Binomial Theorem(11th Grade > Mathematics ) Questions and answers for exam Preparation

11th Grade > Mathematics

BINOMIAL THEOREM MCQs

Total Questions : 30 | Page 3 of 3 pages

The value of $(^{21} C_{1} -^{10} C_{1}) + (^{21} C_{2} -^{10} C_{2}) + (^{21} C_{3} -^{10} C_{3}) + (^{21} C_{4} -^{10} C_{4}) + . . . + (^{21} C_{10} -^{10} C_{10})$ is

$2^{21} - 2^{11}$
$2^{21} - 2^{10}$
$2^{20} - 2^{9}$
$2^{20} - 2^{10}$

Discuss Question

Answer: Option D. ->

2^{20} - 2^{10}

:
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 22.

If the coefficients of $x^{3}$ and $x^{4}$ in the expansion of $(1 + a x + b x^{2}$ ) $(1 - 2 x)^{18}$ in powers of $x$ are both zero, then $(a, b)$ is equal to

$(16, \frac{251}{3})$
$(14, \frac{251}{3})$
$(14, \frac{272}{3})$
$(16, \frac{272}{3})$

Discuss Question

Answer: Option D. ->

(16, \frac{272}{3})

:
D

(1 + a x + b x^{2}) (1 - 2 x)^{18} = 1 (1 - 2 x)^{18} + a x (1 - 2 x)^{18} + b x^{2} (1 - 2 x)^{18}

Coefficient of

x^{3} : (- 2)^{3}^{18} C_{3} + a (- 2)^{2}^{18} C_{2} + b (- 2)^{18} C_{1} = 0

\frac{4 \times (17 \times 16)}{(3 \times 2)} - 2 a \times \frac{17}{2} + b = 0

----- (1)
Coefficient of

x^{4} : (- 2)^{4}^{18} C_{4} + a (- 2)^{3}^{18} C_{3} + b (- 2)^{2}^{18} C_{2} = 0

(4 \times 20) - 2 a \times \frac{16}{3} + b = 0

---- (2)
From equations (1) and (2), we get

4 (\frac{17 \times 8}{3} - 20) + 2 a (\frac{16}{3} - \frac{17}{2}) = 0

\Rightarrow

a = 16

\Rightarrow b = \frac{2 \times 16 \times 16}{3} - 80 = \frac{272}{3}

Question 23.

The term independent of x expansion of ${(\frac{x + 1}{x^{\frac{2}{3}} - x^{\frac{1}{3}} + 1} - \frac{x - 1}{x - x^{\frac{1}{2}}})}^{10}$ is

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 24.

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 25.

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

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

Discuss Question

Answer: Option D. ->

(\begin{matrix} n + 2 r \end{matrix})

:
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 26.

Coefficient of $t^{24} i n (1 + t^{2})^{12} (1 + t^{12}) (1 + t^{24})$ is ?

$^{12} C_{6} + 3$
$^{12} C_{6} + 1$
$^{12} C_{6}$
$^{12} C_{6} + 2$

Discuss Question

Answer: Option D. ->

^{12} C_{6} + 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 27.

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 and 0< 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

Question 28.

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 29.

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 30.

The given expression is ${(x + \sqrt{x^{3} - 1})}^{5} + {(x - \sqrt{x^{3} - 1})}^{5}$ is a

polynomial of fractional degree.
polynomial of degree 7.
polynomial with integer degree.
not a polynomial.

Discuss Question

Answer: Option B. -> polynomial of degree 7.
:
B and C
The given expression is

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

we know that

(x + a)^{n} + (x - a)^{n} = 2 [^{n} C_{0} x^{n} +^{n} C_{2} x^{n - 2} a^{2} +^{n} C^{4} x^{n - 4} a^{4} + \cdot]

Therefore the given expression is equal to

2 [^{5} C_{0} x^{5} +^{5} C_{2} x^{3} (x^{3} - 1) +^{5} C_{4} x (x^{3} - 1)^{2}] .

Maximum power of

x

involved here is 7.
Also only + ve integral powers of

x

are involved.
Therefore, the given expression is a polynomial of degree 7.

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