2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2

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Karnataka 2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2

Part – A

2nd PUC Basic Maths Binomial Theorem Ex 4.2 Two Marks Questions and Answers

1. Find
Question (i).
The 5^th term in \(\left(\frac{4 x}{5}-\frac{5}{2 x}\right)^{8}\)
Answer:
\(\left(\frac{4 x}{5}-\frac{5}{2 x}\right)^{8}\) compare with (x + a)ⁿ
⇒ x → \(\frac{4 x}{5}\), a → \(\frac{4 x}{5}\) n → 8,
To find ^th term put r = 4
T_r+ = ⁿC_r.x^{n – r}.a^r

T₅ = ⁸C₄.(2²).2^-4
8C₄.2^8-4 = ⁸C₄.2⁴ = 1120

Question (ii).
The 8^th term in \(\left(\frac{a}{2}-\frac{3}{b}\right)^{10}\)
Answer:
\(\left(\frac{a}{2}-\frac{3}{b}\right)^{10}\) compare with (x+a)ⁿ
⇒ x → \(\frac{a}{5}\), a → \(\frac{2}{b}\) n = 10,
To find 8^th term put r = 7
T_r+ = ⁿC_r.x^{n – r}.a^r

Question (iii).
The 6^th term in (√x – √y)¹⁷
Answer:
Compare (√x – √y)¹⁷ with (x + a)ⁿ
x → √x a → -√y and n = 17
To find 6^th term put r = 5

Question (iv).
The 7^th term in (3x² – \( \frac{y}{3}\) )⁹
Answer:
Here x → 3x² a → \(-\frac{y}{3}\) nn = 9
Put r = 6
T₆₊₁ =⁹C₆ (3x²)^9-6. ( \(-\frac{y}{3}\) )⁶

Question (v).
the 10^th term in \(\left(\frac{a}{b}-\frac{2 b}{a^{2}}\right)^{12}\)
Answer:

Question (vi).
the 11^th term in \(\left(x+\frac{1}{\sqrt{x}}\right)^{14}\)
Answer:
Here x = x a = \(\frac{1}{\sqrt{x}}\) ,n = 14 and
put r = 10
T₁₀₊₁ = ¹⁴C₁₀ .x^{14 – 10} . \(\left(\frac{1}{\sqrt{x}}\right)^{10}\) ¹⁴C₄.x⁴ \(\frac{1}{x^{5}}\) = \(\frac{1001}{x}\)

Part – B

2nd PUC Basic Maths Binomial Theorem Ex 4.2 Three Marks Questions and Answers

Question 2.
Find the middele term in the expansion of

Question (i).
\(\left(x-\frac{1}{2 y}\right)^{10}\)
Answer:
\(\left(x-\frac{1}{2 y}\right)^{10}\)
Here n = even i.e 10 ∴ we have only one middle term  \(\frac{\mathrm{n}}{2}+1=\frac{10}{2}+1\) = 6^th term

Question (ii).
\(\left(\frac{a}{x}+b x\right)^{12}\)
Answer:
\(\left(\frac{a}{x}+b x\right)^{12}\)
Here
n = 12 (even)
∴ We have only one middle term

Question (iii).
\(\left(\frac{2 a}{3}-\frac{3}{3 a}\right)^{6}\)
Answer:
Here n = 6 (even)

Question (iv).
\(\left(3 x-\frac{1}{6} x^{3}\right)^{8}\)
Answer:

Question (v).
\(\left(\frac{a}{3}+\frac{b}{3}\right)^{8}\)
Answer:
Here n = 8(even)
∴ middle term \frac{\mathrm{n}}{2} + 1 = 4 + 1 = 5^th

3. Find the middle terms in the expansion of

Question (i).
Find the middle term in the expansion of
\(\left(3 x-\frac{2}{x^{2}}\right)^{15}\)
Answer:
Here n = 15 and odd, so we have two middle terms i.e, \(\frac{n+1}{2}=\frac{15+1}{2}=8^{t h}\) and 8 + 1 = 9^th terms to find 8 ^thterm to find 8 ^th term pur r = 7

Question (ii)
\(\left(\frac{x}{2}+\frac{3}{x^{2}}\right)^{19}\)
Answer:
Here n = 19 (odd)∴ We have two middle terms \(\frac{n+1}{2}=\frac{20}{2}\) 10^th and 10 + 1 =11^th term To find 10^th term put r = 9

