Binomial Theorem(11th And 12th > Mathematics ) Questions and answers for exam Preparation

11th And 12th > Mathematics

BINOMIAL THEOREM MCQs

Total Questions : 45 | Page 1 of 5 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 the third term in the binomial expansion of $(1 + x)^{m}$ is - $\frac{1}{8} x^{2}$ , then the rational value of m is

2
$\frac{1}{2}$
3
4

Discuss Question

Answer: Option B. ->

\frac{1}{2}

:
B

$m_{c_{2}} = - \frac{1}{8} \Rightarrow \frac{m (m - 1)}{2} = - \frac{1}{8} \Rightarrow 4 m^{2} - 4 m + 1 = 0 \Rightarrow (2 m - 1)^{2} = 0 \Rightarrow m = \frac{1}{2}$

Question 3.

If the $(r + 1)^{t h}$ term in the expansion of ${(\sqrt[3]{\frac{a}{\sqrt{b}}} + \sqrt{\frac{b}{\sqrt[3]{a}}})}^{21}$ has the same power of a and b, then value of r is

Discuss Question

Answer: Option A. -> 9
:
A

We have $T_{r + 1}$ = $^{21} C_{r} {(\sqrt[3]{\frac{a}{\sqrt{b}}})}^{21 - r} {(\sqrt{\frac{b}{\sqrt[3]{a}}})}^{r}$

= $^{21} C_{r} a^{7 - (\frac{r}{2})} b^{\frac{2}{3} r - (\frac{7}{2})}$

Since the powers of a and b are the same,

therefore

7 - $\frac{r}{2}$ = $\frac{2}{3}$ r - $\frac{7}{2}$ ⇒ r = 9

Question 4.

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 5.

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 6.

$r^{t h}$ 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 7.

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 8.

$6^{t h}$ 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 9.

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 10.

Find the value of
$\frac{(18^{3} + 7^{3} + 3 \cdot 18 \cdot 7 \cdot 25)}{3^{6} + 6 \cdot 243 \cdot 2 + 15 \cdot 81 \cdot 4 + 20 \cdot 27 \cdot 8 + 15 \cdot 9 \cdot 16 + 6 \cdot 3 \cdot 32 + 64}$

Discuss Question

Answer: Option A. -> 1
:
A

The numerator is of the form

$a^{3}$ + $b^{3} + 3 a b (a + b) = (a + b)^{3}$

∴ N' = $(18 + 7)^{3}$ = $25^{3}$

∴ D' = $3^{6} +^{6} C_{1} \cdot 3^{5} \cdot 2^{1} +^{6} C_{2} \cdot 3^{4} \cdot 2^{2} +^{6} C_{3} \cdot 3^{3} \cdot 2^{3} +^{6} C_{4} \cdot 3^{2} \cdot 2^{4} +^{6} C_{5} \cdot 3^{1} \cdot 2^{5} +^{6} C_{6} \cdot 2^{6}$