Bhagya

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 7 Limits Ex 7.7 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 7 Limits Ex 7.7

I. Evaluate the following:

Question 1.
\(\lim _{x \rightarrow \infty}\left[\frac{a x^{3}+b x^{2}+c x+d}{e x^{3}+f x^{2}+g x+h}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow \infty}\left[\frac{x^{3}+3 x+2}{(x+4)(x-6)(x-3)}\right]\)
Solution:

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

II. Evaluate the following:

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

Question 2.
\(\lim _{x \rightarrow \infty}\left[\sqrt{x^{2}+4 x+16}-\sqrt{x^{2}+16}\right]\)
Solution:

Question 3.
\(\lim _{x \rightarrow \infty}\left[\sqrt{x^{4}+4 x^{2}}-x^{2}\right]\)
Solution:

III. Evaluate the following:

Question 1.
\(\lim _{x \rightarrow \infty}\left[\frac{\left(3 x^{2}+4\right)\left(4 x^{2}-6\right)\left(5 x^{2}+2\right)}{4 x^{6}+2 x^{4}-1}\right]\)
Solution:

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

Question 3.
\(\lim _{x \rightarrow \infty}[\sqrt{x}(\sqrt{x+1}-\sqrt{x})]\)
Solution:

Question 4.
\(\lim _{x \rightarrow \infty}\left[\frac{(2 x-1)^{20}(3 x-1)^{30}}{(2 x+1)^{50}}\right]\)
Solution:

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

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 3 Permutations and Combination Ex 3.6 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 3 Permutations and Combination Ex 3.6

Question 1.
Find the value of
(a) ¹⁵C₄
Solution:

(b) ⁸⁰C₂
Solution:

(c) ¹⁵C₄ + ¹⁵C₅
Solution:

(d) ²⁰C₁₆ – ¹⁹C₁₆
Solution:

Question 2.
Find n if
(a) ⁶P₂ = n(⁶C₂)
Solution:

(b) ²ⁿC₃ : ⁿC₂ = 52 : 3
Solution:

(c) ⁿC_n-3 = 84
Solution:

Question 3.
Find r if ¹⁴C_2r : ¹⁰C_2r-4 = 143 : 10.
Solution:

∴ 2r(2r – 1) (2r – 2) (2r – 3) = 14 × 12 × 10
∴ 2r(2r – 1) (2r – 2) (2r – 3) = 8 × 7 × 6 × 5
Comparing on both sides, we get
∴ r = 4

Question 4.
Find n and r if,
(a) ⁿP_r = 720 and ⁿC_n-r = 120
Solution:

(b) ⁿC_r-1 : ⁿC_r : ⁿC_r+1 = 20 : 35 : 42
Solution:

Question 5.
If ⁿP_r = 1814400 and ⁿC_r = 45, find ⁿ⁺⁴C_r+3.
Solution:

Question 6.
If ⁿC_r-1 = 6435, ⁿC_r = 5005, ⁿC_r+1 = 3003, find ^rC₅.
Solution:

Question 7.
Find the number of ways of drawing 9 balls from a bag that has 6 red balls, 8 green balls, and 7 blue balls so that 3 balls of every colour are drawn.
Solution:
9 balls are to be selected from 6 red, 8 green, 7 blue balls such that the selection consists of 3 balls of each colour.
∴ 3 red balls can be selected from 6 red balls in ⁶C₃ ways.
3 reen balls can be selected from 8 green balls in ⁸C₃ ways.
3 blue balls can be selected from 7 blue balls in ⁷C₃ ways.
∴ Number of ways selection can be done if the selection consists of 3 balls of each colour

Question 8.
Find the number of ways of selecting a team of 3 boys and 2 girls from 6 boys and 4 girls.
Solution:
There are 6 boys and 4 girls.
A team of 3 boys and 2 girls is to be selected.
∴ 3 boys can be selected from 6 boys in ⁶C₃ ways.
2 girls can be selected from 4 girls in ⁴C₂ ways.
∴ Number of ways the team can be selected

Question 9.
After a meeting, every participant shakes hands with every other participants. If the number of handshakes is 66, find the number of participants in the meeting.
Solution:
Let there be n participants present in the meeting.
A handshake occurs between 2 persons.
∴ Number of handshakes = ⁿC₂
Given 66 handshakes were exchanged.
66 = ⁿC₂
66 = \(\frac{\mathrm{n} !}{2 !(\mathrm{n}-2) !}\)
66 × 2 = \(\frac{n(n-1)(n-2) !}{(n-2) !}\)
132 = n (n – 1)
n(n – 1) = 12 × 11
Comparing on both sides, we get n = 12
∴ 12 participants were present at the meeting.

Question 10.
If 20 points are marked on a circle, how many chords can be drawn?
Solution:
To draw a chord we need to join two points on the circle.
There are 20 points on a circle.
∴ Total number of chords possible from these points

Question 11.
Find the number of diagonals of an n-sided polygon. In particular, find the number of diagonals when
(i) n = 10
(ii) n = 15
(iii) n = 12
(iv) n = 8
Solution:
In n-sided polygon, there are ‘n’ points and ‘n’ sides.
∴ Through ‘n’ points we can draw ⁿC₂ lines including sides.
∴ Number of diagonals in n sided polygon = ⁿC₂ – n (n = number of sides)

Question 12.
There are 20 straight lines in a plane so that no two lines are parallel and no three lines are concurrent. Determine the number of points of intersection.
Solution:
There are 20 lines such that no two of them are parallel and no three of them are concurrent.
Since no two lines are parallel, they intersect at a point.
∴ Number of points of intersection if no two lines are parallel and no three lines are concurrent = ²⁰C₂
= \(\frac{20 !}{2 ! 18 !}\)
= \(\frac{20 \times 19 \times 18 !}{2 \times 1 \times 18 !}\)
= 190

Question 13.
Ten points are plotted on a plane. Find the number of straight lines obtained by joining these points if (a) no three points are collinear (b) four points are collinear
Solution:
There are 10 points on a plane.
(a) When no three of them are collinear.
A line is obtained by joining 2 points.
∴ Number of lines passing through these points = ¹⁰C₂
= \(\frac{10 !}{2 ! 8 !}\)
= \(\frac{10 \times 9 \times 8 !}{2 \times 1 \times 8 !}\)
= 5 × 9
= 45

(b) When 4 of them are collinear.
If no three points are collinear, we get a total of ¹⁰C₂ = 45 lines by joining them. …..[From (i)]
Since 4 points are collinear, only one line passes through these points instead of ⁴C₂ lines.
∴ ⁴C₂ – 1 extra lines are included in 45 lines.
Number of lines passing through these points
= 45 – (⁴C₂ – 1)
= 45 – \(\frac{4 !}{2 ! 2 !}\) + 1
= 45 – \(\frac{4 \times 3 \times 2 !}{2 \times 2 !}\) + 1
= 45 – 6 + 1
= 40

Question 14.
Find the number of triangles formed by joining 12 points if
(a) no three points are collinear
(b) four points are collinear
Solution:
There are 12 points on the plane.
(a) When no three of them are collinear.
A triangle can be drawn by joining any three non-collinear points.
∴ Number of triangles that can be obtained from these points = ¹²C₃
= \(\frac{12 !}{3 ! 9 !}\)
= \(\frac{12 \times 11 \times 10 \times 9 !}{3 \times 2 \times 1 \times 9 !}\)
= 220

