Binomial Theorem Test

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Binomial Theorem Test - 45

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Weekly Quiz Competition

Question 1
1 / -0

The middle term in the expansion of $$\left( \dfrac { 2x^{ 2 } }{ 3 } +\dfrac { 3 }{ { 2 }x^{ 2 } } \right) ^{ 10 }$$ is

A
$$251$$

B
$$252$$

C
$$250$$

D
None

Solution

$$ (\frac{2x^{2}}{3}+\frac{3}{2x^{2}})^{10} $$
$$ \Rightarrow (a+b)^{n} = ^{n}C_{0}a^{n}+^{n}4a^{n}b+--+^{n}C_{\pi }a^{n-\pi }b^{\pi }+-^{n}C_{n}b^{n} $$
$$ \therefore $$ middle term is $$ 6^{th} $$ here, so :
$$ = 7_{6} \Rightarrow ^{10}C_{5} (\frac{2x^{2}}{3})^{10-5} (\frac{2}{2x^{2}})^{5} $$
$$ \Rightarrow \dfrac{10\times 9\times 8\times 7\times 6}{5\times 4\times 5\times 2\times 1} (\dfrac{2x^{2}}{3^{8}})^{5} (\dfrac{2}{2x^{2}})^{5} $$
$$ \Rightarrow 4\times 63\times 2 (1) $$
$$ \Rightarrow 252\times 1 $$
$$ \Rightarrow 252 $$
(Option B)
Question 2
1 / -0

The sum of rational term in the expansion of $${(3^{\frac{1}{5}}+2^{\frac{1}{3}})}^{15}$$ is

A
$$31$$

B
$$59$$

C
$$51$$

D
$$61$$

Solution

Given that ,

$${{\left( {{3}^{\frac{1}{5}}}+{{2}^{\frac{1}{3}}} \right)}^{15}}$$

General term is

$$ {{t}_{r+1}}{{=}^{15}}{{C}_{r}}{{\left( {{3}^{\frac{1}{5}}} \right)}^{15-r}}{{\left( {{2}^{\frac{1}{3}}} \right)}^{r}} $$

$$ {{=}^{15}}{{C}_{r}}{{.3}^{\frac{15-r}{5}}}{{.2}^{\frac{r}{3}}} $$

For general term r=0,5

$${{t}_{0+1}}{{=}^{15}}{{C}_{0}}{{.3}^{3}}=27$$

$${{t}_{15+1}}={{t}_{16}}{{=}^{15}}{{C}_{15}}{{3}^{0}}.$$$${{2}^{5}}$$ =32

Hence, this is the required value = $$27+32 =59$$
Question 3
1 / -0

The value of $$\displaystyle\sum^{10}_{r=0}$$ $$^{20}C_r$$ is equal to?

A
$$\dfrac{1}{2}(2^{20}+$$ $$^{20}C_{10})$$

B
$$\dfrac{1}{2}(2^{28}+$$ $$^{19}C_{10})$$

C
$$20(2^{18}+$$ $$^{19}C_{11})$$

D
$$10(2^{18}+$$ $$^{19}C_{11})$$

Solution

$$\begin{array}{l} \sum _{ r=0 }^{ 10 }{ { \, ^{ 20 } }{ C_{ r } } } =\, { \, ^{ 20 } }{ C_{ 0 } }+\, { \, ^{ 20 } }{ C_{ 1 } }........{ \, ^{ 20 } }{ C_{ r } }={ 2^{ 20 } } \\ \left( { ^{ 20 }{ C_{ 0 } }{ +^{ 20 } }{ C_{ 1 } }{ { .... }^{ 20 } }{ C_{ 10 } } } \right) +\left( { ^{ 20 }{ C_{ 11 } }{ { ...... }^{ 20 } }{ C_{ 20 } } } \right) ={ 2^{ 20 } } \\ \left( { ^{ 20 }{ C_{ 0 } }{ { ...... }^{ 20 } }{ C_{ 10 } } } \right) +\left( { ^{ 20 }{ C_{ 9 } }{ +^{ 20 } }{ C_{ 8 } }{ { ..... }^{ 20 } }{ C_{ 0 } } } \right) ={ 2^{ 20 } } \\ =2\left( { ^{ 20 }{ C_{ 0 } }{ { ........ }^{ 20 } }{ C_{ 10 } } } \right) ={ 2^{ 20 } }{ +^{ 20 } }{ C_{ 10 } } \\ { =^{ 20 } }{ C_{ 0 } }+{ ....^{ 20 } }{ C_{ 10 } }={ 2^{ 19 } }+{ \frac { 1 }{ 2 } ^{ 20 } }{ C_{ 10 } } \end{array}$$
Question 4
1 / -0

