Binomial Theorem(11th Grade > Mathematics ) Questions and answers for exam Preparation

11th Grade > Mathematics

BINOMIAL THEOREM MCQs

Total Questions : 30 | Page 2 of 3 pages

Question 11.

The value of ${(\sqrt{2} + 1)}^{6}$ + ${(\sqrt{2} - 1)}^{6}$ will be

-198
198
99
-99

Discuss Question

Answer: Option B. -> 198
:
B

$(x + a)^{n}$ + $(x - a)^{n}$ = 2 $[x^{n} +^{n} C_{2} x^{n - 2} a^{2} +^{n} C_{4} x^{n - 4} a^{4} +^{n} C_{6} x^{n - 6} a^{6} + . . . . . . .]$

Here, n = 6, x = $\sqrt{2}$ , a = 1;

$^{6} C_{2}$ = 15, $^{6} C_{4}$ = 15, $^{6} C_{6}$ = 1

∴ ( $\sqrt{2} + 1)^{6}$ + ( $\sqrt{2} - 1)^{6}$ = 2 $[(\sqrt{2})^{6} + 15. (\sqrt{2})^{4} .1 + 15 (\sqrt{2})^{2} .1 + 1.1]$

$= 2 [8 + 15 \times 4 + 15 \times 2 + 1] = 198$

Question 12.

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

$\frac{1.3.5....... (5 n - 1)}{n!}$ $x^{n}$
$\frac{2.4.6.......2 n}{n!}$ $x^{2 n + 1}$
$\frac{1.3.5....... (2 n - 1)}{n!}$ $x^{n}$
$\frac{1.3.5....... (2 n - 1)}{n!}$ $2^{n} x^{n}$

Discuss Question

Answer: Option D. ->

\frac{1.3.5....... (2 n - 1)}{n!}

2^{n} x^{n}

:
D

Middle term of $(1 + 2 x)^{2 n}$ is $T_{n + 1}$ = $^{2 n} C_{n}$ $x^{n}$

= $\frac{(2 n)!}{n! n!}$ $x^{n}$ = $\frac{1.3.5....... (2 n - 1)}{n!}$ $2^{n}$ $x^{n}$ .

Question 13.

If the coefficient of $(2 r + 3)^{t h}$ and $(r - 3)^{t h}$ terms in the expansion of ( $1 + x)^{18}$ are equal, then r =

Discuss Question

Answer: Option D. -> 6
:
D

$^{18} C_{2 r + 3}$ = $^{18} C_{r - 3}$ ⇒ 2r + 3 + r - 3 = 18 ⇒ r = 6

Question 14.

If the coefficient of x in the expansion of $(x^{2} + \frac{k}{x})^{5}$ is 270, then k =

Discuss Question

Answer: Option C. -> 3
:
C

$T_{r + 1}$ = $^{5} C_{r}$ $(x^{2})^{5 - r}$ ( $\frac{k}{x})^{r}$

For coefficient of x, 10 - 2r - r = 1 ⇒ r = 3

Hence, $T_{3 + 1}$ = $^{5} C_{3}$ $(x^{2})^{5 - 3}$ ( $\frac{k}{x})^{3}$

According to question, 10 $k^{3}$ = 270 ⇒ k = 3.

Question 15.

If the coefficient of $x^{7}$ in $(a x^{2} + \frac{1}{b x})^{11}$ is equal to the coefficient of $x^{- 7}$ in $(a x - \frac{1}{b x^{2}})^{11}$ , then ab =

1
$\frac{1}{2}$
2
3

Discuss Question

Answer: Option A. -> 1
:
A

In the expansion of $(a x^{2} + \frac{1}{b x})^{11}$ , the general

term is

$T_{r + 1}$ = $^{11} C_{r}$ $(a x^{2})^{11 - r}$ $(\frac{1}{b x})^{r}$ = $^{11} C_{r}$ $(a)^{11 - r}$ $(\frac{1}{b^{r}}) x^{22 - 3 r}$

For $x^{7}$ , we must have 22 - 3r = 7 ⇒ r = 5, and

the coefficient of $x^{7}$ = $^{11} C_{5}$ . $a^{11 - 5} \frac{1}{5}$ ) = $^{11} C_{5}$ $\frac{a^{6}}{b^{5}}$

Similarly, in the expansion of $(a x - \frac{1}{b x^{2}})^{11}$ , the

general term is $T_{r + 1}$ = $^{11} C_{r} (- 1)^{r}$ $\frac{a^{11 - r}}{b^{r}}$ . $x^{11 - 3 r}$

For $x^{- 7}$ we must have, 11 - 3r = -7 ⇒ r = 6, and

the coefficient of $x^{- 7}$ is $^{11} C_{6}$ $\frac{a^{5}}{b^{6}}$ = $^{11} C_{5}$ $\frac{a^{5}}{b^{6}}$ .

As given, $^{11} C_{5}$ $\frac{a^{6}}{b^{5}}$ = $^{11} C_{5}$ $\frac{a^{5}}{b^{6}}$ ⇒ ab = 1.

Question 16.

If the coefficient of the middle term in the expansion of $(1 + x)^{2 n + 2}$ is p and the coefficients of middle terms in the expansion of $(1 + x)^{2 n + 1}$ are q and r, then

p + q = r
p + r = q
p = q + r
p + q + r = 0

Discuss Question

Answer: Option C. -> p = q + r
:
C

Since $(n + 2)^{t h}$ term is the middle term in the

expansion of $(1 + x)^{2 n + 2}$ , therefore p = $^{2 n + 2} C_{n + 1}$ .

Since $(n + 1)^{t h}$ and $(n + 2)^{t h}$ terms are middle terms in the expansion of $(1 + x)^{2 n + 1}$ ,
therefore q = $^{2 n + 1} C_{n}$ and r = $^{2 n + 1} C_{2 n + 1}$ .

But , $^{2 n + 1} C_{n}$ + $^{2 n + 1} C_{n + 1}$ = $^{2 n + 2} C_{n + 1}$

∴ q + r = p

Question 17.

If $(1 + a x)^{n}$ = 1 + 8x + 24 $x^{2}$ + ......, then the value of a and n is

2, 4
2, 3
3, 6
1, 2

Discuss Question

Answer: Option A. -> 2, 4
:
A

As given $(1 + a x)^{n}$ = 1 + 8x + 24 $x^{2}$ + .........

⇒ 1 + $\frac{n}{1}$ ax + $\frac{n (n - 1)}{1.2}$ $a^{2} x^{2}$ + ......... = 1 + 8x + 24 $x^{2}$ + ...........

⇒na = 8, $\frac{n (n - 1)}{1.2}$ $a^{2}$ = 24 ⇒ na(n-1)a = 48

⇒8(8 - a) = 48 ⇒ 8 - a = 6 ⇒ a = 2 ⇒ n = 4.

Question 18.

$C_{0} - C_{1} + C_{2} - C_{3} + . . . . . . . . + (- 1)^{n} C_{n}$ is equal to

$2^{n}$
$2^{n} - 1$
0
$2^{n - 1}$

Discuss Question

Answer: Option C. -> 0
:
C

We know that

$(1 + x)^{n} =^{n} C_{0} +^{n} C_{1} x +^{n} C_{2} x^{2} + . . . . . . +^{n} C_{n} x^{n}$

Putting x = -1, we get

$(1 - 1)^{n}$ = $^{n} C_{0}$ - $^{n} C_{1}$ + $^{n} C_{2}$ - ....... $(- 1)^{n}$ $^{n} C_{n}$

Therefore $C_{0} - C_{1} + C_{2} - C_{3} + . . . . . . (- 1)^{n} C_{n}$ = 0

Question 19.

In the polynomial (x - 1)(x - 2)(x - 3)............... .........(x - 100), the coefficient of $x^{99}$ is

5050
-5050
100
99

Discuss Question

Answer: Option B. -> -5050
:
B

(x - 1)(x - 2)(x - 3)..............(x - 100)

Number of terms = 100;

∴ Coefficient of $x^{99}$

= (x - 1)(x - 2)(x - 3)........(x -100)

= (-1 - 2 - 3 - .......... - 100) = -(1 + 2 +.........+ 100)

= - $\frac{100 \times 101}{2}$ = - 5050.

Question 20.

In the expansion of ( $\frac{a}{x} + b x)^{12}$ , the coefficient of $x^{- 10}$ will be

$12 a^{11}$
$12 b^{11} a$
$12 a^{11} b$
$12 a^{11} b^{11}$

Discuss Question

Answer: Option C. ->

12 a^{11} b

:
C

$T_{r + 1} = 12_{c_{r}} {(\frac{a}{x})}^{12 - r} {(b x)}^{r} = 12_{c_{r}} . a^{12 - r} . b^{r} . x^{2 r - 12} ∴ 2 x - 12 = - 10 \Rightarrow r = 1 ∴ coefficient = 12. a^{11} . b$