(b) When 4 of these points are collinear.
If no three points are collinear, total we get ¹²C₃ = 220 triangles by joining them. ……[From (i)]
Since 4 points are collinear, no triangle can be formed by joining these four points.
∴ ⁴C₃ extra triangles are included in 220 triangles.
∴ Number of triangles that can be obtained from these points = ¹²C₃ – ⁴C₃
= 220 – \(\frac{4 !}{3 ! \times 1 !}\)
= 220 – \(\frac{4 \times 3 !}{3 !}\)
= 220 – 4
= 216

Question 15.
A word has 8 consonants and 3 vowels. How many distinct words can be formed if 4 consonants and 2 vowels are chosen?
Solution:
There are 8 consonants and 3 vowels.
From 8 consonants, 4 can be selected in ⁸C₄
= \(\frac{8 !}{4 ! 4 !}\)
= \(\frac{8 \times 7 \times 6 \times 5 \times 4 !}{4 \times 3 \times 2 \times 1 \times 4 !}\)
= 70 ways.
From 3 vowels, 2 can be selected in ³C₂
= \(\frac{3 !}{2 ! 1 !}\)
= \(\frac{3 \times 2 !}{2 !}\)
= 3 ways.
Now, to form a word, these 6 ietters (i.e., 4 consonants and 2 vowels) can be arranged in ⁶P₆ = 6! ways.
∴ Total number of words that can be formed = 70 × 3 × 6!
= 70 × 3 × 720
= 151200
∴ 151200 words of 4 consonants and 2 vowels can be formed.

Question 16.
Find n if,
(i) ⁿC₈ = ⁿC₁₂
Solution:
ⁿC₈ = ⁿC₁₂
If ⁿC_x = ⁿC_y, then either x = y or x = n – y
∴ 8 = 12 or 8 = n – 12
But 8 = 12 is not possible
∴ 8 = n – 12
∴ n = 20

(ii) ²³C_3n = ²³C_2n+3
Solution:
²³C_3n = ²³C_2n+3
If ⁿC_x = ⁿC_y, then either x = y or x = n – y
∴ 3n = 2n + 3 or 3n = 23 – 2n – 3
∴ n = 3 or n = 4

(iii) ²¹C_6n = \({ }^{21} \mathrm{C}_{\left(\mathrm{n}^{2}+5\right)}\)
Solution:
²¹C_6n = \({ }^{21} \mathrm{C}_{\left(\mathrm{n}^{2}+5\right)}\)
If ⁿC_x = ⁿC_y, then either x = y or x = n – y
∴ 6n = n² + 5 or 6n = 21 – (n² + 5)
∴ n² – 6n + 5 = 0 or 6n = 21 – n² – 5
∴ n² – 6n + 5 = 0 or n² + 6n – 16 = 0
If n² – 6n + 5 = 0, then (n – 1)(n – 5) = 0
∴ n = 1 or n = 5
If n = 5 then
n² + 5 = 30 > 21
∴ n ≠ 5
∴ n = 1
If n² + 6n – 16 = 0, then (n + 8)(n – 2) = 0
n = -8 or n = 2
n ≠ -8
∴ n = 2
∴ n = 1 or n = 2

Check:
n = 2
∴ n² + 5 = 2² + 5 = 9
²¹C_6n = ²¹C₁₂
and \({ }^{21} \mathrm{C}_{\left(\mathrm{n}^{2}+5\right)}\) = ²¹C₉
∴ ⁿC_r = ⁿC_n-r
∴ ²¹C₁₂ = ²¹C₉
∴ n = 2 is a right answer.

(iv) ²ⁿC_r-1 = ²ⁿC_r+1
Solution:
²ⁿC_r-1 = ²ⁿC_r+1
If ⁿC_x = ⁿC_y, then either x = y or x = n – y
∴ r – 1 = r + 1 or r – 1 = 2n – (r + 1)
But r – 1 = r + 1 is not possible
∴ r – 1 = 2n – (r + 1)
∴ r + r = 2n
∴ r = n

Check:
²ⁿC_r-1 = ²ⁿC_n-1
and ²ⁿC_r+1 = ²ⁿC_n+1
using ⁿC_r = ⁿC_n-r, we have
²ⁿC_n+1 = ²ⁿC_2n-(n+1) = ²ⁿC_n-1
∴ ²ⁿC_r-1 = ²ⁿC_r+1

(v) ⁿC_n-2 = 15
Solution:
ⁿC_n-2 = 15
∴ ⁿC₂ = 15 …..[∵ ⁿC_r = ⁿC_n-r]
∴ \(\frac{n !}{(n-2) ! 2 !}=15\)
∴ \(\frac{n(n-1)(n-2) !}{(n-2) ! \times 2 \times 1}=15\)
∴ n(n – 1) = 30
∴ n(n – 1) = 6 × 5
Equating both sides, we get
∴ n = 6

Question 17.
Find x if ⁿP_r = x ⁿC_r.
Solution:

Question 18.
Find r if ¹¹C₄ + ¹¹C₅ + ¹²C₆ + ¹³C₇ = ¹⁴C_r.
Solution:

Question 19.
Find the value of \(\sum_{r=1}^{4}{ }^{(21-r)} \mathrm{C}_{4}\).
Solution:

Question 20.
Find the differences between the greatest values in the following:
(a) ¹⁴C_r and ¹²C_r
Solution:
Greatest value of ¹⁴C_r.
Here, n = 14, which is even.
Greatest value of nCr occurs at r = \(\frac{n}{2}\) if n is even.
∴ r = \(\frac{n}{2}\)
∴ r = \(\frac{14}{2}\) = 7

∴ Difference between the greatest values of ¹⁴C_r and ¹²C_r = ¹⁴C_r – ¹²C_r
= 3432 – 924
= 2508

(b) ¹³C_r and ⁸C_r
Solution:
Greatest value of ¹³C_r.
Here n = 13, which is odd.
Greatest value of nCr occurs at r = \(\frac{n-1}{2}\) if n is odd.
∴ r = \(\frac{\mathrm{n}-1}{2}\)
∴ r = \(\frac{13-1}{2}\) = 6

∴ Difference between the greatest values of ¹³C_r and ⁸C_r = ¹³C_r – ⁸C_r
= 1716 – 70
= 1646

(c) ¹⁵C_r and ¹¹C_r
Solution:
Greatest value of ¹⁵C_r.
Here n = 15, which is odd.
Greatest value of nCr occurs at r = \(\frac{n-1}{2}\) if n is odd.
∴ r = \(\frac{n-1}{2}\)
∴ r = \(\frac{15-1}{2}\) = 7

∴ Difference between the greatest values of ¹⁵C_r and ¹¹C_r = ¹⁵C_r – ¹¹C_r
= 6435 – 462
= 5973

Question 21.
In how many ways can a boy invite his 5 friends to a party so that at least three join the party?
Solution:
Boy can invite = (3 or 4 or 5 friends)
Consider the following table:

∴ Number of ways a boy can invite his friends to a party so that three or more of join the party = 10 + 5 + 1 = 16

Question 22.
A group consists of 9 men and 6 women. A team of 6 is to be selected. How many of possible selections will have at least 3 women?
Solution:
There are 9 men and 6 women.
A team of 6 persons is to be formed such that it consist of at least 3 women.
Consider the following table:

∴ No. of ways this can be done = 1680 + 540 + 54 + 1 = 2275
∴ 2275 teams can be formed if team consists of at least 3 women.

