Class 11

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 9 Differentiation Ex 9.1 Questions and Answers.

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

Question 1.
Find the derivatives of the following w.r.t. x by using the method of the first principle.
(a) x² + 3x – 1
Solution:
Let f(x) = x² + 3x – 1
∴ f(x + h) = (x + h)² + 3(x + h) – 1
= x² + 2xh + h² + 3x + 3h – 1
By first principle, we get

(b) sin(3x)
Solution:
Let f(x) = sin 3x
f(x + h) = sin3(x + h) = sin(3x + 3h)
By first principle, we get

(c) e^2x+1
Solution:

(d) 3^x
Solution:

(e) log(2x + 5)
Solution:
Let f(x) = log(2x + 5)
∴ f(x + h) = log[2(x + h) + 5] = log (2x + 2h + 5)
By first principle, we get

(f) tan(2x + 3)
Solution:
Let f(x) = tan(2x + 3)
∴ f(x + h) = tan[2(x + h) + 3] = tan(2x + 2h + 3)
By first principle, we get

(g) sec(5x – 2)
Solution:
Let f(x) = sec(5x – 2)
f(x + h) = sec[5(x + h) – 2] = sec(5x + 5h – 2)
By first principle, we get

(h) x√x
Solution:

Question 2.
Find the derivatives of the following w.r.t. x. at the points indicated against them by using the method of the first principle.
(i) \(\sqrt{2 x+5}\) at x = 2
Solution:

(ii) tan x at x = \(\frac{\pi}{4}\)
Solution:

(iii) 2^3x+1 at x = 2
Solution:

(iv) log(2x + 1) at x = 2
Solution:
Let f(x) = log(2x + 1)
∴ f(2) = log [2(2) + 1] = log 5 and
f(2 + h) = log [2(2 + h) + 1] = log(2h + 5)
By first principle, we get

(v) e^3x-4 at x = 2
Solution:

(vi) cos x at x = \(\frac{5 \pi}{4}\)
Solution:

Question 3.
Show that the function f is not differentiable at x = -3,
where f(x) = x² + 2 for x < -3
= 2 – 3x for x ≥ -3
Solution:


∴ L f'(-3) ≠ R f'(-3)
∴ f is not differentiable at x = -3.

Question 4.
Show that f(x) = x² is continuous and differentiable at x = 0.
Solution:

Question 5.
Discuss the continuity and differentiability of
(i) f(x) = x |x| at x = 0
Solution:

(ii) f(x) = (2x + 3) |2x + 3| at x = \(-\frac{3}{2}\)
Solution:

Question 6.
Discuss the continuity and differentiability of f(x) at x = 2.
f(x) = [x] if x ∈ [0, 4). [where [ ] is a greatest integer (floor) function]
Solution:
Explanation:
x ∈ [0, 4)
∴ 0 ≤ x < 4
We will plot graph for 0 ≤ x < 4
not for x < 0 and upto x = 4 making on X-axis.

f(x) = [x]
∴ Greatest integer function is discontinuous at all integer values of x and hence not differentiable at all integers.
∴ f is not continuous at x = 2.
∵ f(x) = 1, x < 2
= 2, x ≥ 2
x ∈ neighbourhood of x = 2.
∴ L.H.L. = 1, R.H.L. = 2
∴ f is not continuous at x = 2.
∴ f is not differentiable at x = 2.

Question 7.
Test the continuity and differentiability of
f(x) = 3x + 2 if x > 2
= 12 – x² if x ≤ 2 at x = 2.
Solution:

Question 8.
If f(x) = sin x – cos x if x ≤ \(\frac{\pi}{2}\)
= 2x – π + 1 if x > \(\frac{\pi}{2}\)
Test the continuity and differentiability of f at x = \(\frac{\pi}{2}\).
Solution:

Question 9.
Examine the function
f(x) = x² cos(\(\frac{1}{x}\)), for x ≠ 0
= 0, for x = 0
for continuity and differentiability at x = 0.
Solution:

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 4 Methods of Induction and Binomial Theorem Miscellaneous Exercise 4 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Methods of Induction and Binomial Theorem Miscellaneous Exercise 4

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

Question 1.
The total number of terms in the expression of (x + y)¹⁰⁰ + (x – y)¹⁰⁰ after simplification is:
(A) 50
(B) 51
(C) 100
(D) 202
Answer:
(B) 51
Hint:

Question 2.
The middle term in the expansion of (1 + x)²ⁿ will be:
(A) (n – 1)^th
(B) n^th
(C) (n + 1)^th
(D) (n + 2)^th
Answer:
(C) (n + 1)^th
Hint:
(1 + x)²ⁿ has (2n + 1) terms.
∴ (n + 1 )^th term is the middle term.

Question 3.
In the expansion of (x² – 2x)¹⁰, the coefficient of x¹⁶ is
(A) -1680
(B) 1680
(C) 3360
(D) 6720
Answer:
(C) 3360
Hint:
(x² – 2x)¹⁰ = x¹⁰ (x – 2)¹⁰
To get the coefficient of x¹⁶ in (x² – 2x)¹⁰,
we need to check coefficient of x⁶ in (x – 2)¹⁰
∴ Required coefficient = ¹⁰C₆ (-2)⁴
= 210 × 16
= 3360

Question 4.
The term not containing x in expansion of \((1-x)^{2}\left(x+\frac{1}{x}\right)^{10}\) is
(A) ¹¹C₅
(B) ¹⁰C₅
(C) ¹⁰C₄
(D) ¹⁰C₇
Answer:
(A) ¹¹C₅
Hint:

Question 5.
The number of terms in expansion of (4y + x)⁸ – (4y – x)⁸ is
(A) 4
(B) 5
(C) 8
(D) 9
Answer:
(A) 4
Hint:

Question 6.
The value of ¹⁴C₁ + ¹⁴C₃ + ¹⁴C₅ + …. + ¹⁴C₁₁ is
(A) 2¹⁴ – 1
(B) 2¹⁴ – 14
(C) 2¹²
(D) 2¹³ – 14
Answer:
(D) 2¹³ – 14
Hint:

Question 7.
The value of ¹¹C₂ + ¹¹C₄ + ¹¹C₆ + ¹¹C₈ is equal to
(A) 2¹⁰ – 1
(B) 2¹⁰ – 11
(C) 2¹⁰ + 12
(D) 2¹⁰ – 12
Answer:
(D) 2¹⁰ – 12
Hint:

Question 8.
In the expansion of (3x + 2)⁴, the coefficient of the middle term is
(A) 36
(B) 54
(C) 81
(D) 216
Answer:
(D) 216
Hint:
(3x + 2)⁴ has 5 terms.
∴ (3x + 2)⁴ has 3rd term as the middle term.
The coefficient of the middle term

= 6 × 9 × 4
= 216

Question 9.
The coefficient of the 8th term in the expansion of (1 + x)¹⁰ is:
(A) 7
(B) 120
(C) ¹⁰C₈
(D) 210
Answer:
(B) 120
Hint:
r = 7
t₈ = ¹⁰C₇ x⁷ = ¹⁰C₃ x⁷
∴ Coefficient of 8th term = ¹⁰C₃ = 120

Question 10.
If the coefficients of x² and x³ in the expansion of (3 + ax)⁹ are the same, then the value of a is
(A) \(-\frac{7}{9}\)
(B) \(-\frac{9}{7}\)
(C) \(\frac{7}{9}\)
(D) \(\frac{9}{7}\)
Answer:
(D) \(\frac{9}{7}\)
Hint:

(II) Answer the following.

Question 1.
Prove by the method of induction, for all n ∈ N.
(i) 8 + 17 + 26 + ….. + (9n – 1) = \(\frac{n}{2}\) (9n + 7)
Solution:
Let P(n) ≡ 8 + 17 + 26 +…..+(9n – 1) = \(\frac{n}{2}\) (9n + 7), for all n ∈ N.
Step I:
Put n = 1
L.H.S. = 8
R.H.S. = \(\frac{1}{2}\) [9(1) + 7] = 8
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 8 + 17 + 26 +…..+ (9k – 1) = \(\frac{k}{2}\) (9k + 7) ……(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., 8 + 17 + 26 + …… + [9(k + 1) – 1]

∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 8 + 17 + 26 +…..+ (9n – 1) = \(\frac{n}{2}\) (9n + 7) for all n ∈ N.

(ii) 1² + 4² + 7² + …… + (3n – 2)² = \(\frac{n}{2}\) (6n² – 3n – 1)
Solution:
Let P(n) = 1² + 4² + 7² + ….. + (3n – 2)² = \(\frac{n}{2}\) (6n² – 3n – 1), for all n ∈ N.
Step I:
Put n = 1
L.H.S.= 1² = 1
R.H.S.= \(\frac{1}{2}\) [6(1)² – 3(1) – 1] = 1
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 1² + 4² + 7² +…..+ (3k – 2)² = \(\frac{k}{2}\) (6k² – 3k – 1) ……(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that

∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 1² + 4² + 7² + … + (3n – 2)² = \(\frac{n}{2}\) (6n² – 3n – 1) for all n ∈ N.

(iii) 2 + 3.2 + 4.2² + …… + (n + 1) 2^n-1 = n . 2ⁿ
Solution:
Let P(n) ≡ 2 + 3.2 + 4.2² +…..+ (n + 1) 2^n-1 = n.2ⁿ, for all n ∈ N.
Step I:
Put n = 1
L.H.S. = 2
R.H.S. = 1(2¹) = 2
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 2 + 3.2 + 4.2² + ….. + (k + 1) 2^k-1 = k.2^k …..(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that
2 + 3.2 + 4.2² +….+ (k + 2) 2^k = (k + 1) 2^k+1

∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 2 + 3.2 + 4.2² +……+ (n + 1) 2^n-1 = n.2ⁿ for all n ∈ N.

(iv) \(\frac{1}{3.4 .5}+\frac{2}{4.5 .6}+\frac{3}{5.6 .7}+\ldots+\frac{n}{(n+2)(n+3)(n+4)}\) = \(\frac{n(n+1)}{6(n+3)(n+4)}\)
Solution:

Question 2.
Given that t_n+1 = 5t_n – 8, t₁ = 3, prove by method of induction that t_n = 5^n-1 + 2.
Solution:
Let the statement P(n) has L.H.S. a recurrence relation t_n+1 = 5t_n – 8, t₁ = 3
and R.H.S. a general statement t_n = 5^n-1 + 2.
Step I:
Put n = 1
L.H.S. = 3
R.H.S. = 5^1-1 + 2 = 1 + 2 = 3
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.
Put n = 2
L.H.S = t₂ = 5t₁ – 8 = 5(3) – 8 = 7
R.H.S. = t₂ = 5^2-1 + 2 = 5 + 2 = 7
∴ L.H.S. = R.H.S.
∴ P(n) is tme for n = 2.

Step II:
Let us assume that P(n) is true for n = k.
∴ t_k+1 = 5t_k – 8 and t_k = 5^k-1 + 2

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that
t_k+1 = 5^k+1-1 + 2 = 5^k + 2
t_k+1 = 5t_k – 8 and t_k = 5^k-1 + 2 ……[From Step II]
∴ t_k+1 = 5(5^k-1 + 2) – 8 = 5^k + 2
∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ t_n = 5^n-1 + 2, for all n ∈ N.

Question 3.
Prove by method of induction
\(\left(\begin{array}{cc}
3 & -4 \\
1 & -1
\end{array}\right)^{n}=\left(\begin{array}{cc}
2 n+1 & -4 n \\
n & -2 n+1
\end{array}\right)\), ∀ n ∈ N.
Solution:


Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ \(\left(\begin{array}{cc}
3 & -4 \\
1 & -1
\end{array}\right)^{n}=\left(\begin{array}{cc}
2 n+1 & -4 n \\
n & -2 n+1
\end{array}\right)\), ∀ n ∈ N.

Question 4.
Expand (3x² + 2y)⁵
Solution:
Here, a = 3x², b = 2y, n = 5.
Using binomial theorem,

Question 5.
Expand \(\left(\frac{2 x}{3}-\frac{3}{2 x}\right)^{4}\)
Solution:

Question 6.
Find third term in the expansion of \(\left(9 x^{2}-\frac{y^{3}}{6}\right)^{4}\)
Solution:

Question 7.
Find tenth term in the expansion of \(\left(2 x^{2}+\frac{1}{x}\right)^{12}\)
Solution:

Question 8.
Find the middle term(s) in the expansion of
(i) \(\left(\frac{2 a}{3}-\frac{3}{2 a}\right)^{6}\)
Solution:
Here, a = \(\frac{2 a}{3}\), b = \(\frac{-3}{2 a}\), n = 6.
Now, n is even.
∴ \(\frac{\mathrm{n}+2}{2}=\frac{6+2}{2}=4\)
∴ Middle term is t4, for which r = 3.

∴ The Middle term is -20.

(ii) \(\left(x-\frac{1}{2 y}\right)^{10}\)
Solution:
Here, a = x, b = \(-\frac{1}{2 y}\), n = 10.
Now, n is even.
∴ \(\frac{\mathrm{n}+2}{2}=\frac{10+2}{2}=6\)
∴ Middle term is t6, for which r = 5

(iii) (x² + 2y²)⁷
Solution:
Here, a = x², b = 2y², n = 7.
Now, n is odd.
∴ \(\frac{\mathrm{n}+1}{2}=\frac{7+1}{2}=4, \frac{\mathrm{n}+3}{2}=\frac{7+3}{2}=5\)
∴ Middle terms are t₄ and t₅, for which r = 3 and r = 4 respectively.

∴ Middle terms are 280x⁸y⁶ and 560x⁶y⁸.

(iv) \(\left(\frac{3 x^{2}}{2}-\frac{1}{3 x}\right)^{9}\)
Solution:

Question 9.
Find the coefficients of
(i) x⁶ in the expantion of \(\left(3 x^{2}-\frac{1}{3 x}\right)^{9}\)
Solution:

(ii) x⁶⁰ in the expansion of \(\left(\frac{1}{x^{2}}+x^{4}\right)^{18}\)
Solution:

Question 10.
Find the constant term in the expansion of
(i) \(\left(\frac{4 x^{2}}{3}+\frac{3}{2 x}\right)^{9}\)
Solution:

(ii) \(\left(2 x^{2}-\frac{1}{x}\right)^{12}\)
Solution:

Question 11.
Prove by method of induction
(i) log_a xⁿ = n log_a x, x > 0, n ∈ N
Solution:

(ii) 15^2n-1 + 1 is divisible by 16, for all n ∈ N.
Solution:
15^2n-1 + 1 is divisible by 16, if and only if (15^2n-1 + 1) is is a multiple of 16.
Let P(n) ≡ 15^2n-1 + 1 = 16m, where m ∈ N.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 15^2n-1 + 1 is divisible by 16, for all n ∈ N.

(iii) 5²ⁿ – 2²ⁿ is divisible by 3, for all n ∈ N.
Solution:

Question 12.
If the coefficient of x¹⁶ in the expansion of (x² + ax)¹⁰ is 3360, find a.
Solution:

Question 13.
If the middle term in the expansion of \(\left(x+\frac{b}{x}\right)^{6}\) is 160, find b.
Solution:

∴ 160 = \(\frac{6 \times 5 \times 4 \times 3 !}{3 \times 2 \times 1 \times 3 !} \times b^{3}\)
∴ 160 = 20b³
∴ 8 = b³
∴ b = 2

Question 14.
If the coefficients of x² and x³ in theexpansion of (3 + kx)⁹ are equal, find k.
Solution:

Question 15.
If the constant term in the expansion of \(\left(x^{3}+\frac{\mathrm{k}}{x^{8}}\right)^{11}\) is 1320, find k.
Solution:

Question 16.
Show that there is no term containing x⁶ in the expansion of \(\left(x^{2}-\frac{3}{x}\right)^{11}\).
Solution:

Question 17.
Show that there is no constant term in the expansion of \(\left(2 x-\frac{x^{2}}{4}\right)^{9}\)
Solution:

Question 18.
State, first four terms in the expansion of \(\left(1-\frac{2 x}{3}\right)^{-1 / 2}\)
Solution:

Question 19.
State, first four terms in the expansion of \((1-x)^{-1 / 4}\).
Solution:

Question 20.
State, first three terms in the expansion of \((5+4 x)^{-1 / 2}\)
Solution:

Question 21.
Using the binomial theorem, find the value of \(\sqrt[3]{995}\) upto four places of decimals.
Solution:

Question 22.
Find approximate value of \(\frac{1}{4.08}\) upto four places of decimals.
Solution:

Question 23.
Find the term independent of x in the expansion of (1 – x²) \(\left(x+\frac{2}{x}\right)^{6}\).
Solution:

Question 24.
(a + bx) (1 – x)⁶ = 3 – 20x + cx² + …, then find a, b, c.
Solution:

Question 25.
The 3rd term of (1 + x)ⁿ is 36x². Find 5th term.
Solution:

Question 26.
Suppose (1 + kx)ⁿ = 1 – 12x + 60x² – …… find k and n.
Solution:

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 5 Sets and Relations Miscellaneous Exercise 5 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 5 Sets and Relations Miscellaneous Exercise 5

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

Question 1.
For the set A = {a, b, c, d, e} the correct statement is
(A) {a, b} ∈ A
(B) {a} ∈ A
(C) a ∈ A
(D) a ∉ A
Answer:
(C) a ∈ A

Question 2.
If aN = {ax : x ∈ N}, then set 6N ∩ 8N =
(A) 8N
(B) 48N
(C) 12N
(D) 24N
Answer:
(D) 24N
Hint:
6N = {6x : x ∈ N} = {6, 12, 18, 24, 30, ……}
8N = {8x : x ∈ N} = {8, 16, 24, 32, ……}
∴ 6N ∩ 8N = {24, 48, 72, …..}
= {24x : x ∈ N}
= 24N

Question 3.
If set A is empty set then n[P[P[P(A)]]] is
(A) 6
(B) 16
(C) 2
(D) 4
Answer:
(D) 4
Hint:
A = Φ
∴ n(A) = 0
∴ n[P(A)] = 2^n(A) = 2⁰ = 1
∴ n[P[P(A)]] = 2^n[P(A)] = 2¹ = 2
∴ n[P[P[P(A)]]] = 2^n[P[P(A)]] = 2² = 4

Question 4.
In a city 20% of the population travels by car, 50% travels by bus and 10% travels by both car and bus. Then, persons travelling by car or bus are
(A) 80%
(B) 40%
(C) 60%
(D) 70%
Answer:
(C) 60%
Hint:
Let C = Population travels by car
B = Population travels by bus
n(C) = 20%, n(B) = 50%, n(C ∩ B) = 10%
n(C ∪ B) = n(C) + n(B) – n(C ∩ B)
= 20% + 50% – 10%
= 60%

Question 5.
If the two sets A and B are having 43 elements in common, then the number of elements common to each of the sets A × B and B × A is
(A) 43²
(B) 2⁴³
(C) 43⁴³
(D) 2⁸⁶
Answer:
(A) 43²

Question 6.
Let R be a relation on the set N be defied by {(x, y) / x, y ∈ N, 2x + y = 41} Then R is
(A) Reflexive
(B) Symmetric
(C) Transitive
(D) None of these
Answer:
(D) None of these

Question 7.
The relation “>” in the set of N (Natural number) is
(A) Symmetric
(B) Reflexive
(C) Transitive
(D) Equivalent relation
Answer:
(C) Transitive
Hint:
For any a ∈ N, a ≯ a
∴ (a, a) ∉ R
∴ > is not reflexive.
For any a, b ∈ N, if a > b, then b ≯ a.
∴ > is not symmetric.
For any a, b, c ∈ N,
if a > b and b > c, then a > c
∴ > is transitive.

Question 8.
A relation between A and B is
(A) only A × B
(B) An Universal set of A × B
(C) An equivalent set of A × B
(D) A subset of A × B
Answer:
(D) A subset of A × B

Question 9.
If (x, y) ∈ N × N, then xy = x² is a relation that is
(A) Symmetric
(B) Reflexive
(C) Transitive
(D) Equivalence
Answer:
(D) Equivalence
Hint:
Let x ∈ R, then xx = x²
∴ x is related to x.
∴ Given relation is reflexive.
Letx = 0 and y = 2,
then xy = 0 × 2 = 0 = x²
∴ x is related to y.
Consider, yx = 2 × 0 = 0 ≠ y²
∴ y is not related to x.
∴ Given relation is not symmetric.
Let x be related to y and y be related to z.
∴ xy = x² and yz = y²
∴ x = \(\frac{x^{2}}{y}\) and z = \(\frac{y^{2}}{y}\) = y …..[if y ≠ 0]
Consider, xz = \(\frac{x^{2}}{y}\) × y = x²
∴ x is related to z.
∴ Given relation is transitive.

Question 10.
If A = {a, b, c}, The total no. of distinct relations in A × A is
(A) 3
(B) 9
(C) 8
(D) 29
Answer:
(D) 29

(II) Answer the following.

Question 1.
Write down the following sets in set builder form:
(i) {10, 20, 30, 40, 50}
(ii) {a, e, i, o, u}
(iii) {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}
Solution:
(i) Let A = {10, 20, 30, 40, 50}
∴ A = {x/x = 10n, n ∈ N and n ≤ 5}

(ii) Let B = {a, e, i, o, u}
∴ B = {x/x is a vowel of English alphabets}

(iii) Let C = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}
∴ C = {x/x is a day of a week}

Question 2.
If U = {x/x ∈ N, 1 ≤ x ≤ 12}, A = {1,4, 7,10}, B = {2, 4, 6, 7, 11}, C = {3, 5, 8, 9, 12}. Write down the sets.
(i) A ∪ B
(ii) B ∩ C
(iii) A – B
(iv) B ∩ C’
(v) A ∪ B ∪ C
(vi) A ∩ (B ∪ C)
Solution:
U = {x/x ∈ N, 1 ≤ x ≤ 12} = {1, 2, 3, …., 12}
A = {1, 4, 7, 10}, B = {2, 4, 6, 7, 11}, C = {3, 5, 8, 9, 12}
(i) A ∪ B = {1, 2, 4, 6, 7, 10, 11}

(ii) B ∩ C = {}

(iii) A – B = {1, 10}

(iv) C’ = {1, 2, 4, 6, 7, 10, 11}
∴ B ∩ C’ = {2, 4, 6, 7, 11}

(v) A ∪ B ∪ C = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

(vi) B ∪ C = {2, 3, 4, 5, 6, 7, 8, 9, 11, 12}
∴ A ∩ (B ∪ C) = {4, 7}

Question 3.
In a survey of 425 students in a school, it was found that 115 drink apple juice, 160 drink orange juice, and 80 drink both apple as well as orange juice. How many drinks neither apple juice nor orange juice?
Solution:
Let A = set of students who drink apple juice
B = set of students who drink orange juice
X = set of all students
∴ n(X) = 425, n(A) = 115, n(B) = 160, n(A ∩ B) = 80

No. of students who neither drink apple juice nor orange juice = n(A’ ∩ B’) = n(A ∪ B)’
= n(X) – n(A ∪ B)
= 425 – [n(A) + n(B) – n(A ∩ B)]
= 425 – (115 + 160 – 80)
= 230

Question 4.
In a school, there are 20 teachers who teach Mathematics or Physics. Of these, 12 teach Mathematics and 4 teach both Physics and Mathematics. How many teachers teach Physics?
Solution:
Let A = set of teachers who teach Mathematics
B = set of teachers who teach Physics
∴ n(A ∪ B) = 20, n(A) = 12, n(A ∩ B) = 4

Since n(A ∪ B) = n(A) + n(B) – n(A ∩ B),
20 = 12 + n(B) – 4
∴ n(B) = 12
∴ Number of teachers who teach physics = 12

Question 5.
(i) If A = {1, 2, 3} and B = {2, 4}, state the elements of A × A, A × B, B × A, B × B, (A × B) ∩ (B × A).
(ii) If A = {-1, 1}, find A × A × A.
Solution:
(i) A = {1, 2, 3} and B = {2, 4}
A × A = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}
A × B = {(1, 2), (1, 4), (2, 2), (2, 4), (3, 2), (3, 4)}
B × A = {(2, 1), (2, 2), (2, 3), (4, 1), (4, 2), (4, 3)}
B × B = {(2, 2), (2, 4), (4, 2), (4, 4)}
∴ (A × B) ∩ (B × A) = {(2, 2)}

(ii) A = {-1, 1}
∴ A × A × A = {(-1, -1, -1), (-1, -1, 1), (-1, 1, -1), (-1, 1, 1), (1, -1, -1), (1, -1, 1), (1, 1, -1), (1, 1, 1)}

Question 6.
If A = {1, 2, 3}, B = {4, 5, 6}, check if the following are relations from A to B. Also, write its domain and range.
(i) R₁ = {(1, 4), (1, 5), (1, 6)}
(ii) R₂ = {(1, 5), (2, 4), (3, 6)}
(iii) R₃ = {(1, 4), (1, 5), (3, 6), (2, 6), (3, 4)}
(iv) R₄ = {(4, 2), (2, 6), (5, 1), (2, 4)}
Solution:
A = {1, 2, 3}, B = {4, 5, 6}
∴ A × B = {(1, 4), (1, 5), (1, 6), (2,4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
(i) R₁ = {(1, 4), (1, 5), (1, 6)}
Since R₁ ⊆ A × B,
R₁ is a relation from A to B.
Domain (R₁) = Set of first components of R₁ = {1}
Range (R₁) = Set of second components of R₁ = {4, 5, 6}

(ii) R₂ = {(1, 5),(2, 4),(3, 6)}
Since R₂ ⊆ A × B,
R₂ is a relation from A to B.
Domain (R₂) = Set of first components of R₂ = {1, 2, 3}
Range (R₂) = Set of second components of R₂ = {4, 5, 6}

(iii) R₃ = {(1, 4), (1, 5), (3, 6), (2, 6), (3, 4)}
Since R₃ ⊆ A × B,
R₃ is a relation from A to B.
Domain (R₃) = Set of first components of R₃ = {1, 2, 3}
Range (R₃) = Set of second components of R₃ = {4, 5, 6}

(iv) R₄ = {(4, 2), (2, 6), (5, 1), (2, 4)}
Since (4, 2) ∈ R₄, but (4, 2) ∉ A × B,
R₄ ⊄ A × B
∴ R₄ is not a relation from A to B.

Question 7.
Determine the domain and range of the following relations.
(i) R = {(a, b) / a ∈ N, a < 5, b = 4}
(ii) R = {(a, b) / b = |a – 1|, a ∈ Z, |a| < 3}
Solution:
(i) R = {(a, b) / a ∈ N, a < 5, b = 4}
∴ Domain (R) = {a / a ∈ N, a < 5} = {1, 2, 3, 4}
Range (R) = {b / b = 4} = {4}

(ii) R = {(a, b) / b = |a – 1|, a ∈ Z, |a| < 3}
Since a ∈ Z and |a| < 3,
a < 3 and a > -3
∴ -3 < a < 3
∴ a = -2, -1, 0, 1, 2
b = |a – 1|
When a = -2, b = 3
When a = -1, b = 2
When a = 0, b = 1
When a = 1, b = 0
When a = 2, b = 1
Domain (R) = {-2, -1, 0, 1, 2}
Range (R) = {0, 1, 2, 3}

Question 8.
Find R : A → A when A = {1, 2, 3, 4} such that
(i) R = {(a, b) / a – b = 10}
(ii) R = {(a, b) / |a – b| ≥ 0}
Solution:
R : A → A, A = {1, 2, 3,4}
(i) R = {(a, b)/a – b = 10} = { }

(ii) R = {(a, b) / |a – b| ≥ 0}
= {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)}
A × A = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)}
∴ R = A × A

Question 9.
R : {1, 2, 3} → {1, 2, 3} given by R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)}. Check if R is
(i) reflexive
(ii) symmetric
(iii) transitive
Solution:
R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)}
(i) Here, (x, x) ∈ R, for x ∈ {1, 2, 3}
∴ R is reflexive.

