2020
Q301
TS-EAMCET
MCQ
iCON Education HYD, 79930 92826, 73309 72826
20 May 2026
Let $x \in \mathbf{R}$ be so small that the powers of $x$ beyond two are insignificant and negligibly small. For such $x$, if $(1-x)^3(2+x)^6$ is approximated by $a+b x+c x^2$, then $a+b+c=$
Show Answer
Practice Quiz
Correct Answer: A
Explanation:
We have,
$(1-x)^3=1-3 x+3 x^2$ [neglecting
higher term]
$ (2+x)^6={ }^6 C_0 2^6 x^0+{ }^6 C_1 2^5 x^1+{ }^6 C_2 2^4 x^2 $
[neglecting higher term]
$ =64+192 x+240 x^2 $
Now, $(1-x)^3(2+x)^6=\left(1-3 x+3 x^2\right)$
$ \begin{array}{r} \left(64+192 x+240 x^2\right) \\ =64+192 x+240 x^2-192 x \\ -576 x^2+192 x^2 \end{array} $
(neglecting higher term)
$ \begin{aligned} & =-144 x^2+64 \\ \therefore \quad & a=64, b=0, c=-144 \\ \therefore & a+b+c=64+0-144=-80 \end{aligned} $
2020
Q302
TS-EAMCET
MCQ
iCON Education HYD, 79930 92826, 73309 72826
20 May 2026
For $0 < x < 1$, the expansion of $\left(1+\frac{1}{x}\right)^{\frac{1}{2}}$ is
A.
$1+\frac{1}{2 x}-\frac{1}{2!}\left(\frac{1}{2 x}\right)^2+\frac{1 \cdot 3}{3!}\left(\frac{1}{2 x}\right)^3-\frac{1 \cdot 3 \cdot 5}{4!}\left(\frac{1}{2 x}\right)^4+\ldots \infty$
B.
$\frac{1}{\sqrt{x}}+\frac{1}{2} \sqrt{x}-\frac{1}{2!} \frac{x \sqrt{x}}{2^2}+\frac{1 \cdot 3}{3!} \frac{x^2 \sqrt{x}}{2^3}-\ldots . \infty$
C.
$1+\frac{1}{\sqrt{x}}+\frac{1}{2} x \sqrt{x}+\frac{1}{2!} \frac{x^2 \sqrt{x}}{2^3}+\frac{1 \cdot 3}{3!} \frac{x^3 \sqrt{x}}{2^4}+\ldots . \infty$
D.
$\frac{1}{\sqrt{x}}+\frac{1}{2 x \sqrt{x}}-\frac{1}{2!}\left(\frac{1}{2 x}\right)^2 \frac{1}{\sqrt{x}}+\frac{1 \cdot 3}{3!}\left(\frac{1}{2 x}\right)^3 \frac{1}{\sqrt{x}}-\ldots \ldots \infty$
Show Answer
Practice Quiz
Correct Answer: A
Explanation:
$ \begin{aligned} &\\ &\begin{aligned} & \text { Given, } \\ & \left(1+\frac{1}{x}\right)^{1 / 2} \\ & =1+\frac{1}{2}\left(\frac{1}{x}\right)+\frac{\left(\frac{1}{2}\right)\left(\frac{1}{2}-1\right)}{2!}\left(\frac{1}{x}\right)^2 \\ & +\frac{\left(\frac{1}{2}\right)\left(\frac{1}{2}-1\right)\left(\frac{1}{2}-2\right)}{3!}\left(\frac{1}{x}\right)^3 \\ & +\frac{\left(\frac{1}{2}\right)\left(\frac{1}{2}-1\right)\left(\frac{1}{2}-2\right)\left(\frac{1}{2}-3\right)}{4!}\left(\frac{1}{x}\right)^4+\ldots \\ & =1+\frac{1}{2 x}-\frac{1}{2!}\left(\frac{1}{2 x}\right)^2+\frac{1 \cdot 3}{3!}\left(\frac{1}{2 x}\right)^3 \\ & -\frac{1 \cdot 3 \cdot 5}{4!}\left(\frac{1}{2 x}\right)^4+\ldots . \infty \end{aligned} \end{aligned} $
2020
Q303
BITSAT
MCQ
iCON Education HYD, 79930 92826, 73309 72826
11 Jun 2026
The value of ${}^{47}{C_4} + \sum\limits_{r = 1}^5 {{}^{52 - r}{C_3}} $ is equal to
Show Answer
Practice Quiz
Correct Answer: C
Explanation:
${}^{47}{C_4} + \sum\limits_{r = 1}^5 {{}^{52 - r}{C_3} = {}^{47}{C_4} + {}^{51}{C_3} + {}^{50}{C_3} + {}^{49}{C_3} + {}^{48}{C_3} + {}^{47}{C_3}} $
$ = {}^{51}{C_3} + {}^{50}{C_3} + {}^{49}{C_3} + {}^{48}{C_3} + ({}^{47}{C_3} + {}^{47}{C_4}) = {}^{52}{C_4}$
2020
Q304
BITSAT
MCQ
iCON Education HYD, 79930 92826, 73309 72826
11 Jun 2026
The coefficient of x8 in the polynomial (x $-$ 1) (x $-$ 2) ..... (x $-$ 10)
Show Answer
Practice Quiz
Correct Answer: B
Explanation:
We have,
$(x - 1)(x - 2)(x - 3)....(x - 10)$
Coefficient of x8 = Sum of the terms taken two at a time i.e.,
$\sum\limits_{1 \le i < j \le 10}^{10} {{x_i}\,.\,{x_j} = {1 \over 2}\left[ {{{\left( {\sum\limits_{i = 1}^{10} {{x_i}} } \right)}^2} - \left( {\sum\limits_{i = 1}^{10} {{x_i}^2} } \right)} \right]} $
$ = {1 \over 2}[{(1 + 2 + 3 + ... + 10)^2} - {({1^2} + {2^2} + {3^2} + ... + 10)^2}]$
$ = {1 \over 2}\left[ {{{\left( {{{10 \times 11} \over 2}} \right)}^2} - \left( {{{10 \times 11 \times 21} \over 6}} \right)} \right]$
$ = {1 \over 2}({(55)^2} - 385) = {1 \over 2}(3025 - 385) = {{2640} \over 2} = 1320$
2019
Q305
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
If 20 C1 + (22 ) 20 C2 + (32 ) 20 C3 + ..... + (202
)
20 C20 = A(2$\beta $ ), then the ordered pair (A, $\beta $) is equal to :
Show Answer
Practice Quiz
Correct Answer: B
Explanation:
S = ${1^2}\,{}^{20}{C_1} + {2^2}\,{}^{20}{C_2} + {3^2}\,{}^{20}{C_3} + .......... + {20^2}\,{}^{20}{C_{20}}$
$ \Rightarrow $ $\sum\limits_{r = 1}^{20} {{r^2}\,{}^{20}{C_r}} $
$ \Rightarrow $ $\sum\limits_{r = 1}^{20} {r\,.\left( {r.{}^{20}{C_r}} \right)} $
$ \Rightarrow $ $20\sum\limits_{r = 1}^{20} {r\,.} {}^{19}{C_{r - 1}}$
$ \Rightarrow $ $20\sum\limits_{r = 1}^{20} {(r - 1 + 1)\,.} {}^{19}{C_{r - 1}}$
$ \Rightarrow $ $20\sum\limits_{r = 1}^{20} {(r - 1)\,.} {}^{19}{C_{r - 1}} + 20\sum\limits_{r = 1}^{20} {{}^{19}{C_{r - 1}}} $
$ \Rightarrow $ $20 \times 19\sum\limits_{r = 2}^{20} {{}^{18}{C_{r - 2}}} + 20 \times {2^{19}}$
$ \Rightarrow $ 20 $ \times $ 19 $ \times $ 218 + 20 $ \times $ 219
$ \Rightarrow $ $20 \times {2^{18}}\left( {19 + 2} \right) = 20 \times 21 \times {2^{18}}$
$ \Rightarrow $ $420 \times {2^{18}}$
2019
Q306
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The term independent of x in the expansion of
$\left( {{1 \over {60}} - {{{x^8}} \over {81}}} \right).{\left( {2{x^2} - {3 \over {{x^2}}}} \right)^6}$ is equal to :
Show Answer
Practice Quiz
Correct Answer: C
Explanation:
Given expression = $\left( {{1 \over {60}} - {{{x^8}} \over {81}}} \right).{\left( {2{x^2} - {3 \over {{x^2}}}} \right)^6}$
= ${1 \over {60}}{\left( {2{x^3} - {3 \over {{x^2}}}} \right)^6} - {{{x^8}} \over {81}}{\left( {2{x^2} - {3 \over {{x^2}}}} \right)^6}$
So its general term is
Tr + 1 = ${1 \over {60}}{}^6{C_r}{\left( {2{x^2}} \right)^{6 - r}}{\left( { - {3 \over {{x^2}}}} \right)^r} - {{{x^8}} \over {81}}{}^6{C_r}{\left( {2{x^2}} \right)^{6 - r}}{\left( { - {3 \over {{x^2}}}} \right)^r}$
= ${1 \over {60}}{}^6{C_r}{\left( 2 \right)^{6 - r}}{\left( { - 3} \right)^r}{x^{12 - 4r}} - {1 \over {81}}{}^6{C_r}{\left( 2 \right)^{6 - r}}{\left( { - 3} \right)^r}{x^{20 - 4r}}$ .....(i)
For this term to be independent of x, put r = 3 in
1st part and r = 5 in 2nd part.