Question 23.
A committee of 10 persons is to be formed from a group of 10 women and 8 men. How many possible committees will have at least 5 women? How many possible committees will have men in majority?
Solution:
(i) A committee of 10 persons is to be formed from 10 women and 8 men such that the committee contains at least 5 women.
Consider the following table:

∴ Number of committees with at least 5 women
= 14112 + 14700 + 6720 + 1260 + 81
= 36873

(ii) Number of committees with men in majority = Total number of committees – (Number of committees with women in majority + women and men equal in number)
= ¹⁸C₁₀ – 36873
= ¹⁸C₈ – 36873
= 43758 – 36873
= 6885

Question 24.
A question paper has two sections. Section I has 5 questions and section II has 6 questions. A student must answer at least two questions from each section among 6 questions he answers. How many different choices does the student have in choosing questions?
Solution:
There are 11 questions, out of which 5 questions are from section I and 6 questions are from section II.
The student has to select 6 questions taking at least 2 questions from each section.
Consider the following table:

∴ Number of choices = 150 + 200 + 75 = 425
∴ In 425 ways students can select 6 questions, taking at least 2 questions from each section.

Question 25.
There are 3 wicketkeepers and 5 bowlers among 22 cricket players. A team of 11 player is to be selected so that there is exactly one wicketkeeper and at least 4 bowlers in the team. How many different teams can be formed?
Solution:
There are 22 cricket players, of which 3 are wicketkeepers and 5 are bowlers.
A team of 11 players is to be chosen such that exactly one wicket keeper and at least 4 bowlers are to be included in the team.
Consider the following table:

∴ Number of ways a team of 11 players can be selected
= 45045 + 6006
= 51051

Question 26.
Five students are selected from 11. How many ways can these students be selected if
(a) two specified students are selected?
(b) two specified students are not selected?
Solution:
5 students are to be selected from 11 students.
(a) When 2 specified students are included,
then remaining 3 students can be selected from (11 – 2) = 9 students.
∴ Number of ways of selecting 3 students from 9 students = ⁹C₃
= \(\frac{9 !}{3 ! \times 6 !}\)
= \(\frac{9 \times 8 \times 7 \times 6 !}{3 \times 2 \times 1 \times 6 !}\)
= 84
∴ Selection of students is done in 84 ways when 2 specified students are included.

(b) When 2 specified students are not included, then 5 students can be selected from the remaining (11 – 2) = 9 students.
∴ Number of ways of selecting 5 students from 9 students = ⁹C₅
= \(\frac{9 !}{5 ! 4 !}\)
= \(\frac{9 \times 8 \times 7 \times 6 \times 5 !}{5 ! \times 4 \times 3 \times 2 \times 1}\)
= 126
∴ Selection of students is done in 126 ways when 2 specified students are not included.

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 7 Limits Ex 7.6 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 7 Limits Ex 7.6

I. Evaluate the following limits:

Question 1.
\(\lim _{x \rightarrow 0}\left[\frac{9^{x}-5^{x}}{4^{x}-1}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow 0}\left[\frac{5^{x}+3^{x}-2^{x}-1}{x}\right]\)
Solution:

Question 3.
\(\lim _{x \rightarrow 0}\left(\frac{a^{x}+b^{x}+c^{x}-3}{\sin x}\right)\)
Solution:

Question 4.
\(\lim _{x \rightarrow 0}\left(\frac{6^{x}+5^{x}+4^{x}-3^{x+1}}{\sin x}\right)\)
Solution:

Question 5.
\(\lim _{x \rightarrow 0}\left(\frac{8^{\sin x}-2^{\tan x}}{e^{2 x}-1}\right)\)
Solution:

II. Evaluate the following limits:

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

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

Question 3.
\(\lim _{x \rightarrow 0}\left[\frac{5 x+3}{3-2 x}\right]^{\frac{2}{x}}\)
Solution:

Question 4.
\(\lim _{x \rightarrow 0}\left[\frac{\log (3-x)-\log (3+x)}{x}\right]\)
Solution:

Question 5.
\(\lim _{x \rightarrow 0}\left[\frac{4 x+1}{1-4 x}\right]^{\frac{1}{x}}\)
Solution:

Question 6.
\(\lim _{x \rightarrow 0}\left[\frac{5+7 x}{5-3 x}\right]^{\frac{1}{3 x}}\)
Solution:

III. Evaluate the following limits:

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

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

Question 3.
\(\lim _{x \rightarrow 0}\left[\frac{15^{x}-5^{x}-3^{x}+1}{x \cdot \sin x}\right]\)
Solution:

Question 4.
\(\lim _{x \rightarrow 0}\left[\frac{(25)^{x}-2(5)^{x}+1}{x \cdot \sin x}\right]\)
Solution:

Question 5.
\(\lim _{x \rightarrow 0}\left[\frac{(49)^{x}-2(35)^{x}+(25)^{x}}{\sin x \cdot \log (1+2 x)}\right]\)
Solution:

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 7 Limits Ex 7.5 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 7 Limits Ex 7.5

I. Evaluate the following:

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

Question 2.
\(\lim _{x \rightarrow a} \frac{\sin x-\sin a}{\sqrt[5]{x}-\sqrt[5]{a}}\)
Solution:

Question 3.
\(\lim _{x \rightarrow \pi}\left[\frac{\sqrt{5+\cos x}-2}{(\pi-x)^{2}}\right]\)
Solution:

Question 4.
\(\lim _{x \rightarrow \frac{\pi}{6}}\left[\frac{\cos x-\sqrt{3} \sin x}{\pi-6 x}\right]\)
Solution:

Question 5.
\(\lim _{x \rightarrow 1}\left[\frac{1-x^{2}}{\sin \pi x}\right]\)
Solution:

II. Evaluate the following:

Question 1.
\(\lim _{x \rightarrow \frac{\pi}{6}}\left[\frac{2 \sin x-1}{\pi-6 x}\right]\)
Solution:

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

Question 3.
\(\lim _{x \rightarrow \frac{\pi}{6}}\left[\frac{2-\sqrt{3} \cos x-\sin x}{(6 x-\pi)^{2}}\right]\)
Solution:

Question 4.
\(\lim _{x \rightarrow a}\left[\frac{\sin (\sqrt{x})-\sin (\sqrt{a})}{x-a}\right]\)
Solution:

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

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 1 Complex Numbers Ex 1.1 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.1

Question 1.
Simplify:
(i) √-16 + 3√-25 + √-36 – √-625
Solution:

= 4i + 3(5i) + 6i – 25i
= 25i – 25i
= 0

(ii) 4√-4 + 5√-9 – 3√-16
Solution:

Question 2.
Write the conjugates of the following complex numbers
(i) 3 + i
Solution:
Conjugate of (3 + i) is (3 – i).

(ii) 3 – i
Solution:
Conjugate of (3 – i) is (3 + i).

(iii) √-5 – √7 i
Solution:
Conjugate of (√-5 – √7 i) is (√-5 + √7 i).

(iv) -√-5
Solution:
-√-5 = -√5 × √-1 = -√5 i
Conjugate of (-√-5) is √5 i

(v) 5i
Solution:
Conjugate of (5i) is (-5i).

(vi) √5 – i
Solution:
Conjugate of (√5 – i) is (√5 + i).

(vii) √2 + √3 i
Solution:
Conjugate of (√2 + √3 i) is (√2 – √3 i)

(viii) cos θ + i sin θ
Solution:
Conjugate of (cos θ + i sin θ) is (cos θ – i sin θ)

Question 3.
Find a and b if
(i) a + 2b + 2ai = 4 + 6i
Solution:
a + 2b + 2ai = 4 + 6i
Equating real and imaginary parts, we get
a + 2b = 4 …..(i)
2a = 6 ……(ii)
∴ a = 3
Substituting, a = 3 in (i), we get
3 + 2b = 4
∴ b = \(\frac{1}{2}\)
∴ a = 3 and b = \(\frac{1}{2}\)

Check:
For a = 3 and b = \(\frac{1}{2}\)
Consider, L.H.S. = a + 2b + 2ai
= 3 + 2(\(\frac{1}{2}\)) + 2(3)i
= 4 + 6i
= R.H.S.