Question (iii)
\(\left(2 x^{2}+\frac{1}{\sqrt{x}}\right)^{11}\)
Answer:
Here n = 11 (odd)
∴ the two middle terms are \(\frac{13+1}{2}\) = 7^th and 6 + 1 = 7^th terms
To find 6^th term put r = 5

Question (iv).
\(\left(\sqrt{x}-\frac{4}{x^{2}}\right)^{11}\)
Answer:
Here n = 11(odd) we have two middle term
i.e,  \( \frac{1+1}{2}\) = 6^th and 6 + 1 = 7^th terms
To find 6^th term put r = 5.

Question (v).
\(\left(\sqrt{x}-\frac{3}{x^{2}}\right)^{13}\)
Answer:
\(\left(\sqrt{x}-\frac{3}{x^{2}}\right)^{13}\)
Here n = 13 (odd)
∴ the two middle terms are \(\frac{13+1}{2}\) = 7^th and 7 + 1 = 8^th terms
To find 7^th term put r = 6

Part – C

2nd PUC Basic Maths Binomial Theorem Ex 4.2 Five Marks Questions and Answers

4.
Question (i).
Find the coefficent of xⁿ in \(\left(x+\frac{2}{x^{2}}\right)^{17}\)
Answer:
Here x = x  a  = \(\frac{2}{x^{2}}\) and  n = 17
T_r+1 = ⁿC_r . x ^n-r.a^r
= ¹⁷C_r .x ^17-r.( \(\frac{2}{x^{2}}\) )^r = ¹⁷C_r . 2^r.x ^17-r-2r
= ¹⁷C_r 2 ^r.x ^17-3r
To find coefficient of x¹¹ equate the power of x to 11
⇒ 17 – 3r = 11 ⇒ 17 – 11 = 3r
⇒ 3r = 6 ⇒ r = 2
T₂₊₁ = ¹⁷C₂ . 2².x¹¹
∴ Coefficient of x” is ¹⁷C₂ .2² = \(\frac{17 \times 16 \times 2}{2 \times 1}\) = 544

Question (ii).
Y³ in \(\left(7 y^{2}-\frac{2}{y}\right)^{12}\)
Answer:
Here x =7y², a = \(-\frac{2}{y}\) and n = 12
T_r+1 = 12C^r(7Y²)^12-r.\(\left(\frac{-2}{y}\right)^{r}\)
= ¹²C_r.7^12-r .y^24-2r. y^-r(-2)^r
T_r+1 = ¹²C_r.7^12-r.(-2)^r.y^24-3r
To find the coefficient of y³ equate the power of y ro 3
i.e., 24 – 3r = 3 ⇒ 21 = 3r ⇒ r= 7
∴T₇₊₁ = ¹²C₇ .7^12-7. (-2)⁷ y³
= ^-12C₇ .7⁵ 2⁷ . y³
∴Coefficient of y³ is ^-12C₇.7⁵.2⁷

Question (iii).
x¹¹ in \(\left(\sqrt{x}-\frac{2}{x}\right)^{17}\)
Answer:
lere, x → √x, a →\(\frac{-2}{x}\) and n = 17

To find the coefficient of x2 equate the power of x to -11
∴ \(\frac{17-3 r}{2}\) = -11 ⇒ 17 – 3r = – 22
⇒ 17+22 = 3r ⇒ 39 = 3.r ⇒ r = 13
T₁₃₊₁ = ¹⁷C₁₃(-2)¹³.x^-11
Coefficient of x^-11 is ^-17C₁₃.2¹³

Question (iv).
X¹⁸ in \(\left(x^{2}-\frac{6}{x}\right)^{15}\)
lere, x → x², a →\(\frac{-6}{x}\) and r = 15
T_r+1 = ¹⁵C_r.(x²)^15-r\(\left(\frac{-6}{x}\right)^{r}\)
T_r+1 = ¹⁵C_r.x^30-2r.(-6)^r.x^-r
= ¹⁵C_r.(-6)^r.x^30-2r-r
= ¹⁵C_r(-6)^r.x^30-3r.
To find the coefficient of x¹⁸,equate the power of x to 18
∴ 30 – 3r = 18 ⇒ 30 – 18 = 3r ⇒ 3r = 12 ⇒ r = 4
T₄₊₁ = ¹⁵C₄(-6) ⁴x¹⁸
∴ Coefficient of x¹⁸ is ¹⁵C₄. (6)⁴

