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

12th Grade > Mathematics

BINOMIAL THEOREM MCQs

Binomial Theorem

Total Questions : 74 | Page 7 of 8 pages

Question 61. The value of

(^{21} C_{1} -^{10} C_{1}) + (^{21} C_{2} -^{10} C_{2}) + (^{21} C_{3} -^{10} C_{3}) + (^{21} C_{4} -^{10} C_{4}) + . . . + (^{21} C_{10} -^{10} C_{10})

221−211
221−210
220−29
220−210

Discuss Question

Answer: Option D. -> 220−210
:
D

(^{21} C_{1} -^{10} C_{1}) + (^{21} C_{2} -^{10} C_{2}) + (^{21} C_{3} -^{10} C_{3}) + . . . + (^{21} C_{10} -^{10} C_{10})

= (^{21} C_{1} +^{21} C_{2} + \dots +^{21} C_{10}) - (^{10} C_{1} +^{10} C_{2} + \dots +^{10} C_{10}) = (^{21} C_{1} +^{21} C_{2} + \dots +^{21} C_{10}) - (2^{10} - 1) = \frac{1}{2} (^{21} C_{1} +^{21} C_{2} + \dots^{21} C_{21} - 1) - (2^{10} - 1) = \frac{1}{2} (2^{21} - 2) - (2^{10} - 1) = 2^{20} - 1 - 2^{10} + 1 = 2^{20} - 2^{10}

Question 62. The coefficient of

x^{7}

in the expansion of

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

is?

132
144
-132
-144

Discuss Question

Answer: Option D. -> -144
:
D

(1 - x - x^{2} + x^{3})^{6} = (1 - x)^{6} (1 - x^{2})^{6} = (1 -^{6} C_{1} x +^{6} C_{2} x^{2} -^{6} C_{3} x^{3} +^{6} C_{4} x^{4} -^{6} C_{5} x^{5} +^{6} C_{6} x^{6}) (1 -^{6} C_{1} x^{2} +^{6} C_{2} x^{4} -^{6} C_{3} x^{6} + \dots)

\Rightarrow

coefficient of

x^{7}

^{6} C_{1} .^{6} C_{3} -^{6} C_{3} .^{6} C_{2} +^{6} C_{5} .^{6} C_{1} = 6 \times 20 - 20 \times 15 + 6 \times 6 = - 144

Question 63. Coefficient of

t^{24} i n (1 + t^{2})^{12} (1 + t^{12}) (1 + t^{24})

is ?

12C6+3
12C6+1
12C6
12C6+2

Discuss Question

Answer: Option D. -> 12C6+2
:
D
Here, Coefficient of

t^{24} i n {(1 + t^{2})^{12} (1 + t^{12}) (1 + t^{24})}

Coefficient of

t^{24} i n {(1 + t^{2})^{12} . (1 + t^{12} + t^{24} + t^{36})}

Coefficient of

t^{24} i n {(1 + t^{2})^{12} + t^{12} (1 + t^{2})^{12} + t^{24} (1 + t^{2})^{12}}

[Neglecting

t^{36} (1 + t^{2})^{12}]

Coefficient of

t^{24} = (^{12} C_{12} +^{12} C_{6} +^{12} C_{0}) = 2 +^{12} C_{6}

Question 64. The number of integral terms in the expansion of

(\sqrt{3} +^{8} \sqrt{5})^{256}

is ?

Discuss Question

Answer: Option B. -> 33
:
B

(\sqrt{3} +^{8} \sqrt{5})^{256} ∴ T_{r + 1} =^{256} C_{r} (\sqrt{3})^{256 - r} (^{8} \sqrt{5})^{r} =^{256} C_{r} (3)^{\frac{256 - r}{2}} (5)^{\frac{r}{8}}

Terms would be integeral if

\frac{(256 - r)}{2}

And

\frac{r}{8}

are possitive integers.
As

0 \leq r \leq 256

r = 0, 8, 16, 24 \dots 256

For the above values of r,

\frac{(256 - r)}{2}

is also an integer. Hence, the total number fo values of r is 33.

Question 65. The term independent of x expansion of

{(\frac{x + 1}{x^{\frac{2}{3}} - x^{\frac{1}{3}} + 1} - \frac{x - 1}{x - x^{\frac{1}{2}}})}^{10}

Discuss Question

Answer: Option C. -> 210
:
C

{[\frac{x + 1}{x^{\frac{2}{3}} - x^{\frac{1}{3}} + 1} - \frac{x - 1}{x - x^{\frac{1}{2}}}]}^{10}

= {⎡ ⎢⎢⎢ ⎣ \frac{{(x^{\frac{1}{3}})}^{3} + 1^{3}}{x^{\frac{2}{3}} - x^{\frac{1}{3}} + 1} - \frac{{(\sqrt{x})^{2} - 1}}{\sqrt{x} (\sqrt{x} - 1)} ⎤ ⎥⎥⎥ ⎦}^{10}