(ii) (a – b) + (a + b)i = a + 5i
Solution:
(a – b) + (a + b)i = a + 5i
Equating real and imaginary parts, we get
a – b = a ……(i)
a + b = 5 ……(ii)
From (i), b = 0
Substituting b = 0 in (ii), we get
a + 0 = 5
∴ a = 5
∴ a = 5 and b = 0

(iii) (a + b) (2 + i) = b + 1 + (10 + 2a)i
Solution:
(a + b) (2 + i) = b + 1 + (10 + 2a)i
2(a + b) + (a + b)i = (b + 1) + (10 + 2a)i
Equating real and imaginary parts, we get
2(a + b) = b + 1
∴ 2a + b = 1 ……(i)
and a + b = 10 + 2a
-a + b = 10 …….(ii)
Subtracting equation (ii) from (i), we get
3a = -9
∴ a = -3
Substituting a = – 3 in (ii), we get
-(-3) + b = 10
∴ b = 7
∴ a = -3 and b = 7

(iv) abi = 3a – b + 12i
Solution:
abi = 3a – b + 12i
∴ 0 + abi = (3a – b) + 12i
Equating real and imaginary parts, we get
3a – b = 0
∴ 3a = b …..(i)
and ab = 12
∴ b = \(\frac{12}{a}\) ……..(ii)
Substituting b = \(\frac{12}{a}\) in (i), we get
3a = \(\frac{12}{a}\)
3a² = 12
a² = 4
a = ±2
When a = 2, b = \(\frac{12}{a}\) = \(\frac{12}{2}\) = 6
When a = -2, b = \(\frac{12}{a}\) = \(\frac{12}{-2}\) = -6
∴ a = 2 and b = 6 or a = -2 and b = -6

(v) \(\frac{1}{a+i b}\) = 3 – 2i
Solution:

(vi) (a + ib) (1 + i) = 2 + i
Solution:
(a + ib)(1 + i) = 2 + i
a + ai + bi + bi² = 2 + i
a + (a + b)i + b(-1) = 2 + i ……(∵ i² = -1)
(a – b) + (a + b)i = 2 + i
Equating real and imaginary parts, we get
a – b = 2 ……(i)
a + b = 1 …….(ii)
Adding equations (i) and (ii), we get
2a = 3
∴ a = \(\frac{3}{2}\)
Substituting a = \(\frac{3}{2}\) in (ii), we get
\(\frac{3}{2}\) + b = 1
∴ b = 1 – \(\frac{3}{2}\) = \(\frac{-1}{2}\)
∴ a = \(\frac{3}{2}\) and b = \(\frac{-1}{2}\)

Question 4.
Express the following in the form of a + ib, a, b ∈ R, i = √-1. State the values of a and b:
(i) (1 + 2i)(-2 + i)
Solution:
(1 + 2i)(-2 + i) = -2 + i – 4i + 2i²
= -2 – 3i + 2(-1) ……[∵ i² = -1]
∴ (1 + 2i)(-2 + i) = -4 – 3i
∴ a = -4 and b = -3

(ii) (1 + i)(1 – i)-1
Solution:

(iii) \(\frac{i(4+3 i)}{1-i}\)
Solution:

(iv) \(\frac{(2+i)}{(3-i)(1+2 i)}\)
Solution:

(v) \(\left(\frac{1+i}{1-1}\right)^{2}\)
Solution:

(vi) \(\frac{3+2 i}{2-5 i}+\frac{3-2 i}{2+5 i}\)
Solution:

(vii) (1 + i)^-3
Solution:

(viii) \(\frac{2+\sqrt{-3}}{4+\sqrt{-3}}\)
Solution:

(ix) (-√5 + 2√-4 ) + (1 – √-9 ) + (2 + 3i)(2 – 3i)
Solution:
(-√5 + 2√-4) + (1 – √-9) + (2 + 3i)(2 – 3i)
= (-√5 + 2√4.√-1) + (1 – √9.√-1) + 4 – 9i²
= [-√5 + 2(2)i] + (1 – 3i) + 4 – 9i²
= -√5 + 4i + 1 – 3i + 4 – 9(-1) ……[∵ i² = -1]
= (14 – √5) + i
∴ a = 14 – √5 and b = 1

(x) (2 + 3i)(2 – 3i)
Solution:
(2 + 3i)(2 – 3i)
= 4 – 9i²
= 4 – 9(-1) …[∵ i² = -1]
= 4 + 9
= 13
∴ (2 + 3i)(2 – 3i) = 13 + 0i
∴ a = 13 and b = 0

(xi) \(\frac{4 i^{8}-3 i^{9}+3}{3 i^{11}-4 i^{10}-2}\)
Solution:

Question 5.
Show that (-1 + √3i)³ is a real number.
Solution:

Question 6.
Find the value of (3 + \(\frac{2}{\mathrm{i}}\)) (i⁶ – i⁷) (1 + i¹¹).
Solution:

Question 7.
Evaluate the following:
(i) i³⁵
(ii) i⁸⁸⁸
(iii) i⁹³
(iv) i¹¹⁶
(v) i⁴⁰³
(vi) \(\frac{1}{i^{58}}\)
(vii) i^-888
(viii) i³⁰ + i⁴⁰ + i⁵⁰ + i⁶⁰
Solution:

Question 8.
Show that 1 + i¹⁰ + i²⁰ + i³⁰ is a real number.
Solution:

= 1 – 1 + 1 – 1
= 0, which is a real number.

Question 9.
Find the value of
(i) i⁴⁹ + i⁶⁸ + i⁸⁹ + i¹¹⁰
(ii) i + i² + i³ + i⁴
Solution:

Question 10.
Simplify: \(\frac{\mathbf{i}^{592}+\mathbf{i}^{590}+\mathbf{i}^{588}+\mathbf{i}^{586}+\mathbf{i}^{584}}{\mathbf{i}^{582}+\mathbf{i}^{580}+\mathbf{i}^{578}+\mathbf{i}^{576}+\mathbf{i}^{574}}\)
Solution:

Question 11.
Find the value of 1 + i² + i⁴ + i⁶ + i⁸ + …… + i²⁰.
Solution:

Question 12.
Show that 1 + i¹⁰ + i¹⁰⁰ – i¹⁰⁰⁰ = 0.
Solution:

Question 13.
Is (1 + i¹⁴ + i¹⁸ + i²²) a real number? Justify your answer.
Solution:

Question 14.
Evaluate: \(\left(\mathbf{i}^{37}+\frac{1}{\mathbf{i}^{67}}\right)\)
Solution:

Question 15.
Prove that: (1 + i)⁴ × \(\left(1+\frac{1}{\mathrm{i}}\right)^{4}\) = 16
Solution:

Question 16.
Find the value of \(\frac{\mathbf{i}^{6}+\mathbf{i}^{7}+\mathbf{i}^{8}+\mathbf{i}^{9}}{\mathbf{i}^{2}+\mathbf{i}^{3}}\)
Solution:

Question 17.
If a = \(\frac{-1+\sqrt{3} i}{2}\), b = \(\frac{-1-\sqrt{3} i}{2}\), then show that a² = b and b² = a.
Solution:

Question 18.
If x + iy = (a + ib)³, show that \(\frac{x}{a}+\frac{y}{b}\) = 4(a² – b²)
Solution:
x + iy = (a + ib)³
x + iy = a³ + 3a²bi + 3ab²i² + b³i³
x + iy = a³ + 3a²bi – 3ab² – b³i ……[∵ i² = -1, i³ = -i]
x + iy = (a³ – 3ab²) + (3a²b – b³)i
Equating real and imaginary parts, we get
x = a³ – 3ab² and y = 3a²b – b³
\(\frac{x}{a}\) = a² – 3b² and \(\frac{y}{b}\) = 3a² – b²
\(\frac{x}{a}+\frac{y}{b}\) = a² – 3b + 3a² – b²
\(\frac{x}{a}+\frac{y}{b}\) = 4a² – 4b²
\(\frac{x}{a}+\frac{y}{b}\) = 4(a² – b²)

Alternate Method:
x + iy = (a + ib)³
x + iy = a³ + 3a²bi + 3ab²i² + b³i³
x + iy = a³ + 3a²bi – 3ab² – b³i …..[∵ i² = -1, i³ = -i]
x + iy = (a³ – 3ab²) + (3a²b – b³)i
Equating real and imaginary parts, we get
x = a³ – 3ab² and y = 3a²b – b³
Consider

Question 19.
If \(\frac{a+3 i}{2+i b}\) = 1 – i, show that (5a – 7b) = 0.
Solution:
\(\frac{a+3 i}{2+i b}\) = 1 – i
a + 3i = (1 – i)(2 + ib)
= 2 + bi – 2i – bi²
= 2 + (b – 2)i – b(-1) ……[∵ i² = -1]
a + 3i = (2 + b) + (b – 2)i
Equating real and imaginary parts, we get
a = 2 + b and 3 = b – 2
a = 2 + b and b = 5
a = 2 + 5 = 7
5a – 7b = 5(7) – 7(5) = 35 – 35 = 0

Question 20.
If x + iy = \(\sqrt{\frac{a+i b}{c+i d}}\), prove that \(\left(x^{2}+y^{2}\right)^{2}=\frac{a^{2}+b^{2}}{c^{2}+d^{2}}\)
Solution:

Question 21.
If (a + ib) = \(\frac{1+i}{1-i}\), then prove that a² + b² = 1.
Solution:

∴ a + bi = 0 + i
Equating real and imaginary parts, we get
a = 0 and b = 1
a² + b² = 0² + 1² = 1

Question 22.
Show that \(\left(\frac{\sqrt{7}+i \sqrt{3}}{\sqrt{7}-i \sqrt{3}}+\frac{\sqrt{7}-i \sqrt{3}}{\sqrt{7}+i \sqrt{3}}\right)\) is real.
Solution:

Question 23.
If (x + iy)³ = y + vi, then show that \(\frac{y}{x}+\frac{v}{y}\) = 4(x² – y²).
Solution:

Question 24.
Find the values of x and y which satisfy the following equations (x, y ∈ R)
(i) (x + 2y) + (2x – 3y)i + 4i = 5
Solution:
(x + 2y) + (2x – 3y)i + 4i = 5
(x + 2y) + (2x – 3y)i = 5 – 4i
Equating real and imaginary parts, we get
x + 2y = 5 ……(i)
and 2x – 3y = -4 …..(ii)
Equation (i) x 2 – equation (ii) gives
7y = 14
∴ y = 2
Substituting y = 2 in (i), we get
x + 2(2) = 5
x + 4 = 5
∴ x = 1
∴ x = 1 and y = 2

Check:
For x = 1 and y = 2
Consider, L.H.S. = (x + 2y) + (2x – 3y)i + 4i
= (1 + 4) + (2 – 6)i + 4i
= 5 – 4i + 4i
= 5
= R.H.S.

(ii) \(\frac{x+1}{1+i}+\frac{y-1}{1-i}=i\)
Solution:

(x + y) + (y – x – 2)i = 2i
(x + y) + (y – x – 2)i = 0 + 2i
Equating real and imaginary parts, we get
x + y = 0 and y – x – 2 = 2
∴ x + y = 0 …….(i)
and -x + y = 4 …….(ii)
Adding (i) and (ii), we get
2y = 4
∴ y = 2
Substituting y = 2 in (i), we get
x + 2 = 0
∴ x = -2
∴ x = -2 and y = 2

(iii) \(\frac{x+i y}{2+3 i}+\frac{2+i}{2-3 i}=\frac{9}{13}(1+i)\)
Solution:

(2x + 3y + 1) + (8 – 3x + 2y)i = 9 + 9i
Equating real and imaginary parts, we get
2x + 3y + 1 = 9 and 8 – 3x + 2y = 9
2x + 3y = 8 ……(i)
and 3x – 2y = – 1 ……(ii)
Equation (i) × 2 + equation (ii) × 3 gives
13x = 13
∴ x = 1
Substituting x = 1 in (i), we get
2(1) + 3y = 8
3y = 6
∴ y = 2
∴ x = 1 and y = 2

(iv) If x(1 + 3i) + y(2 – i) – 5 + i³ = 0, find x + y
Solution:
x(1 + 3i) + y(2 – i) – 5 + i³ = 0
x + 3xi + 2y – yi – 5 – i = 0 ……[∵ i³ = -i]
(x + 2y – 5) + (3x – y – 1)i = 0 + 0i
Equating real and imaginary parts, we get
x + 2y – 5 = 0 …..(i)
and 3x – y – 1 = 0 ……(ii)
Equation (i) + equation (ii) × 2 gives
7x – 7 = 0
7x = 1
∴ x = 1
Substituting x = 1 in (i), we get
1 + 2y – 5 = 0
2y = 4
y = 2
∴ x = 1 and y = 2
∴ x + y = 1 + 2 = 3

(v) If x + 2i + 15i⁶y = 7x + i³(y + 4), find x + y
Solution:
x + 2i + 15i⁶y = 7x + i³(y + 4)
x + 2i + 15(i²)³ y = 7x + i³(y + 4)
x + 2i + 15(-1)³ y = 7x – i(y + 4) ……[∵ i² = -1, i³ = -i]
x + 2i – 15y – 7x + iy + 4i = 0
(-6x – 15y) + i(y + 6) = 0 + 0i
Equating real and imaginary parts, we get
-6x – 15y = 0 and y + 6 = 0
-6x – 15y = 0 and y = -6
-6x – 15(-6) = 0
-6x + 90 = 0
∴ x = 15
∴ x + y = 15 – 6 = 9

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 2 Sequences and Series Ex 2.6 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.6

Question 1.
Find the sum \(\sum_{r=1}^{n}(r+1)(2 r-1)\).
Solution:

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

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

= \(\frac{n}{4}\) [(n + 1) + 2]
= \(\frac{n}{4}\) (n + 3)

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

Question 5.
Find the sum 5 × 7 + 9 × 11 + 13 × 15 + ….. upto n terms.
Solution:
5 × 7 + 9 × 11 + 13 × 15 + ….. upto n terms
Now, 5, 9, 13, … are in A.P. with
rth term = 5 + (r – 1) (4) = 4r + 1
7, 11, 15, ….. are in A.P. with
rth term = 7 + (r – 1) (4) = 4r + 3
∴ 5 × 7 + 9 × 11 + 13 × 15 + …… upto n terms

Question 6.
Find the sum 2² + 4² + 6² + 8² + ….. upto n terms.
Solution:

Question 7.
Find (70² – 69²) + (68² – 67²) + (66² – 65²) + …… + (2² – 1²)
Solution:
Let S = (70² – 69²) + (68² – 67²) + …… + (2² – 1²)
∴ S = (2² – 1²) + (4² – 3²) + ….. + (70² – 69²)
Here, 2, 4, 6,…, 70 are in A.P. with rth term = 2r
and 1, 3, 5, …,69 are in A.P. with rth term = 2r – 1