So from (i) the term independent of x = ${1 \over {60}} \times {2^3} \times {\left( { - 3} \right)^3} \times {}^6{C_3} + \left( { - {1 \over {81}}} \right)(2){( - 3)^5} \times {}^6{C_5}$
= -72 + 36 = -36
2019
Q307
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The coefficient of x18 in the product
(1 + x) (1 – x)10 (1 + x + x2 )9
is :
Show Answer
Practice Quiz
Correct Answer: D
Explanation:
Coefficient of x18 in (1 + x) (1 - x)10 (1 + x + x2 )9
$ \Rightarrow $ Coefficient of x18 in {(1 - x) (1 - x2 ) (1 + x + x2 )}9
$ \Rightarrow $ Coefficient of x18 in (1 - x2 ) (1 - x3 )9
$ \Rightarrow $ 9 C6 - 0 = 84
2019
Q308
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The smallest natural number n, such that the coefficient of x in the expansion of ${\left( {{x^2} + {1 \over {{x^3}}}} \right)^n}$ is n C23 , is :
Show Answer
Practice Quiz
Correct Answer: C
Explanation:
General term
${T_{r + 1}} = {}^n{C_r}{x^{2n - 2r}}.{x^{ - 3r}}$
$ \therefore $ $2n - 5r = 1 \Rightarrow 2n = 5r + 1$
$ \therefore $ $r = {{2n - 1} \over 5}$
$ \Rightarrow $ Coefficient of x = ${}^n{C_{\left( {{{2n - 1} \over 5}} \right)}} = {}^n{C_{23}}$
$ \Rightarrow $ ${{2n - 1} \over 5} = 23\,\,or\,\,n - \left( {{{2n - 1} \over 5}} \right) = 23$
$ \Rightarrow $ 2n - 1 = 115 $ \Rightarrow $ n = 58
and n = 38
$ \therefore $ smallest n = 38
2019
Q309
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
If the coefficients of x2
and x3
are both zero, in the expansion of the expression (1 + ax + bx2
) (1 – 3x)15 in
powers of x, then the ordered pair (a,b) is equal to :
Show Answer
Practice Quiz
Correct Answer: B
Explanation:
(1 + ax + bx2 )(1 – 3x)15
Co-eff. of x2 = 1.15 C2 (–3)2 + a.15 C1 (–3) + b.15 C0
$ = {{15 \times 14} \over 2} \times 9 - 15 \times 3a + b = 0$ (given)
$ \Rightarrow $ 945 – 45a + b = 0 ...(i)
Now co-eff. of x3 = 0
$ \Rightarrow $ 15 C3 (–3)3 + a.15 C2 (–3)2 + b.15 C1 (–3) = 0
$ \Rightarrow {{15 \times 14 \times 13} \over {3 \times 2}} \times ( - 3 \times 3 \times 3) + a \times {{15 \times 14 \times 9} \over 2 } – b × 3 × 15 = 0$
$ \Rightarrow $ 15 × 3[–3 × 7 × 13 + a × 7 × 3 – b] = 0
$ \Rightarrow $ 21a – b = 273 ...(ii)
From (i) and (ii)
a = +28, b = 315 $ \equiv $ (a, b) $ \equiv $ (28, 315)
2019
Q310
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
If some three consecutive in the binomial
expansion of (x + 1)n is powers of x are in the ratio
2 : 15 : 70, then the average of these three
coefficient is :-
Show Answer
Practice Quiz
Correct Answer: D
Explanation:
Given ${}^n{C_{r - 1}}:{}^n{C_r}:{}^n{C_{r + 1}} = 2:15:70$
${{{}^n{C_{r - 1}}} \over {{}^n{C_r}}} = {2 \over {15}}$
$ \Rightarrow {r \over {n - r + 1}} = {2 \over {15}}$
$ \Rightarrow 15r = 2n - 2r + 2$
$ \Rightarrow 17r = 2n + 2$ .... (i)
Now ${{{}^n{C_r}} \over {{}^n{C_{r + 1}}}} = {{15} \over {70}}$
$ \Rightarrow {{r + 1} \over {n - r}} = {3 \over {14}}$
$ \Rightarrow $ 14r + 14 = 3n – 3r
$ \Rightarrow $ 17r = 3n – 14 ... (ii)
Now From (i) and (ii) equation
2n + 2 = 3n - 14 $ \Rightarrow $ n = 16
By putting n = 16 in equation (i)
$ \Rightarrow $ r = 2
$ \therefore $ Average of coefficient = ${{{}^{16}{C_1} + {}^{16}{C_2} + {}^{16}{C_3}} \over 3} $
= ${{16 + 120 + 560} \over 3}$
= ${{696} \over 3} = 232$
2019
Q311
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
If the fourth term in the binomial expansion of ${\left( {{2 \over x} + {x^{{{\log }_8}x}}} \right)^6}$
(x > 0) is 20 × 87 , then a value of
x is :
Show Answer
Practice Quiz
Correct Answer: B
Explanation:
${\left( {{2 \over x} + {x^{{{\log }_8}x}}} \right)^6}$
Given T4 = 20 × 87
$ \Rightarrow $ ${}^6{C_3}{\left( {{2 \over x}} \right)^3}{\left( {{x^{{{\log }_8}x}}} \right)^3}$ = 20 × 87
$ \Rightarrow $ 20$ \times $${8 \over {{x^3}}} \times {x^{3{{\log }_8}x}}$ = 20 × 87
$ \Rightarrow $ ${x^{3{{\log }_8}x - 3}}$ = 86
Taking ${{{\log }_8}}$ both side
$ \Rightarrow $ (${3{{\log }_8}x - 3}$) $ \times $ ${{{\log }_8}x}$ = 6
$ \Rightarrow $ $3{\left( {{{\log }_8}x} \right)^2}$ - 3${{{\log }_8}x}$ = 6
$ \Rightarrow $ ${\left( {{{\log }_8}x} \right)^2}$ - ${{{\log }_8}x}$ = 2
$ \Rightarrow $ (${{{\log }_8}x}$ - 2)(${{{\log }_8}x}$ + 1) = 0
$ \Rightarrow $ ${{{\log }_8}x}$ = 2 or ${{{\log }_8}x}$ = -1
$ \Rightarrow $ x = 82 or x = ${1 \over 8}$
2019
Q312
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
If the fourth term in the binomial expansion of
${\left( {\sqrt {{x^{\left( {{1 \over {1 + {{\log }_{10}}x}}} \right)}}} + {x^{{1 \over {12}}}}} \right)^6}$ is equal to 200, and x > 1,
then the value of x is :
Show Answer
Practice Quiz
Correct Answer: C
Explanation:
Fourth term (T4 )
= ${}^6{C_3}{\left( {\sqrt {{x^{\left( {{1 \over {1 + {{\log }_{10}}x}}} \right)}}} } \right)^3}{\left( {{x^{{1 \over {12}}}}} \right)^3}$
= $20{\left( {{x^{\left( {{1 \over {1 + {{\log }_{10}}x}}} \right)}}} \right)^{{3 \over 2}}}\left( {{x^{{1 \over 4}}}} \right)$
= $20 \times {x^{\left( {{1 \over {1 + {{\log }_{10}}x}}} \right){3 \over 2}}} \times {x^{{1 \over 4}}}$
= $20 \times {x^{\left( {{3 \over {2\left( {1 + {{\log }_{10}}x} \right)}} + {1 \over 4}} \right)}}$
Given, T4 = 200
$ \therefore $ $20 \times {x^{\left( {{3 \over {2\left( {1 + {{\log }_{10}}x} \right)}} + {1 \over 4}} \right)}}$ = 200
$ \Rightarrow $ ${x^{\left( {{3 \over {2\left( {1 + {{\log }_{10}}x} \right)}} + {1 \over 4}} \right)}}$ = 10
Taking log10 on both sides
$\left( {{3 \over {2\left( {1 + {{\log }_{10}}x} \right)}} + {1 \over 4}} \right){\log _{10}}x$ = 1
put log10 x = t
$\left( {{3 \over {2\left( {1 + t} \right)}} + {1 \over 4}} \right)t$ = 1
$ \Rightarrow $ $\left( {{{\left( {1 + t} \right) + 6} \over {4\left( {1 + t} \right)}}} \right) \times t$ = 1
$ \Rightarrow $ t2 + 7t = 4 + 4t
$ \Rightarrow $ t2 + 3t - 4 = 0
$ \Rightarrow $ (t + 4)(t - 1) = 0
$ \Rightarrow $ t = 1 or t = - 4
$ \therefore $ log10 x = 1
$ \Rightarrow $ x = 10
or log10 x = - 4
$ \Rightarrow $ x = 10-4
But as x > 1 so x $ \ne $ 10-4
$ \therefore $ x = 10
2019
Q313
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The sum of the co-efficients of all even
degree terms in x in the expansion of
${\left( {x + \sqrt {{x^3} - 1} } \right)^6}$ + ${\left( {x - \sqrt {{x^3} - 1} } \right)^6}$, (x > 1) is equal to:
Show Answer
Practice Quiz
Correct Answer: D
Explanation:
