Question

If $f(x)=\left|[\sin x-2 \sqrt{2} \cos x+2]+\frac{1}{1+t^2}\right|$ where $x, t \in R$ and $g(t)$ is the minimum value of $f(x)$ for $x \in R$, then the value of $\int\limits_{-\infty}^{\infty} g(t) d(t)$ is
[Note: $[ k ]$ denotes greatest integer less than or equal to $k$.]

• A. 1
• B. 2
• C. $\frac{\pi}{4}$
• D. $\frac{\pi}{2}$

Answer 1

Answer: (B)

$f(x)=\left|[\sin x-2 \sqrt{2} \cos x+2]+\frac{1}{1+t^2}\right|, x, t \in R$
$-1 \leq \sin x-2 \sqrt{2} \cos x+2 \leq 5$
$\therefore[\sin x-2 \sqrt{2} \cos x+2]=-1,0,1,2,3,4,5$
$g(t)=\operatorname{Min} .( f ( x ))=\operatorname{Min} .\left(\left|-1+\frac{1}{1+ t ^2}\right|, \frac{1}{1+ t ^2}, 1+\frac{1}{1+ t ^2}, \ldots .5+\frac{1}{1+ t ^2}\right)$

$\therefore \int\limits_{-\infty}^{\infty} g(t) d t=2 \int\limits_0^{\infty} g(t) d t=2\left(\int\limits_0^1\left(1-\frac{1}{1+t^2}\right) d t+\int\limits_1^{\infty} \frac{1}{1+t^2} d t\right)=2 .$