Question 8.
Find the sum 1 × 3 × 5 + 3 × 5 × 7 + 5 × 7 × 9 + …… + (2n – 1) (2n + 1) (2n + 3)
Solution:
Let S = 1 × 3 × 5 + 3 × 5 × 7 + ….. upto n terms
Here, 1, 3, 5, 7 … are in A.P. with rth term = 2r – 1,
3, 5, 7, 9,… are in A.P. with rth term = 2r + 1,
5, 7, 9, 11,… are in A.P. with rth term = 2r + 3

Question 9.
If \(\frac{1 \times 2+2 \times 3+3 \times 4+4 \times 5+\ldots \text { upto } n \text { terms }}{1+2+3+4+\ldots \text { upto } n \text { terms }}\) = \(\frac{100}{3}\), find n.
Solution:

Question 10.
If S₁, S₂ and S₃ are the sums of first n natural numbers, their squares and their cubes respectively, then show that 9\(\mathrm{S}_{2}{ }^{2}\) = S₃(1 + 8S₁).
Solution:

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 7 Limits Ex 7.4 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 7 Limits Ex 7.4

I. Evaluate the following limits:

Question 1.
\(\lim _{\theta \rightarrow 0}\left[\frac{\sin (m \theta)}{\tan (n \theta)}\right]\)
Solution:

Question 2.
\(\lim _{\theta \rightarrow 0}\left[\frac{1-\cos 2 \theta}{\theta^{2}}\right]\)
Solution:

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

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

II. Evaluate the following limits:

Question 1.
\(\lim _{x \rightarrow 0}\left[\frac{1-\cos (n x)}{1-\cos (m x)}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow \frac{\pi}{6}}\left[\frac{2-{cosec} x}{\cot ^{2} x-3}\right]\)
Solution:

Question 3.
\(\lim _{x \rightarrow \frac{\pi}{4}}\left[\frac{\cos x-\sin x}{\cos 2 x}\right]\)
Solution:

III. Evaluate the following limits:

Question 1.
\(\lim _{x \rightarrow 0}\left[\frac{\cos (a x)-\cos (b x)}{\cos (c x)-1}\right]\)
Solution:

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

Question 3.
\(\lim _{x \rightarrow \frac{\pi}{4}}\left[\frac{\tan ^{2} x-\cot ^{2} x}{\sec x-{cosec} x}\right]\)
Solution:

Question 4.
\(\lim _{x \rightarrow \frac{\pi}{6}}\left[\frac{2 \sin ^{2} x+\sin x-1}{2 \sin ^{2} x-3 \sin x+1}\right]\)
Solution:

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 3 Permutations and Combination Ex 3.2 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 3 Permutations and Combination Ex 3.2

Question 1.
Evaluate:
(i) 8!
Solution:
8!
= 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
= 40320

(ii) 10!
Solution:
10!
= 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
= 3628800

(iii) 10! – 6!
Solution:
10! – 6!
= 10 × 9 × 8 × 7 × 6! – 6!
= 6! (10 × 9 × 8 × 7 – 1)
= 6! (5040 – 1)
= 6 × 5 × 4 × 3 × 2 × 1 × 5039
= 3628080

(iv) (10 – 6)!
Solution:
(10 – 6)!
= 4!
= 4 × 3 × 2 × 1
= 24

Question 2.
Compute:
(i) \(\frac{12 !}{6 !}\)
Solution:
\(\frac{12 !}{6 !}=\frac{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 !}{6 !}\)
= 12 × 11 × 10 × 9 × 8 × 7
= 665280

(ii) \(\left(\frac{12}{6}\right) !\)
Solution:
\(\left(\frac{12}{6}\right) !\)
= 2!
= 2 × 1
= 2

(iii) (3 × 2)!
Solution:
(3 × 2)!
= 6!
= 6 × 5 × 4 × 3 × 2 × 1
= 720

(iv) 3! × 2!
Solution:
3! × 2!
= 3 × 2 × 1 × 2 × 1
= 12

(v) \(\frac{9 !}{3 ! 6 !}\)
Solution:
\(\frac{9 !}{3 ! 6 !}=\frac{9 \times 8 \times 7 \times 6 !}{(3 \times 2 \times 1) \times 6 !}=84\)

(vi) \(\frac{6 !-4 !}{4 !}\)
Solution:
\(\frac{6 !-4 !}{4 !}=\frac{6 \times 5 \times 4 !-4 !}{4 !}=\frac{4 !(6 \times 5-1)}{4 !}=29\)

(vii) \(\frac{8 !}{6 !-4 !}\)
Solution:
\(\frac{8 !}{6 !-4 !}=\frac{8 \times 7 \times 6 \times 5 \times 4 !}{6 \times 5 \times 4 !-4 !}\)
= \(\frac{8 \times 7 \times 6 \times 5 \times 4 !}{4 !(6 \times 5-1)}\)
= \(\frac{1680}{29}\)
= 57.93

(viii) \(\frac{8 !}{(6-4) !}\)
Solution:
\(\frac{8 !}{(6-4) !}=\frac{8 !}{2 !}\)
= \(\frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 !}{2 !}\)
= 20160

Question 3.
Write in terms of factorials
(i) 5 × 6 × 7 × 8 × 9 × 10
Solution:
5 × 6 × 7 × 8 × 9 × 10 = 10 × 9 × 8 × 7 × 6 × 5
Multiplying and dividing by 4!, we get
= \(\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 !}{4 !}\)
= \(\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{4 !}\)
= \(\frac{10 !}{4 !}\)

(ii) 3 × 6 × 9 × 12 × 15
Solution:
3 × 6 × 9 × 12 × 15
= 3 × (3 × 2) × (3 × 3) × (3 × 4) × (3 × 5)
= (3⁵) (5 × 4 × 3 × 2 × 1)
= 3⁵ (5!)

(iii) 6 × 7 × 8 × 9
Solution:
6 × 7 × 8 × 9 = 9 × 8 × 7 × 6
Multiplying and dividing by 5!, we get
= \(\frac{9 \times 8 \times 7 \times 6 \times 5 !}{5 !}\)
= \(\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{5 !}\)
= \(\frac{9 !}{5 !}\)

(iv) 5 × 10 × 15 × 20
Solution:
5 × 10 × 15 × 20
= (5 × 1) × (5 × 2) × (5 × 3) × (5 × 4)
= (5⁴) (4 × 3 × 2 × 1)
= (5⁴) (4!)