Question (v).
X^-2 in \(\left(x+\frac{1}{x^{2}}\right)^{17}\)
Answer:
Here, x → x, a → \(\frac{1}{x^{2}}\) and n = 17
∴ T_r+1 = ¹⁷C_r.x^17-r\(\left(\frac{1}{x^{2}}\right)^{r}\)
T_r+1 = ¹⁷C_r.x^17-r-2r
= ¹⁷C_rx^17-3r
To find the coefficient of x^-2,equate the power of x to -2
17 – 3r = -2 ⇒ 17 + 2 = 3r ⇒ 3r ⇒ r = \(\frac{19}{3}\)
Since r is a fraction the coefficient of x^-2 is 0.

Question (vi).
X⁵ in \(\left(x+\frac{1}{x^{2}}\right)^{17}\)
Answer:
Here, x → x, a → \(\frac{1}{x^{2}}\) and n = 17
T_r+1 = ¹⁷C_r.x^17-r\(\left(\frac{1}{x^{2}}\right)^{r}\)
= ¹⁷C_rx^17-3r
To find the coefficient of x⁵, equate the power of x to 5
∴ 17 – 3r = 5
12 = 23 ⇒ r = 4
T₄₊₁ = T₅ = ¹⁷C₄.x⁵
∴ the coefficient of x⁵ is ¹⁷C₄

Question (vii).
X¹⁸ in \(\left(x^{2}+\frac{3 a}{x}\right)^{15}\)
Answer:
Here x → x² a → \(\frac{3 \mathrm{a}}{\mathrm{x}}\) , n=15
∴ T_r+1 = ¹⁵C_r.(x²)^15-r \(\left(\frac{3 a}{x}\right)^{r}\)
= ¹⁵C_rx^30-2r.(3a)^rx^r
=¹⁵C_r .3^r.a^r.x^30-3r
To find the coefficient of x¹⁸, we get 30 – 3r = 18
12 = 3r ⇒ r =4
∴ T₄₊₁ = ¹⁵C₄.3⁴.a⁴.x¹⁸
∴ coefficient of x¹⁸ is ¹⁵C₄.(3a)⁴

5. Find the term independent of x in

Question (i).
\(\left(\frac{4 x^{2}}{3}+\frac{3}{2 x}\right)^{9}\)
Here x → \(\frac{4 x^{2}}{3}\) , a = \(\frac{3}{2 x}\) and n = 9

To find the term independent of x, equate the power of x to zero.
e., 18 – 3r = 0 ⇒ r = 6

Question (ii).
\(\left(x^{3}-\frac{3}{x^{2}}\right)^{15}\)
Answer:
Here x → x³, a = \(\frac{-3}{x^{2}}\) and n = 15
T_r+1 = ¹⁵C_r.(x³)^15-r.\(\left(\frac{-3}{x^{2}}\right)^{r}\)
= ¹⁵C_r.x^45-r.x^-2r (-3)^r
= ¹⁵C_r.(-3)^r.x^45-5r
To find the term independent of x we have 45 – 5r = O
:. 45 = 5r ⇒ r = 9
T₉₊₁ = ¹⁵C₉.(-3)⁹.x⁰
T₁₀ = ^-15C₉.(3)⁹ is the term independent of x.

Question (iii).
\(\left(\sqrt{x}+\frac{1}{3 x^{2}}\right)^{10}\)
Answer:

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

Question (v).
\(\left(3 x-\frac{2}{x^{2}}\right)^{15}\)
Answer:
Here x → 3x a → \(\frac{-2}{x^{2}}\) and n =15
T_r+1 = ¹⁵C_r.(3x)^15-r.\(\left(\frac{-2}{x^{2}}\right)^{r}\)
= ¹⁵C_r.3^15-r.(-2)^r . x^15-r
We have x^15-3r. = x⁰ ⇒ 15 = 3r ⇒ r = 5
T = ¹⁵C₅.3^15-5.(-2)⁵ = ^-15C₅ .3¹⁰ 2⁵
∴ The term independent of x is ^-15C₅.3¹⁰.2⁵

