^{2}}{3}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a² = 5 and b² = 3
Equations of tangents to the ellipse \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\) having slope m are
y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\)
y = mx ± \(\sqrt{5 m^{2}+3}\)
y – mx =±\(\sqrt{5 m^{2}+3}\)
Squaring both the sides, we get
y² – 2mxy + m²x² = 5m² + 3
(x² – 5) m² – 2xym + (y² – 3) = 0
The roots m₁ and m₂ of this quadratic equation are the slopes of the tangents.
m₁m₂ = \(\frac{y^{2}-3}{x^{2}-5}\)
Since the tangents are at right angles,
m₁m₂ = -1
\(\frac{y^{2}-3}{x^{2}-5}=-1\)
y² – 3 = -x² + 5
x² + y² = 8, which is the required equation of the locus.

Alternate method:
The locus of the point of intersection of perpendicular tangents is the director circle of an ellipse.
The equation of the director circle of an ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) is x² + y² = a² + b²
Here, a² = 5 and b² = 3
x² + y² = 5 + 3
x² + y² = 8, which is the required equation of the locus.

Question 13.
Tangents are drawn through a point P to the ellipse 4x² + 5y² = 20 having inclinations θ₁ and θ₂ such that tan θ₁ + tan θ₂ = 2. Find the equation of the locus of P.
Solution:
Given equation of the ellipse is 4x² + 5y² = 20.
\(\frac{x^{2}}{5}+\frac{y^{2}}{4}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get
a² = 5 and b² = 4
Since inclinations of tangents are θ₁ and θ₂,
m₁ = tan θ₁ and m₂ = tan θ₂
Equation of tangents to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) having slope m are
y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\)
y = mx ± \(\sqrt{5 \mathrm{~m}^{2}+4}\)
y – mx = ± \(\sqrt{5 \mathrm{~m}^{2}+4}\)
Squaring both the sides, we get
y² – 2mxy + m²x² = 5m² + 4
(x² – 5)m² – 2xym + (y² – 4) = 0
The roots m₁ and m₂ of this quadratic equation are the slopes of the tangents.
m₁ + m₂ = \(\frac{-(-2 x y)}{x^{2}-5}=\frac{2 x y}{x^{2}-5}\)
Given, tan θ₁ + tan θ₂ = 2
m₁ + m₂ = 2
\(\frac{2 x y}{x^{2}-5}\) = 2
xy = x² – 5
x² – xy – 5 = 0, which is the required equation of the locus of P.

Question 14.
Show that the locus of the point of intersection of tangents at two points on an ellipse, whose eccentric angles differ by a constant, is an ellipse.
Solution:
Let P(θ₁) and Q(θ₂) be any two points on the given ellipse such that θ₁ – θ₂ = k, where k is a constant.
The equation of the tangent at point P(θ₁) is
\(\frac{x \cos \theta_{1}}{\mathrm{a}}+\frac{y \sin \theta_{1}}{\mathrm{~b}}=1\) ……(i)
The equation of the tangent at point Q(θ₂) is
\(\frac{x \cos \theta_{2}}{\mathrm{a}}+\frac{y \sin \theta_{2}}{\mathrm{~b}}=1\) ……(ii)
Multiplying equation (i) by cos θ₂ and equation (ii) by cos θ₁ and subtracting, we get
\(\frac{y}{b}\) (sin θ₁ cos θ₂ – sin θ₂ cos θ₁) = cos θ₂ – cos θ₁

Question 15.
P and Q are two points on the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) with eccentric angles θ₁ and θ₂. Find the equation of the locus of the point of intersection of the tangents at P and Q if θ₁ + θ₂ = \(\frac{\pi}{2}\).
Solution:
Given equation of the ellipse is \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\).
θ₁ and θ₂ are the eccentric angles of a tangent.
Equation of tangent at point P is
\(\frac{x}{a} \cos \theta_{1}+\frac{y}{b} \sin \theta_{1}=1\) ……(i)
Equation of tangent at point Q is
\(\frac{x}{a} \cos \theta_{2}+\frac{y}{b} \sin \theta_{2}=1\) ………(ii)
θ₁ + θ₂ = \(\frac{\pi}{2}\) …..[Given]

i.e., points P and Q coincide, which is not possible, as P and Q are two different points.
cos θ₁ – sin θ₁ ≠ 0
Dividing equation (iii) by (cos θ₁ – sin θ₁), we get
\(\frac{x_{1}}{a}=\frac{y_{1}}{b}\)
bx₁ – ay₁ = 0
bx – ay = 0, which is the required equation of locus of point M.

Question 16.
The eccentric angles of two points P and Q of the ellipse 4x² + y² = 4 differ by \(\frac{2 \pi}{3}\). Show that the locus of the point of intersection of the tangents at P and Q is the ellipse 4x² + y² = 16.
Solution:
Given equation of the ellipse is 4x² + y² = 4.
\(\frac{x^{2}}{1}+\frac{y^{2}}{4}=1\)
Let P(θ₁) and Q(θ₂) be any two points on the given ellipse such that
θ₁ – θ₂ = \(\frac{2 \pi}{3}\)
The equation of the tangent at point P(θ₁) is
\(\frac{x \cos \theta_{1}}{1}+\frac{y \sin \theta_{1}}{2}=1\) ……(i)
The equation of the tangent at point Q(θ₂) is
\(\frac{x \cos \theta_{2}}{1}+\frac{y \sin \theta_{2}}{2}=1\)
Multiplying equation (i) by cos θ₂ and equation (ii) by cos θ₁ and subtracting, we get

Question 17.
Find the equations of the tangents to the ellipse \(\frac{x^{2}}{16}+\frac{y^{2}}{9}=1\), making equal intercepts on co-ordinate axes.
Solution:
Given equation of the ellipse is \(\frac{x^{2}}{16}+\frac{y^{2}}{9}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get
a² = 16 and b² = 9
Since the tangents make equal intercepts on the co-ordinate axes,
m = -1.
Equations of tangents to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) having slope m are
y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\)
y = -x ± \(\sqrt{16(-1)^{2}+9}\)
y = -x ± \(\sqrt{25}\)
x + y = ±5

Question 18.
A tangent having slope \(\left(-\frac{1}{2}\right)\) to the ellipse 3x² + 4y² = 12 intersects the X and Y axes in the points A and B respectively. If O is the origin, find the area of the triangle AOB.
Solution:
The equation of the ellipse is 3x² + 4y² = 12
\(\frac{x^{2}}{4}+\frac{y^{2}}{3}=1\)
Comparing with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get
a² = 4, b² = 3
The equation of tangent with slope m is

It meets X axis at A
∴ for A, put y = 0 in equation (1), we get,
x = ±4
∴ A = (±4, 0)
Similarly, B = (0, ±2)
∴ OA = 4, OB = 2
∴ Area of ΔOAB = \(\frac{1}{2}\) × OA × OB
= \(\frac{1}{2}\) × 4 × 2
= 4 sq. units