(ii) Here, (1, 2) ∈ R, but (2, 1) ∉ R.
∴ R is not symmetric.

(iii) Here, (1, 2), (2, 3) ∈ R,
But (1, 3) ∉ R.
∴ R is not transitive.

Question 10.
Check if R : Z → Z, R = {(a, b) | 2 divides a – b} is an equivalence relation.
Solution:
(i) Since 2 divides a – a,
(a, a) ∈ R
∴ R is reflexive. .

(ii) Let (a, b) ∈ R
Then 2 divides a – b
∴ 2 divides b – a
∴ (b, a) ∈ R
∴ R is symmetric.

(iii) Let (a, b) ∈ R, (b, c) ∈ R
Then a – b = 2m, b – c = 2n,
∴ a – c = 2(m + n), where m, n are integers.
∴ 2 divides a – c
∴ (a, c) ∈ R
∴ R is transitive.
Thus, R is an equivalence relation.

Question 11.
Show that the relation R in the set A = {1, 2, 3, 4, 5} Given by R = {(a, b) / |a – b| is even} is an equivalence relation.
Solution:
(i) Since |a – a| is even,
∴ (a, a) ∈ R
∴ R is reflexive.

(ii) Let (a, b) ∈ R
Then |a – b| is even
∴ |b – a| is even
∴ (b, a) ∈ R
∴ R is symmetric.

(iii) Let (a, b), (b, c) ∈ R
Then a – b = ±2m, b – c = ±2n
∴ a – c = ±2(m + n), where m, n are integers.
∴ (a, c) ∈ R
∴ R is transitive
Thus, R is an equivalence relation.

Question 12.
Show that the following are equivalence relations:
(i) R in A is set of all books given by R = {(x, y) / x and y have same number of pages}
(ii) R in A = {x ∈ Z | 0 ≤ x ≤ 12} given by R = {(a, b) / |a – b| is a multiple of 4}
(iii) R in A = (x ∈ N/x ≤ 10} given by R = {(a, b) | a = b}
Solution:
(i) a. Clearly (x, x) ∈ R
∴ R is reflexive.

b. If (x, y) ∈ R then (y, x) ∈ R.
∴ R is symmetric.

c. Let (x, y) ∈ R, (y, x) ∈ R.
Then x, y, and z are 3 books having the same number of pages.
∴ (x, z) ∈ R as x, z has the same number of pages.
∴ R is transitive.
Thus, R is an equivalence relation.

(ii) a. Since |a – a| is a multiple of 4,
(a, a) ∈ R
∴ R is reflexive.

b. Let (a, b) ∈ R
Then a – b = ±4m,
∴ b – a = ±4m, where m is an integer
∴ (b, a) ∈ R
∴ R is symmetric.

c. Let (a, b), (b, c) ∈ R
a – b = ± 4m, b – c = ± 4n,
∴ a – c = ±4(m + n), where m, n are integers
∴ (a, c) ∈ R
∴ R is transitive
Thus, R is an equivalence relation.

(iii) a. Since a = a
∴ (a, a) ∈ R
∴ R is reflexive.

b. Let (a, b) ∈ R Then a = b
∴ b = a
∴ (b, a) ∈ R
∴ R is symmetric.

c. Let (a, b), (b, c) ∈ R
Then, a = b, b = c
∴ a = c
∴ (a, c) ∈ R
∴ R is transitive.
Thus, R is an equivalence relation.

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 8 Continuity Miscellaneous Exercise 8 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8

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

Question 1.
f(x) = \(\frac{2^{\cot x}-1}{\pi-2 x}\), for x ≠ \(\frac{\pi}{2}\)
= log √2, for x = \(\frac{\pi}{2}\)
(A) f is continuous at x = \(\frac{\pi}{2}\)
(B) f has a jump discontinuity at x = \(\frac{\pi}{2}\)
(C) f has a removable discontinuity
(D) \(\lim _{x \rightarrow \frac{\pi}{2}} f(x)=2 \log 3\)
Answer:
(A) f is continuous at x = \(\frac{\pi}{2}\)
Hint:

Question 2.
If f(x) = \(\frac{1-\sqrt{2} \sin x}{\pi-4 x}\), for x ≠ \(\frac{\pi}{4}\) is continuous at x = \(\frac{\pi}{4}\), then f(\(\frac{\pi}{4}\)) =
(A) \(\frac{1}{\sqrt{2}}\)
(B) \(-\frac{1}{\sqrt{2}}\)
(C) \(-\frac{1}{4}\)
(D) \(\frac{1}{4}\)
Answer:
(D) \(\frac{1}{4}\)
Hint:

Question 3.
If f(x) = \(\frac{(\sin 2 x) \tan 5 x}{\left(e^{2 x}-1\right)^{2}}\), for x ≠ 0 is continuous at x = 0, then f(0) is
(A) \(\frac{10}{e^{2}}\)
(B) \(\frac{10}{e^{4}}\)
(C) \(\frac{5}{4}\)
(D) \(\frac{5}{2}\)
Answer:
(D) \(\frac{5}{2}\)
Hint:

Question 4.
f(x) = \(\frac{x^{2}-7 x+10}{x^{2}+2 x-8}\), for x ∈ [-6, -3]
(A) f is discontinuous at x = 2
(B) f is discontinuous at x = -4
(C) f is discontinuous at x = 0
(D) f is discontinuous at x = 2 and x = -4
Answer:
(B) f is discontinuous at x = -4
Hint:
f(x) = \(\frac{x^{2}-7 x+10}{x^{2}+2 x-8}\), for x ∈ [-6, -3]
= \(\frac{x^{2}-7 x+10}{(x+4)(x-2)}\)
Here f(x) is a rational function and is continuous everywhere except at the points Where denominator becomes zero.
Here, denominator becomes zero when x = -4 or x = 2
But x = 2 does not lie in the given interval.
∴ x = -4 is the point of discontinuity.

Question 5.
If f(x) = ax² + bx + 1, for |x – 1| ≥ 3 and
= 4x + 5, for -2 < x < 4
is continuous everywhere then,
(A) a = \(\frac{1}{2}\), b = 3
(B) a = \(-\frac{1}{2}\), b = -3
(C) a = \(-\frac{1}{2}\), b = 3
(D) a = \(\frac{1}{2}\), b = -3
Answer:
(A) a = \(\frac{1}{2}\), b = 3
Hint:
f(x) = ax² + bx + 1, |x – 1| ≥ 3
= 4x + 5; -2 < x < 4
The first interval is
|x – 1| ≥ 3
∴ x – 1 ≥ 3 or x – 1 ≤ -3
∴ x ≥ 4 or x ≤ -2
∴ f(x) is same for x ≤ -2 as well as x ≥ 4.
∴ f(x) is defined as:
f(x) = ax² + bx + 1; x ≤ -2
= 4x + 5; -2 < x < 4
= ax² + bx + 1; x ≥ 4
f(x) is continuous everywhere.
∴ f(x) is continuous at x = -2 and x = 4.
As f(x) is continuous at x = -2,
\(\lim _{x \rightarrow-2^{-}} f(x)=\lim _{x \rightarrow-2^{+}} f(x)\)
∴ \(\lim _{x \rightarrow-2}\left(a x^{2}+b x+1\right)=\lim _{x \rightarrow-2}(4 x+5)\)
∴ a(-2)² + b(-2) + 1 = 4(-2) + 5
∴ 4a – 2b + 1 = -3
∴ 4a – 2b = -4
∴ 2a – b = -2 …..(i)
∵ f(x) is continuous at x = 4,
\(\lim _{x \rightarrow 4^{-}} \mathrm{f}(x)=\lim _{x \rightarrow 4^{+}} \mathrm{f}(x)\)
∴ \(\lim _{x \rightarrow 4}(4 x+5)=\lim _{x \rightarrow 4}\left(a x^{2}+b x+1\right)\)
4(4) + 5 = a(4)² + b(4) + 1
16a + 4b + 1 = 21
16a + 4b = 20
4a + b = 5 …..(ii)
Adding (i) and (ii), we get
6a = 3
∴ a = \(\frac{1}{2}\)
Substituting a = \(\frac{1}{2}\) in (ii), we get
4(\(\frac{1}{2}\)) + b = 5
∴ 2 + b = 5
∴ b = 3
∴ a = \(\frac{1}{2}\), b = 3

Question 6.
f(x) = \(\frac{\left(16^{x}-1\right)\left(9^{x}-1\right)}{\left(27^{x}-1\right)\left(32^{x}-1\right)}\), for x ≠ 0
= k, for x = 0
is continuous at x = 0, then ‘k’ =
(A) \(\frac{8}{3}\)
(B) \(\frac{8}{15}\)
(C) \(-\frac{8}{15}\)
(D) \(\frac{20}{3}\)
Answer:
(B) \(\frac{8}{15}\)
Hint:

Question 7.
f(x) = \(\frac{32^{x}-8^{x}-4^{x}+1}{4^{x}-2^{x+1}+1}\), for x ≠ 0
= k, for x = 0,
is continuous at x = 0, then value of ‘k’ is
(A) 6
(B) 4
(C) (log 2) (log 4)
(D) 3 log 4
Answer:
(A) 6
Hint:

Question 8.
If f(x) = \(\frac{12^{x}-4^{x}-3^{x}+1}{1-\cos 2 x}\), for x ≠ 0 is continuous at x = 0 then the value of f(0) is
(A) \(\frac{\log 12}{2}\)
(B) log 2 . log 3
(C) \(\frac{\log 2 \cdot \log 3}{2}\)
(D) None of these
Answer:
(B) log 2 . log 3
Hint:

Question 9.
If f(x) = \(\left(\frac{4+5 x}{4-7 x}\right)^{\frac{4}{x}}\), for x ≠ 0 and f(0) = k, is continuous at x = 0, then k is
(A) e⁷
(B) e³
(C) e¹²
(D) \(e^{\frac{3}{4}}\)
Answer:
(C) e¹²
Hint:

Question 10.
If f(x) = \(\lfloor x\rfloor\) for x ∈ (-1, 2), then f is discontinuous at
(A) x = -1, 0, 1, 2
(B) x = -1, 0, 1
(C) x = 0, 1
(D) x = 2
Answer:
(C) x = 0, 1
Hint:
f(x) = \(\lfloor x\rfloor\), x ∈ (-1, 2)
This function is discontinuous at all integer values of x between -1 and 2.
∴ f(x) is discontinuous at x = 0 and x = 1.

II. Discuss the continuity of the following functions at the point(s) or on the interval indicated against them.

Question 1.
f(x) = \(\frac{x^{2}-3 x-10}{x-5}\), for 3 ≤ x ≤ 6, x ≠ 5
= 10, for x = 5
= \(\frac{x^{2}-3 x-10}{x-5}\), for 6 < x ≤ 9
Solution:

Question 2.
f(x) = 2x² – 2x + 5, for 0 ≤ x ≤ 2
= \(\frac{1-3 x-x^{2}}{1-x}\), for 2 < x < 4
= \(\frac{x^{2}-25}{x-5}\), for 4 ≤ x ≤ 7 and x ≠ 5
= 7, for x = 5
Solution:
The domain of f(x) is [0, 7].
(i) For 0 ≤ x ≤ 2
f(x) = 2x² – 2x + 5
It is a polynomial function and is Continuous at all point in [0, 2).

(ii) For 2 < x < 4
f(x) = \(\frac{1-3 x-x^{2}}{1-x}\)
It is a rational function and is continuous everwhere except at points where its denominator becomes zero.
Denominator becomes zero at x = 1
But x = 1 does not lie in the interval.
f(x) is continuous at all points in (2, 4).

(iii) For 4 ≤ x ≤ 7, x ≤ 5
f(x) = \(\frac{x^{2}-25}{x-5}\)
It is a rational function and is continuous everywhere except at points where its denominator becomes zero.
Denominator becomes zero at x = 5
But x = 5 does not lie in the interval.
∴ f(x) is continuous at all points in (4, 7] – {5}.

(iv) For continuity at x = 2:

∴ f(x) is continuous at x = 2.

(v) For continuity at x = 4:

∴ f(x) is continuous at x = 4.

(vi) For continuity at x = 5.
f(5) = 7

∴ f(x) is discontinuous at x = 5.
Thus, f(x) is continuous at all points on its domain except at x = 5.

Question 3.
f(x) = \(\frac{\cos 4 x-\cos 9 x}{1-\cos x}\), for x ≠ 0
f(0) = \(\frac{68}{15}\), at x = 0 on \(-\frac{\pi}{2}\) ≤ x ≤ \(\frac{\pi}{2}\)
Solution:
The domain of f(x) is [\(-\frac{\pi}{2}\), \(\frac{\pi}{2}\)]
(i) For [\(-\frac{\pi}{2}\), \(\frac{\pi}{2}\)] – {0}:
f(x) = \(\frac{\cos 4 x-\cos 9 x}{1-\cos x}\)
It is a rational function and is continuous everywhere except at points where its denominator becomes zero.
Denominator becomes zero when cos x = 1,
i.e., x = 0
But x = 0 does not lie in the interval.
∴ f(x) is continuous at all points in [\(-\frac{\pi}{2}\), \(\frac{\pi}{2}\)] – {0}

(ii) For continuity at x = 0:


∴ \(\lim _{x \rightarrow 0} f(x) \neq f(0)\)
∴ f(x) is discontinuous at x = 0.

