JEE Advance - Mathematics (2021 - Paper 2 Online - No. 16)
Which of the following statements is TRUE?
$${\psi _1}(x) \le 1$$, for all x > 0
$${\psi _2}(x) \le 0$$, for all x > 0
$$f(x) \ge 1 - {e^{ - {x^2}}} - {2 \over 3}{x^3} + {2 \over 5}{x^5}$$, for all $$x \in \left( {0,{1 \over 2}} \right)$$
$$g(x) \le {2 \over 3}{x^3} - {2 \over 5}{x^5} + {1 \over 7}{x^7}$$, 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.
$${\psi _1}(x) = {e^{ - x}} + x$$
$$\therefore$$ $$\psi {'_1}(x) = 1 - {e^{ - x}} \Rightarrow \psi {'_1}(x) = 0$$
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.
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