NCERT Solutions For Class 11 Math Chapter – 8 Exercise – 8.2

NCERT  Solutions For Class 11 Math Chapter – 8 Exercise – 8.2

Find the coefficient of

1. x⁵ in (x + 3)⁸

Solution:

The general term T_r+1 in the binomial expansion is given by T_r+1 = ⁿ C _r a^n-r b^r

Here x⁵ is the T_r+1 term so a= x, b = 3 and n =8

T_r+1 = ⁸C_r x^8-r 3^r…………… (i)

For finding out x⁵

We have to equate x⁵= x^8-r

⇒ r= 3

Putting value of r in (i) we get

= 1512 x⁵

Hence the coefficient of x⁵= 1512

2. a⁵b⁷ in (a – 2b)¹² .

Solution:

The general term T_r+1 in the binomial expansion is given by T_r+1 = ⁿ C _r a^n-r b^r

Here a = a, b = -2b & n =12

Substituting the values, we get

T_r+1 = ¹²C_r a^12-r (-2b)^r………. (i)

To find a⁵

We equate a^12-r =a⁵

r = 7

Putting r = 7 in (i)

T₈ = ¹²C₇ a⁵ (-2b)⁷

= -101376 a⁵ b⁷

Hence the coefficient of a⁵b⁷= -101376

Write the general term in the expansion of

3. (x² – y)⁶

Solution:

The general term T_r+1 in the binomial expansion is given by

T_r+1 = ⁿ C _r a^n-r b^r…….. (i)

Here a = x² , n = 6 and b = -y

Putting values in (i)

T_r+1 = ⁶C_r x ^2(6-r) (-1)^r y^r

= -1^{r 6}c_r .x^{12 – 2r}. y^r

4. (x² – y x)¹², x ≠ 0.

Solution:

The general term T_r+1 in the binomial expansion is given by T_r+1 = ⁿ C _r a^n-r b^r

Here n = 12, a= x² and b = -y x

Substituting the values we get

T_n+1 =¹²C_r × x^2(12-r) (-1)^r y^r x^r

= -1^{r 12}c_r .x^{24 –2r}. y^r

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

Solution:

The general term T_r+1 in the binomial expansion is given by T_r+1 = ⁿ C _r a^n-r b^r

Here a= x, n =12, r= 3 and b = -2y

By substituting the values we get

T₄ = ¹²C₃ x⁹ (-2y)³

= -1760 x⁹ y³

6. Find the 13^th term in the expansion of 

Solution:

Find the middle terms in the expansions of

Solution:

Solution:

9. In the expansion of (1 + a)^m+n, prove that coefficients of a^m and aⁿ are equal.

Solution:

We know that the general term T_r+1 in the binomial expansion is given by T_r+1 = ⁿC_r a^n-r b^r

Here n= m+n, a = 1 and b= a

Substituting the values in the general form

T_r+1 = ^m+n C_r 1^m+n-r a^r

= ^m+n C_r a^r…………. (i)

Now we have that the general term for the expression is,

T_r+1 =  ^m+n C_r a^r

Now, For coefficient of a^m

T_m+1 =  ^m+n C_m a^m

Hence, for coefficient of a^m, value of r = m

So, the coefficient is ^m+n C _m

Similarly, Coefficient of aⁿ is ^m+n C _n

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

Solution:

The general term T_r+1 in the binomial expansion is given by T_r+1 = ⁿC_r a^n-r b^r

Here the binomial is (1+x)ⁿ with a = 1 , b = x and n = n

The (r+1)^th term is given by

T_(r+1) = ⁿC_r 1^n-r x^r

T_(r+1) = ⁿC_r x^r

The coefficient of (r+1)^th term is ⁿC_r

The r^th term is given by (r-1)^th term

T_(r+1-1) = ⁿC_r-1 x^r-1

T_r = ⁿC_r-1 x^r-1

∴ the coefficient of r^th term is ⁿC_r-1

For (r-1)^th term we will take (r-2)^th term

T_r-2+1 = ⁿC_r-2 x^r-2

T_r-1 = ⁿC_r-2 x^r-2

∴ the coefficient of (r-1)^th term is ⁿC_r-2

Given that the coefficient of (r-1)^th, r^th and r+1^th term are in ratio 1:3:5

Therefore,

⇒ 5r = 3n – 3r + 3

⇒ 8r – 3n – 3 =0………….2

We have 1 and 2 as

n – 4r ± 5 =0…………1

8r – 3n – 3 =0…………….2

Multiplying equation 1 by number 2

2n -8r +10 =0……………….3

Adding equation 2 and 3

2n -8r +10 =0

-3n – 8r – 3 =0

⇒ -n = -7

n =7 and r = 3

11. Prove that the coefficient of xⁿ in the expansion of (1 + x)²ⁿ is twice the coefficient of xⁿ in the expansion of (1 + x)^{2n – 1}.

Solution:

The general term T_r+1 in the binomial expansion is given by T_r+1 = ⁿC_r a^n-r b^r

The general term for binomial (1+x)²ⁿ is

T_r+1 = ²ⁿC_r x^r …………………..1

To find the coefficient of xⁿ

r = n

T_n+1 = ²ⁿC_n xⁿ

The coefficient of xⁿ = ²ⁿC_n

The general term for binomial (1+x)^2n-1 is

T_r+1 = ^2n-1C_r x^r

To find the coefficient of xⁿ

Putting n = r

T_r+1 = ^2n-1C_r xⁿ

The coefficient of xⁿ = ^2n-1C_n

We have to prove

Coefficient of xⁿ in (1+x)²ⁿ = 2 coefficient of xⁿ in (1+x)^2n-1

Consider LHS = ²ⁿC_n

12. Find a positive value of m for which the coefficient of x² in the expansion (1 + x)^m is 6.

Solution:

The general term T_r+1 in the binomial expansion is given by T_r+1 = ⁿC_r a^n-r b^r

Here a = 1, b = x and n = m

Putting the value

T_r+1 = ^m C_r 1^m-r x^r

= ^m C_r x^r

We need coefficient of x²

∴ putting r = 2

T₂₊₁ = ^mC₂ x²

The coefficient of x² = ^mC₂

Given that coefficient of x² = ^mC₂ = 6

⇒ m (m – 1) = 12

⇒ m²– m – 12 =0

⇒ m²– 4m + 3m – 12 =0

⇒ m (m – 4) + 3 (m – 4) = 0

⇒ (m+3) (m – 4) = 0

⇒ m = – 3, 4

We need positive value of m so m = 4