Question

If $g\left(x\right)={\int }_0^x\cos 4tdt$, then $g(x+\pi )$ equals:

• A. $\frac{g\left(x\right)}{g\left(\pi \right)}$
• B. $g\left(x\right)+g\left(\pi \right)$
• C. $g\left(x\right)-g\left(\pi \right)$
• D. $g\left(x\right).g\left(\pi \right)$

Answer 1

Answer: (B), (C)

$g\left(x\right)+g\left(\pi \right)$
$g\left(x\right)-g\left(\pi \right)$

$g\left(x\right)={\int }_0^x\cos 4tdt$
$g^{\prime }\left(x\right)=cos4x$ $\Rightarrow g\left(x\right)=\frac{sin4x}{4}+k$
Since $g\left(0\right)=0$,
$g\left(x\right)=\frac{sin4x}{4}$
$g(x+\pi )=g\left(x\right)+g\left(\pi \right)=g\left(x\right)-g\left(\pi \right)\left(sinceg\right(\pi )=0)$
Hence, options 'B' and 'C' are correct.