Let ${\left( {a + x} \right)^n}$ = Odd trems(A) + Even terms(B)
So ${\left( {a - x} \right)^n}$ = Odd terms(A) - Even terms(B)
$\therefore$ ${\left( {a + x} \right)^n} - {\left( {a - x} \right)^n}$
= (A + B) + (A - B)
= 2A
= 2[odd terms]
= 2[ T1 + T3 + T5 + ....... ]
So in case of
${\left( {x + \sqrt {{x^3} - 1} } \right)^6}$ + ${\left( {x - \sqrt {{x^3} - 1} } \right)^6}$
= 2[ T1 + T3 + T5 + T5 ]
= 2[ ${}^6{C_0}{x^6} + {}^6{C_2}{x^4}\left( {{x^3} - 1} \right)$
$ + {}^6{C_4}{x^2}{\left( {{x^3} - 1} \right)^2} + {}^6{C_6}{\left( {{x^3} - 1} \right)^3}$]
= 2[ ${x^6} + 15{x^4}\left( {{x^3} - 1} \right) + 15{x^2}{\left( {{x^3} - 1} \right)^2} + {\left( {{x^3} - 1} \right)^3}$ ]
= 2[ ${x^6} + 15\left( {{x^7} - {x^4}} \right)$
$ + 15{x^2}\left( {{x^6} - 2{x^3} + 1} \right) + \left( {{x^9} - 3{x^6} + 3{x^3} - 1} \right)$ ]
$ \therefore $ Sum of coefficient of all even degree terms
= 2[ 1 - 15 + 15 + 15 - 3 - 1 ] = 24
2019
Q314
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The sum of the series
2.20 C0
+ 5.20 C1 + 8.20 C2 + 11.20 C3 + ... +62.20 C20 is equal to :
Show Answer
Practice Quiz
Correct Answer: A
Explanation:
Here general term = (3r + 2)20 Cr
$ \therefore $ Sum of the series = $\sum\limits_{r = 0}^{20} {\left( {3r + 2} \right)} {}^{20}{C_r}$
= $3\sum\limits_{r = 0}^{20} {r.} {}^{20}{C_r} + 2\sum\limits_{r = 0}^{20} {{}^{20}{C_r}} $
= 3 $ \times $ 20$ \times $220 - 1 + 2$ \times $220
= 60$ \times $219 + 221
= 221 [15 + 1]
= 221 $ \times $ 16
= 225
2019
Q315
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The total number of irrational terms in the binomial expansion of (71/5 – 31/10 )60 is :
Show Answer
Practice Quiz
Correct Answer: A
Explanation:
General term Tr+1 = 60 Cr , ${7^{{{60 - r} \over 5}}}{3^{{r \over {10}}}}$
$ \therefore $ for rational term, r = 0, 10, 20, 30, 40, 50, 60
$ \Rightarrow $ no of rational terms = 7
$ \therefore $ number of irrational terms = 54
2019
Q316
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
A ratio of the 5th term from the beginning to the 5th term from the end in the binomial expansion of ${\left( {{2^{1/3}} + {1 \over {2{{\left( 3 \right)}^{1/3}}}}} \right)^{10}}$ is :
Show Answer
Practice Quiz
Correct Answer: D
Explanation:
${{{T_5}} \over {T_5^1}} = {{{}^{10}{C_4}{{\left( {{2^{1/3}}} \right)}^{10 - 4}}{{\left( {{1 \over {2{{\left( 3 \right)}^{1/3}}}}} \right)}^4}} \over {{}^{10}{C_4}{{\left( {{1 \over {2\left( {{3^{1/3}}} \right)}}} \right)}^{10 - 4}}{{\left( {{2^{1/3}}} \right)}^4}}} = 4.{\left( {36} \right)^{1/3}}$
2019
Q317
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
Let (x + 10)50 + (x $-$ 10)50 = a0 + a1 x + a2 x2 + . . . . + a50 x50 , for all x $ \in $ R; then ${{{a_2}} \over {{a_0}}}$ is equal to
Show Answer
Practice Quiz
Correct Answer: A
Explanation:
(10 + x)50 + (10 $-$ x)50
$ \Rightarrow $ a2 = 2.50 C2 1048 , a0 = 2.1050
${{{a_2}} \over {{a_0}}} = {{^{50}{C_2}} \over {{{10}^2}}} = 12.25$
2019
Q318
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
Let Sn = 1 + q + q2 + . . . . . + qn and Tn = 1 + $\left( {{{q + 1} \over 2}} \right) + {\left( {{{q + 1} \over 2}} \right)^2}$ + . . . . . .+ ${\left( {{{q + 1} \over 2}} \right)^n}$ where q is a real number and q $ \ne $ 1. If 101 C1 + 101 C2 . S1 + .... + 101 C101 . S100 = $\alpha $T100 then $\alpha $ is equal to
Show Answer
Practice Quiz
Correct Answer: C
Explanation:
101 C1 + 101 C2 S1 + . . . . . . . + 101 C101 S100
$=$ $\alpha $T100
101 C1 + 101 C2 (1 + q) + 101 C3 (1 + q + q2 ) +
. . . . . .+101 C101 (1 + q + . . . . . + q100 )
$ = 2\alpha {{\left( {1 - {{\left( {{{1 + q} \over 2}} \right)}^{101}}} \right)} \over {\left( {1 - q} \right)}}$
$ \Rightarrow $ 101 C1 (1 $-$ q) + 101 C2 (1 $-$ q2 ) +
. . . . . . + 101 C101 (1 $-$ q101 )
$ = 2\alpha \left( {1 - {{\left( {{{1 + q} \over 2}} \right)}^{101}}} \right)$
$ \Rightarrow $ (2101 $-$ 1) $-$ ((1 + q)101 $-$ 1)
$ = 2\alpha \left( {1 - {{\left( {{{1 + q} \over 2}} \right)}^{101}}} \right)$
$ \Rightarrow $ ${2^{101}}\left( {1 - {{\left( {{{1 + q} \over 2}} \right)}^{101}}} \right) = 2\alpha \left( {1 - {{\left( {{{1 + q} \over 2}} \right)}^{101}}} \right)$
$ \Rightarrow $ $\alpha = {2^{100}}$
2019
Q319
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The sum of the real values of x for which the middle term in the binomial expansion of ${\left( {{{{x^3}} \over 3} + {3 \over x}} \right)^8}$ equals 5670 is :
Show Answer
Practice Quiz
Correct Answer: A
Explanation:
${T_5} = {}^8{C_4}{{{x^{12}}} \over {81}} \times {{81} \over {{x^4}}} = 5670$
$ \Rightarrow 70{x^8} = 5670$
$ \Rightarrow x = \pm \sqrt 3 $
2019
Q320
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The value of r for which 20 Cr 20 C0 + 20 Cr$-$1 20 C1 + 20 Cr$-$2 20 C2 + . . . . .+ 20 C0 20 Cr is maximum, is
Show Answer
Practice Quiz
Correct Answer: A
Explanation:
Given sum = coefficient of xr in the expansion of
(1 + x)20 (1 + x)20 ,
Which is equal to 40 Cr
It is maximum when r = 20
2019
Q321
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The positive value of $\lambda $ for which the co-efficient of x2
in the expression x2 ${\left( {\sqrt x + {\lambda \over {{x^2}}}} \right)^{10}}$ is 720, is -
Show Answer
Practice Quiz
Correct Answer: A
Explanation:
The general term in the expansion of the binomial expression $(a+b)^n$ is
$
T_{r+1}={ }^n C_r a^{n-r} b^r
$
Therefore, the general term in the expansion of the binomial expression $x^2\left(\sqrt{x}+\frac{\lambda}{x^2}\right)^{10}$ is
$
\begin{aligned}
T_{r+1} & =x^2\left({ }^{10} C_r(\sqrt{x})^{10-r}\left(\frac{\lambda}{x^2}\right)^r\right) \\\\
& ={ }^{10} C_r x^2 \cdot x^{\frac{10-r}{2}} \lambda^r x^{-2 r} \\\\
& ={ }^{10} C_r \lambda^r x^{2+\frac{10-r}{2}-2 r}
\end{aligned}
$
Now, for the coefficient of $x^2$,
$
\begin{aligned}
2+\frac{10-r}{2}-2 r =2 \\\\
\Rightarrow \frac{10-r}{2}-2 r =0 \\\\
\Rightarrow 10-r =4 r \Rightarrow r=2
\end{aligned}
$
So, the coefficient of $x^2$ is ${ }^{10} C_2 \lambda^2=720$
$
\begin{aligned}
\Rightarrow & \frac{10 !}{2 ! 8 !} \lambda^2 =720 \\\\
\Rightarrow & \frac{10 \cdot 9 \cdot 8 !}{2 \cdot 8 !} \lambda^2 =720 \\\\
\Rightarrow & 45 \lambda^2 =720 \\\\
\Rightarrow & \lambda^2 =16 \\\\
\Rightarrow & \lambda = \pm 4
\end{aligned}
$
Given the problem asks for the positive value of $\lambda$, $\lambda = 4$.