Question 4.
f(x) = \(\frac{\sin ^{2} \pi x}{3(1-x)^{2}}\), for x ≠ 1
= \(\frac{\pi^{2} \sin ^{2}\left(\frac{\pi x}{2}\right)}{3+4 \cos ^{2}\left(\frac{\pi x}{2}\right)}\), for x = 1, at x = 1
Solution:

Question 5.
f(x) = \(\frac{|x+1|}{2 x^{2}+x-1}\), for x ≠ -1
= 0, for x = -1, at x = -1
Solution:

Question 6.
f(x) = [x + 1] for x ∈ [-2, 2)
Where [*] is greatest integer function.
Solution:
f(x) = [x + 1], x ∈ [-2, 2)
∴ f(x) = -1, x ∈ [-2, -1)
= 0, x ∈ [-1, 0)
= 1, x ∈ [0, 1)
= 2, x ∈ [1, 2)

∴ \(\lim _{x \rightarrow-1^{-}} \mathrm{f}(x)=\lim _{x \rightarrow-1^{+}} \mathrm{f}(x)\)
∴ f(x) is discontinuous at x = -1.
Similarly, f(x) is discontinuous at the points x = 0 and x = 1.

Question 7.
f(x) = 2x² + x + 1, for |x – 3| ≥ 2
= x² + 3, for 1 < x < 5
Solution:
|x – 3| ≥ 2
∴ x – 3 ≥ 2 or x – 3 ≤ -2
∴ x ≥ 5 or x ≤ 1
∴ f(x) = 2x² + x + 1, x ≤ 1
= x² + 3, 1 < x < 5
= 2x² + x + 1, x ≥ 5
Consider the intervals
x < 1 , i.e., (-∞, 1)
1 < x < 5, i.e., (1, 5) x > 5, i.e., (5, ∞)
In all these intervals, f(x) is a polynomial function and hence is continuous at all points.
For continuity at x = 1:

∴ f(x) is discontinuous at x = 5.
∴ f(x) is continuous for all x ∈ R, except at x = 5.

III. Identify discontinuities if any for the following functions as either a jump or a removable discontinuity on their respective domains.

Question 1.
f(x) = x² + x – 3, for x ∈ [-5, -2)
= x² – 5, for x ∈ (-2, 5]
Solution:
f(-2) has not been defined.
\(\lim _{x \rightarrow-2^{-}} f(x)=\lim _{x \rightarrow-2^{-}}\left(x^{2}+x-3\right)\)
= (-2)² + (-2) – 3
= 4 – 2 – 3
= -1
\(\lim _{x \rightarrow-2^{+}} f(x)=\lim _{x \rightarrow-2^{+}}\left(x^{2}-5\right)\)
= (-2)² – 5
= 4 – 5
= -1
∴ \(\lim _{x \rightarrow-2^{-}} f(x)=\lim _{x \rightarrow-2^{+}} f(x)\)
∴ \(\lim _{x \rightarrow-2} f(x) \text { exists. }\)
But f(-2) has not been defined.
∴ f(x) has a removable discontinuity at x = -2.

Question 2.
f(x) = x² + 5x + 1, for 0 ≤ x ≤ 3
= x³ + x + 5, for 3 < x ≤ 6
Solution:
\(\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{-}}\left(x^{2}+5 x+1\right)\)
= (3)² + 5(3) + 1
= 9 + 15 + 1
= 25
\(\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3^{+}}\left(x^{3}+x+5\right)\)
= (3)³ + 3 + 5
= 27 + 3 + 5
= 35
∴ \(\lim _{x \rightarrow 3^{-}} f(x) \neq \lim _{x \rightarrow 3^{+}} f(x)\)
∴ \(\lim _{x \rightarrow 3} f(x)\) does not exist.
∴ f(x) is discontinuous at x = 3.
∴ f(x) has a jump discontinuity at x = 3.

Question 3.
f(x) = \(\frac{x^{2}+x+1}{x+1}\), for x ∈ [0, 3)
= \(\frac{3 x+4}{x^{2}-5}\), for x ∈ [3, 6]
Solution:


∴ f(x) is continuous at x = 3.

IV. Discuss the continuity of the following functions at the point or on the interval indicated against them. If the function is discontinuous, identify the type of discontinuity and state whether the discontinuity is removable. If it has a removable discontinuity, redefine the function so that it becomes continuous.

Question 1.
f(x) = \(\frac{(x+3)\left(x^{2}-6 x+8\right)}{x^{2}-x-12}\)
Solution:
f(x) = \(\frac{(x+3)\left(x^{2}-6 x+8\right)}{x^{2}-x-12}\)
= \(\frac{(x+3)(x-2)(x-4)}{(x-4)(x+3)}\)
∴ f(x) is not defined at x = 4 and x = -3.
∴ The domain of function f = R – {-3, 4}.
For x ≠ -3 and 4,

f(x) is discontinuous at x = 4 and x = -3.
This discontinuity is removable.
∴ f(x) can be redefined as
f(x) = \(\frac{(x+3)\left(x^{2}-6 x+8\right)}{x^{2}-x-12}\), for x ≠ 4, x ≠ -3
= -5, for x ∈ R – {-3, 4}, x = -3
= 2, for x ∈ R – {-3, 4}, x = 4

Question 2.
f(x) = x² + 2x + 5, for x ≤ 3
= x³ – 2x² – 5, for x > 3
Solution:

∴ f(x) is discontinuous at x = 3.
This discontinuity is irremovable.

V. Find k if the following functions are continuous at the points indicated against them.

Question 1.
f(x) = \(\left(\frac{5 x-8}{8-3 x}\right)^{\frac{3}{2 x-4}}\), for x ≠ 2
= k, for x = 2 at x = 2.
Solution:
f(x) is continuous at x = 2. …..(given)

Question 2.
f(x) = \(\frac{45^{x}-9^{x}-5^{x}+1}{\left(k^{x}-1\right)\left(3^{x}-1\right)}\), for x ≠ 0
= \(\frac{2}{3}\), for x = 0, at x = 0
Solution:
f(x) is continuous at x = 0 …..(given)

VI. Find a and b if the following functions are continuous at the points or on the interval indicated against them.

Question 1.
f(x) = \(\frac{4 \tan x+5 \sin x}{a^{x}-1}\), for x < 0
= \(\frac{9}{\log 2}\), for x = 0
= \(\frac{11 x+7 x \cdot \cos x}{b^{x}-1}\), for x < 0
Solution:

Question 2.
f(x) = ax² + bx + 1, for |2x – 3| ≥ 2
= 3x + 2, for \(\frac{1}{2}\) < x < \(\frac{5}{2}\)
Solution:

VII. Find f(a), if f is continuous at x = a where,

Question 1.
f(x) = \(\frac{1+\cos (\pi x)}{\pi(1-x)^{2}}\), for x ≠ 1 and at a = 1.
Solution:

Question 2.
f(x) = \(\frac{1-\cos [7(x-\pi)]}{5(x-\pi)^{2}}\), for x ≠ π and at a = π.
Solution:

VIII. Solve using intermediate value theorem.

Question 1.
Show that 5^x – 6x = 0 has a root in [1, 2].
Solution:
Let f(x) = 5^x – 6x
5^x and 6x are continuous functions for all x ∈ R.
∴ 5^x – 6x is also continuous for all x ∈ R.
i.e., f(x) is continuous for all x ∈ R.
A root of f(x) exists, if f(x) = 0 for at least one value of x.
f(1) = 5¹ – 6(1) = -1 < 0
f(2) = (5)² – 6(2) = 13 > 0
∴ f(1) < 0 and f(2) > 0
∴ By intermediate value theorem, there has to be a point ‘c’ between 1 and 2 such that f(c) = 0.
∴ There is a root of the given equation in [1, 2].

Question 2.
Show that x³ – 5x² + 3x + 6 = 0 has at least two real roots between x = 1 and x = 5.
Solution:
Let f(x) = x³ – 5x² + 3x + 6
f(x) is a polynomial function and hence it is continuous for all x ∈ R.
A root of f(x) exists, if f(x) = 0 for at least one value of x.
Here, we have been asked to show that f(x) has at least two roots between x = 1 and x = 5.
f(1) = (1)³ – 5(1)² + 3(1) + 6
= 5 > 0
f(2) = (2)³ – 5(2)² + 3(2) + 6
= 8 – 20 + 6 + 6
= 0
∴ x = 2 is a root of f(x).
Also, f(3) = (3)³ – 5(3)² + 3(3) + 6
= 27 – 45 + 9 + 6
= -3 < 0
f(4) = (4)³ – 5(4)² + 3(4) + 6
= 64 – 80 + 12 + 6
= 2 > 0
∴ f(3) < 0 and f(4) > 0
∴ By intermediate value theorem, there has to be a point ‘c’ between 3 and 4 such that f(c) = 0.
∴ There are two roots, x = 2 and a root between x = 3 and x = 4.
Thus, there are at least two roots of the given equation between x = 1 and x = 5.

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 3 Permutations and Combination Ex 3.5 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 3 Permutations and Combination Ex 3.5

Question 1.
In how many different ways can 8 friends sit around a table?
Solution:
We know that ‘n’ persons can sit around a table in (n – 1)! ways.
∴ 8 friends can sit around a table in 7!
= 7 × 6 × 5 × 4 × 3 × 2 × 1
= 5040 ways.

Question 2.
A party has 20 participants. Find the number of distinct ways for the host to sit with them around a circular table. How many of these ways have two specified persons on either side of the host?
Solution:
A party has 20 participants.
All of them and the host (i.e., 21 persons) can be seated at a circular table in (21 – 1)! = 20! ways.
When two particular participants are seated on either side of the host.
The host takes the chair in 1 way.
These 2 persons can sit on either side of the host in 2! ways.
Once the host occupies his chair, it is not circular permutation more.
The remaining 18 people occupy their chairs in 18! ways.
∴ A total number of arrangements possible if two particular participants are seated on either side of the host = 2! × 18! = 2 × 18!

Question 3.
Delegates from 24 countries participate in a round table discussion. Find the number of seating arrangements where two specified delegates are (a) always together. (b) never together.
Solution:
(a) Delegates of 24 countries are to participate in a round table discussion such that two specified delegates are always together.
Let us consider these 2 delegates as one unit. They can be arranged among themselves in 2! ways.
Also, these two delegates are to be seated with 22 other delegates (i.e. total of 23) which can be done in (23 – 1)! = 22! ways.
∴ Required number of arrangements = 2! × 22!

(b) When 2 specified delegates are never together then, other 22 delegates can be participate in a round table discussion in (22 – 1)! = 21! ways.
∴ There are 22 places of which any 2 places can be filled by those 2 delegates so that they are never together.
∴ Two specified delegates can be arranged in ²²P₂ ways.
∴ Required number of arrangements = ²²P₂ × 21!
= \(\frac{22 !}{(22-2) !} \times 21 !\)
= \(\frac{22 !}{20 !}\) × 21!
= 22 × 21 × 21!
= 21 × 22 × 21!
= 21 × 22!

Question 4.
Find the number of ways for 15 people to sit around the table so that no two arrangements have the same neighbours.
Solution:
There are 15 people to sit around a table.
∴ They can be arranged in(15 – 1)! = 14! ways.
But, they should not have the same neighbour in any two arrangements.
Around the table, arrangements (i.e., clockwise and anticlockwise) coincide.

∴ Required number of arrangements = \(\frac{14 !}{2}\)

Question 5.
A committee of 10 members sits around a table. Find the number of arrangements that have the President and the Vice-president together.
Solution:
A committee of 10 members sits around a table.
But, President and Vice-president sit together.
Let us consider President and Vice-president as one unit.
They can be arranged among themselves in 2! ways.
Now, this unit with the other 8 members of the committee is to be arranged around a table, which can be done in (9 – 1)! = 8! ways.
∴ Required number of arrangements = 8! × 2! = 2 × 8!

Question 6.
Five men, two women, and a child sit around a table. Find the number of arrangements where the child is seated (a) between the two women. (b) between two men.
Solution:
5 men, 2 women, and a child sit around a table.
(a) When the child is seated between two women.
5 men, 2 women, and a child are to be seated around a round table such that the child is seated between two women.
∴ the two women can be seated on either side of the child in 2! ways.
Let us consider these 3 (two women and a child) as one unit.
Now, this one unit is to be arranged with the remaining 5 men,
i.e., a total of 6 units are to be arranged around a round table, which can be done in (6 – 1)! = 5! ways.
∴ Required number of arrangements = 5! × 2!
= 120 × 2
= 240

(b) Two men can be selected from 5 men in
⁵C₂ = \(\frac{5 !}{2 !(5-2) !}=\frac{5 \times 4 \times 3 !}{2 \times 3 !}\) = 10 ways.
Also, these two men can sit on either side of the child in 2! ways.
Let us take two men and a child as one unit.
Now, this one unit is to be arranged with the remaining 3 men and 2 women,
i.e., a total of 6 units (3 + 2 + 1) are to be arranged around a round table, which can be done in (6 – 1)! = 5! ways.
∴ Required number of arrangements = 10 × 2! × 5!
= 10 × 2 × 120
= 2400

Question 7.
Eight men and six women sit around a table. How many sitting arrangements will have no two women together?
Solution:
8 men can be seated around a table in (8 – 1)! = 7! ways.
No two women should sit together.
There are 8 gaps created by 8 men’s seats.
∴ Women can be seated in 8 gaps in ⁸P₆ ways.
∴ Required number of arrangements = 7! × ⁸P₆

Question 8.
Find the number of seating arrangements for 3 men and 3 women to sit around a table so that exactly two women are together.
Solution:
2 women (who wish to sit together) can be selected from 3 in
³C₂ = \(\frac{3 !}{2 !(3-2) !}=\frac{3 \times 2 !}{2 ! \times 1 !}\) = 3 ways.
Also, these two women can sit together in 2! ways.
Let us take two women as one unit.
Now, this one unit is to be arranged with the remaining 3 men and 1 woman,
i.e., a total of 5 units are to be arranged around a round table, which can be done in (5 – 1)! = 4! ways.
∴ Required number of arrangements = 3 × 2! × 4!
= 3 × 2 × 24
= 144

Question 9.
Four objects in a set of ten objects are alike. Find the number of ways of arranging them in a circular order.
Solution:
Ten things can be arranged in a circular order of which 4 are alike in \(\frac{9 !}{4 !}\) ways.
∴ Required number of arrangements = \(\frac{9 !}{4 !}\)

Question 10.
Fifteen persons sit around a table. Find the number of arrangements that have two specified persons not sitting side by side.
Solution:
Since 2 particular persons can’t be sitting side by side,
the other 13 persons can be arranged around the table in (13 – 1)! = 12! ways.
The two persons who are not sitting side by side may take 13 positions created by 3 persons in ¹³P₂ ways.
∴ Required number of arrangements = 12! × ¹³P₂
= 12! × 13 × 12
= 13 × 12! × 12
= 12 × 13!