Question 4.
Evaluate: \(\frac{n !}{r !(n-r) !}\) for
(i) n = 8, r = 6
(ii) n = 12, r = 12
(iii) n = 15, r = 10
(iv) n = 15, r = 8
Solution:

Question 5.
Find n, if
(i) \(\frac{n}{8 !}=\frac{3}{6 !}+\frac{1 !}{4 !}\)
Solution:

(ii) \(\frac{n}{6 !}=\frac{4}{8 !}+\frac{3}{6 !}\)
Solution:

(iii) \(\frac{1 !}{n !}=\frac{1 !}{4 !}-\frac{4}{5 !}\)
Solution:

(iv) (n + 1)! = 42 × (n -1)!
Solution:
(n + 1)! = 42(n – 1)!
∴ (n + 1) n (n – 1)! = 42(n – 1)!
∴ n² + n = 42
∴ n² + n – 42 = 0
∴ (n + 7)(n – 6) = 0
∴ n = -7 or n = 6
But n ≠ -7 as n ∈ N
∴ n = 6

(v) (n + 3)! = 110 × (n + 1)!
Solution:
(n + 3)! = (110) (n + 1)!
∴ (n + 3)(n + 2)(n + 1)! = 110(n + 1)!
∴ (n + 3) (n + 2) = (11) (10)
Comparing on both sides, we get
n + 3 = 11
∴ n = 8

Question 6.
Find n, if:
(i) \(\frac{(17-n) !}{(14-n) !}=5 !\)
Solution:

∴ (17 – n) (16 – n) (15 – n) = 6 × 5 × 4
Comparing on both sides, we get
17 – n = 6
∴ n = 11

(ii) \(\frac{(15-n) !}{(13-n) !}=12\)
Solution:
\(\frac{(15-n) !}{(13-n) !}=12\)
∴ \(\frac{(15-n)(14-n)(13-n) !}{(13-n) !}=12\)
∴ (15 – n) (14 – n) = 4 × 3
Comparing on both sides, we get
∴ 15 – n = 4
∴ n = 11

(iii) \(\frac{n !}{3 !(n-3) !}: \frac{n !}{5 !(n-5) !}=5: 3\)
Solution:

∴ 12 = (n – 3)(n – 4)
(n – 3)(n – 4) = 4 × 3
Comparing on both sides, we get
n – 3 = 4
∴ n = 7

(iv) \(\frac{n !}{3 !(n-3) !}: \frac{n !}{5 !(n-7) !}=1: 6\)
Solution:

∴ 120 = (n – 3)(n – 4) (n – 5)(n – 6)
∴ (n – 3)(n – 4) (n – 5)(n – 6) = 5 × 4 × 3 × 2
Comparing on both sides, we get
n – 3 = 5
∴ n = 8

(v) \(\frac{(2 n) !}{7 !(2 n-7) !}: \frac{n !}{4 !(n-4) !}=24: 1\)
Solution:


(2n – 1)(2n – 3)(2n – 5) = \(\frac{24 \times 7 \times 6 \times 5}{16}\)
∴ (2n – 1)(2n – 3)(2n – 5) = 9 × 7 × 5
Comparing on both sides. We get
∴ 2n – 1 = 9
∴ n = 5

Question 7.
Show that \(\frac{n !}{r !(n-r) !}+\frac{n !}{(r-1) !(n-r+1) !}=\frac{(n+1) !}{r !(n-r+1) !}\)
Solution:

Question 8.
Show that \(\frac{9 !}{3 ! 6 !}+\frac{9 !}{4 ! 5 !}=\frac{10 !}{4 ! 6 !}\)
Solution:

Question 9.
Show that \(\frac{(2 n) !}{n !}\) = 2n (2n – 1)(2n – 3)…5.3.1
Solution:

Question 10.
Simplify
(i) \(\frac{(2 n+2) !}{(2 n) !}\)
Solution:

(ii) \(\frac{(n+3) !}{\left(n^{2}-4\right)(n+1) !}\)
Solution:

(iii) \(\frac{1}{n !}-\frac{1}{(n-1) !}-\frac{1}{(n-2) !}\)
Solution:

(iv) n[n! + (n – 1)!] + n²(n – 1)! + (n + 1)!
Solution:

(v) \(\frac{n+2}{n !}-\frac{3 n+1}{(n+1) !}\)
Solution:

(vi) \(\frac{1}{(n-1) !}+\frac{1-n}{(n+1) !}\)
Solution:

(vii) \(\frac{1}{n !}-\frac{3}{(n+1) !}-\frac{n^{2}-4}{(n+2) !}\)
Solution:

(viii) \(\frac{n^{2}-9}{(n+3) !}+\frac{6}{(n+2) !}-\frac{1}{(n+1) !}\)
Solution:

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 7 Limits Ex 7.3 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 7 Limits Ex 7.3

I. Evaluate the following limits:

Question 1.
\(\lim _{x \rightarrow 0}\left[\frac{\sqrt{6+x+x^{2}}-\sqrt{6}}{x}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow 3}\left[\frac{\sqrt{2 x+3}-\sqrt{4 x-3}}{x^{2}-9}\right]\)
Solution:

Question 3.
\(\lim _{y \rightarrow 0}\left[\frac{\sqrt{1-y^{2}}-\sqrt{1+y^{2}}}{y^{2}}\right]\)
Solution:

Question 4.
\(\lim _{x \rightarrow 2}\left[\frac{\sqrt{2+x}-\sqrt{6-x}}{\sqrt{x}-\sqrt{2}}\right]\)
Solution:

II. Evaluate the following limits:

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

Question 2.
\(\lim _{x \rightarrow 2}\left[\frac{x^{2}-4}{\sqrt{x+2}-\sqrt{3 x-2}}\right]\)
Solution:

Question 3.
\(\lim _{x \rightarrow 2}\left[\frac{\sqrt{1+\sqrt{2+x}}-\sqrt{3}}{x-2}\right]\)
Solution:

Question 4.
\(\lim _{y \rightarrow 0}\left[\frac{\sqrt{a+y}-\sqrt{a}}{y \sqrt{a+y}}\right]\)
Solution:

Question 5.
\(\lim _{x \rightarrow 0}\left(\frac{\sqrt{x^{2}+9}-\sqrt{2 x^{2}+9}}{\sqrt{3 x^{2}+4}-\sqrt{2 x^{2}+4}}\right)\)
Solution:

III. Evaluate the following limits:

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

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

Question 3.
\(\lim _{x \rightarrow 4}\left[\frac{x^{2}+x-20}{\sqrt{3 x+4}-4}\right]\)
Solution:

Question 4.
\(\lim _{z \rightarrow 4}\left[\frac{3-\sqrt{5+z}}{1-\sqrt{5-z}}\right]\)
Solution:

Question 5.
\(\lim _{x \rightarrow 0}\left(\frac{3}{x \sqrt{9-x}}-\frac{1}{x}\right)\)
Solution:

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 1 Complex Numbers Ex 1.4 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.4

Question 1.
Find the value of
(i) ω¹⁸
(ii) ω²¹
(iii) ω^-30
(iv) ω^-105
Solution:

Question 2.
If ω is the complex cube root of unity, show that
(i) (2 – ω)(2 – ω²) = 7
Solution:
ω is the complex cube root of unity.
ω³ = 1 and 1 + ω + ω² = 0
Also, 1 + ω² = -ω, 1 + ω = -ω² and ω + ω² = -1
L.H.S. = (2 – ω)(2 – ω²)
= 4 – 2ω² – 2ω + ω³
= 4 – 2(ω² + ω) + 1
= 4 – 2(-1) + 1
= 4 + 2 + 1
= 7
= R.H.S.