So, the correct option is (A) 4.
2019
Q322
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
If ${\sum\limits_{i = 1}^{20} {\left( {{{{}^{20}{C_{i - 1}}} \over {{}^{20}{C_i} + {}^{20}{C_{i - 1}}}}} \right)} ^3} = {k \over {21}}$ then k is equal to
Show Answer
Practice Quiz
Correct Answer: A
Explanation:
${\sum\limits_{i = 1}^{20} {\left( {{{^{20}{C_{i - 1}}} \over {^{20}{C_i}{ + ^{20}}{C_{i - 1}}}}} \right)} ^3} = {k \over {21}}$
$ \Rightarrow \,\,\sum\limits_{i = 1}^{20} {{{\left( {{{{}^{20}{C_{i - 1}}} \over {{}^{21}{C_i}}}} \right)}^3}} = {k \over {21}}$
$ \Rightarrow \,\,\sum\limits_{i = 1}^{20} {{{\left( {{i \over {21}}} \right)}^3}} = {k \over {21}}$
$ \Rightarrow \,\,{1 \over {{{\left( {21} \right)}^3}}}{\left[ {{{20\left( {21} \right)} \over 2}} \right]^2} = {k \over {21}}$
$ \Rightarrow 100 = k$
2019
Q323
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
If the third term in the binomial expansion of ${\left( {1 + {x^{{{\log }_2}x}}} \right)^5}$ equals 2560, then a possible value of x is -
Show Answer
Practice Quiz
Correct Answer: D
Explanation:
${\left( {1 + {x^{{{\log }_2}x}}} \right)^5}$
${T_3} = {}^5{C_2}.{\left( {{x^{{{\log }_2}x}}} \right)^2} = 2560$
$ \Rightarrow \,\,10.{x^{2{{\log }_2}x}} = 2560$
$ \Rightarrow \,\,{x^{2\log 2x}} = 256$
$ \Rightarrow \,\,2{({\log _2}x)^2} = {\log _2}256$
$ \Rightarrow 2{({\log _2}x)^2} = 8$
$ \Rightarrow \,\,{({\log _2}x)^2} = 4$
$ \Rightarrow \,\,{\log _2}x = 2$ or $-$ 2
$x = 4$ or ${1 \over 4}$
2019
Q324
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The coefficient of t4 in the expansion of ${\left( {{{1 - {t^6}} \over {1 - t}}} \right)^3}$ is :
Show Answer
Practice Quiz
Correct Answer: B
Explanation:
${\left( {{{1 - {t^6}} \over {1 - t}}} \right)^3}$
= (1 $-$ t6 )3 (1 $-$ t)$-$3
= (1 $-$ 3 C1 t6 + 3 C2 t12 $-$ 3 C3 t18 ) $ \times $ (1 $-$ t)$-$3
coefficient of t4 is 1 $ \times $ coefficient of t4 in (1 $-$ t)$-$3
= 1 $ \times $ 3+4$-$1 C4 (By multinomial theorem)
= 6 C4 = 15
2019
Q325
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
If the fractional part of the number $\left\{ {{{{2^{403}}} \over {15}}} \right\} is \, {k \over {15}}$, then k is equal to :
Show Answer
Practice Quiz
Correct Answer: A
Explanation:
${{{2^{403}}} \over {15}}$
$ = {{{2^3}\, \cdot \,{2^{400}}} \over {15}}$
$ = {8 \over {15}}{\left( {16} \right)^{100}}$
$ = {8 \over {15}}{\left( {15 + 1} \right)^{100}}$
$ = {8 \over {15}}\left( {{}^{100}{C_0} + {}^{100}{C_1}\,15 + {}^{100}{C_2}{{\left( {15} \right)}^2} + .....{{\left( {15} \right)}^{100}}} \right)$
$ = {8 \over {15}} + 8\left( {{}^{100}{C_1} + {}^{100}{C_2}\,\left( {15} \right) + ..... + {{\left( {15} \right)}^{99}}} \right)$
$ = {8 \over {15}} + 8$ (integer)
$ \therefore $ Fractional part $ = {8 \over {15}}$
According to the question,
${k \over {15}} = {8 \over {15}}$
$ \Rightarrow $ K $=$ 8
2018
Q326
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The coefficient of x2 in the expansion of the product
(2$-$x2 ) .((1 + 2x + 3x2 )6 + (1 $-$ 4x2 )6 ) is :
Show Answer
Practice Quiz
Correct Answer: B
Explanation:
Given,
(2 $-$ x2 ) . (1 + 2x + 3x2 ) 6 + (1 $-$ 4x2 )6 )
Let, a = ((1 + 2x + 3x2 )6 + (1 $-$ 4x2 )6 )
$\therefore\,\,\,\,$ Given statement becomes,
(2 $-$ x2 ) . (a)
= 2a $-$ x2 (a)
Here coefficients of x2 is
= 2 (coefficient of x2 in a ) $-$ 1 (constant in a)
(1 + 2x + 3x2 ) 6 = 6 C0 + 6 C1 (2x + 3x2 ) + 6 C2 (2x + 3x2 )2
+ . . . . . .+ (2x + 3x2 )6
(1 $-$ 4x2 )6 = 6 C0 $-$ 6 C1 (4x2 ) + 6 C2 (4x2 )2 + . . . . .+ (4x2 )6
Coefficient of x2 in (1 + 2x + 3x2 )6
= 6 C1 $ \times $ 3 + 6 C2 $ \times $ 4
= 18 + 60
Coefficient of x2 in (1 $-$ 4x2 )6
= $-$ 6 C1 $ \times $ 4
= $-$ 24
Coefficient of x2 in ((1 + 2x + 3x2 )6 + (1 $-$ 4x2 )6 )
= 60 + 18 $-$ 24
= 54
Constant term in (1 + 2x + 3x2 )6 = 6 C0 = 1
Constant term in (1 $-$ 4x)6 = 6 C0 = 1
$\therefore\,\,\,\,$ Constant term in ((1 + 2x + 3x2 )6 + (1 $-$ 4x)6 )
= 1 + 1 = 2
$\therefore\,\,\,\,$ Coefficient of x2 in (2 $-$ x2 ) ((1 + 2x + 3x2 )6 + (1 $-$ 4x2 )6 )
= 2 (54) $-$ 1 (2)
= 108 $-$ 2
= 106
2018
Q327
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The sum of the co-efficients of all odd degree terms in the expansion of
${\left( {x + \sqrt {{x^3} - 1} } \right)^5} + {\left( {x - \sqrt {{x^3} - 1} } \right)^5}$, $\left( {x > 1} \right)$ is
Show Answer
Practice Quiz
2018
Q328
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The coefficien of x10 in the expansion of (1 + x)2 (1 + x2 )3 (1 + x3 )4 is equal to :
Show Answer
Practice Quiz
Correct Answer: A
Explanation:
$ \because $$\,\,\,$ (1 + x)2 = 1 + 2x + x2 ,
(1 + x2 )3 = 1 + 3x2 + 3x4 + x6
and (1 + x3 )4 = 1 + 4x3 + 6x6 + 4x9 + x12
So, the possible combination for x10 are :
x . x9 , x . x6 . x3 , x2 . x2 . x6 , x4 . x6
Corresponding coefficients are 2 $ \times $ 4, 2$ \times $ 1 $ \times $4, 1 $ \times $3$ \times $ 6,
3 $ \times $ 6 or 8, 8, 18, 18.
$\therefore\,\,\,$ Sum of the coefficient is 8 + 8 + 18 + 18 = 52
Therefore, the coefficient of x10 in the expansion of
(1 + x)2 (1 + x2 )3 (1 + x3 )4 is 52.
