Test Index

Definite Integration

Section: Mathematics

Question : 10 of 46

Marks: +1, -0

$\psi_{1}(x) \le 1$ , for all $x>0$
$\psi_{2}(x) \le 0$ , for all $x>0$
$f(x) \ge 1-e^{-x^{2}}-\;\frac{2}{3}x^{3}+\;\frac{2}{5}x^{5}$ , for all $x \in \left(0, \;\frac{1}{2}\right)$
$g(x) \le \;\frac{2}{3}x^{3}-\;\frac{2}{5}x^{5}+\;\frac{1}{7}x^{7}$ , for all $x \in \left(0, \;\frac{1}{2}\right)$

Solution:

\psi_{1}(x)=e^{-x}+x, x \ge 0

\psi_{1}'(x)=1-e^{-x}>0

\Rightarrow \psi_{1}(x)

is increasing

\psi_{1}(x) \ge \psi_{1}(0) \forall x \ge 0 \Rightarrow \psi_{1}(x) \ge 1

Option (a) is incorrect.

(b)

\;\; \psi_{2}(x)=x^{2}-2x+2-2e^{-x} \;\; x \ge 0

\psi_{2}'(x)=2x-2+2e^{-x}=2\psi_{1}(x)-2 \ge 0, \;\; \forall x \ge 0

\Rightarrow \psi_{2}(x)

is increasing

\Rightarrow \;\; \psi_{2}^{-}(x) \ge \psi_{2}(0) \;\; \Rightarrow \;\; \psi_{2}(x) \ge 0

So, option (b) is incorrect.

(c)

f(x)=2\int\limits_{0}^{x}(t-t^{2})e^{-t^{2}}dt \;\; x \in \left(0, \;\frac{1}{2}\right)

=\int\limits_{0}^{x}2te^{-t^{2}}dt-\int\limits_{0}^{x}2t^{2}e^{-t^{2}}dt

=-e^{-x^{2}}\big|_{0}^{x} - \;\; \int\limits_{0}^{x}2t^{2}e^{-t^{2}}dt

Let

H(x)=f(x)-1+e^{-x^{2}}+\;\frac{2}{3}x^{3}-\;\frac{2}{5}x^{5}, x \in \left(0, \;\frac{1}{2}\right)

H(0)=0

H'(x)=2(x-x^{2})e^{-x^{2}}-2xe^{-x^{2}}+2x^{2}-2x^{4}

=-2x^{2}e^{-x^{2}}+2x^{2}-2x^{4}=2x^{2}(1-x^{2}-e^{-x^{2}})

\because \;\; e^{-x} \ge 1-x \forall x \ge 0

\Rightarrow H'(x) \le 0

\Rightarrow \;\; H(x)

is decreasing

\Rightarrow \;\; 1-1(x) < 0 \forall x \in \left(0, \;\frac{1}{2}\right)

Let

P(x)=g(x)-\;\frac{2}{3}x^{3}+\;\frac{2}{5}x^{5}-\;\frac{1}{7}x^{7}, x \in \left(0, \;\frac{1}{2}\right)

P'(x)=2x^{2}e^{-x^{2}}-2x^{2}+2x^{4}-x^{6}

=2x^{2}\left(1-\;\frac{x^{2}}{1}+\;\frac{x^{4}}{2}-\;\frac{x^{6}}{3}+\dots\right)-2x^{2}+2x^{4}-x^{6}

=-\;\frac{x^{8}}{3}+\;\frac{x^{10}}{12} \dots

\Rightarrow P'(x) \le 0 \Rightarrow P(x)

is decreasing

\Rightarrow P(x) \le 0

So, option

(d)

is correct.

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