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

12th Grade > Mathematics

BINOMIAL THEOREM MCQs

Binomial Theorem

Total Questions : 74 | Page 4 of 8 pages

Question 31. The greatest integer less than or equal to

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

Discuss Question

Answer: Option B. -> 197
:
B
Let

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

= k + f, where k is integral part and f
the fraction (0 ≤ f < 1).
Let

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

= f, (0 < f < 1),
Since 0 < (

\sqrt{2}

- 1) < 1
Now, k + 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}

} = 198 ..........(i)
∴ f + f' = 198 - k = an integer
But, 0≤ f < 1 and 0 < f' < 1 ⇒ 0 < (f + f') < 2
⇒ f + f' = 1, ( ∵ f + f' is an integer)
∴ By (i), l = 198 - (f + f') = 198 - 1 = 197.

Question 32.

r^{t h}

term in the expansion of

(a + 2 x)^{n}

n(n+1).....(n−r+1)r!an−r+1(2x)r
n(n−1).....(n−r+2)(r−1)!an−r+1(2x)r−1
n(n+1).....(n−r)(r+1)!an−r(x)r
None of these

Discuss Question

Answer: Option B. -> n(n−1).....(n−r+2)(r−1)!an−r+1(2x)r−1
:
B

r^{t h}

term of

(a + 2 x)^{n}

^{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 33. 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 34. 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 35. The middle term in the expansion of

(1 + x)^{2 n}

1.3.5.......(5n−1)n!xn
2.4.6.......2nn!x2n+1
1.3.5.......(2n−1)n!xn
1.3.5.......(2n−1)n!2nxn

Discuss Question

Answer: Option D. -> 1.3.5.......(2n−1)n!2nxn
:
D
Middle term of

(1 + 2 x)^{2 n}

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 36. Middle term in the expansion of

(1 + 3 x + 3 x^{2} + x^{3})^{6}

4th
3rd
10th
None of these

Discuss Question

Answer: Option C. -> 10th
:
C

(1 + 3 x + 3 x^{2} + x^{3})^{6}

((1 + x)^{3})^{6}

(1 + x)^{18}

Hence the middle term is

10^{t h}

Question 37. The value of x in the expression

[x + x^{l o g_{10} (x)}]^{5}

, if the third term in the expansion is 10,00,000

10
11
12
None of these

Discuss Question

Answer: Option A. -> 10
:
A

T_{3}

^{5} C_{2}

x^{2}

(x^{l o g_{10} x})^{3}

10^{6}

Put

^{5} C_{2}

= 10 [ ∵

l o g_{10} 10 = 1

].
If x = 10, then

10^{3}

10^{2.1}

10^{5}

is satisfied.
Hence x = 10.

Question 38. If the coefficient of

x^{7}

(a x^{2} + \frac{1}{b x})^{11}

is equal to the coefficient of

x^{- 7}

(a x - \frac{1}{b x^{2}})^{11}

, then ab =

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}

^{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 39. 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}

,
thereforeq =

^{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 40. If P is a prime number, then

n^{p}

- n is divisible by p when n is a

Natural number greater than 1
Irrational number
Complex number
Odd number

Discuss Question

Answer: Option A. -> Natural number greater than 1
:
A

n^{p}

- n is divisible by p for any natural number
greater than 1. It is Fermet's theorem.
Trick:Let n = 4 and p = 2
Then

(4)^{2}

- 4 = 16 - 4 = 12, it is divisible by 2.
So,it is true for any natural number greater than 1.

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