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1

JEE Advanced 2021 Paper 2 Online

MCQ (Single Correct Answer)
Let $${\psi _1}:[0,\infty ) \to R$$, $${\psi _2}:[0,\infty ) \to R$$, f : (0, $$\infty$$) $$\to$$ R and g : [0, $$\infty$$) $$\to$$ R be functions such that f(0) = g(0) = 0,

$${\psi _1}(x) = {e^{ - x}} + x,x \ge 0$$,

$${\psi _2}(x) = {x^2} - 2x - 2{e^{ - x}} + 2,x \ge 0$$,

$$f(x) = \int_{ - x}^x {(|t| - {t^2}){e^{ - {t^2}}}dt,x > 0} $$ and

$$g(x) = \int_0^{{x^2}} {\sqrt t {e^{ - t}}dt,x > 0} $$.
Which of the following statements is TRUE?
A
$${\psi _1}(x) \le 1$$, for all x > 0
B
$${\psi _2}(x) \le 0$$, for all x > 0
C
$$f(x) \ge 1 - {e^{ - {x^2}}} - {2 \over 3}{x^3} + {2 \over 5}{x^5}$$, for all $$x \in \left( {0,{1 \over 2}} \right)$$
D
$$g(x) \le {2 \over 3}{x^3} - {2 \over 5}{x^5} + {1 \over 7}{x^7}$$, for all $$x \in \left( {0,{1 \over 2}} \right)$$

Explanation

For option (a)

$${\psi _1}(x) = {e^{ - x}} + x$$

$$\therefore$$ $$\psi {'_1}(x) = 1 - {e^{ - x}} \Rightarrow \psi {'_1}(x) = 0$$

Here, $${\psi _1}(x)$$ is always increasing.

Now, $${\psi _1}(x) > 1$$

($$\because$$ $${\psi _1}(0) = 1$$)

Thus, option (a) is incorrect.

For option (b)

$${\psi _2}(x) = {x^2} - 2x - 2{e^{ - x}} + 2$$

$$ \Rightarrow \psi {'_2}(x) = 2x - 2 + 2{e^{ - x}}$$

$$ \Rightarrow \psi '{'_2}(x) = 2 - 2{e^{ - x}} \ge 0,\forall x > 0$$

$$ \Rightarrow \psi {'_2}(x)$$ is increasing $$ \Rightarrow \psi {'_2}(x) > \psi {'_2}(0)$$

$$ \Rightarrow \psi {'_2}(x) > 0,\forall x > 0$$

Thus, option (b) is incorrect.

For option (c)

$$f(x) = 2\int_0^x {(t - {t^2}){e^{ - {t^2}}}dt = ( - {e^{ - {t^2}}})_0^x - 2\int_0^x {{t^2}.{e^{ - {t^2}}}} dt} $$

$$ = 1 - {e^{ - {x^2}}} - 2\int_0^x {{t^2}\left( {1 - {t^2} + {{{t^4}} \over {2!}} + ...} \right)dt} $$

$$ = 1 - {e^{ - {x^2}}} - {{2{x^3}} \over 3} + {{2{x^5}} \over 5} - {{2{x^7}} \over {2.7}} + {{2{x^9}} \over {6.9}}...$$

$$ \Rightarrow f(x) - 1 + {e^{ - {x^2}}} + {{2{x^3}} \over 3} - {{2{x^5}} \over 5} < 0$$ in $$\left( {0,{1 \over 2}} \right)$$

Hence, option (c) is also incorrect.

For option (d)

$$g(x) = \int_0^{{x^2}} {\sqrt t .{e^{ - t}}dt} $$

Put $$t = {u^2} \Rightarrow dt = 2u\,du$$

$$\therefore$$ $$g(x) = \int_x^x {2.{u^2}{e^{ - {u^2}}}du = \int_0^x {2{u^2}\left( {1 - {u^2} + {{{u^4}} \over {2!}}....} \right)du} } $$

$$g(x) = {{2{x^3}} \over 3} - {{2{x^5}} \over 5} + {{2{x^7}} \over {2.7}} - {{2{x^9}} \over {9.6}}...$$

$$ \Rightarrow g(x) - {{2{x^3}} \over 3} + {{2{x^5}} \over 5} - {{2{x^7}} \over {2.7}} \le 0,\forall x \in \left( {0,{1 \over 2}} \right)$$

$$ \Rightarrow g(x) \le {{2{x^3}} \over 3} - {{2{x^5}} \over 5} + {1 \over 7}{x^7},\forall x \in \left( {0,{1 \over 2}} \right)$$