2018
Q329
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
If n is the degree of the polynomial,
${\left[ {{2 \over {\sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1} }}} \right]^8} + $ ${\left[ {{2 \over {\sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1} }}} \right]^8}$
and m is the coefficient of xn in it, then the ordered pair (n, m) is equal to :
Show Answer
Practice Quiz
Correct Answer: C
Explanation:
Given,
${\left[ {{2 \over {\sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1} }}} \right]^8}$ + ${\left[ {{2 \over {\sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1} }}} \right]^8}$
= ${\left[ {{2 \over {\sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1} }} \times {{\sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1} } \over {\sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1} }}} \right]^8}$
+ ${\left[ {{2 \over {\sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1} }} \times {{\sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1} } \over {\sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1} }}} \right]^8}$
= ${\left[ {{{2(\sqrt {5x^3 + 1} + \sqrt {5x^3 - 1} } \over 2}} \right]^8}$ + ${\left[ {{{2(\sqrt {5x^3 - 1} - \sqrt {5x^3 - 1} } \over 2}} \right]^8}$
= (${\sqrt {5{x^3} + 1} }$ + ${\sqrt {5{x^3} - 1} }$)8 + (${\sqrt {5{x^3} - 1} }$ - ${\sqrt {5{x^3} - 1} }$)8
= 2 $\left[ {{}^8{C_0}} \right.{(\sqrt {5{x^3} + 1} )^8}$ +
+ ${}^8{C_2}{(\sqrt {5{x^3} + 1} )^6}\sqrt {5{x^3} - 1} $
+ ${}^8{C_4}{(\sqrt {5{x^3} + 1} )^4}{(5{x^3} - 1)^2}$
+ ${}^8{C_6}{(\sqrt {5{x^3} + 1} )^2}{(5{x^3} - 1)^3}$
+ $\left. {{}^8{C_8}{{(5{x^3} - 1)}^4}} \right]$
= 2$[{}^8{C_0}{(5{x^3} + 1)^4}$
+ ${}^8{C_2}{(5{x^3} + 1)^3}(5{x^3} - 1)$
+ ${}^8{C_4}(5{x^3} + 1)^2{(5{x^3} - 1)^2}$
+ ${}^8{C_6}(5{x^3} + 1){(5{x^3} - 1)^3}$
+ ${}^8{C_8}{(5{x^3} - 1)^4}]$
Here maximum power of x is 12
$ \therefore $ Degree of polynomial = 12
Coefficient of x12
= 2 [8 C0 54 + 8 C2
$\times$ 53 $\times$ 5 + 8 C4
$\times$ 52 $\times$ 52 + 8 C6 $\times$ 5 $\times$ 53 + 8 C8 $\times$ 54 ]
= 160000 = (20)4
2017
Q330
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The coefficient of x−5 in the binomial expansion of
${\left( {{{x + 1} \over {{x^{{2 \over 3}}} - {x^{{1 \over 3}}} + 1}} - {{x - 1} \over {x - {x^{{1 \over 2}}}}}} \right)^{10}},$ where x $ \ne $ 0, 1, is :
Show Answer
Practice Quiz
Correct Answer: A
Explanation:
${\left( {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right)^{10}}$
= ${\left( {{{{{\left( {{x^{1/3}}} \right)}^3} + {{\left( {{1^{1/3}}} \right)}^3}} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{{{\left( {\sqrt x } \right)}^2} - {{\left( 1 \right)}^2}} \over {x - {x^{1/2}}}}} \right)^{10}}$
= ${\left( {{{\left( {{x^{1/3}} + 1} \right)\left( {{x^{2/3}} - {x^{1/3}} + 1} \right)} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)} \over {\sqrt x \left( {\sqrt x - 1} \right)}}} \right)^{10}}$
= ${\left( {\left( {{x^{1/3}} + 1} \right) - {{\left( {\sqrt x + 1} \right)} \over {\sqrt x }}} \right)^{10}}$
= ${\left( {\left( {{x^{1/3}} + 1} \right) - \left( {1 + {1 \over {\sqrt x }}} \right)} \right)^{10}}$
= ${\left( {{x^{1/3}} - {1 \over {{x^{1/2}}}}} \right)^{10}}$
[Note:
For ${\left( {{x^\alpha } \pm {1 \over {{x^\beta }}}} \right)^n}$ the $\left( {r + 1} \right)$th term with power m of x is
$r = {{n\alpha - m} \over {\alpha + \beta }}$]
Here $\alpha = {1 \over 3}$, $\beta = {1 \over 2}$ and m = -5
then $r = {{10 \times {1 \over 3} - (-5)} \over {{1 \over 3} + {1 \over 2}}}$ = ${{25} \over 3} \times {6 \over 5}$ = 10
$\therefore$ T11 is the term with x-5 .
$\therefore$ T11 = ${}^{10}{C_{10}}$ = 1
2017
Q331
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
If (27)999 is divided by 7, then the remainder is :
Show Answer
Practice Quiz
Correct Answer: D
Explanation:
We have,
${{{{\left( {27} \right)}^{999}}} \over 7}$
= ${{{{\left( {28 - 1} \right)}^{999}}} \over 7}$
= ${{28\,\lambda - 1} \over 7}$
= ${{28\,\lambda - 7 + 7 - 1} \over \lambda }$
= ${{7\left( {4\lambda - 1} \right) + 6} \over 7}$
$\therefore\,\,\,$ Remainder = 6
2017
Q332
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The value of $\left( {{}^{21}{C_1} - {}^{10}{C_1}} \right) + \left( {{}^{21}{C_2} - {}^{10}{C_2}} \right) + \left( {{}^{21}{C_3} - {}^{10}{C_3}} \right)$
$\left( {{}^{21}{C_4} - {}^{10}{C_4}} \right)$$ + .... + \left( {{}^{21}{C_{10}} - {}^{10}{C_{10}}} \right)$ is
Show Answer
Practice Quiz
2016
Q333
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
If the coefficients of x−2 and x−4 in the expansion of ${\left( {{x^{{1 \over 3}}} + {1 \over {2{x^{{1 \over 3}}}}}} \right)^{18}},\left( {x > 0} \right),$ are m and n respectively, then ${m \over n}$ is equal to :
Show Answer
Practice Quiz
Correct Answer: A
Explanation:
Tr+1 = 18 Cr ${\left( {{x^{{1 \over 3}}}} \right)^{18 - r}}$ . ${\left( {{1 \over {2{x^{{1 \over 3}}}}}} \right)^r}$
= 18 Cr ${\left( {{1 \over 2}} \right)^r}\,\,.\,\,{x^{{{18 - 2r} \over 3}}}$
For coefficient of x$-$2 ,
${{18 - 2r} \over 3}$ = $-$2
$ \Rightarrow $ r = 12
$ \therefore $ Coefficient of x$-$2 is (m) = 18 C12 ${\left( {{1 \over 2}} \right)^{12}}$
For coefficient of x$-$4 ,
${{18 - 2r} \over 3}$ = $-$ 4
$ \Rightarrow $ r = 15
$ \therefore $ Coefficient of x$-$4 is (n) = 18 C15 $\left( {{1 \over {2}}} \right)$15
$ \therefore $ ${m \over n} = {{^{18}{C_{12}}{{\left( {{1 \over 2}} \right)}^{12}}} \over {^{18}{C_{15}}{{\left( {{1 \over 2}} \right)}^{15}}}}$
= ${{{}^{18}{C_6} \times {{\left( 2 \right)}^3}} \over {{}^{18}{C_3}}}$
= 182
2016
Q334
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
For x $ \in $ R , x $ \ne $ -1,
if (1 + x)2016 + x(1 + x)2015 + x2 (1 + x)2014 + . . . . + x2016 =
$\sum\limits_{i = 0}^{2016} {{a_i}} \,{x^i},\,\,$ then a17 is equal to :
A.
${{2017!} \over {17!\,\,\,2000!}}$
B.
${{2016!} \over {17!\,\,\,1999!}}$
C.
${{2017!} \over {2000!}}$
D.
${{2016!} \over {16!}}$
Show Answer
Practice Quiz
Correct Answer: A
Explanation:
Assume,
P = (1 + x)2016 + x(1 + x)2015 + . . . . .+ x2015 . (1 + x) + x2016 . . . . .(1)
Multiply this with $\left( {{x \over {1 + x}}} \right),$
$\left( {{x \over {1 + x}}} \right)P = $ x(1 + x)2015 + x2 (1 + x)2014 +
. . . . . . + x2016 + ${{{x^{2017}}} \over {1 + x}}$ . . . . . (2)
Performing (1) $-$ (2), we get
${P \over {1 + x}} = $ (1 + x)2016 $-$ ${{{x^{2017}}} \over {1 + x}}$
$ \Rightarrow $ P = (1 + x)2017 $-$ x2017
$ \therefore $ a17 = coefficient of x17 $=$ 2017 C17 $=$ ${{2017!} \over {17!\,\,2000!}}$
2016
Q335
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
If the number of terms in the expansion of ${\left( {1 - {2 \over x} + {4 \over {{x^2}}}} \right)^n},\,x \ne 0,$ is 28, then the sum of the coefficients of all the terms in this expansion, is :
Show Answer
Practice Quiz
Correct Answer: B
Explanation:
Total no of terms in ${\left( {1 - {2 \over x} + {4 \over {{x^2}}}} \right)^n}$ = ${}^{n + 2}{C_2}$ = 28
(n+2)(n+1) = 56
$ \Rightarrow n = 6$
Sum of coefficient = (1 - 2 + 4)6 = 36 = 729
2015
Q336
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The sum of coefficients of integral power of $x$ in the binomial expansion ${\left( {1 - 2\sqrt x } \right)^{50}}$ is :
A.
