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

12th Grade > Mathematics

BINOMIAL THEOREM MCQs

Binomial Theorem

Total Questions : 74 | Page 5 of 8 pages

Question 41. 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 42. If

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

, then

C_{0}^{2} + C_{1}^{2} + C_{2}^{2} + C_{3}^{2} + . . . . . . . . . . + C_{n}^{2}

n!n!n!
(2n)!n!n!
(2n)!n!
None of these

Discuss Question

Answer: Option B. -> (2n)!n!n!
:
B

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

...........(i)
and

(1 + \frac{1}{x})^{n}

C_{0} + C_{1} \frac{1}{x} + C_{2} (\frac{1}{x})^{2} + . . . . . . . . + C_{n} (\frac{1}{x})^{n}

..........(ii)
If we multiply (i) and (ii), we get

C_{0}^{2} + C_{1}^{2} + C_{2}^{2} + . . . . . . . . + C_{n}^{2}

is the term independent of x and hence it is equal to the term independent of x in the product

(1 + x)^{n} (1 + \frac{1}{x})^{n}

or in

\frac{1}{x^{n}}

(1 + x)^{2 n}

or term
containing

x^{n}

(1 + x)^{2 n}

. Clearly the coefficient of

x^{n}

(1 + x)^{2 x}

T_{n + 1}

and equal to

^{2 n} C_{n}

\frac{(2 n)!}{n! n!}

Trick:Solving conversely.
Put n = 1, n = 2,.........then we get

S_{1}

^{1} C_{0}^{2} +^{1} C_{1}^{2}

= 2,

S_{2}

^{2} C_{0}^{2} +^{2} C_{1}^{2} +^{2} C_{2}^{2} = 1^{2} + 2^{2} + 1^{2}

= 6
Now check the options
(a) Does not hold given condition,
(b) (i) Put n = 1, then

\frac{2!}{1! 1!} = 2

(ii) Put n = 2, then

\frac{4!}{2! 2!} = \frac{4 \times 3 \times 2 \times 1}{2 \times 1 \times 2 \times 1} = 6

Note: Students should remember this question as identity

Question 43. In the expansion of

(\frac{1 + x}{1 - x})^{2}

, the coefficient of

x^{n}

4n
4n-3
4n+1
None of these

Discuss Question

Answer: Option A. -> 4n
:
A
Given term can be written as

(1 + x)^{2}

(1 - x)^{- 2}

(1 + 2 x + x^{2}) [1 + 2 x + 3 x^{2} + . . . . . . . + (n - 1) x^{n - 2} + n x^{n - 1} + (n + 1) x^{n} + . . . . . . . . . . .]

x^{n} (n + 1 + 2 n + n - 1)

+..........
Therefore coefficient of

x^{n}

is 4n.

Question 44.

(\frac{a}{a + x})^{\frac{1}{2}}

(\frac{a}{a - x})^{\frac{1}{2}}

2 + 3x24a2 + .......
1 + 3x28a2 + .....
2 + xa + 3x24a2 + ......
2 - xa + 3x24a2 + .......

Discuss Question

Answer: Option A. -> 2 + 3x24a2 + .......
:
A

(\frac{a + x}{a})^{\frac{- 1}{2}}

(\frac{a - x}{a})^{\frac{- 1}{2}}

(1 + \frac{x}{a})^{\frac{- 1}{2}} + (1 - \frac{x}{a})^{\frac{- 1}{2}}

=[1 + (-

\frac{1}{2}

)(

\frac{x}{a}

) +

\frac{(- \frac{1}{2}) (- \frac{3}{2})}{2.1}

(\frac{x}{a})^{2}

+ ........]
+

[1 + (- \frac{1}{2}) (- \frac{x}{a}) + \frac{(- \frac{1}{2}) (\frac{3}{2})}{2.1} (- \frac{x}{a})^{2} + . . . . . . . .]

= 2 +

\frac{3 x^{2}}{4 a^{2}}

+ ......
Here odd terms cancel each other.

Question 45. The fourth term in the expansion of

(1 - 2 x)^{\frac{3}{2}}

will be

- 34x4
x32
-x32
34x4

Discuss Question

Answer: Option B. -> x32
:
B
Expansion of

(1 - 2 x)^{\frac{3}{2}}

1 + \frac{3}{2} (- 2 x) + \frac{3}{2} . \frac{1}{2} . \frac{1}{2} (- 2 x)^{2} + \frac{3}{2} . \frac{1}{2} (- \frac{1}{2}) \frac{1}{6} (- 2 x)^{3} + . . . . . . . . .

Hence

4^{t h}

term is

\frac{x^{3}}{2}

Question 46.

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

is equal to

2n
2n−1
0
2n−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 47. If

\frac{1}{(x^{2} + \frac{1}{x})^{\frac{4}{3}}}

can be expanded by binomial theorem if

x < 1
|x| < 1
x > 1
|x| > 1

Discuss Question

Answer: Option D. -> |x| > 1
:
D
The expansion is

x^{- \frac{8}{3}} (1 + \frac{1}{x^{3}})^{-} \frac{4}{3}

Hence |

\frac{1}{x^{3}}

| < 1 ⇒ |x| > 1

Question 48.

\frac{C_{0}}{1}

\frac{C_{2}}{3}

\frac{C_{4}}{5}

\frac{C_{6}}{7}

+..........=

2n+1n+1
2n+1−1n+1
2nn+1
None of these

Discuss Question

Answer: Option C. -> 2nn+1
:
C
Putting the values of

C_{0}, C_{2}, C_{4}

........., we get
= 1 +

\frac{n (n - 1)}{3.2!}

\frac{n (n - 1) (n - 2) (n - 3)}{5.4!}

+ .........
=

\frac{1}{n + 1}

[(n+1) +

\frac{(n + 1) n (n - 1)}{3!}

\frac{(n + 1) n (n - 1) (n - 2) (n - 3)}{5!}

+ ..........]
Put n+1 = N
=

\frac{1}{N}

[N +

\frac{N (N - 1) (N - 2)}{3!}

\frac{N (N - 1) (N - 2) (N - 3) (N - 4)}{5!}

+ .........]
=

\frac{1}{N}

{

^{N} C_{1}

^{N} C_{3}

^{N} C_{5}

+..........}
=

\frac{1}{N}

{

2^{N - 1}

} =

\frac{2^{n}}{n + 1}

{ ∵ N = n+1}
Trick:Put n = 1, then

S_{1}

\frac{^{1} C_{0}}{1}

\frac{1}{1}

= 1
At n = 2,

S_{2}

\frac{^{2} C_{0}}{1}

\frac{^{2} C_{2}}{3}

= 1 +

\frac{1}{3}

\frac{4}{3}

Also (c) ⇒

S_{1}

= 1,

S_{2}

\frac{4}{3}

Question 49. If a and d are two complex numbers, then the sum
to (n+1) terms of the following series
a

C_{0}

- (a + d)

C_{1}

+ (a + 2d)

C_{2}

- ........... is

a2n
na
0
None of these

Discuss Question

Answer: Option C. -> 0
:
C
We can write
a

C_{0}

- (a + d)

C_{1}

+ (a + 2d)

C_{2}

- ........ upto (n+1) terms

= a (C_{0} - C_{1} + C_{2} - . . . . . . . .) + d (- C_{1} + 2 C_{2} - 3 C_{3} + . . . . . .)

.............(i)
Again,

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

..........(ii)
Differentiating with respect to x,

- n (1 - x)^{n - 1}

= -

C_{1}

+ 2

C_{2}

x - .......... +

(- 1)^{n} C_{n} n x^{n - 1}

......................(iii)
Putting x = 1 in (ii) and (iii), we get

C_{0}

C_{1}

C_{2}

- ........ + (-1)

^{n} C_{n}

= 0
and -

C_{1}

+ 2

C_{2}

- ........+

(- 1)^{n} n . C_{n}

= 0
Thus the required sum to (n+1) terms, by (i)
= a.0 + d.0 = 0.

Question 50. Cube root of 217 is

6.01
6.04
6.02
None of these

Discuss Question

Answer: Option A. -> 6.01
:
A

(217)^{\frac{1}{3}}

(6^{3} + 1)^{\frac{1}{3}}

6 (1 + \frac{1}{6^{3}})^{\frac{1}{3}}

On expansion by binomial theorem
=

6 (1 + \frac{1}{3 \times 216} - \frac{1 \times 2}{3 \times 3 \times 2} (\frac{1}{216})^{2} + . . . . . . . . .)

= 6.01

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