Hence, option (d) is correct.
2

JEE Advanced 2021 Paper 2 Online

MCQ (Single Correct Answer)
Let $${\psi _1}:[0,\infty ) \to R$$, $${\psi _2}:[0,\infty ) \to R$$, f : (0, $$\infty$$) $$\to$$ R and g : [0, $$\infty$$) $$\to$$ R be functions such that f(0) = g(0) = 0,

$${\psi _1}(x) = {e^{ - x}} + x,x \ge 0$$,

$${\psi _2}(x) = {x^2} - 2x - 2{e^{ - x}} + 2,x \ge 0$$,

$$f(x) = \int_{ - x}^x {(|t| - {t^2}){e^{ - {t^2}}}dt,x > 0} $$ and

$$g(x) = \int_0^{{x^2}} {\sqrt t {e^{ - t}}dt,x > 0} $$.
Which of the following statements is TRUE?
A
$$f(\sqrt {\ln 3} ) + g(\sqrt {\ln 3} ) = {1 \over 3}$$
B
For every x > 1, there exists an $$\alpha$$ $$\in$$ (1, x) such that $${\psi _1}(x) = 1 + \alpha x$$
C
For every x > 0, there exists a $$\beta$$ $$\in$$ (0, x) such that $${\psi _2}(x) = 2x({\psi _1}(\beta ) - 1)$$
D
f is an increasing function on the interval $$\left[ {0,{3 \over 2}} \right]$$

Explanation

$${\psi _1}(x) = {e^{ - x}} + x,x \ge 0$$

$${\psi _2}(x) = {x^2} - 2x - 2{e^{ - x}} + 2,x \ge 0$$

$$f(x) = \int_{ - x}^x {(\left| t \right| - {t^2}){e^{ - {t^2}}}dt,\,x > 0} $$

$$ = 2\int_0^x {(t - {t^2}){e^{ - {t^2}}}dt} $$ .... (i)

$$g(x) = \int_0^{{x^2}} {\sqrt t \,.\,{e^{ - t}}dt,x > 0} $$

Put $$t = {z^2} \Rightarrow dt = 2zdz$$

$$\therefore$$ $$g(x) = \int_0^x {z.{e^{ - {z^2}}}.\,2zdz} $$

$$ = 2\int_0^x {{z^2}.{e^{ - {z^2}}}dz = 2\int_0^x {{t^2}.{e^{ - {t^2}}}dt} } $$

$$f'(x) = 2(x - {x^2}){e^{ - {x^2}}} = 2x(1 - x){e^{ - {x^2}}}$$

$$\therefore$$ f is increasing for x$$\in$$(0, 1) and f is decreasing for x$$\in$$(1, $$\infty$$). Hence, option (d) is incorrect.

Now, $$f(x) + g(x) = 2\int_0^x {t.{e^{ - {t^2}}}dt} $$

$$ = [ - {e^{ - {t^2}}}]_0^x = ( - {e^{ - {x^2}}}) - ( - 1) = 1 - {e^{ - {x^2}}}$$

$$ \Rightarrow f(x) + g(x) = 1 - {e^{ - {x^2}}}$$

$$\therefore$$ $$f(\sqrt {\ln 3} ) + g(\sqrt {\ln 3} ) = 1 - {e^{ - {{(\sqrt {\ln 3} )}^2}}} = 1 - {e^{ - \ln 3}} = 1 - {1 \over 3} = {2 \over 3}$$

Hence, option (a) is incorrect.

$$\because$$ $${\psi _1}(x) = {e^{ - x}} + x$$ .... (iii)

$$\because$$ $$ \Rightarrow \psi {'_1}(x) = 1 - {e^{ - x}} < 1$$

Then, for $$\alpha \in (1,x),{\psi _1}(x) = 1 + \alpha x$$ does not true for $$\alpha$$ > 1. So, option (b) is incorrect.

Now, $${\psi _2}(x) = {x^2} - 2x - 2{e^{ - x}} + 2$$

$$ \Rightarrow \psi {'_2}(x) = 2x - 2 + 2{e^{ - x}} = 2(x + {e^{ - x}}) - 2$$

$$ = \psi {'_2}(x) = 2{\psi _1}(x) - 2$$ [from Eq. (iii)]

By LMVT,

$$\psi {'_2}(\beta ) = {{{\psi _2}(x) - {\psi _2}(0)} \over {x - 0}}$$, for $$\beta \in (0,x)$$

$$ \Rightarrow {{{\psi _2}(x) - 0} \over {x - 0}} = \psi {'_2}(\beta ) \Rightarrow {\psi _2}(x) = x\,.\,\psi {'_2}(\beta ) = 2x({\psi _1}(\beta ) - 1)$$

Hence, option (c) is correct.
3

JEE Advanced 2021 Paper 1 Online

MCQ (Single Correct Answer)
The area of the region

$$\left\{ {\matrix{ {(x,y):0 \le x \le {9 \over 4},} & {0 \le y \le 1,} & {x \ge 3y,} & {x + y \ge 2} \cr } } \right\}$$ is
A
$${{11} \over {32}}$$
B
$${{35} \over {96}}$$
C
$${{37} \over {96}}$$
D
$${{13} \over {32}}$$

Explanation



Required area = Shaded region

On solving x + y = 2 and x = 3y, we get

$$A \equiv \left( {{3 \over 2},{1 \over 2}} \right)$$

On solving y = 0 and x + y = 2, we get

B ≡ (2, 0)

On solving x = $${9 \over 4}$$ and x = 3y, we get

D ≡ $$\left( {{9 \over 4},{3 \over 4}} \right)$$

and C ≡ $$\left( {{9 \over 4},0} \right)$$

Required area = Area of ∆OCD − Area of ∆OBA

= $${1 \over 2} \times \left( {{9 \over 4} - 0} \right) \times {3 \over 4} - {1 \over 2} \times \left( {2 - 0} \right) \times {1 \over 2}$$

= $${{27} \over {32}} - {1 \over 2}$$

= $${{11} \over {32}}$$

4

JEE Advanced 2020 Paper 1 Offline

MCQ (Single Correct Answer)
Let the functions f : R $$ \to $$ R and g : R $$ \to $$ R be defined by

f(x) = ex $$-$$ 1 $$-$$ e$$-$$|x $$-$$ 1|

and g(x) = $${1 \over 2}$$(ex $$-$$ 1 + e1 $$-$$ x).

The the area of the region in the first quadrant bounded by the curves y = f(x), y = g(x) and x = 0 is
A
$$(2 - \sqrt 3 ) + {1 \over 2}(e - {e^{ - 1}})$$
B
$$(2 + \sqrt 3 ) + {1 \over 2}(e - {e^{ - 1}})$$
C
$$(2 - \sqrt 3 ) + {1 \over 2}(e + {e^{ - 1}})$$
D
$$(2 + \sqrt 3 ) + {1 \over 2}(e + {e^{ - 1}})$$

Explanation

The given functions f : R $$ \to $$ R and g : R $$ \to $$ R be defined by

$$f(x) = {e^{x - 1}} - {e^{ - |x - 1|}}$$

$$ = \left| \matrix{ {e^{x - 1}} - {e^{1 - x}},\,x\, \ge \,1 \hfill \cr 0,\,x < 1 \hfill \cr} \right.$$

and $$g(x) = {1 \over 2}({e^{x - 1}} + {e^{1 - x}})$$

For point of intersection of curves f(x) and g(x) put f(x) = g(x)

for $$x\, \ge \,1,\,{e^{x - 1}} - {e^{1 - x}} = {1 \over 2}({e^{x - 1}} + {e^{1 - x}})$$

$$ \Rightarrow {e^{x - 1}} = 3{e^{1 - x}} \Rightarrow {e^{2x}} = 3{e^2}$$

$$ \Rightarrow x = {1 \over 2}\log _e^3 + 1$$

So, required area is

$$\int_0^{1/2\log _e^3 + 1} {(g(x) - f(x))dx} $$

$$ = \int_0^{1/2\log _e^3 + 1} {g(x)dx - \int_1^{1/2\log _e^3 + 1} {f(x)dx} } $$

$$ = {1 \over 2}\int_0^{1/2\log _e^3 + 1} {({e^{x - 1}} + {e^{1 - x}})dx - } \int_1^{1/2\log _e^3 + 1} {({e^{x - 1}} - {e^{1 - x}})dx} $$



$$ = {1 \over 2}[{e^{x - 1}} - {e^{1 - x}}]_0^{{1 \over 2}\log _e^3 + 1} - [{e^{x - 1}} + {e^{1 - x}}]_1^{{1 \over 2}\log _e^3 + 1}$$

$$ = {1 \over 2}\left[ {\sqrt 3 - {1 \over {\sqrt 3 }} - {e^{ - 1}} + e} \right] - \left[ {\sqrt 3 + {1 \over {\sqrt 3 }} - 1 - 1} \right]$$

$$ = {1 \over {\sqrt 3 }} + {1 \over 2}(e - {e^{ - 1}}) - {4 \over {\sqrt 3 }} + 2$$

$$ = (2 - \sqrt 3 ) + {1 \over 2}(e - {e^{ - 1}})$$

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