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 2 Sequences and Series Ex 2.4 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.4

Question 1.
Verify whether the following sequences are H.P.
(i) \(\frac{1}{3}, \frac{1}{5}, \frac{1}{7}, \frac{1}{9}, \ldots\)
Solution:
\(\frac{1}{3}, \frac{1}{5}, \frac{1}{7}, \frac{1}{9}, \ldots\)
Here, the reciprocal sequence is 3, 5, 7, 9,…
t₁ = 3, t₂ = 5, t₃ = 7, t₄ = 9, …..
t₂ – t₁ = t₃ – t₂ = t₄ – t₃ = 2 = constant
∴ The reciprocal sequence is an A.P.
∴ The given sequence is a H.P.

(ii) \(\frac{1}{3}, \frac{1}{6}, \frac{1}{12}, \frac{1}{24}, \ldots\)
Solution:
\(\frac{1}{3}, \frac{1}{6}, \frac{1}{12}, \frac{1}{24}, \ldots\)
Here, the reciprocal sequence is 3, 6, 12, 24,…
t₁ = 3, t₂ = 6, t₃ = 12, ……
t₂ – t₁ = 3, t₃ – t₂ = 6
t₂ – t₁ ≠ t₃ – t₂
∴ The reciprocal sequence is not an A.P.
∴ The given sequence is not a H.P.

(iii) \(5, \frac{10}{17}, \frac{10}{32}, \frac{10}{47}, \ldots\)
Solution:

∴ The reciprocal sequence is an A.P.
∴ The given sequence is a H.P.

Question 2.
Find the nth term and hence find the 8th term of the following HPs.
(i) \(\frac{1}{2}, \frac{1}{5}, \frac{1}{8}, \frac{1}{11}, \ldots\)
Solution:
\(\frac{1}{2}, \frac{1}{5}, \frac{1}{8}, \frac{1}{11}, \ldots\) are in H.P.
∴ 2, 5, 8, 11,… are in A.P.
∴ a = 2, d = 3
t_n = a + (n – 1)d
= 2 + (n – 1)(3)
= 3n – 1
∴ nth term of H.P. = \(\frac{1}{3 n-1}\)
∴ 8th term of H.P. = \(\frac{1}{3(8)-1}\) = \(\frac{1}{23}\)

(ii) \(\frac{1}{4}, \frac{1}{6}, \frac{1}{8}, \frac{1}{10}, \ldots\)
Solution:
\(\frac{1}{4}, \frac{1}{6}, \frac{1}{8}, \frac{1}{10}, \ldots\) are in H.P.
∴ 4, 6, 8, 10, … are in A.P.
∴ a = 4, d = 2
t_n = a + (n – 1)d
= 4 + (n – 1) (2)
= 2n + 2
∴ nth term of H.P. = \(\frac{1}{2 n+2}\)
∴ 8th term of H.P. = \(\frac{1}{2(8)+2}\) = \(\frac{1}{18}\)

(iii) \(\frac{1}{5}, \frac{1}{10}, \frac{1}{15}, \frac{1}{20}, \ldots\)
Solution:
\(\frac{1}{5}, \frac{1}{10}, \frac{1}{15}, \frac{1}{20}, \ldots\) are in H.P.
∴ 5, 10, 15, 20, … are in A.P.
∴ a = 5, d = 5
t_n = a + (n – 1)d
= 5 + (n – 1) (5)
= 5n
∴ nth term of H.P. = \(\frac{1}{5 n}\)
∴ 8th term of H.P. = \(\frac{1}{5(8)}\) = \(\frac{1}{40}\)

Question 3.
Find A.M. of two positive numbers whose G.M. and H.M. are 4 and \(\frac{16}{5}\) respectively.
Solution:
G.M. = 4, H.M. = \(\frac{16}{5}\)
Now, (G.M.)² = (A.M.) (H.M.)
∴ 4² = A.M. × \(\frac{16}{5}\)
∴ A.M. = 16 × \(\frac{5}{16}\)
∴ A.M. = 5

Question 4.
Find H.M. of two positive numbers whose A.M. and G.M. are \(\frac{15}{2}\) and 6.
Solution:
A.M. = \(\frac{15}{2}\), G.M. = 6
Now, (G.M.)² = (A.M.) (H.M.)
∴ 6² = \(\frac{15}{2}\) × H.M.
∴ H.M. = 36 × \(\frac{2}{15}\)
∴ H.M. = \(\frac{24}{5}\)

Question 5.
Find G.M. of two positive numbers whose A.M. and H.M. are 75 and 48.
Solution:
A.M. = 75, H.M. = 48
Now, (G.M.)² = (A.M.) (H.M.)
∴ (G.M.)² = 75 × 48
∴ (G.M.)² = 25 × 3 × 16 × 3
∴ (G.M.)² = 5² × 4² × 3²
∴ G.M. = 5 × 4 × 3
∴ G.M. = 60

Question 6.
Insert two numbers between \(\frac{1}{4}\) and \(\frac{1}{3}\) so that the resulting sequence is a H.P.
Solution:
Let the required numbers be \(\frac{1}{\mathrm{H}_{1}}\) and \(\frac{1}{\mathrm{H}_{2}}\).
∴ \(\frac{1}{4}, \frac{1}{\mathrm{H}_{1}}, \frac{1}{\mathrm{H}_{2}}, \frac{1}{3}\) are in H.P.
∴ 4, H₁, H₂, 3 are in A.P.
t₁ = 4, t₂ = H₁, t₃ = H₂, t₄ = 3
∴ t₁ = a = 4, t₄ = 3
t_n = a + (n – 1)d
t₄ = 4 + (4 – 1)d
3 = 4 + 3d
3d = -1
∴ d = \(\frac{-1}{3}\)
H₁ = t₂ = a + d = 4 – \(\frac{1}{3}\) = \(\frac{11}{3}\)
H₂ = t₃ = a + 2d = 4 – \(\frac{2}{3}\) = \(\frac{10}{3}\)
∴ For resulting sequence to be H.P. we need to insert numbers \(\frac{3}{11}\) and \(\frac{3}{10}\).

Question 7.
Insert two numbers between 1 and -27 so that the resulting sequence is a G.P.
Solution:
Let the required numbers be G₁ and G₂.
∴ 1, G₁, G₂, -27 are in G.P.
t₁ = 1, t₂ = G₁, t₃ = G₂, t₄ = -27
∴ t₁ = a = 1
t_n = ar^n-1
t₄ = (1) r^4-1
-27 = r³
r³ = (-3)³
∴ r = -3
∴ G₁ = t₂ = ar = 1(-3) = -3
G₂ = t₃ = ar² = 1(-3)² = 9
∴ For resulting sequence to be G.P. we need to insert numbers -3 and 9.

Question 8.
If the A.M. of two numbers exceeds their G.M. by 2 and their H.M. by \(\frac{18}{5}\), find the numbers.
Solution:
Let a and b be the two numbers.
A = \(\frac{a+b}{2}\), G = \(\sqrt{a b}\), H = \(\frac{2 a b}{a+b}\)
According to the given conditions,

Consider, G = A – 2 = 10 – 2 = 8
\(\sqrt{a b}\) = 8
ab = 64
a(20 – a) = 64 …..[From (i)]
a² – 20a + 64 = 0
(a – 4)(a – 16) = 0
∴ a = 4 or a = 16
When a = 4, b = 20 – 4 = 16
When a = 16, b = 20 – 16 = 4
∴ The two numbers are 4 and 16.

Question 9.
Find two numbers whose A.M. exceeds their G.M. by \(\frac{1}{2}\) and their H.M. by \(\frac{25}{26}\).
Solution:
Let a and b be the two numbers.
A = \(\frac{a+b}{2}\), G = \(\sqrt{a b}\), H = \(\frac{2 a b}{a+b}\)
According to the given conditions,

\(\sqrt{a b}\) = 6
ab = 36
a(13 – a) = 36 ……[From (i)]
a² – 13a + 36 = 0
(a – 4)(a – 9) = 0
∴ a = 4 or a = 9
When a = 4, b = 13 – 4 = 9
When a = 9, b = 13 – 9 = 4
∴ The two numbers are 4 and 9.

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 5 Sets and Relations Ex 5.2 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 5 Sets and Relations Ex 5.2

Question 1.
If (x – 1, y + 4) = (1, 2), find the values of x and y.
Solution:
(x – 1, y + 4) = (1, 2)
By the definition of equality of ordered pairs, we have
x – 1 = 1 and y + 4 = 2
∴ x = 2 and y = -2

Question 2.
If \(\left(x+\frac{1}{3}, \frac{y}{3}-1\right)=\left(\frac{1}{2}, \frac{3}{2}\right)\), find x and y.
Solution:
\(\left(x+\frac{1}{3}, \frac{y}{3}-1\right)=\left(\frac{1}{2}, \frac{3}{2}\right)\)
By the definition of equality of ordered pairs, we have
x + \(\frac{1}{3}\) = \(\frac{1}{2}\) and \(\frac{y}{3}\) – 1 = \(\frac{3}{2}\)
∴ x = \(\frac{1}{2}\) – \(\frac{1}{3}\) and \(\frac{y}{3}\) = \(\frac{3}{2}\) + 1
∴ x = \(\frac{1}{6}\) and y = \(\frac{15}{2}\)

Question 3.
If A = {a, b, c}, B = {x, y}, find A × B, B × A, A × A, B × B.
Solution:
A = (a, b, c}, B = {x, y}
A × B = {(a, x), (a, y), (b, x), (b, y), (c, x), (c, y)}
B × A = {(x, a), (x, b), (x, c), (y, a), (y, b), (y, c)}
A × A = {(a, a), (a, b), (a, c), (b, a), (b, b), (b, c), (c, a), (c, b), (c, c)}
B × B = {(x, x), (x, y), (y, x), (y, y)}

Question 4.
If P = {1, 2, 3} and Q = {1, 4}, find sets P × Q and Q × P.
Solution:
P = {1, 2, 3}, Q = {1, 4}
∴ P × Q = {(1, 1), (1, 4), (2, 1), (2, 4), (3, 1), (3, 4)}
and Q × P = {(1, 1), (1, 2), (1, 3), (4, 1), (4, 2), (4, 3)}

Question 5.
Let A = {1, 2, 3, 4}, B = {4, 5, 6}, C = {5, 6}. Verify,
(i) A × (B ∩ C) = (A × B) ∩ (A × C)
(ii) A × (B ∪ C) = (A × B) ∪ (A × C)
Solution:
A = {1, 2, 3, 4}, B = {4, 5, 6}, C = {5, 6}
(i) B ∩ C = {5, 6}
A × (B ∩ C) = = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6)}
A × B = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}
A × C = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6)}
∴ (A × B) ∩ (A × C) = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6)}
∴ A × (B ∩ C) = (A × B) ∩ (A × C)

(ii) B ∪ C = {4, 5, 6}
A × (B ∪ C) = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3,4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}
A × B = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}
A × C = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6)}
∴ (A × B) ∪ (A × C) = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}
∴ A × (B ∪ C) = (A × B) ∪ (A × C)

Question 6.
Express {(x, y) / x² + y² = 100, where x, y ∈ W} as a set of ordered pairs.
Solution:
{(x, y) / x² + y² = 100, where x, y ∈ W}
We have, x² + y² = 100
When x = 0 and y = 10,
x² + y² = 0² + 10² = 100
When x = 6 and y = 8,
x² + y² = 6² + 8² = 100
When x = 8 and y = 6,
x² + y² = 8² + 6² = 100
When x = 10 and y = 0,
x² + y² = 10² + 0² = 100
∴ Set of ordered pairs = {(0, 10), (6, 8), (8, 6), (10, 0)}

Question 7.
Let A = {6, 8} and B = {1, 3, 5}. Show that R₁ = {(a, b) / a ∈ A, b ∈ B, a – b is an even number} is a null relation, R₂ = {(a, b) / a ∈ A, b ∈ B, a + b is an odd number} is a universal relation.
Solution:
A = {6, 8}, B = {1, 3, 5}
R₁ = {(a, b)/ a ∈ A, b ∈ B, a – b is an even number}
a ∈ A
∴ a = 6, 8
b ∈ B
∴ b = 1, 3, 5
When a = 6 and b = 1, a – b = 5, which is odd
When a = 6 and b = 3, a – b = 3, which is odd
When a = 6 and b = 5, a – b = 1, which is odd
When a = 8 and b = 1, a – b = 7, which is odd
When a = 8 and b = 3, a – b = 5, which is odd
When a = 8 and b = 5, a – b = 3, which is odd
Thus, no set of values of a and b gives a – b as even.
∴ R₁ has a null relation from A to B.
A × B = {(6, 1), (6, 3), (6, 5), (8, 1), (8, 3), (8, 5)}
When a = 6 and b = 1, a + b = 7, which is odd
When a = 6 and b = 3, a + b = 9, which is odd
When a = 6 and b = 5, a + b = 11, which is odd
When a = 8 and b = 1, a + b = 9, which is odd
When a = 8 and b = 3, a + b = 11, which is odd
When a = 8 and b = 5, a + b = 13, which is odd
∴ R₂ = {(6, 1), (6, 3), (6, 5), (8, 1), (8, 3), (8, 5)}
Here, R₂ = A × B
∴ R₂ has a universal relation from A to B.