(ii) (1 + ω – ω²)⁶ = 64
Solution:
ω is the complex cube root of unity.
ω³ = 1 and 1 + ω + ω² = 0
Also, 1 + ω² = -ω, 1 + ω = -ω² and ω + ω² = -1

(iii) (1 + ω)³ – (1 + ω²)³ = 0
Solution:
ω is the complex cube root of unity.
ω³ = 1 and 1 + ω + ω² = 0
Also, 1 + ω² = -ω, 1 + ω = -ω² and ω + ω² = -1

(iv) (2 + ω + ω²)³ – (1 – 3ω + ω²)³ = 65
Solution:
ω is the complex cube root of unity.
ω³ = 1 and 1 + ω + ω² = 0
Also, 1 + ω² = -ω, 1 + ω = -ω² and ω + ω² = -1

(v) (3 + 3ω + 5ω²)⁶ – (2 + 6ω + 2ω²)³ = 0
Solution:
ω is the complex cube root of unity.
ω³ = 1 and 1 + ω + ω² = 0
Also, 1 + ω² = -ω, 1 + ω = -ω² and ω + ω² = -1

(vi) \(\frac{a+b \omega+c \omega^{2}}{c+a \omega+b \omega^{2}}\) = ω²
Solution:
ω is the complex cube root of unity.
ω³ = 1 and 1 + ω + ω² = 0
Also, 1 + ω² = -ω, 1 + ω = -ω² and ω + ω² = -1

(vii) (a + b) + (aω + bω²) + (aω² + bω) = 0
Solution:
ω is the complex cube root of unity.
ω³ = 1 and 1 + ω + ω² = 0
Also, 1 + ω² = -ω, 1 + ω = -ω² and ω + ω² = -1

(viii) (a – b)(a – bω)(a – bω²) = a³ – b³
Solution:
ω is the complex cube root of unity.
ω³ = 1 and 1 + ω + ω² = 0
Also, 1 + ω² = -ω, 1 + ω = -ω² and ω + ω² = -1

(ix) (a + b)² + (aω + bω²)² + (aω²+ bω)² = 6ab
Solution:
ω is the complex cube root of unity.
ω³ = 1 and 1 + ω + ω² = 0
Also, 1 + ω² = -ω, 1 + ω = -ω² and ω + ω² = -1

Question 3.
If ω is the complex cube root of unity, find the value of
(i) ω + \(\frac{1}{\omega}\)
Solution:
ω is the complex cube root of unity.
ω³ = 1 and 1 + ω + ω² = 0
Also, 1 + ω² = -ω, 1 + ω = -ω² and ω + ω² = -1
\(\omega+\frac{1}{\omega}=\frac{\omega^{2}+1}{\omega}=\frac{-\omega}{\omega}=-1\)

(ii) ω² + ω³ + ω⁴
Solution:
ω is the complex cube root of unity.
ω³ = 1 and 1 + ω + ω² = 0
Also, 1 + ω² = -ω, 1 + ω = -ω² and ω + ω² = -1
ω² + ω³ + ω⁴
= ω²(1 + ω + ω²)
= ω²(0)
= 0

(iii) (1 + ω²)³
Solution:
ω is the complex cube root of unity.
ω³ = 1 and 1 + ω + ω² = 0
Also, 1 + ω² = -ω, 1 + ω = -ω² and ω + ω² = -1
(1 + ω²)³
= (-ω)³
= -ω³
= -1

(iv) (1 – ω – ω²)³ + (1 – ω + ω²)³
Solution:
ω is the complex cube root of unity.
ω³ = 1 and 1 + ω + ω² = 0
Also, 1 + ω² = -ω, 1 + ω = -ω² and ω + ω² = -1
(1 – ω – ω²)³ + (1 – ω + ω²)³
= [1 – (ω + ω²)]³ + [(1 + ω²) – ω]³
= [1 – (-1)]² + (-ω – ω)³
= 2³ + (-2ω)³
= 8 – 8ω³
= 8 – 8(1)
= 0

(v) (1 + ω)(1 + ω²)(1 + ω⁴)(1 + ω⁸)
Solution:
ω is the complex cube root of unity.
ω³ = 1 and 1 + ω + ω² = 0
Also, 1 + ω² = -ω, 1 + ω = -ω² and ω + ω² = -1
(1 + ω)(1 + ω²)(1 + ω⁴)(1 + ω⁸)
= (1 + ω)(1 + ω²)(1 + ω)(1 + ω²) …..[∵ ω³ = 1, ω⁴ = ω]
= (-ω²)(-ω)(-ω²)(-ω)
= ω⁶
= (ω³)²
= (1)²
= 1

Question 4.
If α and β are the complex cube roots of unity, show that
(i) α² + β² + αβ = 0
(ii) α⁴ + β⁴ + α^-1β^-1 = 0
Solution:
α and β are the complex cube roots of unity.

Question 5.
If x = a + b, y = αa + βb and z = aβ + bα, where α and β are complex cube roots of unity, show that xyz = a³ + b³.
Solution:
x = a + b, y = αa + βb, z = aβ + bα
α and β are the complex cube roots of unity.
∴ α = \(\frac{-1+i \sqrt{3}}{2}\) and β = \(\frac{-1-i \sqrt{3}}{2}\)

Question 6.
Find the equation in cartesian coordinates of the locus of z if
(i) |z| = 10
Solution:
Let z = x + iy
|z| = 10
|x + iy| = 10
\(\sqrt{x^{2}+y^{2}}\) = 10
∴ x² + y² = 100

(ii) |z – 3| = 2
Solution:
Let z = x + iy
|z – 3| = 2
|x + iy – 3| = 2
|(x – 3) + iy| = 2
\(\sqrt{(x-3)^{2}+y^{2}}\) = 2
∴ (x – 3)² + y² = 4

(iii) |z – 5 + 6i| = 5
Solution:
Let z = x + iy
|z – 5 + 6i| = 5
|x + iy – 5 + 6i| = 5
|(x – 5) + i(y + 6)| = 5
\(\sqrt{(x-5)^{2}+(y+6)^{2}}\) = 5
∴ (x – 5)² + (y + 6)² = 25

(iv) |z + 8| = |z – 4|
Solution:
Let z = x + iy
|z + 8| = |z – 4|
|x + iy + 8| = |x + iy – 4|
|(x + 8) + iy | = |(x – 4) + iy|
\(\sqrt{(x+8)^{2}+y^{2}}=\sqrt{(x-4)^{2}+y^{2}}\)
(x + 8)² + y² = (x – 4)² + y²
x² + 16x + 64 + y² = x² – 8x + 16 + y²
16x + 64 = -8x + 16
24x + 48 = 0
∴ x + 2 = 0

(v) |z – 2 – 2i | = |z + 2 + 2i|
Solution:
Let z = x + iy
|z – 2 – 2i| = |z + 2 + 2i|
|x + iy – 2 – 2i | = |x + iy + 2 + 2i |
|(x – 2) + i(y – 2)| = |(x + 2) + i(y + 2)|
\(\sqrt{(x-2)^{2}+(y-2)^{2}}=\sqrt{(x+2)^{2}+(y+2)^{2}}\)
(x – 2)² + (y – 2)² = (x + 2)² + (y + 2)²
x²– 4x + 4 + y² – 4y + 4 = x² + 4x + 4 + y² + 4y + 4
-4x – 4y = 4x + 4y
8x + 8y = 0
x + y = 0
y = -x

(vi) \(\frac{|z+3 i|}{|z-6 i|}=1\)
Solution:
Let z = x + iy

x² + (y + 3)² = x² + (y – 6)²
y² + 6y + 9 = y² – 12y + 36
18y – 27 = 0
2y – 3 = 0

Question 7.
Use De Moivre’s theorem and simplify the following:
(i) \(\frac{(\cos 2 \theta+i \sin 2 \theta)^{7}}{(\cos 4 \theta+i \sin 4 \theta)^{3}}\)
Solution:

(ii) \(\frac{\cos 5 \theta+i \sin 5 \theta}{(\cos 3 \theta-i \sin 3 \theta)^{2}}\)
Solution:

(iii) \(\frac{\left(\cos \frac{7 \pi}{13}+i \sin \frac{7 \pi}{13}\right)^{4}}{\left(\cos \frac{4 \pi}{13}-i \sin \frac{4 \pi}{13}\right)^{6}}\)
Solution:

Question 8.
Express the following in the form a + ib, a, b ∈ R, using De Moivre’s theorem.
(i) (1 – i)⁵
Solution:

(ii) (1 + i)⁶
Solution:

(iii) (1 – √3 i)⁴
Solution:

(iv) (-2√3 – 2i)⁵
Solution: