Binomial Theorem Test

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Binomial Theorem Test - 16

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Weekly Quiz Competition

Question 1
1 / -0

The number of terms in the expansion of $$\left [ (a+4b)^{3}(a-4b)^{3} \right ]^{2}$$ are

A
$$6$$

B
$$7$$

C
$$8$$

D
$$32$$

Solution

$$[(a+4b)\:^3(a-4b)\:^3]\:^2$$
$$=[(a+4b)(a-4b)]\:^6$$
$$=[a^2-16b^2]\:^6$$
Hence total number of terms is $$n+1$$
Here $$n=6$$
Therefore, total number of terms is $$7.$$
Question 2
1 / -0

The number of terms in the expansion of

$$(a_{1}+b_{1})(a_{2}+b_{2}).....(a_{n}+b_{n})$$

A
$$2^{\mathrm{n}}$$

B
$$3^{\mathrm{n}}$$

C
$$3^{2\mathrm{n}}$$

D
$$2^{2n}$$

Solution

Each bracket in the above expansion contains 2 elements
Therefore
$$2$$ brackets will have $$2^2=4 $$ elements
$$3$$ brackets will have $$2^3=8$$ elements
$$4$$ brackets will have $$2^4=16 $$ elements
:
:
$$n$$ brackets will have $$2^n$$ elements
Hence, there will be $$2^n$$ elements
Question 3
1 / -0

(i) The no.of distinct terms in the expansion of $$({x}_{1}+{x}_{2}+\ldots.+{x}_{{n}})^{3}$$ is $$n+2{c}_{3}$$

(ii) The no. of irrational terms in the expansion $$(2^{1/5}+3^{1/10})^{55}$$ is $$55$$

A
(i) is true (ii) is false

B
both (i) and (ii) are true

C
both (i) and (ii) are false

D
(i) is false (ii) is true

Solution

Basic questions. $$^{n+r-1}C_r$$ is $$1^{st}$$ one and $$2^{nd}$$ is $$\dfrac {55}{l}l, (5,10) + 1$$ which is $$6$$, hence is wrong.
Question 4
1 / -0

The sum of the coefficients of the middle terms of $$(1+{x})^{2{n}-1}$$ is

A
$$^{2n-1}C_{n}$$

B
$$^{2n-1}C_{n+1}$$

C
$$^{2n}C_{n-1}$$

D
$$^{2n}C_{n}$$

Solution

Consider: $$(1+x)^{2n-1}$$
Since $$2n-1$$ is odd, the middle terms are $$\left(\dfrac{2n-1+1}{2}\right)^{th}$$ and $$\left(\dfrac{2n-1+1}{2}+1\right)^{th}$$ terms
Now, consider the following
$$T_{r+1}=\:^nC_{r}a^{n-r}b^r$$...(i)
Where $$T$$ represents the term
Here the middle terms are the $$n^{th}$$ term and $$(n+1)^{th}$$ term.
$$T_{n+1}=\ ^{2n-1}C_{n}1^{n-1}x^{n}$$ $$=\:^{2n-1}C_{n}x^{n}$$
Hence, the coefficient is $$\:^{2n-1}C_{n}$$ ...(1)

$$T_{n}=\ ^{2n-1}C_{n-1}1^{n}x^{n-1}$$$$=\:^{2n-1}C_{n-1}x^{n-1}$$
Hence, $$\:^{2n-1}C_{n-1}$$ is the coefficient ...(2)

Therefore, sum of the coefficients of the middle terms is
$$\ ^{2n-1}C_{n}+\ ^{2n-1}C_{n-1}$$
$$=\ ^{2n}C_{n}$$

Hence, option D.
Question 5
1 / -0

If $$x + y = 1$$, then $$\displaystyle \sum_{r=0}^{n}r^{n}C_{r}x^{r}.y^{n-r}=$$

A
$$1$$

B
$$n$$

C
$$nx$$

D
$$ny$$

Solution

$$\displaystyle \sum_{r=0} ^{n}r\:^{n}C_{r}x^ry^{n-r}$$
$$=\displaystyle \sum_{r=0} ^{n}r\frac{n!}{(n-r)!r!}x^ry^{n-r}$$
$$=nx(y+x)^{n-1}$$
$$=nx(1)^{n-1}$$ ... given in the above question
$$=nx$$
Hence, the answer is Option C
Question 6
1 / -0

In the expansion of $$\left ( 2.2^{\dfrac{2x}{3}}+2^{\dfrac{-x}{3}} \right )^{n}$$ the sum of the last four binomial coefficients exceeds the sum of the first three binomial coefficients by 20 and if the middle term is-the numerically largest term, then x belongs to