${1 \over 2}\left( {{3^{50}} - 1} \right)$
B.
${1 \over 2}\left( {{2^{50}} + 1} \right)$
C.
${1 \over 2}\left( {{3^{50}} + 1} \right)$
D.
${1 \over 2}\left( {{3^{50}}} \right)$
Show Answer
Practice Quiz
Correct Answer: C
Explanation:
${\left( {1 - 2\sqrt x } \right)^{50}}$
= ${}^{50}{C_0} + {}^{50}{C_1}.\left( { - 2\sqrt x } \right) + {}^{50}{C_2}.{\left( { - 2\sqrt x } \right)^2} + ....$
Now we need to find out those coefficient where degree of x is integer and you can see at odd terms power of x is integer.
Let ${\left( {1 - 2\sqrt x } \right)^{50}}$ = Odd(A) - Even(B)
So ${\left( {1 + 2\sqrt x } \right)^{50}}$ = A + B
$\therefore$ 2A = ${\left( {1 + 2\sqrt x } \right)^{50}}$ + ${\left( {1 - 2\sqrt x } \right)^{50}}$
$ \Rightarrow A = {1 \over 2}\left[ {{{\left( {1 + 2\sqrt x } \right)}^{50}} + {{\left( {1 - 2\sqrt x } \right)}^{50}}} \right]$
Now to find sum of coefficient of A, put x = 1.
$\therefore$ Sum of coefficient of A = ${1 \over 2}\left[ {{{\left( {1 + 2} \right)}^{50}} + {{\left( {1 - 2} \right)}^{50}}} \right]$
= ${1 \over 2}\left[ {{{\left( 3 \right)}^{50}} + 1} \right]$
2014
Q337
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
If the coefficints of ${x^3}$ and ${x^4}$ in the expansion of $\left( {1 + ax + b{x^2}} \right){\left( {1 - 2x} \right)^{18}}$ in powers of $x$ are both zero, then $\left( {a,\,b} \right)$ is equal to:
A.
$\left( {14,{{272} \over 3}} \right)$
B.
$\left( {16,{{272} \over 3}} \right)$
C.
$\left( {16,{{251} \over 3}} \right)$
D.
$\left( {14,{{251} \over 3}} \right)$
Show Answer
Practice Quiz
Correct Answer: B
Explanation:
$\left( {1 + ax + b{x^2}} \right){\left( {1 - 2x} \right)^{18}}$
= ${\left( {1 - 2x} \right)^{18}} + ax{\left( {1 - 2x} \right)^{18}} + b{x^2}{\left( {1 - 2x} \right)^{18}}$
= $\left( {1 + ax + b{x^2}} \right)\left[ {{}^{18}{C_0} - {}^{18}{C_1}\left( {2x} \right) + {}^{18}{C_2}{{\left( {2x} \right)}^2} - {}^{18}{C_3}{{\left( {2x} \right)}^3} + ....} \right]$
Coefficient of x3 is
${\left( { - 2} \right)^3}.{}^{18}{C_3} + a{\left( { - 2} \right)^2}.{}^{18}{C_2} + b\left( { - 2} \right).{}^{18}{C_1}$ = 0
$ \Rightarrow 153a - 9b = 1632$ ....... (1)
Coefficient of x4 is
${\left( { - 2} \right)^4}.{}^{18}{C_4} + a{\left( { - 2} \right)^3}.{}^{18}{C_3} + b{\left( { - 2} \right)^2}.{}^{18}{C_2}$ = 0
$ \Rightarrow 3b - 32a = -240$ ....... (2)
Solving (1) and (2), we get $a$ = 16, b = ${{172} \over 3}$
2013
Q338
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The term independent of $x$ in expansion of
${\left( {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right)^{10}}$ is
Show Answer
Practice Quiz
Correct Answer: C
Explanation:
${\left( {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right)^{10}}$
= ${\left( {{{{{\left( {{x^{1/3}}} \right)}^3} + {{\left( {{1^{1/3}}} \right)}^3}} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{{{\left( {\sqrt x } \right)}^2} - {{\left( 1 \right)}^2}} \over {x - {x^{1/2}}}}} \right)^{10}}$
= ${\left( {{{\left( {{x^{1/3}} + 1} \right)\left( {{x^{2/3}} - {x^{1/3}} + 1} \right)} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)} \over {\sqrt x \left( {\sqrt x - 1} \right)}}} \right)^{10}}$
= ${\left( {\left( {{x^{1/3}} + 1} \right) - {{\left( {\sqrt x + 1} \right)} \over {\sqrt x }}} \right)^{10}}$
= ${\left( {\left( {{x^{1/3}} + 1} \right) - \left( {1 + {1 \over {\sqrt x }}} \right)} \right)^{10}}$
= ${\left( {{x^{1/3}} - {1 \over {{x^{1/2}}}}} \right)^{10}}$
[Note:
For ${\left( {{x^\alpha } \pm {1 \over {{x^\beta }}}} \right)^n}$ the $\left( {r + 1} \right)$th term with power m of x is
$r = {{n\alpha - m} \over {\alpha + \beta }}$]
Here $\alpha = {1 \over 3}$, $\beta = {1 \over 2}$ and m = 0
then $r = {{10 \times {1 \over 3} - 0} \over {{1 \over 3} + {1 \over 2}}}$ = ${{10} \over 3} \times {6 \over 5}$ = 4
$\therefore$ T5 is the term independent of x.
$\therefore$ T5 = ${}^{10}{C_4}$ = 210
2012
Q339
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
If $n$ is a positive integer, then ${\left( {\sqrt 3 + 1} \right)^{2n}} - {\left( {\sqrt 3 - 1} \right)^{2n}}$ is :
B.
an odd positive integer
C.
an even positive integer
D.
a rational number other than positive integers
Show Answer
Practice Quiz
Correct Answer: A
Explanation:
Let ${\left( {a + x} \right)^n}$ = Odd trems(A) + Even terms(B)
So ${\left( {a - x} \right)^n}$ = Odd terms(A) - Even terms(B)
$\therefore$ ${\left( {a + x} \right)^n} - {\left( {a - x} \right)^n}$
= (A + B) - (A - B)
= 2B
= 2[even terms]
= 2[ T2 + T4 + T6 + ....... ]
So in case of ${\left( {\sqrt 3 + 1} \right)^{2n}} - {\left( {\sqrt 3 - 1} \right)^{2n}}$
= 2[ T2 + T4 + T6 + ....... ]
= 2[ ${}^{2n}{C_1}.{\left( {\sqrt 3 } \right)^{2n - 1}} + {}^{2n}{C_3}.{\left( {\sqrt 3 } \right)^{2n - 3}} + ...$
Here in ${\left( {\sqrt 3 } \right)^{2n - 1}}$, 2n - 1 is odd number. So there will be always ${\sqrt 3 }$ in ${\left( {\sqrt 3 } \right)^{2n - 1}}$.
So ${\left( {\sqrt 3 + 1} \right)^{2n}} - {\left( {\sqrt 3 - 1} \right)^{2n}}$ will be always irrational number.
2011
Q340
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The coefficient of ${x^7}$ in the expansion of ${\left( {1 - x - {x^2} + {x^3}} \right)^6}$ is
Show Answer
Practice Quiz
Correct Answer: B
Explanation:
Given,
${\left( {1 - x - {x^2} + {x^3}} \right)^6}$
= ${\left[ {\left( {1 - x} \right) - {x^2}\left( {1 - x} \right)} \right]^6}$
= ${\left( {1 - x} \right)^6}{\left( {1 - {x^2}} \right)^6}$
= $\left( {1 + {}^6{C_1}( - x) + {}^6{C_2}{{( - x)}^2} + {}^6{C_3}{{( - x)}^3} + .......} \right)\times$
$\left( {1 + {}^6{C_1}( - {x^2}) + {}^6{C_2}{{( - {x^2})}^2} + {}^6{C_3}{{( - {x^2})}^3} + .......} \right)$
$\therefore$ Coefficient of x7 = $ - {}^6{C_1} \times - {}^6{C_3} + \left( { - {}^6{C_3}} \right) \times {}^6{C_2} + \left( { - {}^6{C_5}} \right) \times - {}^6{C_1}$
= 120 - 300 + 36 = - 144
2010
Q341
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
Let ${s_1} = \sum\limits_{j = 1}^{10} {j\left( {j - 1} \right){}^{10}} {C_j}$,
${{s_2} = \sum\limits_{j = 1}^{10} {} } j.{}^{10}{C_j}$ and
${{s_3} = \sum\limits_{j = 1}^{10} {{j^2}.{}^{10}{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}}$.
