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

12th Grade > Mathematics

BINOMIAL THEOREM MCQs

Binomial Theorem

Total Questions : 74 | Page 3 of 8 pages

Question 21. 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 22. If n is positive integer and three consecutive coefficients in the expansion of

(1 + x)^{n}

are in the ratio 6 : 33 : 110, then n =

Discuss Question

Answer: Option C. -> 12
:
C
Let the consecutive coefficient of

(1 + x)^{n}

are

^{n} C_{r - 1}

^{n} C_{r}

^{n} C_{r + 1}

By condition,

^{n} C_{r - 1}

^{n} C_{r}

^{n} C_{r + 1}

= 6 : 33 : 110
Now

^{n} C_{r - 1}

^{n} C_{r}

= 6 : 33

\Rightarrow

2n - 13r + 2 = 0 ..............(i)
and

^{n} C_{r}

^{n} C_{r + 1}

= 33: 110

\Rightarrow

3n - 13r - 10 = 0 ...............(ii)
Solving (i) and (ii), we get n = 12 and r = 2.
Aliter:We take first n = 4 [By alternate (a)] but
(a) does not hold. Similarity (b).
So alternate (c), n = 12 gives

(1 + x)^{12}

= [1 + 12 x + \frac{12.11}{2.1}

x^{2}

\frac{12.11.10}{3.2.1} x^{3} + . . . . . . . . .]

So coefficient of II, III and IV terms are

12, 6 \times 11, 2 \times 11 \times 10.

So,

12 : 6 \times 11 : 2 \times 11 \times 10 = 6 : 33 : 110.

Question 23. 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 24. If n is positive integer and three consecutive coefficients in the expansion of

(1 + x)^{n}

are in the ratio 6 : 33 : 110, then n =

Discuss Question

Answer: Option C. -> 12
:
C
Let the consecutive coefficient of

(1 + x)^{n}

are

^{n} C_{r - 1}

^{n} C_{r}

^{n} C_{r + 1}

By condition,

^{n} C_{r - 1}

^{n} C_{r}

^{n} C_{r + 1}

= 6 : 33 : 110
Now

^{n} C_{r - 1}

^{n} C_{r}

= 6 : 33

\Rightarrow

2n - 13r + 2 = 0 ..............(i)
and

^{n} C_{r}

^{n} C_{r + 1}

= 33: 110

\Rightarrow

(1 + x)^{12}

= [1 + 12 x + \frac{12.11}{2.1}

x^{2}

\frac{12.11.10}{3.2.1} x^{3} + . . . . . . . . .]

So coefficient of II, III and IV terms are

12, 6 \times 11, 2 \times 11 \times 10.

So,

12 : 6 \times 11 : 2 \times 11 \times 10 = 6 : 33 : 110.

Question 25. 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 26. If number of terms in the expansion of

(x - 2 y + 3 z)^{n}

are 45, then n =

7
8
9
None of these

Discuss Question

Answer: Option B. -> 8
:
B

\frac{(n + 1) (n + 2)}{2}

= 45 or

n^{2}

+ 3n - 88 = 0

\Rightarrow

n = 8 .

Question 27. 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 28. 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 29.

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 30. In the expansion of

(5^{\frac{1}{2}} + 7^{\frac{1}{8}})^{1024}

, the number of integral terms is

Discuss Question

Answer: Option B. -> 129
:
B
Here n = 1024 =

2^{10}

, a power of 2, where as the
power of 7 is

\frac{1}{8}

2^{- 3}

Now first term

^{1024} C_{0}

(

5^{\frac{1}{2}})^{1024}

5^{512}

(integer)
And after 8 terms, the

9^{t h}

term =

^{1024} C_{8}

(

5^{\frac{1}{2}})^{1016}

(

7^{\frac{1}{8}})^{8}

= an integer
Again,

17^{t h}

term =

^{1024} C_{16}

(

5^{\frac{1}{2}})^{1008}

(

7^{\frac{1}{8}})^{16}

= An integer.
Continuing like this, we get an A.P. 1, 9, 17,..........., 1025,
because

1025^{t h}

term = the last term in the expansion
=

^{1024} C_{1024}

(

7^{\frac{1}{8}})^{1024}

7^{128}

(an integer)
If n is the number of terms of above A.P. we have
1025 =

T_{n}

= 1 + (n - 1)8 ⇒ n = 129.

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