= {⎡ ⎢⎢⎢ ⎣ \frac{(x^{\frac{1}{3}} + 1) (x^{\frac{2}{3}} + 1 - x^{\frac{1}{3}})}{x^{\frac{2}{3}} - x^{\frac{1}{3}} + 1} - \frac{{(\sqrt{x})^{2} - 1}}{\sqrt{x} (\sqrt{x} - 1)} ⎤ ⎥⎥⎥ ⎦}^{10}

{[(x^{\frac{1}{3}} + 1) - \frac{(\sqrt{x} + 1)}{\sqrt{x}}]}^{10} = {(x^{\frac{1}{3}} - x^{- \frac{1}{2}})}^{10}

∴

The general term is

T_{r + 1} =^{10} C_{r} {(x^{\frac{1}{3}})}^{10 - r} {(- x^{- \frac{1}{2}})}^{r} =^{10} C_{r} (- 1)^{r} x^{\frac{10 - r}{3} - \frac{r}{2}}

For independent of x, put

\frac{10 - r}{3} - \frac{r}{2} = 0

\Rightarrow

20 - 2r - 3r = 0

\Rightarrow

20 = 5r

\Rightarrow

r = 4

∴ T_{5} =^{10} C_{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210

Question 66. For

2 \leq r \leq n, (\begin{matrix} n r \end{matrix}) + 2 (\begin{matrix} n r - 1 \end{matrix}) + (\begin{matrix} n r - 2 \end{matrix})

is equal to

(n+1r−1)
(n+1r+1)
2(n+2r)
(n+2r)

Discuss Question

Answer: Option D. -> (n+2r)
:
D

(\begin{matrix} n r \end{matrix}) + 2 (\begin{matrix} n r - 1 \end{matrix}) + (\begin{matrix} n r - 2 \end{matrix}) = [(\begin{matrix} n r \end{matrix}) + (\begin{matrix} n r - 1 \end{matrix})]

[(\begin{matrix} n r - 1 \end{matrix}) + (\begin{matrix} n r - 2 \end{matrix})] = (\begin{matrix} n + 1 r \end{matrix}) + (\begin{matrix} n + 1 r - 1 \end{matrix}) = (\begin{matrix} n + 2 r \end{matrix})

[∵^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}]

Question 67. The coefficients of three consecutive terms of

(1 + x)^{n + 5}

are in the ratio 5 : 10 : 14. Then, n is equal to

Discuss Question

Answer: Option D. -> 6
:
D
Let the three consecutive terms in

(1 + x)^{n + 5}

t_{r}, t_{r + 1}, t_{r} + 2

having coefficients

^{n + 5} C_{r - 1},^{n + 5} C_{r},^{n + 5} C_{r + 1}

Given,

^{n + 5} C_{r - 1},^{n + 5} C_{r},^{n + 5} C_{r + 1} = 5 : 10 : 14

∴ \frac{^{n + 5} C_{r}}{^{n + 5} C_{r - 1}} = \frac{10}{5} a n d \frac{^{n + 5} C_{r + 1}}{^{n + 5} C_{r}} = \frac{14}{10} \Rightarrow \frac{n + 5 - (r - 1)}{r} = 2 a n d \frac{n - r + 5}{r + 1} = \frac{7}{5}

\Rightarrow

n – r + 6 = 2r and 5n – 5r + 25 = 7r + 7

\Rightarrow

n + 6 = 3r and 5n + 18 = 12r

∴ \frac{n + 6}{3} = \frac{5 n + 18}{12}

\Rightarrow

4n + 24 = 5n + 18

\Rightarrow

n = 6

Question 68. 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
23 and 9
32 and 6

Discuss Question

Answer: Option C. -> 23 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 69.

6^{t h}

term in expansion of

(2 x^{2} - \frac{1}{3 x^{2}})^{10}

458017
- 89627
558017
None of these

Discuss Question

Answer: Option B. -> - 89627
:
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 70.

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}

← Previous
1
2
3
4
5
6
7
8
Next →

Latest Videos

Chapter 1 - GLOBAL STEEL SCENARIO & INDIAN STEEL INDUSTRY Part 1 : (13-04-2024) INDUSTRY AND COMPANY AWARENESS (ICA)

Direction Sense Test Part 1 Reasoning (Hindi)

Chapter 1 - RMHP / OHP / OB & BP Part 1 : (14-02-2024) GPOE

Cube & Cuboid Part 1 Reasoning (Hindi)

Data Interpretation (DI) Basic Concept Reasoning (Hindi)

Counting Figures Part 1 Counting Of Straight Lines Reasoning (Hindi)

Sail E0 free webinar Webinars

Real Numbers Part 7 Class 10 Maths

Real Numbers Part 1 Class 10 Maths

Polynomials Part 1 Class 10 Maths

Latest Test Papers

Pie Chart - Test Paper 1 Data Interpretation Reasoning & DI

Tabulation - Test Paper 2 Data Interpretation Reasoning & DI

Test Paper 1 Direction Sense Test Reasoning & DI

ICA , RDIC,GFM Free Modal Test Paper - Non - Technical Branch Old Syllabus 2022 - SAIL Junior Officer E0

GFM Combined 1 Free Revision Test SAIL JO E-0