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
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Good questions but too short
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questions are awesome but we need more questions
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NYC questions
Please add more questions in binomial therom
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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
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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
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Please add some more questions
Pls add some more questions
Very helpful!
please upload more question.
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Easy but few are lengthy
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Chil bro
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