Question 8.
Write the relation in the Roster form. State its domain and range.
(i) R₁ = {(a, a²) / a is a prime number less than 15}
(ii) R₂ = {(a, \(\frac{1}{a}\)) / 0 < a ≤ 5, a ∈ N}
(iii) R₃ = {(x, y / y = 3x, y ∈ {3, 6, 9, 12}, x ∈ {1, 2, 3}}
(iv) R₄ = {(x, y) / y > x + 1, x = 1, 2 and y = 2, 4, 6}
(v) R₅ = {(x, y) / x + y = 3, x, y ∈ {0, 1, 2, 3}}
(vi) R₆ = {(a, b) / a ∈ N, a < 6 and b = 4}
(vii) R₇ = {(a, b) / a, b ∈ N, a + b = 6}
(viii) R₈ = {(a, b)/ b = a + 2, a ∈ Z, 0 < a < 5}
Solution:
(i) R₁ = {(a, a²) / a is a prime number less than 15}
∴ a = 2, 3, 5, 7, 11, 13
∴ a² = 4, 9, 25, 49, 121, 169
∴ R₁ = {(2, 4), (3, 9), (5, 25), (7, 49), (11, 121), (13, 169)}
∴ Domain (R₁) = {a/a is a prime number less than 15}
= {2, 3, 5, 7, 11, 13}
Range (R₁) = {a²/a is a prime number less than 15}
= {4, 9, 25, 49, 121, 169}

(iii) R₃ = {(x, y) / y = 3x, x ∈ {1, 2, 3}, y ∈ {3, 6, 9, 12}}
Here y = 3x
When x = 1, y = 3(1) = 3
When x = 2, y = 3(2) = 6
When x = 3, y = 3(3) = 9
∴ R₃ = {(1, 3), (2, 6), (3, 9)}
∴ Domain (R₃) ={1, 2, 3}
∴ Range (R₃) = {3, 6, 9}

(iv) R₄ = {(x, y) / y > x + 1, x = 1, 2 and y = 2, 4, 6}
Here, y > x + 1
When x = 1 and y = 2, 2 ≯  1 + 1
When x = 1 and y = 4, 4 > 1 + 1
When x = 1 and y = 6, 6 > 1 + 1
When x = 2 and y = 2, 2 ≯  2 + 1
When x = 2 and y = 4, 4 > 2 + 1
When x = 2 and y = 6, 6 > 2 + 1
∴ R₄ = {(1, 4), (1, 6), (2, 4), (2, 6)}
Domain (R₄) = {1, 2}
Range (R₄) = {4, 6}

(v) R₅ = {{x, y) / x + y = 3, x, y ∈ (0, 1, 2, 3)}
Here, x + y = 3
When x = 0, y = 3
When x = 1, y = 2
When x = 2, y = 1
When x = 3, y = 0
∴ R₅ = {(0, 3), (1, 2), (2, 1), (3, 0)}
Domain (R₅) = {0, 1, 2, 3}
Range (R₅) = {3, 2, 1, 0}

(vi) R₆ = {(a, b)/ a ∈ N, a < 6 and b = 4}
a ∈ N and a < 6
∴ a = 1, 2, 3, 4, 5 and b = 4
R₆ = {(1, 4), (2, 4), (3, 4), (4, 4), (5, 4)}
Domain (R₆) = {1, 2, 3, 4, 5}
Range (R₆) = {4}

(vii) R₇ = {(a, b) / a, b ∈ N, a + b = 6}
Here, a + b = 6
When a = 1, b = 5
When a = 2, b = 4
When a = 3, b = 3
When a = 4, b = 2
When a = 5, b = 1
∴ R₇ = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}
Domain (R₇) = {1, 2, 3, 4, 5}
Range (R₇) = {5, 4, 3, 2, 1}

(viii) R₈ = {(a, b) / b = a + 2, a ∈ Z, 0 < a < 5}
Here, b = a + 2
When a = 1, b = 3
When a = 2, b = 4
When a = 3, b = 5
When a = 4, b = 6
∴ R₈ = {(1, 3), (2, 4), (3, 5), (4, 6)}
Domain (R₈) = {1, 2, 3, 4}
Range (R₈) = {3, 4, 5, 6}

Question 9.
Identify which of the following relations are reflexive, symmetric, and transitive.

Solution:

(i) Given, R = {(a, b): a, b ∈ Z, a – b is an integer}
Let a ∈ Z, then a – a ∈ Z
∴ (a, a) ∈ R
∴ R is reflexive.
Let (a, b) ∈ R
∴ a – b ∈ Z
∴ -(a – b) ∈ Z, i.e., b – a ∈ Z
∴ (b, a) ∈ R
∴ R is symmetric.
Let (a, b) and (b, c) ∈ R
∴ a – b ∈ Z and b – c ∈ Z
∴ (a – b) + (b – c) ∈ Z
∴ a – c ∈ Z
∴ (a, c) ∈ R
∴ R is transitive.

(ii) Given, R = {(a, b) : a, b ∈ N, a + b is even}
Let a ∈ N, then a + a = 2a, which is even.
∴ (a, a) ∈ R
∴ R is reflexive.
Let (a, b) ∈ R
∴ a + b is even
∴ b + a is even
∴ (b, a) ∈ R
∴ R is symmetric.
Let (a, b) and (b, c) ∈ R
∴ a + b and b + c is even
Let a + b = 2x and b + c = 2y for x, y ∈ N
∴ (a + b) + (b + c) = 2x + 2y
∴ a + 2b + c = 2(x + y)
∴ a + c = 2(x + y) – 2b = 2(x + y – b)
∴ a + c is even ……..[∵ x, y, b ∈ N, x + y – b ∈ N]
∴ (a, c) ∈ R
∴ R is transitive.

(iii) Given, R = {(a, b) : a, b ∈ N, a divides b}
Let a ∈ N, then a divides a.
∴ (a, a) ∈ R
∴ R is reflexive.
Let a = 2 and b = 8, then 2 divides 8
∴ (a, b) ∈ R
But 8 does not divide 2.
∴ (b, a) ∉ R
∴ R is not symmetric.
Let (a, b) and (b, c) ∈ R
∴ a divides b and b divides c.
Let b = ax and c = by for x, y ∈ N.
∴ c = (ax) y = a(xy)
i.e., a divides c.
∴ (a, c) ∈ R
∴ R is transitive.

(iv) Given, R = {(a, b) : a, b ∈ N, a² – 4ab + 3b² = 0}
Let a ∈ N, then a² – 4aa + 3a² = a² – 4a² + 3a² = 0
∴ (a, a) ∈ R
∴ R is reflexive.
Let a = 3 and b = 1,
then a² – 4ab + 3b² = 9 – 12 + 3 = 0
∴ (a, b) ∈ R
Consider, b² – 4ba + 3a² = 1 – 12 + 9 = -2 ≠ 0
∴ (b, a) ∉ R
∴ R is not symmetric.
Let a = 3, b = 1 and c = \(\frac{1}{3}\),
then a² – 4ab + 3b² = 9 – 12 + 3 = 0 and
b² – 4bc + 3c² = 1 – \(\frac{4}{3}\) + \(\frac{1}{3}\) = 1 – 1 = 0
∴ we get (a, b) and (b, c) ∈ R.
Consider, a² – 4ac + 3c² = 9 – 4 + \(\frac{1}{3}\) = \(\frac{16}{3}\) ≠ 0
∴ (a, c) ∉ R
∴ R is not transitive.

(v) Given, R = {(a, b) : a is sister of b and a, b ∈ G = Set of girls}
Let a ∈ G, then ‘a’ cannot be a sister of herself.
∴ (a, a) ∉ R
∴ R is not reflexive.
Let (a, b) ∈ R
∴ ‘a’ is a sister of ‘b’.
∴ ‘b’ is a sister of ‘a’.
∴ (b, c) ∈ R
∴ R is symmetric.
Let (a, b) and (b, c) ∈ R
∴ ‘a’ is a sister of ‘b’ and ‘b’ is a sister of ‘c’
∴ ‘a’ is a sister of ‘c’.
∴ (a, c) ∈ R
∴ R is transitive.

(vi) Given, R = {(a, b) : Line a is perpendicular to line b in a plane}
Let a be any line in the plane, then a cannot be perpendicular to itself.
∴ (a, a) ∉ R
∴ R is not reflexive.
Let (a, b) ∈ R
∴ a is perpendicular to b.
∴ b is perpendicular to a.
∴ (b, a) ∈ R.
∴ R is symmetric.
Let (a, b) and (b, c) ∈ R.
∴ a is perpendicular to b and b is perpendicular to c.
∴ a is parallel to c.
∴ (a, c) ∉ R
∴ R is not transitive.

(vii) Given, R = {(a, b) : a, b ∈ R, a < b}
Let a ∈ R, then a ≮ a.
∴ (a, a) ∉ R
∴ R is not reflexive.
Let a = 1 and b = 2, then 1 < 2
∴ (a, b) ∈ R
But 2 ≮ 1
∴ (b, a) ∉ R
∴ R is not symmetric.
Let (a, b) and (b, c) ∈ R
∴ a < b and b < c
∴ a < c
∴ (a, c) ∈ R
∴ R is transitive.

(viii) Given, R = {(a, b) : a, b ∈ R, a ≤ b³}
Let a = -3, then a³ = -27.
Here, a ≮ a
∴ (a, a) ∉ R
∴ R is not reflexive.
Let a = 2 and b = 9, then b³ = 729
Here, a < b³
∴ (a, b) ∈ R
Consider, a3 = 8
Here, b ≮ a³
∴ (b, a) ∉ R
∴ R is not symmetric.
Let a = 10, b = 3, c = 2,
then b³ = 27 and c³ = 8
Here, a < b³ and b < c³.
∴ (a, b) and (b, c) ∈ R
But a ≮ c³
∴ (a, c) ∉ R.
∴ R is not transitive.

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 2 Sequences and Series Ex 2.3 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.3

Question 1.
Determine whether the sum to infinity of the following G.P.s exist, if exists find them.
(i) \(\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \ldots\)
Solution:

(ii) \(2, \frac{4}{3}, \frac{8}{9}, \frac{16}{27}, \ldots\)
Solution:

(iii) \(-3,1, \frac{-1}{3}, \frac{1}{9}, \ldots\)
Solution:

(iv) \(\frac{1}{5}, \frac{-2}{5}, \frac{4}{5}, \frac{-8}{5}, \frac{16}{5}, \ldots\)
Solution:

(v) 9, 8.1, 7.29, ……
Solution:
9, 8.1, 7.29, …..
Here, a = 9, r = \(\frac{8.1}{9}\) = 0.9, |r| < 1
∴ Sum to infinity exists.
∴ Sum to infinity = \(\frac{\mathrm{a}}{1-\mathrm{r}}\)
= \(\frac{9}{1-0.9}\)
= \(\frac{9}{0.1}\)
= 90

Question 2.
Express the following recurring decimals as rational numbers.
(i) \(0 . \overline{7}\)
(ii) \(2 . \overline{4}\)
(iii) \(2.3 \overline{5}\)
(iv) \(51.0 \overline{2}\)
Solution:
(i) \(0 . \overline{7}\) = 0.7777… = 0.7 + 0.07 + 0.007 + ….
The terms 0.7, 0.07, 0.007,… are in G.P.
∴ a = 0.7, r = \(\frac{0.07}{0.7}\) = 0.1, |r| = |0.1| < 1
∴ Sum to infinity exists.
∴ Sum to infinity

(ii) \(2 . \overline{4}\) = 2.444 … = 2 + 0.4 + 0.04 + 0.004 + …
The terms 0.4, 0.04, 0.004,… are in G.P.
∴ a = 0.4, r = \(\frac{0.07}{0.7}\) = 0.1, |r| = 10.11 < 1
∴ Sum to infinity exists.
∴ Sum to infinity

(iii) \(2.3 \overline{5}\) = 2.3555… = 2.3 + 0.05 + 0.005 + 0.0005 + …
The terms 0.05,0.005,0.0005,… are in G.P.
∴ a = 0.05, r = \(\frac{0.005}{0.05}\) = 0.1, |r| = |0.1| < 1
∴ Sum to infinity exists.
∴ Sum to infinity

(iv) \(51.0 \overline{2}\) = 51.0222 …. = 51 + 0.02 + 0.002 + 0.0002 + …..
The terms 0.02, 0.002, 0.0002,… are in G.P.
∴ a = 0.02, r = \(\frac{0.002}{0.02}\) = 0.1, |r| = |0.1| < 1
∴ Sum to infinity exists.
∴ Sum to infinity

Question 3.
If the common ratio of a G.P. is \(\frac{2}{3}\) and the sum to infinity is 12, find the first term.
Solution:
r = \(\frac{2}{3}\), sum to infinity = 12 ….. [Given]
Sum to infinity = \(\frac{\mathrm{a}}{1-\mathrm{r}}\)
12 = \(\frac{a}{1-\frac{2}{3}}\)
a = 12 × \(\frac{1}{3}\)
∴ a = 4

Question 4.
If the first term of the G.P. is 6 and its sum to infinity is \(\frac{96}{17}\), find the common ratio.
Solution:
a = 6, sum to infinity = \(\frac{96}{17}\) …..[Given]
Sum to infinity = \(\frac{\mathrm{a}}{1-\mathrm{r}}\)

Question 5.
The sum of an infinite G.P. is 5 and the sum of the squares of these terms is 15, find the G.P.
Solution:
Let the required G.P. be a, ar, ar², ar³, …..
Sum to infinity of this G.P. = 5

Question 6.
Find
(i) \(\sum_{r=1}^{\infty} 4(0.5)^{r}\)
Solution:

