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Q. The remainder left out when $8^{2 n}-(62)^{2 n+1}$ is divided by 9 is :-
(1) 7 (2) 8 (3) 0 (4) 2
[AIEEE 2009]
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Sol. (4) $8^{2 \mathrm{n}}-(62)^{2 \mathrm{n}+1}$ $=(63+1)^{\mathrm{n}}-(63-1)^{2 \mathrm{n}+1}$ $=\left(63 \mathrm{I}_{1}+1\right)-\left(63 \mathrm{I}_{2}-1\right)$ $=63 \mathrm{I}_{3}+2=9 \mathrm{I}+2$
Q. Let $S_{1}=\sum_{j=1}^{10} j(j-1)^{10} C_{j}, S_{2}=\sum_{j=1}^{10} j^{10} C_{j}$ and $S_{3}=\sum_{j=1}^{10} j^{20} C_{j}$
Statement-1: $S_{3}=55 \times 2^{9}$.
Statement-2 : $S_{1}=90 \times 2^{8}$ and $S_{2}=10 \times 2^{8}$.
(1) Statement–1 is true, Statement–2 is true ; Statement–2 is a correct explanation for Statement–1.
(2) Statement–1 is true, Statement–2 is true ; Statement–2 is not a correct explanation for Statement–1.
`(3) Statement–1 is true, Statement–2 is false.
(4) Statement–1 is false, Statement–2 is true.
[AIEEE-2010]
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Sol. (3)
so, statement-2 is wrong.
Q. The coefficient of $x^{7}$ in the expansion of $\left(1-x-x^{2}+x^{3}\right)^{6}$ is :-
(1) – 144 (2) 132 (3) 144 (4) – 132
[AIEEE 2011]
Q. If $\mathrm{n}$ is a positive integer, then $(\sqrt{3}+1)^{2 \mathrm{n}}-(\sqrt{3}-1)^{2 \mathrm{n}}$ is :
(1) a rational number other than positive integers
(2) an irrational number
(3) an odd positive integer
(4) an even positive integer
[AIEEE 2012]
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Sol. (2) $(\sqrt{3}+1)^{2 n}-(\sqrt{3}-1)^{2 n}$ $=2\left[\mathrm{T}_{2}+\mathrm{T}_{2}+\mathrm{T}_{6}+\ldots \ldots+\mathrm{T}_{2 \mathrm{n}}\right]$ $=2\left[2 \mathrm{n} \mathrm{C}_{1}(\sqrt{3})^{2 n-1}+2 \mathrm{n} \mathrm{C}_{3}(\sqrt{3})^{2 \mathrm{n}-3}+\ldots \ldots \ldots\right]$ $=$ An Irrational Number
Q. The term independent of $x$ in expansion of $\left(\frac{x+1}{x^{2 / 3}-x^{1 / 3}+1}-\frac{x-1}{x-x^{1 / 2}}\right)^{10}$ is :
(1) 4 (2) 120 (3) 210 (4) 310
[JEE-Main 2013]
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Sol. (3) $\left(\frac{x+1}{x^{2 / 3}-x^{1 / 3}+1}-\frac{x-1}{x-\sqrt{x}}\right)^{10}$ $\left(\mathrm{x}^{1 / 3}+1-\left(\frac{\sqrt{\mathrm{x}}+1}{\sqrt{\mathrm{x}}}\right)\right)^{10}$ $\left(\mathrm{x}^{1 / 3}-\mathrm{x}^{-1 / 2}\right)^{10}$ $\mathrm{T}_{\mathrm{r}+1}=^{10} \mathrm{C}_{\mathrm{r}}\left(\mathrm{x}^{1 / 3}\right)^{10-\mathrm{r}}\left(\mathrm{x}^{-1 / 2}\right)^{\mathrm{r}}$ $\frac{10-r}{3}-\frac{r}{2}=0$ 20 – 2r = 3r r = 4 $\mathrm{T}_{5}=\mathrm{T}_{4+1}=^{10} \mathrm{C}_{4}=\frac{10 !}{6 ! .4 !}=210$
Q. If the coefficients of $x^{3}$ and $x^{4}$ in the expansion of $\left(1+a x+b x^{2}\right)(1-2 x)^{18}$ in powers of $x$ are both zero, then $(a, b)$ is equal to :-
( 1)$\left(16, \frac{251}{3}\right)$
(2) $\left(14, \frac{251}{3}\right)$
(3) $\left(14, \frac{272}{3}\right)$
( 4)$\left(16, \frac{272}{3}\right)$
[JEE(Main)-2014]
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Sol. (4) In the expansion of $\left(1+a x+b x^{2}\right)(1-2 x)^{18}$ General term $=\left(1+a x+b x^{2}\right) \cdot 18 C_{r}(-2 x)^{r}$ Cofficinet of $x^{3}=18 C_{3}(-2)^{3}+a \cdot\left(8 C_{2}(-2)^{2}+b \cdot 18 C_{1}(-2)=0\right.$ Cofficinet of $\mathrm{x}^{4}=18 \mathrm{C}_{4}(-2)^{4}+\mathrm{a} \cdot^{18} \mathrm{C}_{3}(-2)^{3}+\mathrm{b} \cdot 18 \mathrm{C}_{2}(-2)^{2}=0$ on solving the equations we get $153 \mathrm{a}-9 \mathrm{b}=1632 \quad \ldots$ (i) $3 \mathrm{b}-32 \mathrm{a}=-240 \quad \ldots$ (ii) on solving we get $a=16 \& b=\frac{272}{3}$
Q. The sum of coefficients of integral powers of $x$ in the binomial expansion of $(1-2 \sqrt{x})^{50}$ is :
(1) $\frac{1}{2}\left(3^{50}-1\right)$
(2) $\frac{1}{2}\left(2^{50}+1\right)$
(3) $\frac{1}{2}\left(3^{50}+1\right)$
( 4)$\frac{1}{2}\left(3^{50}\right)$
[JEE(Main)-2015]
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Sol. (3) $\because(1-2 \sqrt{\mathrm{x}})^{50}=50 \mathrm{C}_{0}-^{50} \mathrm{C}_{1}(2 \sqrt{\mathrm{x}})+50 \mathrm{C}_{2}(2 \sqrt{\mathrm{x}})^{2}+\ldots \ldots+50 \mathrm{C}_{50}(2 \sqrt{\mathrm{x}})^{50}$ $\therefore$ consider $(1+2 \sqrt{\mathrm{x}})^{50}=50 \mathrm{C}_{0}+^{50} \mathrm{C}_{1}(2 \sqrt{\mathrm{x}})+\ldots \ldots+50 \mathrm{C}_{50}(2 \sqrt{\mathrm{x}})^{50}$ add both equations and put $\mathrm{x}=1$ $\Rightarrow \mathrm{sum}$ of coefficients of integral powers of $\mathrm{x}=\frac{1}{2}\left(1+3^{50}\right)$
Q. If the number of terms in the expansion of
$\left(1 \frac{2}{x}+\frac{4}{x^{2}}\right)^{n}, x \neq 0,$ is $28,$ then the sum of the
coefficients of all the terms in this expansion, is :-
(1) 729 (2) 64 (3) 2187 (4) 243
[JEE(Main)-2016]
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Sol. (1 or bonus) Number of terms in the expansion of $\left(1-\frac{2}{x}+\frac{4}{x^{2}}\right)^{n}$ is $^{n}+2 C_{2}$ (considering $\frac{1}{x}$ and $\frac{1}{x^{2}}$ distinct) $\therefore n+2 C_{2}=28 \Rightarrow n=6$ $\therefore$ Sum of coefficients $=(1-2+4)^{6}=729$ But number of dissimilar terms actually will be $2 n+1$ (as $\frac{1}{x}$ and $\frac{1}{x^{2}}$ are functions as same variable) Hence it contains error, so a bonus can be expected.
Q. The value of
$\left(21 \mathrm{C}_{1}-10 \mathrm{C}_{1}\right)+\left(21 \mathrm{C}_{2}-^{10} \mathrm{C}_{2}\right)+$
$\left(^{21} \mathrm{C}_{3}-^{10} \mathrm{C}_{3}\right)+\left(21 \mathrm{C}_{4}-^{10} \mathrm{C}_{4}\right)+\ldots .+$
$\left(^{21} \mathrm{C}_{10}-^{10} \mathrm{C}_{10}\right)$ is
(1) $2^{20}-2^{10}$
(2) $2^{21}-2^{11}$
(3) $2^{21}-2^{10}$
(4) $2^{20}-2^{9}$
[JEE(Main)-2017]
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Sol. (1) $\left(^{21} \mathrm{C}_{1}+^{21} \mathrm{C}_{2} \ldots \ldots \ldots+210\right.$ $-\left(^{10} \mathrm{C}_{1}+^{10} \mathrm{C}_{2} \ldots \ldots \ldots .^{10} \mathrm{C}_{10}\right)$ $=\frac{1}{2}\left[\left(^{(21} \mathrm{C}_{1}+\ldots .+^{21} \mathrm{C}_{10}\right)+\left(^{21} \mathrm{C}_{11}+\ldots .2^{21} \mathrm{C}_{20}\right)\right]-\left(2^{10}-1\right)$ $=\frac{1}{2}\left[2^{21}-2\right]-\left(2^{10}-1\right)$ $=\left(2^{20}-1\right)-\left(2^{10}-1\right)=2^{20}-2^{10}$
Q. The sum of the co-efficients of all odd degree terms in the expansion of $(x+\sqrt{x^{3}-1})^{5}+$ $(x-\sqrt{x^{3}-1})^{5},(x>1)$ is-
(1) 0 (2) 1 (3) 2 (4)
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Sol. (3) using $(x+a)^{5}+(x-a)^{5}$ $=2\left[^{5} C_{0} x^{5}+^{5} C_{2} x^{3} \cdot a^{2}+^{5} C_{4} x \cdot a^{4}\right]$ $(x+\sqrt{x^{3}-1})^{5}+(x-\sqrt{x^{3}-1})^{5}$ $=2\left[^{5} C_{0} x^{5}+^{5} C_{2} x^{3}\left(x^{3}-1\right)+^{5} C_{4} x\left(x^{3}-1\right)^{2}\right]$ $\Rightarrow 2\left[x^{5}+10 x^{6}-10 x^{3}+5 x^{7}-10 x^{4}+5 x\right]$ considering odd degree terms, $\quad 2\left[x^{5}+5 x^{7}-10 x^{3}+5 x\right]$ $\therefore$ Sum of coefficients of odd terms is 2
Good questions but too short
.
questions are awesome but we need more questions
thanks
NYC questions
Please add more questions in binomial therom
Ya
Hello
Super
The answer of the question which came in jee 2016 should 243
no 729 is correct answer
Number of terms =(n+1)(n+2)2=28
⇒n=6
∴a0+a1x+a2x2+….+a2nx2n=(1−2x+4×2)n
Put x=1,n=6,a0+a1+a2+…+a2n=36=729
Pls add some more questions
Hello
It’s really very helpful in last 20 day prepration. In few questions whole concept of a chapter is covered and pattern is also clear about exam.
It’s really beneficial for me.
Very good question but printing is not perfect
Good
EXCELLENT QUESTIONS BRO BUT PLZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ ADD SOME MORE QUESTIONS
Please add some more questions
Pls add some more questions
Very helpful!
please upload more question.
Very helpful and useful regular questions
Very Useful and helpful
Easy but few are lengthy
Mujhse toh ek bhi nahi ho rha ha yarrr
Chil bro
good
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