A.
Statement - 1 is true, Statement- 2 is true; Statement - 2 is not a correct explanation for Statement - 1.
B.
Statement - 1 is true, Statement-2 is false.
C.
Statement - 1 is false, Statement-2 is true.
D.
Statement - 1 is true, Statement-2 is true: -Statement - 2 is a correct explanation for Statement - 1.
Show Answer
Practice Quiz
Correct Answer: B
Explanation:
Note :
$\sum\limits_{r = 0}^n {r.{}^n{C_r}} $ = $ = n{.2^{n - 1}}$
$\sum\limits_{r = 0}^n {{r^2}.{}^n{C_r}} = n\left( {n + 1} \right){2^{n - 2}}$
Given that,
${s_1} = \sum\limits_{j = 1}^{10} {j\left( {j - 1} \right){}^{10}} {C_j}$
=$\sum\limits_{j = 1}^{10} {{j^2}.{}^{10}} {C_j} - \sum\limits_{j = 1}^{10} {j.{}^{10}} {C_j}$
= 10$ \times $11$ \times $${2^{10 - 2}}$ - 10$ \times $${2^{10 - 1}}$
= 10$ \times $${2^{8}}$(11 - 2)
= 10$ \times $9$ \times $${2^{8}}$
= 90$ \times $${2^{8}}$
${{s_2} = \sum\limits_{j = 1}^{10} {} } j.{}^{10}{C_j}$
= 10$ \times $${2^{10-1}}$
= 10$ \times $${2^{9}}$
${{s_3} = \sum\limits_{j = 1}^{10} {{j^2}.{}^{10}{C_j}.} }$
= 10$ \times $11$ \times $${2^{10-2}}$
= ${{110} \over 2} \times {2^9}$
= 55 $ \times $ $ {2^9}$
2009
Q342
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The remainder left out when ${8^{2n}} - {\left( {62} \right)^{2n + 1}}$ is divided by 9 is :
Show Answer
Practice Quiz
Correct Answer: A
Explanation:
${8^{2n}} - {\left( {62} \right)^{2n + 1}}$
= ${\left( {{8^2}} \right)^n} - {\left( {62} \right)^{2n + 1}}$
= ${\left( {1 + 63} \right)^n} - {\left( {1 - 63} \right)^{2n + 1}}$
= $\left( {1 + n.63 + {}^n{C_2}{{.63}^2} + ......} \right)$
+ $\left( {1 + {}^{2n + 1}{C_1}.\left( { - 63} \right) + {}^{2n + 1}{C_2}.{{\left( { - 63} \right)}^2} + ......} \right)$
= 2 + 63$\left[ {\left( {n + {}^n{C_2} + ....} \right) + \left( { - {}^{2n + 1}{C_1} + {}^{2n + 1}{C_2}.63 + ......} \right)} \right]$
= 63$ \times $[Some integral value] + 2
63$ \times $[Some integral value] + 2 by dividing with 9 we will get 2 as remainder as 63 is multiple of 9.
2008
Q343
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
Statement - 1 : $\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{}^n{C_r} = \left( {n + 2} \right){2^{n - 1}}.} $
Statement - 2 : $\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{}^n{C_r}{x^r} = {{\left( {1 + x} \right)}^n} + nx{{\left( {1 + x} \right)}^{n - 1}}.} $
A.
Statement - 1 is false, Statement - 2 is true
B.
Statement - 1 is true, Statement - 2 is true; Statement - 2 is a correct explanation for Statement - 1
C.
Statement - 1 is true, Statement - 2 is true; Statement - 2 is not a correct explanation for Statement - 1
D.
Statement - 1 is true, Statement - 2 is false
Show Answer
Practice Quiz
Correct Answer: B
Explanation:
Check Statement - 1
$\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{}^n{C_r}}$
= $\sum\limits_{r = 0}^n {r.{}^n{C_r}} $ + $\sum\limits_{r = 0}^n {{}^n{C_r}} $
= $\sum\limits_{r = 1}^n {r.{n \over r}{}^{n - 1}{C_{r - 1}}} $ $ + {2^n}$
= $n\sum\limits_{r = 1}^n {{}^{n - 1}{C_{r - 1}}} $ $ + {2^n}$
= $n \times {2^{n - 1}}$$ + {2^n}$
= ${2^{n - 1}}\left[ {n + 2} \right]$
Check Statement 2 :
$\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{}^n{C_r}{x^r}}$
= $\sum\limits_{r = 0}^n r .{}^n{C_r}.{x^r}$ + $\sum\limits_{r = 0}^n {{}^n{C_r}.{x^r}} $
= $n\sum\limits_{r = 1}^n {{}^{n - 1}{C_{r - 1}}.{x^r}} $ $ + {\left( {1 + x} \right)^n}$
= $nx\sum\limits_{r = 1}^n {{}^{n - 1}{C_{r - 1}}.{x^{r - 1}}} + {\left( {1 + x} \right)^n}$
= $nx{\left( {1 + x} \right)^{n - 1}} + {\left( {1 + x} \right)^n}$
Substitude x = 1 in the statement 2 and we get,
$\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{}^n{C_r} = \left( {n + 2} \right){2^{n - 1}}.} $
So Option B is correct.
2007
Q344
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The sum of the series ${}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .....\, - \,.....\, + {}^{20}{C_{10}}$ is
D.
${1 \over 2}{}^{20}{C_{10}}$
Show Answer
Practice Quiz
Correct Answer: D
Explanation:
We know
${}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .... + {}^{20}{C_{10}} - {}^{20}{C_{11}}+ ...... + {}^{20}{C_{20}} = 0$
$ \Rightarrow $ $({}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .... - {}^{20} {C_{9}})$ $+{}^{20}{C_{10}}$ $(-{}^{20}{C_{9}}+ {}^{20}{C_{8}}+...... + {}^{20}{C_{0}})$
(As ${}^{20}{C_{11}} = {}^{20}{C_9}$)
$ \Rightarrow $ $2({}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .... - {}^{20} {C_{9}})$ $+{}^{20}{C_{10}}$ = 0
$ \Rightarrow $ ${}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .... - {}^{20} {C_{9}}$ = $ - {1 \over 2}{}^{20}{C_{10}}$
Adding ${}^{20}{C_{10}}$ both sides,
$ \Rightarrow $ ${}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .... - {}^{20} {C_{9}}$ $ + {}^{20}{C_{10}}$ = $ - {1 \over 2}{}^{20}{C_{10}}$ $ + {}^{20}{C_{10}}$
$ \Rightarrow $ ${}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .... - {}^{20} {C_{9}}$ $ + {}^{20}{C_{10}}$ = $ {1 \over 2}{}^{20}{C_{10}}$
2007
Q345
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
In the binomial expansion of ${\left( {a - b} \right)^n},\,\,\,n \ge 5,$ the sum of ${5^{th}}$ and ${6^{th}}$ terms is zero, then $a/b$ equals
Show Answer
Practice Quiz
Correct Answer: B
Explanation:
According to the question,
t5 + t6 = 0
$\therefore$ ${}^n{C_4}.{a^{n - 4}}.{b^4}$ + $\left( { - {}^n{C_5}.{a^{n - 5}}.{b^5}} \right)$ = 0
By solving we get,
${a \over b} = {{n - 4} \over 5}$
2006
Q346
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
For natural numbers $m$ , $n$, if ${\left( {1 - y} \right)^m}{\left( {1 + y} \right)^n}\,\, = 1 + {a_1}y + {a_2}{y^2} + ..........$ and ${a_1} = {a_2} = 10,$ then $\left( {m,\,n} \right)$ is
A.
$\left( {20,\,45} \right)$
B.
$\left( {35,\,20} \right)$
C.
$\left( {45,\,35} \right)$
D.
$\left( {35,\,45} \right)$
Show Answer
Practice Quiz
Correct Answer: D
Explanation:
${\left( {1 - y} \right)^m}{\left( {1 + y} \right)^n}\,\,$
= $\left( {{}^m{C_0} - {}^m{C_1}y + {}^m{C_2}{y^2} + ....} \right)$ -
$\left( {{}^n{C_0} + {}^n{C_1}y + {}^n{C_2}{y^2} + ....} \right)$
${a_1}$ = Coefficient of y = ${{}^n{C_1}}$ - ${{}^m{C_1}}$ = 10
$ \Rightarrow $ n - m = 10
${a_2}$ = Coefficient of y2
= ${}^n{C_2} + {}^n{C_1} \times {}^m{C_1} + {}^m{C_2} = 10$
$ \Rightarrow {{n\left( {n - 1} \right)} \over 2} - nm + {{m\left( {m - 1} \right)} \over 2} = 10$
$ \Rightarrow n\left( {n - 1} \right) - 2nm + m\left( {m - 1} \right) = 20$
$ \Rightarrow$ (m + 10)(m + 9) - 2(m + 10)m + m(m - 1) = 20
$ \Rightarrow$ 90 + 19m + m2 - 2m2 - 20m + m2 - m - 20 = 0
$ \Rightarrow$ 70 - 2m = 0
$ \Rightarrow$ m = 35
$\therefore$ n = 10 + 35 = 45
2006
Q347
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
If the expansion in powers of $x$ of the function ${1 \over {\left( {1 - ax} \right)\left( {1 - bx} \right)}}$ is ${a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3}.....$ then ${a_n}$ is
A.