(ii) \(\sum_{r=1}^{\infty}\left(-\frac{1}{3}\right)^{r}\)
Solution:

(iii) \(\sum_{r=0}^{\infty}(-8)\left(-\frac{1}{2}\right)^{r}\)
Solution:

(iv) \(\sum_{n=1}^{\infty} 0.4^{n}\)
Solution:

Question 7.
The midpoints of the sides of a square of side 1 are joined to form a new square. This procedure is repeated indefinitely. Find the sum of
(i) the areas of all the squares.
(ii) the perimeters of all the squares.
Solution:

(i) Area of the 1st square = 12
Area of the 2nd square = \(\left(\frac{1}{\sqrt{2}}\right)^{2}\)
Area of the 3rd square = \(\left(\frac{1}{2}\right)^{2}\)
and so on.
∴ Sum of the areas of all the squares

∴ Sum to infinity exists.
∴ Sum of the areas of all the squares = \(\frac{1}{1-\frac{1}{2}}\) = 2

(ii) Perimeter of 1st square = 4
Perimeter of 2nd square = 4\(\left(\frac{1}{\sqrt{2}}\right)\)
Perimeter of 3rd square = 4\(\left(\frac{1}{2}\right)\)
and so on.
Sum of the perimeters of all the squares

Question 8.
A ball is dropped from a height of 10 m. It bounces to a height of 6m, then 3.6 m, and so on. Find the total distance travelled by the ball.
Solution:
Here, on the first bounce, the ball will go 6 m and it will return 6 m.
On the second bounce, the ball will go 3.6 m and it will return 3.6 m, and so on.
Given that, a ball is dropped from a height of 10 m.
∴ Total distance travelled by the ball is = 10 + 2[6 + 3.6 + …]
The terms 6, 3.6 … are in G.P.
a = 6, r = 0.6, |r| = |0.6| < 1
∴ Sum to infinity exists.
∴ Total distance travelled by the ball

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

Maharashtra State Board 11th Maths Solutions Chapter 3 Permutations and Combination Miscellaneous Exercise 3

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

Question 1.
A college offers 5 courses in the morning and 3 in the evening. The number of ways a student can select exactly one course, either in the morning or in the evening is
(A) 5
(B) 3
(C) 8
(D) 15
Answer:
(C) 8
Hint:
Number of ways to select one course from available 8 courses
(i.e., 5 courses in the morning and 3 in the evening) = 5 + 3 = 8

Question 2.
A college has 7 courses in the morning and 3 in the evening. The possible number of choices with the student if he wants to study one course in the morning and one in the evening is
(A) 21
(B) 4
(C) 42
(D) 10
Answer:
(A) 21
Hint:
Number of ways to select one morning and one evening course = ⁷C₁ × ³C₁ = 21

Question 3.
In how many ways can 8 Indians and, 4 American and 4 Englishmen can be seated in a row so that all persons of the same nationality sit together?
(A) 3! 8!
(B) 3! 4! 8! 4!
(C) 4! 4!
(D) 8! 4! 4!
Answer:
(B) 3! 4! 8! 4!
Hint:
8 Indians take their seats in 8! ways, 4 Americans take their seats in 4! ways, 4 Englishmen take their seats in 4! ways.
Three groups of Indians, Americans and Englishmen can be permuted in 3! ways.
Required number = 3! × 8! × 4! × 4!

Question 4.
In how many ways can 10 examination papers be arranged so that the best and the worst papers never come together?
(A) 9 × 8!
(B) 8 × 8!
(C) 9 × 9!
(D) 8 × 9!
Answer:
(D) 8 × 9!
Hint:
Arrange 8 papers in 8! ways and two papers in 9 gaps are arranged in ⁹P₂ ways.
Required number = 8! ⁹P₂
= 8! × 9 × 8
= 9! × 8

Question 5.
In how many ways 4 boys and 3 girls can be seated in a row so that they are alternate.
(A) 12
(B) 288
(C) 144
(D) 256
Answer:
(C) 144
Hint:
B G B G B G B
4 boys take their seats in 4! ways.
3 girls take their seats in 3! ways.
Required number = 4! × 3!
= 24 × 6
= 144

Question 6.
Find the number of triangles which can be formed by joining the angular points of a polygon of 8 sides as vertices.
(A) 16
(B) 56
(C) 24
(D) 8
Answer:
(B) 56
Hint:
A triangle is obtained by joining three vertices.
Number of ways of selecting 3 vertices out of 8 vertices = ⁸C₃
= \(\frac{8 \times 7 \times 6}{1 \times 2 \times 3}\)
= 56

Question 7.
A question paper has two parts, A and B, each containing 10 questions. If a student has to choose 8 from part A and 5 from part B, in how many ways can he choose the questions?
(A) 320
(B) 750
(C) 40
(D) 11340
Answer:
(D) 11340
Hint:
Number of ways to choose 8 questions from Part A and 5 from Part B = ¹⁰C₈ × ¹⁰C₅
= ¹⁰C₂ × ¹⁰C₅
= 45 × 252
= 11340

Question 8.
There are 10 persons among whom two are brothers. The total number of ways in which these persons can be seated around a round table so that exactly one person sits between the brothers is equal to:
(A) 2! × 7!
(B) 2! × 8!
(C) 3! × 7!
(D) 3! × 8!
Answer:
(B) 2! × 8!
Hint:
Select a person from 8 people (i.e., the people excluding two brothers).
This is done in 8 ways.
2 brothers sit adjacent to the selected person on two sides, they may interchange their seats.
Remaining 7 people sit in 7! ways
Required number = 8 × 2 × 7! = 2! × 8!

Question 9.
The number of arrangements of the letters of the word BANANA in which two N’s do not appear adjacently is
(A) 80
(B) 60
(C) 40
(D) 100
Answer:
(C) 40
Hint:
Arrange B, A, A, A in \(\frac{4 !}{3 !}\) ways.
These four letters create 5 gaps in which 2 N are to be filled, this can be done in 5C2 ways, we do not permute those 2N as they are identical.
∴ Required number = \(\frac{4 !}{3 !}\) × ⁵C₂ = 40

Question 10.
The number of ways in which 5 male and 2 female members of a committee can be seated around a round table so that the two females are not seated together is
(A) 840
(B) 600
(C) 720
(D) 480
Answer:
(D) 480
Hint:
5 males take their seats in 4! ways, creating 5 gaps.
In these 5 gaps, 2 females are to be seated.
∴ The number of ways to do this = ⁵C₂ × 2!
Required number = 4! × ⁵C₂ × 2! = 480

(II) Answer the following.

Question 1.
Find the value of r if ⁵⁶P_r+2 : ⁵⁴P_r-1 = 30800 : 1.
Solution:

Question 2.
How many words can be formed by writing letters in the word CROWN in a different order?
Solution:
Five Letters of the word CROWN are to be permuted.
∴ Number of different words = 5! = 120

Question 3.
Find the number of words that can be formed by using all the letters in the word REMAIN. If these words are written in dictionary order, what will be the 40th word?
Solution:
There are 6 letters A, E, I, M, N, R.
Number of words that can be formed by using all these letters = 6! = 720
When a word starts with ‘A’,
‘A’ can be arranged in 1 way and the remaining 5 letters can be arranged among themselves in 5! ways.
The number of words starting with A = 5!
∴ Similarly,
The number of words starting with E = 5!
The number of words starting with I = 5!
The number of words starting with M = 5!
The number of words starting with N = 5!
The number of words starting with R = 5!
Total number of words = 6 × 5! = 720
Number of words starting with AE = 4! = 24
Number of words starting with AIE = 3! = 6
Number of words starting with AIM = 3! = 6
Number of words starting with AINE = 2!
Total words = 24 + 6 + 6 + 2 = 38
39th word is AINMER
40th word is AINMRE

Question 4.
The Capital English alphabet has 11 symmetric letters that appear the same when looked at in a mirror. These letters are A, H, I, M, O, T, U, V, W, X, and Y. How many symmetric three letters passwords can be formed using these letters?
Solution:
There are 11 symmetric letters.
∴ Number of 3 Letter passwords = ¹¹P₃
= 11 × 10 × 9
= 990

Question 5.
How many numbers formed using the digits 3, 2, 0, 4, 3, 2, 3 exceed one million?
Solution:
A number that exceeds one million is to be formed from the digits 3, 2, 0, 4, 3, 2, 3.
Then the numbers should be any number of 7 digits which can be formed from these digits.
Also, among the given numbers 2 is repeated twice and 3 is repeated thrice.
∴ Required number of numbers = Total number of arrangements possible among these digits – number of arrangements of 7 digits which begin with 0.
= \(\frac{7 !}{2 ! 3 !}-\frac{6 !}{2 ! 3 !}\)
= \(\frac{7 \times 6 \times 5 \times 4 \times 3 !}{2 \times 3 !}-\frac{6 \times 5 \times 4 \times 3 !}{2 \times 3 !}\)
= 7 × 6 × 5 × 2 – 6 × 5 × 2
= 6 × 5 × 2(7 – 1)
= 60 × 6
= 360

Question 6.
Ten students are to be selected for a project from a class of 30 students. There are 4 students who want to be together either in the project or not in the project. Find the number of possible selections.
Solution:
Ten students are to be selected for a project from a class of 30 students.
Case I:
If 4 students join the project, then from remaining 26 students, rest of the 6 students are to be selected.
Which can be done in ²⁶C₆
= \(\frac{26 !}{6 !(26-6) !}\)
= \(\frac{26 \times 25 \times 24 \times 23 \times 22 \times 21 \times 20 !}{6 ! \times 20 !}\)
= 230230 ways.

Case II:
If 4 students does not join the project, then from remaining 26 students, all the 10 students are to be selected.
Which can be done in ²⁶C₁₀
= \(\frac{26 !}{10 !(26-10) !}\)
= \(\frac{26 \times 25 \times 24 \times 23 \times 22 \times 21 \times 20 \times 19 \times 18 \times 17 \times 16 !}{10 ! \times 16 !}\)
= 5311735 ways.
∴ Required number of selections = ²⁶C₆ + ²⁶C₁₀
= 230230 + 5311735
= 5541965

Question 7.
A student finds 7 books of his interest but can borrow only three books. He wants to borrow the Chemistry part II book only if Chemistry Part I can also be borrowed. Find the number of ways he can choose three books that he wants to borrow.
Solution:
There are 7 books of student’s interest, but he can borrow only three books.
He wants to borrow the Chemistry part II book only if Chemistry Part I can also be borrowed.
Consider the following table:

Required number of selections = 5 + 10 = 15

Question 8.
30 objects are to be divided into three groups containing 7, 10, 13 objects. Find the number of distinct ways of doing so.
Solution:
First we can select 7 objects out of 30 for the first group in ³⁰C₇ ways.
Now there are 23 objects left out of which we can select 10 objects for the second group in ²³C₁₀ ways.
Remaining 13 objects can be selected for the third group in ⁵C₅ ways.
∴ Required number of ways = ³⁰C₇ × ²³C₁₀ × ¹³C₁₃
= \(\frac{30 !}{23 ! 7 !} \times \frac{23 !}{10 ! 13 !} \times 1\)
= \(\frac{30 !}{7 ! 10 ! 13 !}\)

Question 9.
A student passes an examination if he secures a minimum in each of the 7 subjects. Find the number of ways a student can fail.
Solution:
Every subject a student may pass or fail.
∴ Total number of outcomes = 2⁷ = 128
This number includes one case when the student passes in all subjects.
Required number of ways = 128 – 1 = 127

Question 10.
Nine friends decide to go for a picnic in two groups. One group decides to go by car and the other group decides to go by train. Find the number of different ways of doing so if there must be at least 3 friends in each group.
Solution:
Nine friends decide to go for a picnic in two groups and there must be at least 3 friends in each group.
Consider the following table:

Question 11.
A hall has 12 lamps and every lamp can be switched on independently. Find the number of ways of illuminating the hall.
Solution:
Every lamp is either ON or OFF.
There are 12 lamps
Number of instances = 2¹²
This number includes one case in when all 12 lamps are OFF.
∴ Required Number of ways = 2¹² – 1 = 4095

Question 12.
How many quadratic equations can be formed using numbers from 0, 2, 4, 5 as coefficients if a coefficient can be repeated in an equation?
Solution:
A quadratic equation is to be formed using numbers 0, 2, 4, 5 as coefficients and a coefficient can be repeated.
Let the quadratic equation be ax² + bx + c = 0, a ≠ 0
Consider the following table:

Number of quadratic equations can be formed = 3 × 4 × 4 = 48

Question 13.
How many six-digit telephone numbers can be formed if the first two digits are 45 and no digit can appear more than once?
Solution:
There are total of 10 digits.
Let the telephone number be 45abcd.
There are 8 digits left for the choice of a, b, c, d as repetition is not allowed.
Consider the following table:

∴ Required number of numbers formed = 8 × 7 × 6 × 5 = 1680

Question 14.
A question paper has 6 questions. How many ways does a student have to answer if he wants to solve at least one question?
Solution:
Every question is ‘SOLVED’ or ‘NOT SOLVED’.
There are 6 questions.
Number of outcomes = 2⁶
This number includes one case when the student solves NONE of the questions.
∴ Required number of ways = 2⁶ – 1 = 64 – 1 = 63

Question 15.
Find the number of ways of dividing 20 objects into three groups of sizes 8, 7, and 5.
Solution:
First we can select 8 objects our of 20 for the first group in ²⁰C₈ ways.
Now there are 12 objects left out of which we can select 7 objects for the second group in ¹²C₇ ways.
Remaining 5 objects can be selected for the third group in ⁵C₅ ways.
∴ Required number of ways = ²⁰C₈ × ¹²C₇ × ⁵C₅
= \(\frac{20 !}{8 ! 12 !} \times \frac{12 !}{7 ! 5 !} \times 1\)
= \(\frac{20 !}{8 ! 7 ! 5 !}\)

Question 16.
There are 4 doctors and 8 lawyers in a panel. Find the number of ways for selecting a team of 6 if at least one doctor must be in the team.
Solution:
There are 4 doctors and 8 lawyers in a panel.
A team of 6 with at least one doctor is to be formed.
We count the number by the INDIRECT method of counting.
Number of ways to select a team of 6 people = ¹²C₆
Number of teams with No doctor in any team = ⁸C₆
∴ Required number of ways = ¹²C₆ – ⁸C₆
= 924 – 28
= 896

Question 17.
Four parallel lines intersect another set of five parallel lines. Find the number of distinct parallelograms formed.
Solution:
The first set has 4 parallel lines and another set has 5 parallel lines.
To form a parallelogram, we need 2 lines from each set.
∴ Required number of distinct parallelograms formed = ⁴C₂ × ⁵C₂
= 6 × 10
= 60

Question 18.
There are 12 distinct points A, B, C, …, L, in order, on a circle. Lines are drawn passing through each pair of points.
(i) How many lines are there in total?
(ii) How many lines pass through D?
(iii) How many triangles are determined by lines?
(iv) How many triangles have on vertex C?
Solution:
(i) We need two points to draw a line.
∴ Total number of lines = ¹²C₂ = 66

(ii) Lines are drawn passing through each pair of points.
∴ Lines from point D will pass through all the remaining 11 points.
∴ 11 lines pass through D.

