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 $\underset{-\infty }{\overset{\infty }{\int }}g(t)d(t)$ 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)

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

$\therefore \underset{-\infty }{\overset{\infty }{\int }}g(t)dt=2\underset{0}{\overset{\infty }{\int }}g(t)dt=2\left(\underset{0}{\overset{1}{\int }}\left(1-\frac{1}{1+t^2}\right)dt+\underset{1}{\overset{\infty }{\int }}\frac{1}{1+t^2}dt\right)=2.$