Question

Assertion :The area bounded by $f\left(x\right)=\frac{1}{x^2-2x+2}$ and x-axis is $\pi$ square units. Reason: The function $f\left(x\right)=\frac{1}{x^2-2x+2}$ is continuous $\forall xϵR$ and attains maximum value at x=1, which is also the line of symmetry for f(x). Therefore the area is given by $2{\int }_1^{\infty }f\left(x\right)dx$

• 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 A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

$f\left(x\right)=\frac{1}{x^2-2x+2}=\frac{1}{{(x-1)}^2+1}$ which is define for all values of $xϵR$

So $f\left(x\right)$ is continuous $\forall xϵR$ and $f\left(x\right)$ attains its maximum value for ${(x-1)}^2=0$ i. e. $x=1$
Again $f\left(x\right)=\frac{1}{x^2-2x+2}$ is symmetrical about the line $x=1$ as it satisfies $f(1+x)=f(1-x)$
$\therefore$ Required area $=2{\int }_1^{\infty }f\left(x\right)dx$
$=2{\int }_1^{\infty }\frac{1}{{(x-1)}^2+1^2}=2{\left[{\tan }^{-1}(x-1)\right]}_1^{\infty }$
$=2{\tan }^{-1}\infty -{\tan }^{-1}0=2\times \frac{\pi }{2}=\pi$