Question (vi).
\(\left(x^{2}-\frac{2}{x^{3}}\right)^{5}\)
Answer:
Here x → x² a → \(\frac{-2}{x^{3}}\) = 5_r(x²)^{5 – r} . \(\left(\frac{-2}{x^{3}}\right)^{r}\)= 5_r.x^10-2r.(-2)^r
= 5C_r(-2)^r.x^10-5r
We have 10 – 5r = 0 ⇒ r = 2
T₂₊₁ = T₃ = ⁵C₂(-2)².x⁰ = 4. \(\frac { 5.4}{ 2.1 }\) = 40
∴  The term independent of x is 40

Question (vii).
\(\left(x-\frac{1}{x^{2}}\right)^{21}\)
Answer:
Here x →x and a → \(\frac{-1}{x^{2}}\) and n= 21
T_r+1 = ²¹C_r.(x)^21-r. \(\left(\frac{-1}{x^{2}}\right)^{r}\)
= ²¹C_r.x^21-3r.(-1)^r
We have x⁰= x^21-3r ⇒21 = 3r ⇒ r = 7
∴ T₇₊₁ = T₈ = ²¹C₇ (-1)⁷.x⁰ = ^-21C₇
∴ The term independent of x is -21C₇

Question (viii).
\(\left( \sqrt { 2 } \frac { 2 }{ { x }^{ 2 } } \right)\)²⁰
Answer:

Part – D

2nd PUC Basic Maths Binomial Theorem Ex 4.2 Four Marks Questions and Answers

6. Use binomial theorem to evaluate upt 4 decimals place

Questoin (i)
(102)⁶
(102)⁶ = (100 + 2)⁶
= (100)⁶ + ⁶C₁(1 0O)⁵. 2 + ⁶C₂(100)⁴.2² + ⁶C₃(100)³.2³ + ⁶C₄(100)² 2⁴ + ⁶C₅.100.2⁵ + ⁶C₆2⁶
= 1000000000000 + 120000000000 + 6000000000 + 160000000 + 2400000 + 19200 + 64
= 1,126,162,419,264

Question (ii).
(98)⁴
Answer:
(98)⁴ = (100 – 2 )⁴
= (100) ⁴– ⁴C₁ (100)³.2 + ⁴C₂(100)².2² – ⁴C₃(100).2³ + ⁴C₄.2⁴
= 100000000 – 8000000 + 240000 – 3200 + 16
= 92236816

Question (iii).
(1.0005)⁴
Answer:
(1.0005)⁴ = (1 + 0.0005)⁴
= 1⁴ + ⁴C₁(O.0005) + ⁴C₂(0.0005)² + ⁴C₃(0.0005)³ + ⁴C₄(0.0005)⁴
1 + 0.002 + 0.0000015 + …………….
= 1.00200150 ≈ 1.0020

Question (iv).
(0.99)⁴
Answer:
(0.99)⁴ = (1- 0.01)⁴ = ⁴C₀(0.01) – ⁴C₁ (0.01) + ⁴C₂(0.O1)² – ⁴C₃(0.01)³ + ⁴C₄(0.01)⁴
= 1 – 0.04 + 0.0006 – 0.000004 + 0.00000001
= 0.96059601 ≈ 0.9606

Question 7.
The first three terms in (1 + ax)ⁿ where n is a positive integer are 1, 6x, 16x². Find the vaIue
Answer:
Given . T₁ = lin (1 + ax)ⁿ; T₂ = 6x
ⁿC_{1 .}ax = 6x
nax = 6x
⇒ na = 6 ⇒ a = \(\frac{6}{n}\)
and T₃ = 16x²
\(\frac{n(n-1)}{2}\) a²x² = 16x²

Question 8.
In the expansion of (3 + kx)⁹ the x² and x³ are equal. Find k.
Answer:
Given , T_r+1= ⁹C_r.3^9-r.(kx) ^r
= ⁹C_r.3^9-r.k^rx^r
Coefficient of x² ⇒ x² ⇒ r = 2

36x 3⁷.k² x x²
Coefficient of x³ ⇒ x³ = x^r⇒ r = 3

T₃₊₁ = T₄ = ⁹₃.3^9-3.k₃.x₃ = ⁹C₃.3⁶
k₃.x₃ = 84.3⁶k₃.x₃
84 x 3⁶k₂= 36 x 3₇ x k²

Question 9.
Find the ratio of the coefficient of x⁴ in the two expansions (1+x)⁷ and (1+x)¹⁰?
Answer: