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

12th Grade > Mathematics

BINOMIAL THEOREM MCQs

Binomial Theorem

Total Questions : 74 | Page 2 of 8 pages

Question 11. The number of integral terms in the expansion of

(5^{\frac{1}{2}} + 7^{\frac{1}{6}})^{642}

Discuss Question

Answer: Option B. -> 108
:
B

T_{r + 1}

^{642} C_{r} (5^{\frac{1}{2}})^{642 - r} . (7^{\frac{1}{6}})^{r}

Obviously, r should be a multiple of 6.
Total number of terms=

\frac{642}{6}

= 107,
but first term forr = 0 is also an integer. Hence total terms are 107 + 1 = 108.

Question 12. The value of the expression

(2 + \sqrt{2})^{4}

lies between

134 and 135
135 and 136
136 and 137
None of these

Discuss Question

Answer: Option B. -> 135 and 136
:
B

(2 + \sqrt{(} 2))^{4}

(\sqrt{2})^{4}

(\sqrt{2} + 1)^{4}

= 4[

^{4} C_{0}

^{4} C_{1}

(

\sqrt{2})

^{4} C_{2}

(

\sqrt{2})^{2}

^{4} C_{3}

(

\sqrt{2})^{3}

^{4} C_{4}

(

\sqrt{2})^{4}

]
= 4 [1 + 4

\sqrt{2}

\frac{4.3}{2}

.2 +

\frac{4.3.2}{1.2.3}

\sqrt{2}

+ 4]
= 4[1 + 4

\sqrt{2}

+ 12 + 8

\sqrt{2}

+ 4] = 4[17 + 12

\sqrt{2}

]
= 4[17 + 17] = 4[34] = 136

Question 13. If

\frac{(1 - 3 x)^{\frac{1}{2}} + (1 - x)^{\frac{5}{3}}}{\sqrt{4 - x}}

is approximately equal to a + bx for small values of x, then (a, b) =

(1, 3524)
(1, - 3524)
(2, 3512)
(2, - 3512)

Discuss Question

Answer: Option B. -> (1, - 3524)
:
B

\frac{(1 - 3 x)^{1 / 2} + (1 - x)^{5 / 3}}{2 [1 - \frac{x}{4}]^{1 / 2}} \frac{1}{2} [(1 - \frac{3}{2} x) + (1 - \frac{5 x}{3})] (1 + \frac{x}{8}) \frac{1}{2} [2 - \frac{19}{6} x]] (1 + \frac{x}{8}) = \frac{1}{2} [2 - (- \frac{1}{4} + \frac{19}{6}) x] = 1 - \frac{35}{24} x ∴ (a, b) = (1, - \frac{35}{24})

Question 14. The coefficient of

x^{3}

in the expansion of

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

will be

8
32
50
None of these

Discuss Question

Answer: Option C. -> 50
:
C
=

(1 + 3 x)^{2}

(

1 - 2 x)^{- 1}

= (

1 + 3 x)^{2}

(1 + 2x +

\frac{1.2}{2.1} (- 2 x)^{2} + . . . . . . .

)
=

(1 + 6 x + 9 x^{2}) (1 + 2 x + 4 x^{2} + 8 x^{3} + . . . . . . . .)

Therefore coefficient of

x^{3}

is (8 + 24 + 18) = 50.

Question 15. The value of

\frac{C_{1}}{2}

\frac{C_{3}}{4}

\frac{C_{5}}{6}

+ ...... is equal to

2n−1n+1
n.2n
2nn
2n−1

Discuss Question

Answer: Option A. -> 2n−1n+1
:
A
We know that

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

C_{1} x + C_{3} x^{3} + C_{5} x^{5} + . . . . . . . .

Integrating from x = 0 to x = 1, we get

\frac{1}{2}

\int_{0}^{1} (1 + x)^{n} - (1 - x)^{n} d x

\int_{0}^{1} (C_{1} x + C_{3} x^{3} + C_{5} x^{5} + . . . . . . .) d x

⇒

\frac{1}{2} {\frac{(1 + x)^{n + 1}}{n + 1} + \frac{(1 - x)^{n + 1}}{n + 1}}_{0}^{1} = \frac{C_{1}}{2}

\frac{C_{3}}{4}

\frac{C_{5}}{6}

+ ....
or

\frac{C_{1}}{2}

\frac{C_{3}}{4}

\frac{C_{5}}{6}

+ .......=

\frac{1}{2}

{

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

\frac{0 - 1}{n + 1}

}
=

\frac{1}{2}

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

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

Question 16. 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 17. 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 18.

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 19. In the polynomial (x - 1)(x - 2)(x - 3)............... .........(x - 100), the coefficient of

x^{99}

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. The greatest integer less than or equal to

{(\sqrt{2} + 1)}^{6}

is?

Discuss Question

Answer: Option B. -> 197
:
B
Let

{(\sqrt{2} + 1)}^{6} = I + F,

Where I is an integer and0< F<1.

f = \sqrt{2} - 1 = \frac{1}{\sqrt{2} + 1} \Rightarrow 0 < \sqrt{2} - 1 < 1 \Rightarrow 0 < f < 1

Also

I + F + f = {(\sqrt{2} + 1)}^{6} + {(\sqrt{2} - 1)}^{6}

2 [^{6} C_{0} {.2}^{3} +^{6} C_{2} {.2}^{2} +^{6} C_{4} .2 +^{6} C_{6}]

= 2 (8 + 60 + 30 + 1) = 198

Hence,

F + f = 198 - I

is an integer. But

0 < F + f < 2

∴ F + f = 1

, and thus,

I = 197

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