Test Index

Definite Integration

Section: Mathematics

Question : 9 of 46

Marks: +1, -0

$f(\sqrt{\ln 3})+g(\sqrt{\ln 3})=\frac{1}{3}$
For every $x>1$ , there exists an $\alpha \in (1, x)$ such that $\psi_1(x)=1+ax$
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 $\left[0, \frac{3}{2}\right]$

Solution:

\because \;\; f(x)=\int\limits_{-x}^{x}(|t|-t^{2}) e^{-t^{2}} dt, x>0

\therefore \;\; f(x)=2 \int\limits_{0}^{x}(t-t^{2}) e^{-t^{2}} dt

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

.....(i)

Hence option (d) is wrong.

g'(x)=x e^{-x^{2}} 2x

......(ii)

Add (i) and (ii),

f'(x)+g'(x)=2 x e^{-x^{2}}

f(x)+g(x)=-e^{-x^{2}}+c

\because \;\; f(0)=g(0)=0

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

f(\sqrt{\ell n 3})+g(\sqrt{\ell n 3})=1-\frac{1}{3}=\frac{2}{3}

So, option (a) is wrong.

H(x)=\psi_1(x)-1-\alpha x=e^{-x}+-1-\alpha x

x \ge 1 \; \alpha \in (1, x)

H(1)=e^{-1}+1-1-\alpha<0

H'(x)=-e^{-x}+1-\alpha<0 \Rightarrow H(x)

is decreasing

So option (b) is wrong.

(c)

\psi_2(x)=2(\psi_1(\beta)-1)

Applying L.M.V.T to

\psi_2(x)

[0, x]

\psi_2'(\beta)=\frac{\psi_2(x)-\psi_2(0)}{x}

2\beta-2+2e^{-\beta}=\frac{\psi_2(x)-0}{x}

\Rightarrow \psi_2(x)=2x(\psi_1(\beta-1))

has one solution So, option (c) is correct.

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