Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 4 Methods of Induction and Binomial Theorem Ex 4.3 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 4 Methods of Induction and Binomial Theorem Ex 4.3

Question 1.

In the following expansions, find the indicated term.

(i) \(\left(2 x^{2}+\frac{3}{2 x}\right)^{8}\), 3rd term

Solution:

(ii) \(\left(x^{2}-\frac{4}{x^{3}}\right)^{11}\), 5th term

Solution:

(iii) \(\left(\frac{4 x}{5}-\frac{5}{2 x}\right)^{9}\), 7th term

Solution:

(iv) In \(\left(\frac{1}{3}+a^{2}\right)^{12}\), 9th term

Solution:

(v) In \(\left(3 a+\frac{4}{a}\right)^{13}\), 10th term

Solution:

Question 2.

In the following expansions, find the indicated coefficients.

(i) x^{3} in \(\left(x^{2}+\frac{3 \sqrt{2}}{x}\right)^{9}\)

Solution:

(ii) x^{8} in \(\left(2 x^{5}-\frac{5}{x^{3}}\right)^{8}\)

Solution:

(iii) x^{9} in \(\left(\frac{1}{x}+x^{2}\right)^{18}\)

Solution:

(iv) x^{-3} in \(\left(x-\frac{1}{2 x}\right)^{5}\)

Solution:

(v) x^{-20} in \(\left(x^{3}-\frac{1}{2 x^{2}}\right)^{15}\)

Solution:

Question 3.

Find the constant term (term independent of x) in the expansion of

(i) \(\left(2 x+\frac{1}{3 x^{2}}\right)^{9}\)

Solution:

(ii) \(\left(x-\frac{2}{x^{2}}\right)^{15}\)

Solution:

(iii) \(\left(\sqrt{x}-\frac{3}{x^{2}}\right)^{10}\)

Solution:

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

Solution:

(v) \(\left(2 x^{2}-\frac{5}{x}\right)^{9}\)

Solution:

Question 4.

Find the middle terms in the expansion of

(i) \(\left(\frac{x}{y}+\frac{y}{x}\right)^{12}\)

Solution:

(ii) \(\left(x^{2}+\frac{1}{x}\right)^{7}\)

Solution:

(iii) \(\left(x^{2}-\frac{2}{x}\right)^{8}\)

Solution:

(iv) \(\left(\frac{x}{a}-\frac{a}{x}\right)^{10}\)

Solution:

(v) \(\left(x^{4}-\frac{1}{x^{3}}\right)^{11}\)

Solution:

Question 5.

In the expansion of (k + x)^{8}, the coefficient of x^{5} is 10 times the coefficient of x^{6}. Find the value of k.

Solution:

Question 6.

Find the term containing x^{6} in the expansion of (2 – x) (3x + 1)^{9}.

Solution:

Question 7.

The coefficient of x^{2} in the expansion of (1 + 2x)^{m} is 112. Find m.

Solution:

The coefficient of x^{2} in (1 + 2x)^{m} = ^{m}C_{2} (2^{2})

Given that the coefficient of x^{2} = 112

∴ ^{m}C_{2} (4) = 112

∴ ^{m}C_{2} = 28

∴ \(\frac{\mathrm{m} !}{2 !(\mathrm{m}-2) !}=28\)

∴ \(\frac{m(m-1)(m-2) !}{2 \times(m-2) !}=28\)

∴ m(m – 1) = 56

∴ m^{2} – m – 56 = 0

∴ (m – 8) (m + 7) = 0

As m cannot be negative.

∴ m = 8