${{{b^n} - {a^n}} \over {b - a}}$
B.
${{{a^n} - {b^n}} \over {b - a}}$
C.
${{{a^{n + 1}} - {b^{n + 1}}} \over {b - a}}$
D.
${{{b^{n + 1}} - {a^{n + 1}}} \over {b - a}}$
Show Answer
Practice Quiz
Correct Answer: D
Explanation:
${1 \over {\left( {1 - ax} \right)\left( {1 - bx} \right)}}$
= ${\left( {1 - ax} \right)^{ - 1}}{\left( {1 - bx} \right)^{ - 1}}$
= $\left[ {1 + \left( { - 1} \right)\left( { - ax} \right) + {{\left( { - 1} \right)\left( { - 2} \right)} \over {1.2}}{{\left( { - ax} \right)}^2} + ...} \right]$ -
$\left[ {1 + \left( { - 1} \right)\left( { - bx} \right) + {{\left( { - 1} \right)\left( { - 2} \right)} \over {1.2}}{{\left( { - bx} \right)}^2} + ...} \right]$
= $\left[ {1 + ax + {a^2}{x^2} + ... + {a^{n - 1}}{x^{n - 1}} + {a^n}{x^n}}+.... \right]$ -
$\left[ {1 + bx + {b^2}{x^2} + ... + {b^{n - 1}}{x^{n - 1}} + {b^n}{x^n}}+.... \right]$
Coefficient of xn =
${a^n} + {a^{n - 1}}b + {a^{n - 2}}{b^2} + .... + {b^n}$
= ${a^n}\left[ {1 + {b \over a} + {{{b^2}} \over {{a^2}}} + ..... + {{{b^n}} \over {{a^n}}}} \right]$
= ${a^n}\left[ {{{{{\left( {{b \over a}} \right)}^{n + 1}} - 1} \over {{b \over a} - 1}}} \right]$
= ${a^n}\left[ {{{{b^{n + 1}} - {a^{n + 1}}} \over {{a^{n + 1}}\left( {{{b - a} \over a}} \right)}}} \right]$
= ${{{{b^{n + 1}} - {a^{n + 1}}} \over {b - a}}}$
2005
Q348
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The value of $\,{}^{50}{C_4} + \sum\limits_{r = 1}^6 {^{56 - r}} {C_3}$ is
Show Answer
Practice Quiz
Correct Answer: D
Explanation:
Given, ${}^{50}{C_4} + \sum\limits_{n = 1}^6 {{}^{56 - r}{C_3}} $
$ \Rightarrow $ ${}^{50}{C_4}$ + ${}^{55}{C_3}$ + ${}^{54}{C_3}$ + ${}^{53}{C_3}$ + ${}^{52}{C_3}$ + ${}^{51}{C_3}$ + ${}^{50}{C_3}$
Arrange those this way
$ \Rightarrow $ ${}^{50}{C_4}$ + ${}^{50}{C_3}$ + ${}^{51}{C_3}$ + ${}^{52}{C_3}$ + ${}^{53}{C_3}$ + ${}^{54}{C_3}$ + ${}^{55}{C_3}$
We know this formula [ ${{}^n{C_r}}$ + ${{}^n{C_{r - 1}}}$ = ${{}^{n + 1}{C_r}}$ ] which is used to solve this problem.
$ \Rightarrow $ ${}^{51}{C_4}$ + ${}^{51}{C_3}$ + ${}^{52}{C_3}$ + ${}^{53}{C_3}$ + ${}^{54}{C_3}$ + ${}^{55}{C_3}$
$ \Rightarrow $ ${}^{52}{C_4}$ + ${}^{52}{C_3}$ + ${}^{53}{C_3}$ + ${}^{54}{C_3}$ + ${}^{55}{C_3}$
$ \Rightarrow $${}^{53}{C_4}$ + ${}^{53}{C_3}$ + ${}^{54}{C_3}$ + ${}^{55}{C_3}$
$ \Rightarrow$$ {}^{54}{C_4}$ + ${}^{54}{C_3}$ + ${}^{55}{C_3}$
$ \Rightarrow $${}^{55}{C_4}$ + ${}^{55}{C_3}$
$ \Rightarrow$$ {}^{56}{C_4}$
2005
Q349
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
If the coefficients of rth , (r+1)th , and (r + 2)th terms in the binomial expansion of ${{\rm{(1 + y )}}^m}$ are in A.P., then m and r satisfy the equation
A.
${m^2} - m(4r - 1) + 4\,{r^2} - 2 = 0$
B.
${m^2} - m(4r + 1) + 4\,{r^2} + 2 = 0$
C.
${m^2} - m(4r + 1) + 4\,{r^2} - 2 = 0$
D.
${m^2} - m(4r - 1) + 4\,{r^2} + 2 = 0$
Show Answer
Practice Quiz
Correct Answer: C
Explanation:
Let r = 2
$\therefore$ 2nd, 3rd and 4th terms are in AP.
2nd term = T2 = ${}^m{C_1}.y$
Coefficient of T2 = ${}^m{C_1}$
3rd term = T3 = ${}^m{C_2}.{y^2}$
Coefficient of T3 = ${}^m{C_2}$
4th term = T4 = ${}^m{C_3}.{y^3}$
Coefficient of T2 = ${}^m{C_3}$
$\therefore$ 2.${}^m{C_2}$ = ${}^m{C_1}$ + ${}^m{C_3}$
$ \Rightarrow $ $2.{{m\left( {m - 1} \right)} \over {1.2}}$ = ${m \over 1}$ + ${{m\left( {m - 1} \right)\left( {m - 2} \right)} \over {1.2.3}}$
$ \Rightarrow $ 6m2 - 6m = 6m +m(m2 - 3m + 2)
$ \Rightarrow $ 6m2 - 6m = 6m + m3 - 3m2 + 2m
$ \Rightarrow $ 6m - 6 = 6 + m2 - 3m + 2
$ \Rightarrow $ m2 - 9m + 14 = 0
Now put r = 2 at each option and find answer.
In option C, ${m^2} - m(4r + 1) + 4\,{r^2} - 2 = 0$ putting r = 2 we get
m2 - 9m + 14 = 0. So Option C is correct.
2005
Q350
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
If $x$ is so small that ${x^3}$ and higher powers of $x$ may be neglected, then ${{{{\left( {1 + x} \right)}^{{3 \over 2}}} - {{\left( {1 + {1 \over 2}x} \right)}^3}} \over {{{\left( {1 - x} \right)}^{{1 \over 2}}}}}$ may be approximated as
A.
$1 - {3 \over 8}{x^2}$
B.
$3x + {3 \over 8}{x^2}$
D.
${x \over 2} - {3 \over 8}{x^2}$
Show Answer
Practice Quiz
Correct Answer: C
Explanation:
${\left( {1 + x} \right)^{{3 \over 2}}}$ = 1 + ${3 \over 2}x + {{{3 \over 2}.{1 \over 2}} \over {1.2}}{x^2} + ...$
= 1 + ${3 \over 2}x + {3 \over 8}{x^2}$ (As $x$ is so small, so ${x^3}$ and higher powers of $x$ neglected)
${{{{\left( {1 + x} \right)}^{{3 \over 2}}} - {{\left( {1 + {1 \over 2}x} \right)}^3}} \over {{{\left( {1 - x} \right)}^{{1 \over 2}}}}}$
= ${{\left( {1 + {3 \over 2}x + {3 \over 8}{x^2}} \right) - \left( {1 + {}^3{C_0}.{x \over 2} + {}^3{C_1}.{{\left( {{x \over 2}} \right)}^2}} \right)} \over {{{\left( {1 - x} \right)}^2}}}$
= ${{{3 \over 8}{x^2} - {3 \over 4}{x^2}} \over {{{\left( {1 - x} \right)}^2}}}$
= ${{x^2}\left( {{3 \over 8} - {3 \over 4}} \right){{\left( {1 - x} \right)}^{ - 2}}}$
= ${ - {3 \over 8}{x^2}\left( {1 - {1 \over 2}\left( { - x} \right) + ....} \right)}$
= $ - {3 \over 8}{x^2} - {3 \over {16}}{x^3}$
[ As x3 is so small we can ignore $-{3 \over {16}}{x^3}$]
= $ - {3 \over 8}{x^2}$