(iii) We need three points to draw a triangle.
∴ Number of triangles = ¹²C₃ = 220

(iv) To get the triangles with one vertex as C,
we need two vertices from the remaining 11 vertices.
∴ Number of triangles with vertex at C = ¹¹C₂
= \(\frac{11 \times 10}{2}\)
= 55

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 1 Complex Numbers Ex 1.2 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.2

Question 1.
Find the square root of the following complex numbers:
(i) -8 – 6i
Solution:
Let \(\sqrt{-8-6 i}\) = a + bi, where a, b ∈ R.
Squaring on both sides, we get
-8 – 6i = (a + bi)²
-8 – 6i = a² + b²i² + 2abi
-8 – 6i = (a² – b²) + 2abi …..[∵ i² = -1]
Equating real and imaginary parts, we get
a² – b² = -8 and 2ab = -6

(ii) 7 + 24i
Solution:
Let \(\sqrt{7+24 i}\) = a + bi, where a, b ∈ R.
Squaring on both sides, we get
7 + 24i = (a + bi)²
7 + 24i = a² + b²i² + 2abi
7 + 24i = (a² – b²) + 2abi …..[∵ i² = -1]
Equating real and imaginary parts, we get
a² – b² = 7 and 2ab = 24

(iii) 1 + 4√3 i
Solution:
Let \(\sqrt{1+4 \sqrt{3} i}\) = a + bi, where a, b ∈ R.
Squaring on both sides, we get
1 + 4√3 i = (a + bi)²
1 + 4√3i = a² + b²i² + 2abi
1 + 4√3i = (a² – b²) + 2abi …..[∵ i² = -1]
Equating real arid imaginary parts, we get
a² – b² = 1 and 2ab = 4√3

(iv) 3 + 2√10 i
Solution:
Let \(\sqrt{3+2 \sqrt{10}} i\) = a + bi, where a, b ∈ R.
Squaring on both sides, we get
3 + 2√10 i = a² + b²i² + 2abi
3 + 2√10 i = (a² – b²) + 2abi ……[∵ i² = -1]
Equating real and imaginary parts, we get
a² – b² = 3 and 2ab = 2√10

(v) 2(1 – √3 i)
Solution:
Let \(\sqrt{2(1-\sqrt{3} i)}\) = a + bi, where a, b ∈ R.
Squaring on both sides, we get
2(1 – √3 i) = a² + b²i² + 2abi
2 – 2√3 i = (a² – b²) + 2abi ….[∵ i² = -1]
Equating real and imaginary parts, we get
a² – b² = 2 and 2ab = -2√3
a² – b² = 2 and b = \(-\frac{\sqrt{3}}{a}\)

Question 2.
Solve the following quadratic equations:
(i) 8x² + 2x + 1 = 0
Solution:
Given equation is 8x² + 2x + 1 = 0
Comparing with ax² + bx + c = 0, we get
a = 8, b = 2, c = 1
Discriminant = b² – 4ac
= (2)² – 4 × 8 × 1
= 4 – 32
= -28 < 0
So, the given equation has complex roots.
These roots are given by

(ii) 2x² – √3 x + 1 = 0
Solution:
Given equation is 2x² – √3 x + 1 = 0
Comparing with ax² + bx + c = 0, we get
a = 2, b = -√3, c = 1
Discriminant = b² – 4ac
= (-√3)² – 4 × 2 × 1
= 3 – 8
= -5 < 0
So, the given equation has complex roots.
These roots are given by

(iii) 3x² – 7x + 5 = 0
Solution:
Given equation is 3x² – 7x + 5 = 0
Comparing with ax² + bx + c = 0, we get
a = 3, b = -7, c = 5
Discriminant = b² – 4ac
= (-7)² – 4 × 3 × 5
= 49 – 60
= -11 < 0
So, the given equation has complex roots.
These roots are given by

The roots of the given equation are \(\frac{7+\sqrt{11} \mathrm{i}}{6}\) and \(\frac{7-\sqrt{11} \mathrm{i}}{6}\)

(iv) x² – 4x + 13 = 0
Solution:
Given equation is x² – 4x + 13 = 0
Comparing with ax² + bx + c = 0, we get
a = 1, b = -4, c = 13
Discriminant = b² – 4ac
= (-4)² – 4 × 1 × 13
= 16 – 52
= -36 < 0
So, the given equation has complex roots.
These roots are given by

∴ The roots of the given equation are 2 + 3i and 2 – 3i.

Question 3.
Solve the following quadratic equations:
(i) x² + 3ix + 10 = 0
Solution:
Given equation is x² + 3ix + 10 = 0
Comparing with ax² + bx + c = 0, we get
a = 1, b = 3i, c = 10
Discriminant = b² – 4ac
= (3i)² – 4 × 1 × 10
= 9i² – 40
= -9 – 40 ……[∵ i² = -1]
= -49 < 0
So, the given equation has complex roots.
These roots are given by

∴ x = 2i or x = -5i
∴ The roots of the given equation are 2i and -5i.

(ii) 2x² + 3ix + 2 = 0
Solution:
Given equation is 2x² + 3ix + 2 = 0
Comparing with ax + bx + c = 0, we get
a = 2, b = 3i, c = 2
Discriminant = b² – 4ac
= (3i)² – 4 × 2 × 2
= 9i² – 16
= -9 – 16 …..[∵ i² = -1]
= -25 < 0
So, the given equation has complex roots.
These roots are given by

∴ The roots of the given equation are \(\frac{1}{2}\)i and -2i.

(iii) x² + 4ix – 4 = 0
Solution:
Given equation is x² + 4ix – 4 = 0
Comparing with ax² + bx + c = 0, we get
a = 1, b = 4i, c = -4
Discriminant = b² – 4ac
= (4i)² – 4 × 1 × (-4)
= 16i² + 16
= -16 + 16 …..[∵ i² = -1]
= 0
So, the given equation has equal roots.
These roots are given by

∴ x = -2i
∴ The root of the given equation is -2i.

(iv) ix² – 4x – 4i = 0
Solution:
ix² – 4x – 4i = 0
Multiplying throughout by i, we get
i2x² – 4ix – 4i² = 0
-x² – 4ix + 4 = 0 …[∵ i² = -1]
x² + 4ix – 4 = 0
Comparing with ax² + bx + c = 0, we get
a = 1, b = 4i, c = -4
Discriminant = b² – 4ac
= (4i)² – 4 × 1 × (-4)
= 16i² + 16
= -16 + 16
= 0
So, the given equation has equal roots.
These roots are given by

∴ The root of the given equation is -2i

Question 4.
Solve the following quadratic equations:
(i) x² – (2 + i) x – (1 – 7i) = 0
Solution:
Given equation is x² – (2 + i)x – (1 – 7i) = 0
Comparing with ax² + bx + c = 0, we get
a = 1, b = -(2 + i), c = -(1 – 7i)
Discriminant = b² – 4ac
= [-(2 + i)]² – 4 × 1 × -(1 – 7i)
= 4 + 4i + i² + 4 – 28i
= 4 + 4i – 1 + 4 – 28i …..[∵ i² = – 1]
= 7 – 24i
So, the given equation has complex roots.
These roots are given by

Let \(\sqrt{7-24 i}\) = a + bi, where a, b ∈ R
Squaring on both sides, we get
7 – 24i = a² + i²b² + 2abi
7 – 24i = a² – b² + 2abi
Equating real and imaginary parts, we get
a² – b² = 7 and 2ab = -24

(ii) x² – (3√2 + 2i) x + 6√2 i = 0
Solution:
Given equation is x² – (3√2 + 2i) x + 6√2 i = 0
Comparing with ax² + bx + c = 0, we get
a = 1, b = -(3√2 + 2i), c = 6√2i
Discriminant = b² – 4ac
= [-(3√2 + 2i)]² – 4 × 1 × 6√2 i
= 18 + 12√2i + 4i² – 24√2 i
= 18 – 12√2 i – 4 ……[∵ i² = -1]
= 14 – 12√2 i
So, the given equation has complex roots.
These roots are given by

(iii) x² – (5 – i) x + (18 + i) = 0
Solution:
Given equation is x² – (5 – i)x + (18 + i) = 0
Comparing with ax² + bx + c = 0, we get
a = 1, b = -(5 – i), c = 18 + i
Discriminant = b² – 4ac
= [-(5 – i)]² – 4 × 1 × (18 + i)
= 25 – 10i + i² – 72 – 4i
= 25 – 10i – 1 – 72 – 4i ……[∵ i² = -1]
= -48 – 14i
So, the given equation has complex roots.
These roots are given by

(iv) (2 + i) x² – (5 – i) x + 2(1 – i) = 0
Solution:
Given equation is (2 + i) x² – (5 – i) x + 2(1 – i) = 0
Comparing with ax² + bx + c = 0, we get
a = 2 + i, b = -(5 – i), c = 2(1 – i)
Discriminant = b² – 4ac
= [-(5 – i)]² – 4 × (2 + i) × 2(1 – i)
= 25 – 10i + i² – 8(2 + i) (1 – i)
= 25 – 10i + i² – 8(2 – 2i + i – i²)
= 25 – 10i – 1 – 8(2 – i + 1) …..[∵ i² = -1]
= 25 – 10i – 1 – 16 + 8i – 8
= -2i
So, the given equation has complex roots.
These roots are given by

Let \(\sqrt{-2 i}\) = a + bi, where a, b ∈ R
Squaring on both sides, we get
-2i = a² + b²i² + 2abi
-2i = a² – b² + 2abi
Equating real and imaginary parts, we get
a² – b² = 0 and 2ab = -2
a² – b² = 0 and b = \(-\frac{1}{a}\)

Question 5.
Find the value of
(i) x³ – x² + x + 46, if x = 2 + 3i
Solution:
x = 2 + 3i
x – 2 = 3i
(x – 2)² = 9i²
x² – 4x + 4 = 9(-1) …..[∵ i² = -1]
x² – 4x + 13 = 0 …..(i)

Dividend = Divisor × Quotient + Remainder
∴ x³ – x² + x + 46 = (x² – 4x + 13) (x + 3) + 7
= 0(x + 3) + 7 …..[from(i)]
= 7

Alternate Method:
x = 2 + 3i
α = 2 + 3i, \(\bar{\alpha}\) = 2 – 3i
α\(\bar{\alpha}\) = (2 + 3i)(2 – 3i)
= 4 – 6i + 6i – 9i²
= 4 – 9(-1)
= 4 + 9
= 13
α + \(\bar{\alpha}\) = 2 + 3i + 2 – 3i = 4
∴ Standard form of quadratic equation,
x² – (Sum of roots) x + Product of roots = 0
x² – 4x + 13 = 0

Dividend = Divisor × Quotient + Remainder
∴ x³ – x² + x + 46 = (x² – 4x + 13).(x + 3) + 7
= 0(x + 3) + 7 …..[From (i)]
= 7

(ii) 2x³ – 11x² + 44x + 27, if x = \(\frac{25}{3-4 i}\)
Solution:


Dividend = Divisor × Quotient + Remainder
2x³ – 11x² + 44x + 27 = (x² – 6x + 25)(2x + 1) + 2
= 0.(2x + 1) + 2 …..[From (i)]
= 0 + 2
= 2

(iii) x³ + x² – x + 22, if x = \(\frac{5}{1-2 i}\)
Solution:

Dividend = Divisor × Quotient + Remainder
x³ + x² – x + 22 = (x² – 2x + 5)(x + 3) + 7
= 0.(x + 3) + 7 …..[From (i)]
= 0 + 7
= 7

(iv) x⁴ + 9x³ + 35x² – x + 4, if x = -5 + √-4
Solution:
x = -5 + √-4
x + 5 = √-4
x + 5 = √4 √-1
x + 5 = 2i
(x + 5)² = 4i²
x² + 10x + 25 = 4(-1) ….[∵ i² = -1]
x² + 10x + 29 = 0 …..(i)

Dividend = Divisor × Quotient + Remainder
x⁴ + 9x³ + 35x² – x + 4 = (x² + 10x + 29) (x² – x + 16) – 132x – 460
= 0.(x² – x + 16) – 132x – 460 …..[From (i)]
= -132 (-5 + 2i) – 460
= 660 – 264i – 460
= 200 – 264i

(v) 2x⁴ + 5x³ + 7x² – x + 41, if x = -2 – √3i
Solution:

Dividend = Divisor × Quotient + Remainder
2x⁴ + 5x³ + 7x² – x + 41 = (x² + 4x + 7) (2x² – 3x + 5) + 6
= 0(2x² – 3x + 5) + 6 ……[From (i)]
= 0 + 6
= 6