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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+} = ^{n}C_{r}.x^{n – r}.a^{r}

T_{5} = ^{8}C_{4}.(2^{2}).2^{-4}

^{8}C_{4}.2^{8-4} = ^{8}C_{4}.2^{4} = 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+} = ^{n}C_{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 (3x^{2} – \( \frac{y}{3}\) )^{9
}Answer:

Here x → 3x^{2 }a → \(-\frac{y}{3}\) nn = 9

Put r = 6

T_{6+1 }=^{9}C_{6} (3x^{2})^{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

T_{10+1} = ^{14}C_{10} .x^{14 – 10} . \(\left(\frac{1}{\sqrt{x}}\right)^{10}\) ^{14}C_{4}.x^{4} \(\frac{1}{x^{5}}\) = \(\frac{1001}{x}\)

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 ^{th}term 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 x^{n} in \(\left(x+\frac{2}{x^{2}}\right)^{17}\)

Answer:

Here x = x a = \(\frac{2}{x^{2}}\) and n = 17

T_{r+1} = ^{n}C_{r} . x ^{n-r}.a^{r}

= ^{17}C_{r} .x ^{17-r}.( \(\frac{2}{x^{2}}\) )^{r} = ^{17}C_{r} . 2^{r}.x ^{17-r-2r}

= ^{17}C_{r} 2 ^{r}.x ^{17-3r}

To find coefficient of x^{11} equate the power of x to 11

⇒ 17 – 3r = 11 ⇒ 17 – 11 = 3r

⇒ 3r = 6 ⇒ r = 2

T_{2+1} = ^{17}C_{2} . 2^{2}.x^{11}

∴ Coefficient of x” is ^{17}C_{2} .2^{2} = \(\frac{17 \times 16 \times 2}{2 \times 1}\) = 544

Question (ii).

Y^{3} in \(\left(7 y^{2}-\frac{2}{y}\right)^{12}\)

Answer:

Here x =7y^{2}, a = \(-\frac{2}{y}\) and n = 12

T_{r+1} = 12C^{r}(7Y^{2})^{12-r}.\(\left(\frac{-2}{y}\right)^{r}\)

= ^{12}C_{r}.7^{12-r} .y^{24-2r}. y^{-r}(-2)^{r}

T_{r+1} = ^{12}C_{r}.7^{12-r}.(-2)^{r}.y^{24-3r}

To find the coefficient of y^{3} equate the power of y ro 3

i.e., 24 – 3r = 3 ⇒ 21 = 3r ⇒ r= 7

∴T_{7+1} = ^{12}C_{7} .7^{12-7}. (-2)^{7} y^{3}

= ^{-12}C_{7} .7^{5} 2^{7} . y^{3}

∴Coefficient of y^{3} is ^{-12}C_{7}.7^{5}.2^{7}

Question (iii).

x^{11} 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

T_{13+1} = ^{17}C_{13}(-2)^{13}.x^{-11}

Coefficient of x^{-11} is ^{-17}C_{13}.2^{13}

Question (iv).

X^{18} in \(\left(x^{2}-\frac{6}{x}\right)^{15}\)

lere, x → x^{2}, a →\(\frac{-6}{x}\) and r = 15

T_{r+1} = ^{15}C_{r}.(x^{2})^{15-r}\(\left(\frac{-6}{x}\right)^{r}\)

T_{r+1} = ^{15}C_{r}.x^{30-2r}.(-6)^{r}.x^{-r}

= ^{15}C_{r}.(-6)^{r}.x^{30-2r-r}

= ^{15}C_{r}(-6)^{r}.x^{30-3r}.

To find the coefficient of x^{18},equate the power of x to 18

∴ 30 – 3r = 18 ⇒ 30 – 18 = 3r ⇒ 3r = 12 ⇒ r = 4

T_{4+1} = ^{15}C_{4}(-6) ^{4}x^{18}

∴ Coefficient of x^{18} is ^{15}C_{4}. (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} = ^{17}C_{r}.x^{17-r}\(\left(\frac{1}{x^{2}}\right)^{r}\)

T_{r+1} = ^{17}C_{r}.x^{17-r-2r}

= ^{17}C_{r}x^{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).

X^{5} in \(\left(x+\frac{1}{x^{2}}\right)^{17}\)

Answer:

Here, x → x, a → \(\frac{1}{x^{2}}\) and n = 17

T_{r+1} = ^{17}C_{r}.x^{17-r}\(\left(\frac{1}{x^{2}}\right)^{r}\)

= ^{17}C_{r}x^{17-3r}

To find the coefficient of x^{5}, equate the power of x to 5

∴ 17 – 3r = 5

12 = 23 ⇒ r = 4

T_{4+1} = T_{5} = ^{17}C_{4}.x^{5}

∴ the coefficient of x^{5} is ^{17}C_{4}

Question (vii).

