JEE Advance - Mathematics (2021 - Paper 2 Online - No. 15)
Which of the following statements is TRUE?
$$f(\sqrt {\ln 3} ) + g(\sqrt {\ln 3} ) = {1 \over 3}$$
For every x > 1, there exists an $$\alpha$$ $$\in$$ (1, x) such that $${\psi _1}(x) = 1 + \alpha x$$
For every x > 0, there exists a $$\beta$$ $$\in$$ (0, x) such that $${\psi _2}(x) = 2x({\psi _1}(\beta ) - 1)$$
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.
$${\psi _2}(x) = {x^2} - 2x - 2{e^{ - x}} + 2,x \ge 0$$
$$f(x) = \int_{ - x}^x {(\left| t \right| - {t^2}){e^{ - {t^2}}}dt,\,x > 0} $$
$$ = 2\int_0^x {(t - {t^2}){e^{ - {t^2}}}dt} $$ .... (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.
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