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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 5th term in $$\left(\frac{4 x}{5}-\frac{5}{2 x}\right)^{8}$$
$$\left(\frac{4 x}{5}-\frac{5}{2 x}\right)^{8}$$ compare with (x + a)n
⇒ x → $$\frac{4 x}{5}$$, a → $$\frac{4 x}{5}$$ n → 8,
To find th term put r = 4
Tr+ = nCr.xn – r.ar T5 = 8C4.(22).2-4
8C4.28-4 = 8C4.24 = 1120

Question (ii).
The 8th term in $$\left(\frac{a}{2}-\frac{3}{b}\right)^{10}$$
$$\left(\frac{a}{2}-\frac{3}{b}\right)^{10}$$ compare with (x+a)n
⇒ x → $$\frac{a}{5}$$, a → $$\frac{2}{b}$$ n = 10,
To find 8th term put r = 7
Tr+ = nCr.xn – r.ar  Question (iii).
The 6th term in (√x – √y)17
Compare (√x – √y)17 with (x + a)n
x → √x a → -√y and n = 17
To find 6th term put r = 5 Question (iv).
The 7th term in (3x2 – $$\frac{y}{3}$$ )9
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 10th term in $$\left(\frac{a}{b}-\frac{2 b}{a^{2}}\right)^{12}$$ Question (vi).
the 11th term in $$\left(x+\frac{1}{\sqrt{x}}\right)^{14}$$
Here x = x a = $$\frac{1}{\sqrt{x}}$$ ,n = 14 and
put r = 10
T10+1 = 14C10 .x14 – 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).
$$\left(x-\frac{1}{2 y}\right)^{10}$$
$$\left(x-\frac{1}{2 y}\right)^{10}$$
Here n = even i.e 10 ∴ we have only one middle term  $$\frac{\mathrm{n}}{2}+1=\frac{10}{2}+1$$ = 6th term Question (ii).
$$\left(\frac{a}{x}+b x\right)^{12}$$
$$\left(\frac{a}{x}+b x\right)^{12}$$
Here
n = 12 (even)
∴ We have only one middle term Question (iii).
$$\left(\frac{2 a}{3}-\frac{3}{3 a}\right)^{6}$$
Here n = 6 (even) Question (iv).
$$\left(3 x-\frac{1}{6} x^{3}\right)^{8}$$ Question (v).
$$\left(\frac{a}{3}+\frac{b}{3}\right)^{8}$$
Here n = 8(even)
∴ middle term \frac{\mathrm{n}}{2} + 1 = 4 + 1 = 5th  3. Find the middle terms in the expansion of

Question (i).
Find the middle term in the expansion of
$$\left(3 x-\frac{2}{x^{2}}\right)^{15}$$
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 = 9th terms to find 8 thterm to find 8 th term pur r = 7 Question (ii)
$$\left(\frac{x}{2}+\frac{3}{x^{2}}\right)^{19}$$
Here n = 19 (odd)∴ We have two middle terms $$\frac{n+1}{2}=\frac{20}{2}$$ 10th and 10 + 1 =11th term To find 10th term put r = 9 Question (iii)
$$\left(2 x^{2}+\frac{1}{\sqrt{x}}\right)^{11}$$
Here n = 11 (odd)
∴ the two middle terms are $$\frac{13+1}{2}$$ = 7th and 6 + 1 = 7th terms
To find 6th term put r = 5 Question (iv).
$$\left(\sqrt{x}-\frac{4}{x^{2}}\right)^{11}$$
Here n = 11(odd) we have two middle term
i.e,  $$\frac{1+1}{2}$$ = 6th and 6 + 1 = 7th terms
To find 6th term put r = 5.  Question (v).
$$\left(\sqrt{x}-\frac{3}{x^{2}}\right)^{13}$$
$$\left(\sqrt{x}-\frac{3}{x^{2}}\right)^{13}$$
Here n = 13 (odd)
∴ the two middle terms are $$\frac{13+1}{2}$$ = 7th and 7 + 1 = 8th terms
To find 7th 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 $$\left(x+\frac{2}{x^{2}}\right)^{17}$$
Here x = x  a  = $$\frac{2}{x^{2}}$$ and  n = 17
Tr+1 = nCr . x n-r.ar
= 17Cr .x 17-r.( $$\frac{2}{x^{2}}$$ )r = 17Cr . 2r.x 17-r-2r
= 17Cr 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 $$\left(7 y^{2}-\frac{2}{y}\right)^{12}$$
Here x =7y2, a = $$-\frac{2}{y}$$ and n = 12
Tr+1 = 12Cr(7Y2)12-r.$$\left(\frac{-2}{y}\right)^{r}$$
= 12Cr.712-r .y24-2r. y-r(-2)r
Tr+1 = 12Cr.712-r.(-2)r.y24-3r
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 .712-7. (-2)7 y3
= -12C7 .75 27 . y3
∴Coefficient of y3 is -12C7.75.27

Question (iii).
x11 in $$\left(\sqrt{x}-\frac{2}{x}\right)^{17}$$
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
Tr+1 = 15Cr.(x2)15-r$$\left(\frac{-6}{x}\right)^{r}$$
Tr+1 = 15Cr.x30-2r.(-6)r.x-r
= 15Cr.(-6)r.x30-2r-r
= 15Cr(-6)r.x30-3r.
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 $$\left(x+\frac{1}{x^{2}}\right)^{17}$$
Here, x → x, a → $$\frac{1}{x^{2}}$$ and n = 17
∴ Tr+1 = 17Cr.x17-r$$\left(\frac{1}{x^{2}}\right)^{r}$$
Tr+1 = 17Cr.x17-r-2r
= 17Crx17-3r
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 $$\left(x+\frac{1}{x^{2}}\right)^{17}$$
Here, x → x, a → $$\frac{1}{x^{2}}$$ and n = 17
Tr+1 = 17Cr.x17-r$$\left(\frac{1}{x^{2}}\right)^{r}$$
= 17Crx17-3r
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 $$\left(x^{2}+\frac{3 a}{x}\right)^{15}$$
Here x → x2 a → $$\frac{3 \mathrm{a}}{\mathrm{x}}$$ , n=15
∴ Tr+1 = 15Cr.(x2)15-r $$\left(\frac{3 a}{x}\right)^{r}$$
= 15Crx30-2r.(3a)rxr
=15Cr .3r.ar.x30-3r
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).
$$\left(x^{3}-\frac{3}{x^{2}}\right)^{15}$$
Here x → x3, a = $$\frac{-3}{x^{2}}$$ and n = 15
Tr+1 = 15Cr.(x3)15-r.$$\left(\frac{-3}{x^{2}}\right)^{r}$$
= 15Cr.x45-r.x-2r (-3)r
= 15Cr.(-3)r.x45-5r
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).
$$\left(\sqrt{x}+\frac{1}{3 x^{2}}\right)^{10}$$ Question (iv).
$$\left(\frac{3 x^{2}}{2}-\frac{1}{3 x}\right)^{9}$$ Question (v).
$$\left(3 x-\frac{2}{x^{2}}\right)^{15}$$
Here x → 3x a → $$\frac{-2}{x^{2}}$$ and n =15
Tr+1 = 15Cr.(3x)15-r.$$\left(\frac{-2}{x^{2}}\right)^{r}$$
= 15Cr.315-r.(-2)r . x15-r
We have x15-3r. = x0 ⇒ 15 = 3r ⇒ r = 5
T = 15C5.315-5.(-2)5 = -15C5 .310 25
∴ The term independent of x is -15C5.310.25

Question (vi).
$$\left(x^{2}-\frac{2}{x^{3}}\right)^{5}$$
Here x → x2 a → $$\frac{-2}{x^{3}}$$ = 5r(x2)5 – r . $$\left(\frac{-2}{x^{3}}\right)^{r}$$= 5r.x10-2r.(-2)r
= 5Cr(-2)r.x10-5r
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).
$$\left(x-\frac{1}{x^{2}}\right)^{21}$$
Here x →x and a → $$\frac{-1}{x^{2}}$$ and n= 21
Tr+1 = 21Cr.(x)21-r. $$\left(\frac{-1}{x^{2}}\right)^{r}$$
= 21Cr.x21-3r.(-1)r
We have x0= x21-3r ⇒21 = 3r ⇒ r = 7
∴ T7+1 = T8 = 21C7 (-1)7.x0 = -21C7
∴ The term independent of x is -21C7

Question (viii).
$$\left( \sqrt { 2 } \frac { 2 }{ { x }^{ 2 } } \right)$$<sup>20</sup> 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)3.236C4(100)2 24 + 6C5.100.25 + 6C626
= 1000000000000 + 120000000000 + 6000000000 + 160000000 + 2400000 + 19200 + 64
= 1,126,162,419,264

Question (ii).
(98)4
(98)4 = (100 – 2 )4
= (100) 44C1 (100)3.2 + 4C2(100)2.22 4C3(100).23 + 4C4.24
= 100000000 – 8000000 + 240000 – 3200 + 16
= 92236816 Question (iii).
(1.0005)4
(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
(0.99)4 = (1- 0.01)4 = 4C0(0.01) – 4C1 (0.01) + 4C2(0.O1)24C3(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
Given . T1 = lin (1 + ax)n; T2 = 6x
nC1 .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.
Given , Tr+1= 9Cr.39-r.(kx) r
= 9Cr.39-r.krxr
Coefficient of x2 ⇒ x2 ⇒ r = 2 36x 37.k2 x x2
Coefficient of x3 ⇒ x3 = xr⇒ r = 3 T3+1 = T4 = 93.39-3.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? 