X^{18} in \(\left(x^{2}+\frac{3 a}{x}\right)^{15}\)

Answer:

Here x → x^{2} a → \(\frac{3 \mathrm{a}}{\mathrm{x}}\) , n=15

∴ T_{r+1} = ^{15}C_{r}.(x^{2})^{15-r} \(\left(\frac{3 a}{x}\right)^{r}\)

= ^{15}C_{r}x^{30-2r}.(3a)^{r}x^{r}

=^{15}C_{r} .3^{r}.a^{r}.x^{30-3r}

To find the coefficient of x^{18}, we get 30 – 3r = 18

12 = 3r ⇒ r =4

∴ T_{4+1} = ^{15}C_{4}.3^{4}.a^{4}.x^{18}

∴ coefficient of x^{18} is ^{15}C_{4}.(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 → x^{3}, a = \(\frac{-3}{x^{2}}\) and n = 15

T_{r+1} = ^{15}C_{r}.(x^{3})^{15-r}.\(\left(\frac{-3}{x^{2}}\right)^{r}\)

= ^{15}C_{r}.x^{45-r}.x^{-2r} (-3)^{r}

= ^{15}C_{r}.(-3)^{r}.x^{45-5r}

To find the term independent of x we have 45 – 5r = O

:. 45 = 5r ⇒ r = 9

T_{9+1} = ^{15}C_{9}.(-3)^{9}.x^{0}

T_{10} = ^{-15}C_{9}.(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} = ^{15}C_{r}.(3x)^{15-r}.\(\left(\frac{-2}{x^{2}}\right)^{r}\)

= ^{15}C_{r}.3^{15-r}.(-2)^{r} . x^{15-r}

We have x^{15-3r}. = x^{0} ⇒ 15 = 3r ⇒ r = 5

T = ^{15}C_{5}.3^{15-5}.(-2)^{5} = ^{-15}C_{5} .3^{10} 2^{5}

∴ The term independent of x is ^{-15}C_{5}.3^{10}.2^{5}

Question (vi).

\(\left(x^{2}-\frac{2}{x^{3}}\right)^{5}\)

Answer:

Here x → x^{2} a → \(\frac{-2}{x^{3}}\) = 5_{r}(x^{2})^{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

T_{2+1} = T_{3} = ^{5}C_{2}(-2)^{2}.x^{0} = 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} = ^{21}C_{r}.(x)^{21-r}. \(\left(\frac{-1}{x^{2}}\right)^{r}\)

= ^{21}C_{r}.x^{21-3r}.(-1)^{r}

We have x^{0}= x^{21-3r} ⇒21 = 3r ⇒ r = 7

∴ T_{7+1} = T_{8} = ^{21}C_{7} (-1)^{7}.x^{0} = ^{-21}C_{7}

∴ The term independent of x is -21C_{7}

Question (viii).

\(\left( \sqrt { 2 } \frac { 2 }{ { x }^{ 2 } } \right)\)<sup>20</sup>

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} + ^{6}C_{1}(1 0O)^{5}. 2 + ^{6}C_{2}(100)^{4}.2^{2 }+ ^{6}C_{3}(100)^{3}.2^{3} + ^{6}C_{4}(100)^{2} 2^{4} + ^{6}C_{5}.100.2^{5} + ^{6}C_{6}2^{6}

= 1000000000000 + 120000000000 + 6000000000 + 160000000 + 2400000 + 19200 + 64

= 1,126,162,419,264

Question (ii).

(98)^{4}

Answer:

(98)^{4} = (100 – 2 )^{4}

= (100) ^{4}– ^{4}C_{1} (100)^{3}.2 + ^{4}C_{2}(100)^{2}.2^{2 }– ^{4}C_{3}(100).2^{3} + ^{4}C_{4}.2^{4}

= 100000000 – 8000000 + 240000 – 3200 + 16

= 92236816

Question (iii).

(1.0005)^{4}

Answer:

(1.0005)^{4} = (1 + 0.0005)^{4}

= 1^{4} + ^{4}C_{1}(O.0005) + ^{4}C_{2}(0.0005)^{2 }+ ^{4}C_{3}(0.0005)^{3} + ^{4}C_{4}(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} = ^{4}C_{0}(0.01) – ^{4}C_{1} (0.01) + ^{4}C_{2}(0.O1)^{2} – ^{4}C_{3}(0.01)^{3} + ^{4}C_{4}(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, 16x^{2}. Find the vaIue

Answer:

Given . T_{1} = lin (1 + ax)^{n}; T_{2} = 6x

^{n}C_{1 .}ax = 6x

nax = 6x

⇒ na = 6 ⇒ a = \(\frac{6}{n}\)

and T_{3} = 16x^{2}

\(\frac{n(n-1)}{2}\) a^{2}x^{2} = 16x^{2}

Question 8.

In the expansion of (3 + kx)^{9} the x^{2} and x^{3} are equal. Find k.

Answer:

Given , T_{r+1}= ^{9}C_{r}.3^{9-r}.(kx) ^{r}

= ^{9}C_{r}.3^{9-r}.k^{r}x^{r}

Coefficient of x^{2} ⇒ x^{2} ⇒ r = 2

36x 3^{7}.k^{2} x x^{2}

Coefficient of x^{3} ⇒ x^{3} = x^{r}⇒ r = 3

T_{3+1} = T_{4} = ^{9}_{3}.3^{9-3}.k_{3}.x_{3} = ^{9}C_{3}.3^{6}

k_{3}.x_{3} = 84.3^{6}k_{3}.x_{3}

84 x 3^{6}k_{2}= 36 x 3_{7} x k^{2}

Question 9.

Find the ratio of the coefficient of x^{4} in the two expansions (1+x)^{7} and (1+x)^{10}?

Answer: