Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 8 Binomial Theorem Ex 8.2 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 11 Maths Chapter 8 Binomial Theorem Ex 8.2

Find the coefficient of:

1. x^{5} in (x + 3)^{8}

2. a^{5}b^{7} in (a – 2b)^{12}

Solutions to questions 1 and 2:

1. General term in (x + 3)^{8}

= ^{8}C_{r}x^{8-r}.3^{r}

We have to find the coefficient of x^{5}

8 – r = 5 â‡’ r = 8 – 5 = 3.

Coefficient of x^{5} (putting r = 3)

= ^{8}C_{3}.3^{3} = \(\frac{8.7.6}{1.2.3}\).27 = 56.27 = 1512.

2. (a – 2b)^{12} = [a + (- 2b)]^{12}

General term T_{r+1} = C(12, r)a^{12-r}(- 2b)^{r}

Putting 12 – r = 5 or 12 – 5 = r â‡’ r = 7.

âˆ´ T_{7+1} = C(12, 7)a^{12-7}(- 2b)^{7}

= C(12, 7)a^{5}(- 2)^{7}b^{7}

= C(12, 7)(- 2)^{7}a^{5} b^{7}.

Hence, required coefficient is

= – 11 Ã— 9 Ã— 2^{7} = – 99 Ã— 8 Ã— 128

= – 101376.

3. General term = T_{r+1} = ^{6}C_{r}(x^{2})^{6-r}(- y)^{r}

= (- 1)^{r}\(\frac{6!}{r!(6 – r)!}\).x^{12-2r}.y^{r}.

4. Binomial expression is (x^{2} – yx)^{12}.

âˆ´ General term = T_{r+1} = ^{12}C_{r}(x^{2})^{12-r}.(- yx)^{r}

= \(\frac{12!}{r!(12 – r)!}\).x^{24-2r}(- 1)^{r}y^{r}x^{r}

= \(\frac{(-1)^{r} 12 !}{r !(12-r) !}\).x^{24-r}y^{r}.

5. Find the 4th term in the expansion of (x – 2y)^{12}.

Solution:

4th term = T_{3+1} in the expansion of x + (- 2y)^{12}

= ^{12}C_{3}x^{12-3}(- 2y)^{3}

= \(\frac{12.11.10}{1.2.3}\).x^{9}(- 1)^{3}.2^{3}.y^{3}

= – 220 Ã— 8x^{9}.y^{3} = – 1760x^{9}y^{3}.

6. Find the 13th term in the expansion of (9x – \(\frac{1}{3 \sqrt{x}}\)^{18})

Solution:

13th term = T_{13} = T_{12+1} = ^{18}C_{12}(9x)^{18-12}(- \(\frac{1}{3 \sqrt{x}}\))^{12}

Find the middle term in the expansion of:

7. (3 – \(\frac{x^{3}}{6}\))^{7}

8. (\(\frac{x}{3}\) + 9y)^{10}

Solutions to questions 7 and 8:

7. Number of terms in the expansion is 7 + 1 = 8.

âˆ´ There are two middle terms which are T_{4} and T_{5}.

Hence, we are to find T_{4} and T_{5} in the given expansion.

(3 – \(\frac{x^{3}}{6}\))^{7} = [3 + (- \(\frac{x^{3}}{6}\))^{7}

âˆ´ T_{r+1} = C(7, 3)3^{7-r}(- \(\frac{x^{3}}{6}\))^{r} ………………… (1)

Now, T_{r+1} = T_{4}.

or r + 1 = 4

âˆ´ r = 3.

âˆ´ Putting r = 3 in (1), we have:

Again T_{r+1} = T_{5} or r + 1 = 5.

â‡’ r = 4

Putting r = 4 in (1), we get:

8. Number of terms in the expansion is 10 + 1 = 11.

Middle term of the expansion is \(\frac{11+1}{2}\) = T_{6}.

T_{r+1} = C(10, r)(\(\frac{x}{3}\))^{10-r} (9r)^{r}.

But T_{r+1} = T_{6} or r + 1 = 6.

â‡’ r = 5.

Putting r = 5 in (1), we have:

Question 9.

In the expansion of (1 + a)^{m+n}, prove that coefficients of a^{m} and a^{n} are equal?

Solution:

General term in the expansion of (1 + a)^{m+n}

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

Putting r = m

T_{m+1} = ^{m+n}C_{m}a^{m} ……………….. (1)

âˆ´ Coefficient of a^{m} = ^{m+n}C_{m}.

Again putting r = n, we get

T_{n+1} = ^{m+n}C_{n}a^{n}.

Coefficients of a^{n} = ^{m+n}C_{n}

= ^{m+n}C_{m} ………………….. (2) [âˆµ ^{n}C_{r} = ^{n}C_{n-r}]

From (1) and (2), coefficient of am is equal to coefficient of

a^{m} is equal to coefficient of a^{n}.

Question 10.

The coefficients of the (r – 1)^{th}, rth and (r + 1)^{th} terms in the expansion of (x + 1)^{n} are in the ratio of 1 : 3 : 5. Find both n and r.

Solution:

General term in the expansion of (x + 1)^{n} is

T_{k+1} + 1 = C(n, k)x^{n-k}.

Putting T_{k+1} = T_{r-1} or k + 1 = r – 1 or k = r – 2.

âˆ´ Coefficient of T_{r-1} is C(n, r – 2) ………………….. (1)

Putting T_{k+1} = T_{r} or k + 1 = r, we get

k = r – 1.

âˆ´ Coefficient of T_{r} is C(n, r – 1) ………………….. (2)

Putting T_{r+1} = T_{k+1}, we get

r = k.

âˆ´ Coefficient of T_{r+1} = C(n, r)

According to the problem,

or 5r = 3 (n + 1 – r)

or 8r = 3n + 3 ………………… (5)

From (4) and (5),

2n + 10 = 3n + 3

or 3n – 2n = 10 – 3

From (5), 8r = 21 + 3 = 24

âˆ´ r = 3

âˆ´ n = 7, r = 3.

Question 11.

Prove that the coefficient of x^{n} in (1 + x)^{2nÂ }is twice the coefficient of x^{n} in (1 + x)^{2n-1}.

Solution:

General term in the expansion of (1 + x)^{2n} is

T_{r+1} = C(2n, r)x^{r}.

Putting r = n, we have:

T_{n+1} = C(2n, n)x^{n}.

Coefficient of x^{n} = C(2n, n) ……………………. (1)

Again general term in the expansion of (1 + x)^{2n-1} is

T_{r+1} = C(2n – 1, r)x^{r}.

Putting r = n, we have:

T_{n+1} = C(2n – 1, n)x^{n}.

Coefficient of x^{n} in the expansion of x^{n} is C(2n – 1, n).

According to the problem, we are to prove that

C(2n, n) = 2 Ã— C(2n – 1, n).

Multiplying numerator (N) and denominator (D) by n on R.H.S; we have:

= \(\frac{2n(2n – 1)!}{n!n(n – 1)!}\).

i.e; \(\frac{2n!}{n!n!}\) = \(\frac{2n!}{n!n!}\), which is true. (Proved)

Question 12.

Find a positive value of m for which the coefficient of x^{2} in the expansion of (1 + x)^{m} is 6.

Solution:

Coefficient of x^{2} in the expansion of (1 + x)^{m} is C(m, 2).

According to the problem,

C(m, 2) = 6.

or \(\frac{m(m – 1)}{2!}\) = 6

or m^{2} – m = 12

or m^{2} – m – 12 = 0

or m^{2} – 4m + 3m – 12 = 0

or m(m – 4) + 3(m – 4) = 0

or (m – 4)(m + 3) = 0

m = 4, since m â‰ – 3.