Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Methods of Induction and Binomial Theorem Ex 4.2 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 4 Methods of Induction and Binomial Theorem Ex 4.2

Question 1.

Expand:

(i) (√3 + √2)^{4}

Solution:

Here, a = √3, b = √2 and n = 4.

Using binomial theorem,

∴ (√3 + √2)^{4} = 1(9) (1) + 4(3√3) (√2) + 6(3)(2) + 4(√3) (2√2) + 1(1)(4)

= 9 + 12√6 + 36 + 8√6 + 4

= 49 + 20√6

(ii) (√5 – √2)^{5}

Solution:

Here, a = √5, b = √2 and n = 5.

Using binomial theorem,

Question 2.

Expand:

(i) (2x^{2} + 3)^{4}

Solution:

Here, a = 2x^{2}, b = 3 and n = 4.

Using binomial theorem,

(ii) \(\left(2 x-\frac{1}{x}\right)^{6}\)

Solution:

Here, a = 2x, b = \(\frac{1}{x}\) and n = 6.

Using binomial theorem,

Question 3.

Find the value of

(i) (√3 + 1)^{4} – (√3 – 1)^{4}

Solution:

(ii) (2 + √5)^{5} + (2 – √5)^{5}

Solution:

Adding (i) and (ii), we get

∴ (2 + √5 )^{5} + (2 – √5)^{5} = (32 + 80√5 + 400 + 200√5 + 250 + 25√5) + (32 – 80√5 + 400 – 200√5+ 250 – 25√5 )

= 64 + 800 + 500

= 1364

Question 4.

Prove that:

(i) (√3 + √2)^{6} + (√3 – √2)^{6} = 970

Solution:

(ii) (√5 + 1)^{5} – (√5 – 1)^{5} = 352

Solution:

Question 5.

Using binomial theorem, find the value of

(i) (102)^{4}

Solution:

(ii) (1.1)^{5}

Solution:

Question 6.

Using binomial theorem, find the value of

(i) (9.9)^{3}

Solution:

(ii) (0.9)^{4}

Solution:

Question 7.

Without expanding, find the value of

(i) (x + 1)^{4} – 4(x + 1)^{3} (x – 1) + 6(x + 1)^{2} (x – 1)^{2} – 4(x + 1) (x – 1)^{3} + (x – 1)^{4}

Solution:

(ii) (2x – 1)^{4} + 4(2x – 1)^{3} (3 – 2x) + 6(2x – 1)^{2} (3 – 2x)^{2} + 4(2x – 1)^{1} (3 – 2x)^{3} + (3 – 2x)^{4}

Solution:

Question 8.

Find the value of (1.02)^{6}, correct upto four places of decimals.

Solution:

Question 9.

Find the value of (1.01)^{5}, correct upto three places of decimals.

Solution:

Question 10.

Find the value of (0.9)^{6}, correct upto four places of decimals.

Solution: