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JEE Mains Previous Years Questions with Solutions

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1

JEE Main 2016 (Offline)

MCQ (Single Correct Answer)
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 :
A
243
B
729
C
64
D
2187

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
2

JEE Main 2015 (Offline)

MCQ (Single Correct Answer)
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
$${1 \over 2}\left( {{3^{50}}} \right)$$

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 = $${1 \over 2}\left[ {{{\left( {1 + 2} \right)}^{50}} + {{\left( {1 - 2} \right)}^{50}}} \right]$$

= $${1 \over 2}\left[ {{{\left( 3 \right)}^{50}} + 1} \right]$$
3

JEE Main 2014 (Offline)

MCQ (Single Correct Answer)
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
$$\left( {14,{{251} \over 3}} \right)$$

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 = $${{172} \over 3}$$
4

JEE Main 2013 (Offline)

MCQ (Single Correct Answer)
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
A
4
B
120
C
210
D
310

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

Questions Asked from Mathematical Induction and Binomial Theorem

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