Binomial Theorem Test

Result

Binomial Theorem Test - 51

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Question 1
1 / -0

The number of integral terms in the expansion of $$ (\sqrt{3}+\sqrt[8]{5})^{256} $$ is

A
$$32$$

B
$$33$$

C
$$34$$

D
$$35$$

Solution

In the expansion of $$(\sqrt3 +\sqrt[8]5)^{256}$$,

General term, $$T_{r+1}=^{256}C_r (\sqrt 3)^{256-r}(\sqrt [8]{5})^r$$
$$=\ ^{256}C_r\cdot 3^{(256-r)/2}\cdot 5^{r/8}$$

The terms are integral if $$\dfrac{256-r}{2}$$ and $$\dfrac{r}{8}$$ are both positive integers.
$$\therefore r=0, 8, 16, 24,...., 256$$
Hence, there are $$33$$ integral terns.
Question 2
1 / -0

For a binomial distribution, n = 5.
If P (X = 4) = P (X = 3), then P (X > 2) is

A
0.69

B
0.97

C
0.21

D
0.79

Solution

$$n=5\\P(X=4)=^5C_4p^4q=5p^4q\\P(X=3)=^5C_3p^3q^2=10p^3q^2\\P(X=3)=P(X=4)\implies p=2q\\p+q=1\\q=\dfrac 13,p=\dfrac 23\\P(X>2)=P(X=3)+P(X=4)+P(X=5)\\10\left(\dfrac 23\right)^3\left(\dfrac 13\right)^2+5\left(\dfrac 23\right)^4\left(\dfrac 13\right)+\left(\dfrac 23\right)^5=\dfrac{192}{243}=0.79$$
Question 3
1 / -0

The coefficient of $$t^{50}$$ in $$(1+t)^{41}(1-t+t^2)^{40}$$ is equal to?

A
$$1$$

B
$$50$$

C
$$81$$

D
$$0$$

Solution
Question 4
1 / -0

The coefficient of $$x^{8}$$ in the polynomial $$\left(x-1\right)\left(x-2\right)\left(x-3\right).\left(x-10\right)$$ is:

A
$$2640$$

B
$$1320$$

C
$$1270$$

D
$$2740$$

Solution
Question 5
1 / -0

Find the middle term in the expansion of $$(1-2x+x^2)^n$$

A
$$\dfrac {(2n)!}{(n!)^2}(-1)^n x^n$$

B
$$\dfrac {(2n)!}{(n!)}(-1)^n x^2n$$

C
$$\dfrac {(2n)!}{(n!)^2} x^n$$

D
None of these

Solution

Given expression $$(1-2x+x^2)^n$$

$$\Rightarrow \ (1-2x+x^2)^n =[(1-x)^2]^n =(1-x)^{2n}$$

Middle term $$\left (\dfrac {2n}{2}+1\right)^{th} \Rightarrow (n+1)^{th} $$ term

$$T_{n+1}=^{2n}C_1 (1)^{2n-n} (-x)^n$$

$$=\dfrac {2n!}{(n!)^2}(1)^n (-x)^n$$

Middle term is $$\dfrac {(2n!)}{(n!)^2} (-1)^n x^n$$
Question 6
1 / -0

If $$\left(1+x+x^ {2}+x^ {3}\right)^ {5}=a_{0}+a_{1}x+a_{2}x^ {2}+....+a_{15}x^ {15}$$, then $$a_{10}$$ equals

A
$$99$$

B
$$100$$

C
$$101$$

D
$$110$$

Solution
Question 7
1 / -0

The largest coefficient in the expansion of $${ \left( 1+x \right) }^{ 38 }$$ is

A
$$_{ }^{ 38 }{ { C }_{ 18 } }$$

B
$$_{ }^{ 38 }{ { C }_{ 15 } }$$

C
$$_{ }^{ 38 }{ { C }_{ 20 } }$$

D
$$_{ }^{ 38 }{ { C }_{ 19 } }$$

Solution
Question 8
1 / -0

Given that the term of the expansion $$\displaystyle (x^{1/3}+ x^{-1/2})^{15}$$ which does not contain $$x$$ is $$5$$ m , where m$$\in$$N , m $$=$$

A
1100

B
1010

C
1001

D
None of these

Solution

Given,

$$(x^{\dfrac{1}{3}}+x^{-\dfrac{1}{2}})^{15}$$

$$T_{r+1}=^{15}C_r(x^{\dfrac{1}{3}})^{15-r}(x^{-\dfrac{1}{2}})^r$$

Power of $$x=\dfrac{15-r}{3}-\dfrac{r}{2}=0$$ for $$x^0$$

$$2(15-r)-3r=0$$

$$30-2r-3r=0$$

$$30-5r=0$$

$$r=6$$

$$T_{r+1}=^{15}C_{6}=5005$$

$$5m=5005\Rightarrow m=1001$$
Question 9
1 / -0

If the $$6^{th}$$ term in the expansion of $$\displaystyle \left [ \dfrac{1}{x^{8/3}} + x^2log_{10}x \right ]^8$$ is 5600 , then $$x$$ =

A
10

B
8

C
11

D
9

Solution

From given, we have,

6th term $$=^8C_5 x^{-\frac{8}{3} \times 3}(x^2 \log x)^5$$

$$\Rightarrow \dfrac{ 8 \times 7 \times 6}{3 \times 2} \times x^{-8} \times x^{10}(\log x)^5 =5600$$

$$x^2(\log_{10}x)^5 =100$$

$$\therefore x=10$$
Question 10
1 / -0

If the second term of the expansion $${ [{ a }^{ { 1 }/{ 13 } }+\frac { a }{ \sqrt { { a }^{ -1 } } } ] }^{ n }$$ is $$14{ a }^{ { 5 }/{ 2 } }$$, then the value of $$\frac { ^{ n }{ C }_{ 3 } }{^{ n } { C }_{ 2 } } $$ is .

A
4

B
3

C
12

D
6

Solution