Question

Assertion :Let $I_1={\int }_0^1{e}^{-x}{\cos }^{2}xdx$, $I_2={\int }_0^1{e}^{-x^2}{\cos }^{2}xdx$, $I_3={\int }_0^1{e}^{-x^2}dx$, $I_4={\int }_0^1{e}^{\frac{-x^2}{2}}dx$ then greatest of these integrals is $I_4$ Reason: ${\int }_a^{b}f\left(x\right)dx<{\int }_a^{b}g\left(x\right)\forall f\left(x\right)\ge g\left(x\right)$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is C Assertion is correct but Reason is incorrect

$\because$ $0<x<1$
$\Rightarrow$ $\frac{1}{2}{x}^2<x^2<x$ i.e.$(-\frac{1}{2}{x}^2>-x^2>-x)$
$\Rightarrow$ $-x<-x^2<-\frac{1}{2}{x}^2$
$\Rightarrow$ $e^{-x}<e^{-x^2}<e^{-\frac{1}{2}{x}^2}$
$\Rightarrow$ $e^{-x}<e^{-x^2}$ & $e^{-x^2}{\cos }^{2}x<e^{-x^2}<e^{-\frac{1}{2}{x}^2}$ $(\because {\cos }^{2}x\le 1)$
$\Rightarrow$ $e^{-x}{\cos }^{2}x<e^{-x^2}<{\cos }^{2}x$
$\Rightarrow$ $I_1<I_2$ (A)
Again $e^{-x^2}{\cos }^{2}x<e^{-x^2}<e^{-\frac{1}{2}{x}^2}$
$\Rightarrow$ ${\int }_0^1{e}^{-x^2}{\cos }^{2}x<{\int }_0^1{e}^{-x^2}dx<{\int }_0^1{e}^{-\frac{1}{2}{x}^2}dx$
$\Rightarrow$ $I_2<I_3<I_4$ (B)
from (A) & (B) we have
$I_1<I_2<I_3<I_4$
$\Rightarrow$ $I_4$ is greatest integral so Assertion (A) is true .
Also if $f\left(x\right)\le g\left(x\right)$ in [a, b] then ${\int }_a^{b}f\left(x\right)dx\le {\int }_a^{b}g\left(x\right)dx$
$\therefore$ Reason (R) is false.
Ans: C