Coefficient of $$x^{2}$$ in the expansion of $$\left ( x-\frac{1}{2x} \right )$$is?

A
$$\frac{1}{7}$$

B
$$\frac{2}{7}$$

C
-7

D
7

Solution

Consider the given function:

$$ Cofficient\,\,of\,{{x}^{2}}\,in\,\,the\,\,\exp ansion\,\,of\,\,(x-\dfrac{1}{2x}) $$

$$ =let\,\,the\,\,terms\,{{\,}^{9}}{{C}_{r}}{{(x)}^{9-r}}{{(\dfrac{1}{2x})}^{r}} $$

$$ {{(x)}^{9-r}}{{({{x}^{-1}})}^{r}}={{x}^{2}} $$

$$ \,\,\,\,\,\,\,\,\,\,\,\,\,{{x}^{9-r-r}}={{x}^{2}} $$

$$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,{{x}^{9-2r}}={{x}^{2}} $$

$$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2r=7 $$

$$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,r=\dfrac{2}{7} $$

Hence this is the answer.
Question 5
1 / -0

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

A
$$\dfrac{n-5}{6}$$

B
$$\dfrac{n-4}{5}$$

C
$$\dfrac{5}{n-4}$$

D
$$\dfrac{6}{n-5}$$

Solution

Acc. to Question
$$^{m}C_{4} a^{n-4}(-b)^{4}+ ^{n}C_{5} a^{n-5}(-b)^{5}=0$$
solving we get
$$a/b =\dfrac {n-4}{5}$$
Question 6
1 / -0

If $$x^2+7ax+40$$ and $$x^2+2ax-60$$ have a common factor , then find the value of a,

A
$$\pm1$$

B
$$\pm3$$

C
$$0$$

D
$$\pm2$$

Solution

$$x^{2}+7 a x+40 \text { and } x^{2}+2 a x-60 \text { have a } \\$$
$$\text { common factor }$$

$$x^{2}+7 a x+40 \\$$
$$-x^{2}+2 a x-60 \\$$
$$----------$$
$$5 a x+100$$

$$\text { Therefore, } a x=-20 \text { , Put the value in } \\$$
$$\text { above equations } \\$$

$$x^{2}+ 7(-20)+40=0 \\$$
$$x =\pm 10$$
$$\text { Therefore, } a=\mp 2$$
Question 7
1 / -0

The number of non-zero terms in the expansion of $$(\sqrt{7}+1)^{75}-(\sqrt{7}-1)^{75}$$ is

A
$$36$$

B
$$37$$

C
$$38$$

D
$$39$$

Solution

$$(\sqrt 7+1)^{75}-(\sqrt 7-1)^{75}$$
$$\Rightarrow \sqrt 7\left(1+\dfrac{1}{\sqrt{7}}\right)^{75}-\sqrt 7\left(1-\dfrac{1}{\sqrt{7}}\right)^{75}$$
$$\Rightarrow $$ middle term = no of non zero terms=
$$=36$$
Question 8
1 / -0

The middle term in the expansion of $${ \left( x-\cfrac { 1 }{ 2x } \right) }^{ 10 }$$ is equal to

A
$$\cfrac{105}{32{x}^{2}}$$

B
$$\cfrac{63}{8}$$

C
$$\cfrac{-105}{32{x}^{2}}$$

D
$$\cfrac{-63}{8}$$

Solution

General term
$$T_{r+1}={^{10}C_r}x^{10-r}\left(-\dfrac{1}{2x}\right)^r$$
For middle term, $$r=5$$
$$T_6={^{10}C_5}x^5\left(\dfrac{-1}{2^5x^5}\right)$$
$$=-\dfrac{10\times 9\times 8\times 7\times 6}{5\times 4\times 3\times 2\times 2^5}$$
$$=-\dfrac{3\times 2\times 7\times 3\times 2}{2\times 2\times 2\times 2\times 2}$$
$$T_6=\dfrac{-63}{8}$$.
Question 9
1 / -0

The number of rational terms in the expansion of $$\quad { \left( { 3 }^{ \cfrac { 1 }{ 4 } }+{ 7 }^{ \cfrac { 1 }{ 6 } } \right) }^{ 144 }$$ is

A
$$33$$

B
$$23$$

C
$$12$$

D
$$13$$

Solution

$$(3^{1/4}+7^{1/6})^{144}$$
general term$$={^{144}C_r}3^{\dfrac{1}{4}(144-r)}7^{1/6(r)}$$
$$={^{144}C_r}3^{36-\dfrac{r}{4}}7^{\dfrac{r}{6}}$$
For rational terms
$$r=$$LCM$$(4, 6)=12$$
$$r=0, 12, 24, 36,.....144$$
AP$$\rightarrow$$
$$144=0+(n-1)12$$
$$n-1=\dfrac{144}{2}=12$$
$$n=13$$
$$\therefore$$ $$13$$ rational term.
Question 10
1 / -0

In the expansion of $$(y^{1/5}+x^{1/10})^{55}$$, the number of terms free of a radical sign is

A
$$5$$

B
$$6$$

C
$$50$$

D
$$56$$

Solution

$$\left(y^{1 / 5}+x^{1 / 10}\right)^{55}$$
$$T_{r+1}=55C_r\left(y^{1 / 5}\right)^{r}\left(x^{1 / 10}\right)^{55-r}$$
This will be independent of radicals after making powers of $$x, y$$ integers
So $$\frac{r}{5}$$ and $$\frac{r}{10}$$should be integers
$$\therefore r=0,10,20,30,40,50$$
6 terms will be frel of radicals.

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Re-Attempt Weekly Quiz Competition

Binomial Theorem Test - 45

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Binomial Theorem Test - 45

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SHARING IS CARING

Question 1

$$251$$

$$252$$

$$250$$

None

Solution

Question 2

$$31$$

$$59$$

$$51$$

$$61$$

Solution

Question 3

$$\dfrac{1}{2}(2^{20}+$$ $$^{20}C_{10})$$

$$\dfrac{1}{2}(2^{28}+$$ $$^{19}C_{10})$$

$$20(2^{18}+$$ $$^{19}C_{11})$$

$$10(2^{18}+$$ $$^{19}C_{11})$$

Solution

Question 4

$$\frac{1}{7}$$

$$\frac{2}{7}$$

-7

7

Solution

Question 5

$$\dfrac{n-5}{6}$$

$$\dfrac{n-4}{5}$$

$$\dfrac{5}{n-4}$$

$$\dfrac{6}{n-5}$$

Solution

Question 6

$$\pm1$$

$$\pm3$$

$$0$$

$$\pm2$$

Solution

Question 7

$$36$$

$$37$$

$$38$$

$$39$$

Solution

Question 8

$$\cfrac{105}{32{x}^{2}}$$

$$\cfrac{63}{8}$$

$$\cfrac{-105}{32{x}^{2}}$$

$$\cfrac{-63}{8}$$

Solution

Question 9

$$33$$

$$23$$

$$12$$

$$13$$

Solution

Question 10

$$5$$

$$6$$

$$50$$

$$56$$

Solution

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