A
$$(-2,\infty )$$

B
$$\left ( 2, log _{2} \dfrac{1}{3}\right )$$

C
$$\left ( -2, log _{2} \dfrac{1}{3}\right )$$

D
$$\left ( -2, 2log _{2} \dfrac{2}{3}\right )$$

Solution

Applying the above given condition we get
$$(\:^nC_{n}+\:^nC_{n-1}+\:^nC_{n-2}+\:^nC_{n-3})-(\:^nC_{0}+\:^nC_{1}+\:^nC_{2})=20$$
By using properties of binomial coefficients the above expression is reduced to
$$\:^nC_{n-3}=20$$
$$\:^nC_{3}=20$$
$$\:^nC_{3}=\:^6C_{3}$$
Hence $$n=6$$.
It is given that the middle term is the largest term.
Hence $$T_{3+1}$$
$$=T_{4}=$$largest term.
Therefore $$\dfrac{T_{4}}{T_{3}}>1$$ which implies
$$\dfrac{2}{3}^{2}>2^{x}$$
$$2log2-2log3>xlog2$$
$$2-2log_{2}3>x$$ ...(i)

Now $$\dfrac{T_{4}}{T_{5}}>1$$
Which is simplified as $$2^{x}>\dfrac{1}{4}$$
$$xlog2>-2log2$$
$$x>-2$$...(ii)
From i and ii we get
$$x\epsilon(-2,2-2log_{2}3)$$
Hence answer is Option D
Question 7
1 / -0

If $$x + y = 1$$ then $$\displaystyle \sum_{r=0}^{n}r^{2}$$ $$^{n}C_{r}x^{r}y^{n-r}$$

A
$$nxy$$

B
$$nx(x + yn)$$

C
$$nx(nx + y)$$

D
$$nx$$

Solution

$$\displaystyle \sum_{r=0}^{n}r^{2}$$ $$^{n}C_{r}x^{r}y^{n-r}$$
$$\displaystyle=\sum_{r=0} ^{n}[r(r-1)+r]$$ $$^{n}C_{r}x^ry^{n-r}$$
$$\displaystyle =\sum_{r=0} ^{n}(r(r-1))\cdot\ ^{n}C_{r}x^ry^{n-r}+\sum_{r=0}^{n}r\cdot\ ^{n}C_{r}x^ry^{n-r}$$
$$\displaystyle\sum_{r=0} ^{n}(r(r-1))\cdot\ ^{n}C_{r}x^ry^{n-r} = x^2 \dfrac{d^2}{dx^2}(\sum_{r=0}^{n}$$ $$^{n}C_{r}x^ry^{n-r}) = x^2\dfrac{d^2}{dx^2}(x+y)^n = n(n - 1)x^2$$
$$\displaystyle\sum_{r=0}^{n}r\cdot\ ^{n}C_{r}x^ry^{n-r} = x \dfrac{d}{dx}(\sum_{r=0}^{n}$$ $$^{n}C_{r}x^ry^{n-r}) = x\dfrac{d}{dx}(x+y)^n = nx$$
$$\therefore\displaystyle \sum_{r=0} ^{n}(r(r-1))\cdot\ ^{n}C_{r}x^ry^{n-r}+\sum_{r=0}^{n}r\cdot\ ^{n}C_{r}x^ry^{n-r}$$
$$=n(n-1)x^2+nx$$
$$=nx[x(n-1)+1]$$
$$=nx[nx-x+(x+y)]$$ ...it is given in the question that $$x+y=1$$
$$=nx[nx+y]$$
Hence, the answer is Option C
Question 8
1 / -0

In the binomial expansion of $$(a - b)^{n}, n>5,$$ the sum of 5$$^{th}$$ and 6$$^{th}$$ terms is zero, then $$\displaystyle\frac{a}{b}$$ equals-

A
$$\displaystyle\frac{5}{n-4}$$

B
$$\displaystyle\frac{6}{n-5}$$

C
$$\displaystyle\frac{n-5}{6}$$

D
$$\displaystyle\frac{n-4}{5}$$

Solution

Since, in binomial expansion of $${ \left( a-b \right) }^{ n },n\ge 5$$,
the sum of fifth and sixth term is equal to zero
$$\therefore^{ n }{ { C }_{ 4 } }{ a }^{ n-4 }{ b }^{ 4 }+^{ n }{ { C }_{ 5 } }{ a }^{ n-5 }{ b }^{ 5 }=0$$
$$\displaystyle\Rightarrow\frac { n! }{ \left( n-4 \right) !4! } { a }^{ n-4 }{ b }^{ 4 }-\frac { n! }{ \left( n-5 \right) !5! } { a }^{ n-5 }{ b }^{ 5 }=0$$
$$\displaystyle\Rightarrow\frac { a }{ b } =\frac { n-4 }{ 5 } $$
Question 9
1 / -0

If the fourth term in the expansion of

$$\left(\sqrt{x^\dfrac{1}{\log x+1}}+x^\dfrac{1}{12}\right)^{6}$$ is equal to $$200$$ and $$x>1$$, then $$x$$ is

A
$$10$$

B
$$10^{-4}$$

C
$$1$$

D
$$-4$$

Solution

$$T_{r+1}=\:^nC_{r}a^{n-r}b^r$$
Applying to the above question, we get
$$T_{3+1}=\:^6C_{3}x^{\dfrac {3}{2(\log x+1)}}x^{\dfrac {3}{12}}$$
$$=200$$
$$20x^{\dfrac {3}{2(\log x+1)}}x^{\dfrac {3}{12}}=200$$
$$x^{\dfrac {3}{2(\log x+1)}+\dfrac {1}{4}}=10$$
Taking $$\log$$ to the base $$10$$ ($$\log_{10}$$) on both sides, we get
$$\dfrac{3}{2(\log x+1)}+\dfrac{1}{4}=1$$
$$\dfrac{3}{2(\log x+1)}=\dfrac{3}{4}$$
$$4=2(\log_{10}x+1)$$
$$2=\log_{10}x+1$$
$$1=\log_{10}x$$
Taking anti-logarithm, we get
$$x=10$$
Question 10
1 / -0

If in the expansion of $$(\displaystyle \frac{1}{x}+x\tan x)^{5}$$, the ratio of $$4^{th}$$ term to the $$2^{nd}$$ term is $$\displaystyle \frac{2}{27}\pi^{4}$$, then the value of $${x}$$ can be

A
$$\displaystyle \frac{-\pi}{6}$$

B
$$\displaystyle \frac{-\pi}{3}$$

C
$$\displaystyle \frac{\pi}{3}$$

D
$$\displaystyle \frac{\pi}{12}$$

Solution

From the above given expression by applying binomial theorem, we get.
$$T_{4}=\:^5C_{3}x\tan^3{x}$$
$$=10x\tan^3{x}$$
$$T_{2}=\:^5C_{1}x^{-3}\tan{x}$$
$$=5x^{-3}\tan{x}$$
Therefore $$\displaystyle \frac{T_{4}}{T_{2}}$$
$$=\displaystyle \frac{10x\tan^3{x}}{5x^{-3}\tan{x}}$$
$$=2x^4\tan^2x$$
$$=\displaystyle \frac{2\pi^{4}}{27}$$
$$x^4\tan^2x=\displaystyle \frac{\pi^{4}}{27}$$
Taking root on both sides, we get
$$x^2\tan x=\displaystyle \frac{\pi^2}{3\sqrt3}$$
$$=\displaystyle \frac{\sqrt{3}\pi^2}{9}$$
Therefore $$x^2=\displaystyle \frac{\pi^2}{9}$$ ...(i) and $$\tan x=\sqrt{3}$$...(ii)
Both Eq(i) and Eq (ii) give $$x=\displaystyle \frac{\pi}{3}$$

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Binomial Theorem Test - 16

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Binomial Theorem Test - 16

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Question 1

$$6$$

$$7$$

$$8$$

$$32$$

Solution

Question 2

$$2^{\mathrm{n}}$$

$$3^{\mathrm{n}}$$

$$3^{2\mathrm{n}}$$

$$2^{2n}$$

Solution

Question 3

(i) is true (ii) is false

both (i) and (ii) are true

both (i) and (ii) are false

(i) is false (ii) is true

Solution

Question 4

$$^{2n-1}C_{n}$$

$$^{2n-1}C_{n+1}$$

$$^{2n}C_{n-1}$$

$$^{2n}C_{n}$$

Solution

Question 5

$$1$$

$$n$$

$$nx$$

$$ny$$

Solution

Question 6

$$(-2,\infty )$$

$$\left ( 2, log _{2} \dfrac{1}{3}\right )$$

$$\left ( -2, log _{2} \dfrac{1}{3}\right )$$

$$\left ( -2, 2log _{2} \dfrac{2}{3}\right )$$

Solution

Question 7

$$nxy$$

$$nx(x + yn)$$

$$nx(nx + y)$$

$$nx$$

Solution

Question 8

$$\displaystyle\frac{5}{n-4}$$

$$\displaystyle\frac{6}{n-5}$$

$$\displaystyle\frac{n-5}{6}$$

$$\displaystyle\frac{n-4}{5}$$

Solution

Question 9

$$10$$

$$10^{-4}$$

$$1$$

$$-4$$

Solution

Question 10

$$\displaystyle \frac{-\pi}{6}$$

$$\displaystyle \frac{-\pi}{3}$$

$$\displaystyle \frac{\pi}{3}$$

$$\displaystyle \frac{\pi}{12}$$

Solution

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