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

11th Grade > Mathematics

BINOMIAL THEOREM MCQs

Total Questions : 30 | Page 1 of 3 pages

The first 3 terms in the expansion of $(1 + a x)^{n}$ (n ≠ 0) are 1, 6x and 16 $x^{2}$ . Then the value of a and n are respectively

2 and 9
3 and 2
$\frac{2}{3}$ and 9
$\frac{3}{2}$ and 6

Discuss Question

Answer: Option C. ->

\frac{2}{3}

and 9
:
C

$n_{c_{1}} . a x = 6 x; n_{c_{2}} (a x)^{2} = 16 x^{2} \Rightarrow a n = 6; \frac{n (n - 1)}{2} a^{2} = 16 \Rightarrow a = \frac{2}{3} \Rightarrow n = 9$

Question 2.

If coefficient of $(2 r + 3)^{t h}$ and $(r - 1)^{t h}$ terms in the expansion of $(1 + x)^{15}$ are equal, then value of r is

Discuss Question

Answer: Option A. -> 5
:
A

$^{15} C_{2 r + 2}$ = $^{15} C_{r - 2}$

But $^{15} C_{2 r + 2}$ = $^{15} C_{15 - (2 r + 2)}$ = $^{15} C_{13 - 2 r}$

⇒ $^{15} C_{13 - 2 r}$ = $^{15} C_{r - 2}$ ⇒ r = 5.

Question 3.

If $x^{4}$ occurs in the $r^{t h}$ term in the expansion of $(x^{4} + \frac{1}{x^{3}})^{16}$ , then r =

Discuss Question

Answer: Option C. -> 9
:
C

$T_{r}$ = $^{16} C_{r - 1}$ $(x^{4})^{16 - r}$ $(\frac{1}{x^{3}})^{r - 1}$ = $^{16} C_{r - 1}$ $x^{67 - 7 r}$

$\to 67 - 7 r = 4 \to r = 9$

Question 4.

In $(\sqrt[3]{2} + \frac{1}{\sqrt[3]{3}})^{n}$ if the ratio of $7^{t h}$ term from the beginning to the $7^{t h}$ term from the end is $\frac{1}{6}$ , then n =

7
8
9
None of these

Discuss Question

Answer: Option C. -> 9
:
C

$\frac{1}{6} = \frac{^{n} C_{6} {(2^{\frac{1}{3}})}^{n - 6} {(3^{- \frac{1}{3}})}^{6}}{^{n} C_{n - 6} {(2^{\frac{1}{3}})}^{6} {(3^{- \frac{1}{3}})}^{n - 6}}$
$6^{- 1} = 6^{\frac{1}{3} (n - 12)}$
$n - 12 = - 3$
$n = 9$

Question 5.

$r^{t h}$ term in the expansion of $(a + 2 x)^{n}$ is

$\frac{n (n + 1) . . . . . (n - r + 1)}{r!} a^{n - r + 1} (2 x)^{r}$
$\frac{n (n - 1) . . . . . (n - r + 2)}{(r - 1)!} a^{n - r + 1} (2 x)^{r - 1}$
$\frac{n (n + 1) . . . . . (n - r)}{(r + 1)!} a^{n - r} (x) r$
None of these

Discuss Question

Answer: Option B. ->

\frac{n (n - 1) . . . . . (n - r + 2)}{(r - 1)!} a^{n - r + 1} (2 x)^{r - 1}

:
B

$r^{t h}$ term of $(a + 2 x)^{n}$ is $^{n} C_{r - 1} (a)^{n - r + 1} (2 x)^{r - 1}$

= $\frac{n!}{(n - r + 1)! (r - 1)!} a^{n - r + 1} (2 x)^{r - 1}$

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

Question 6.

$6^{t h}$ term in expansion of $(2 x^{2} - \frac{1}{3 x^{2}})^{10}$ is

$\frac{4580}{17}$
- $\frac{896}{27}$
$\frac{5580}{17}$
None of these

Discuss Question

Answer: Option B. -> -

\frac{896}{27}

:
B

Applying $T_{r + 1}$ = $^{n} C_{r}$ $x^{n - r} d$ for $(x + a)^{n}$

Hence $T_{6}$ = $^{10} C_{5}$ $(2 x^{2})^{5}$ (- $\frac{1}{3 x^{2}})^{5}$

= - $(\frac{10!}{5! 5!})$ (32) $(\frac{1}{243})$ = - $\frac{896}{27}$

Question 7.

If the ratio of the coefficient of third and fourth term in the expansion of $(x - \frac{1}{2 x})^{n}$ is 1:2, then the value of n will be

Discuss Question

Answer: Option D. -> -10
:
D

$T_{3}$ = $^{n} C_{2}$ $(x)^{n - 2}$ $(- \frac{1}{2 x})^{2}$ and $T_{4}$ = $^{n} C_{3}$ $(x)^{n - 3}$ $(- \frac{1}{2 x})^{3}$

But according to the condition,

$\frac{- n (n - 1) \times 3 \times 2 \times 1 \times 8}{n (n - 1) (n - 2) \times 2 \times 1 \times 4}$ = $\frac{1}{2}$ ⇒ n = -10

Question 8.

If n is positive integer and three consecutive coefficients in the expansion of $(1 + x)^{n}$ are in the ratio 6 : 33 : 110, then n =

Discuss Question

Answer: Option C. -> 12
:
C

Let the consecutive coefficient of $(1 + x)^{n}$ are

$^{n} C_{r - 1}$ , $^{n} C_{r}$ , $^{n} C_{r + 1}$

By condition, $^{n} C_{r - 1}$ : $^{n} C_{r}$ : $^{n} C_{r + 1}$ = 6 : 33 : 110

Now $^{n} C_{r - 1}$ : $^{n} C_{r}$ = 6 : 33

$\Rightarrow$ 2n - 13r + 2 = 0 ..............(i)

and $^{n} C_{r}$ : $^{n} C_{r + 1}$ = 33: 110

$\Rightarrow$ 3n - 13r - 10 = 0 ...............(ii)

Solving (i) and (ii), we get n = 12 and r = 2.

Aliter: We take first n = 4 [By alternate (a)] but

(a) does not hold. Similarity (b).

So alternate (c), n = 12 gives

$(1 + x)^{12}$ $= [1 + 12 x + \frac{12.11}{2.1}$ $x^{2}$ + $\frac{12.11.10}{3.2.1} x^{3} + . . . . . . . . .]$

So coefficient of II, III and IV terms are

$12, 6 \times 11, 2 \times 11 \times 10.$

So, $12 : 6 \times 11 : 2 \times 11 \times 10 = 6 : 33 : 110.$

Question 9.

If the coefficients of $p^{t h}$ , $(p + 1)^{t h}$ and $(p + 2)^{t h}$ terms in the expansion of $(1 + x)^{n}$ are in A.P., then

$n^{2} - 2 n p + 4 p^{2}$ = 0
$n^{2} - n (4 p + 1) + 4 p^{2} - 2$ = 0
$n^{2} - n (4 p + 1) + 4 p^{2}$ = 0
None of these

Discuss Question

Answer: Option B. ->

n^{2} - n (4 p + 1) + 4 p^{2} - 2

= 0
:
B

2 $^{n} c_{p} = n_{c_{p - 1}} + p_{c_{p + 1}} \Rightarrow \frac{2}{(n - p)! P!} = \frac{1}{(n - p + 1)! (p - 1)!} + \frac{1}{(n - p - 1)! (p + 1)!} \Rightarrow \frac{2}{(n - p) p} = \frac{1}{(n - p + 1) (n - p)} + \frac{1}{(p + 1) p} o n s i m p l i f i n g w e g e t n^{2} - n (4 p + 1) + 4 p^{2} - 2 = 0$