Question

If $f\left(x\right)=\int \frac{2\sin 6x-5\sin 4x+10\sin 2x}{2\cos 5x-3\cos 3x+7\cos x}dx,f\left(0\right)=1,$ then which of the following option(s) is/are correct ?

• A. $f\left(x\right)+f^{\prime }\left(x\right)+1=0$
• B. $f\left(x\right)+f^{\prime \prime }\left(x\right)=3$
• C. $f_{max}=1$
• D. $f_{min}=1$

Answer 1

Answer: (B), (D)

$f_{min}=1$

Let
$I=\int \frac{2\sin 6x-5\sin 4x+10\sin 2x}{2\cos 5x-3\cos 3x+7\cos x}dx=\int \frac{(2\sin 6x-2\sin 4x)-(3\sin 4x-3\sin 2x)+7\sin 2x}{2\cos 5x-3\cos 3x+7\cos x}dx=\int \frac{4\left(\cos 5x\sin x\right)-6\cos 3x\sin x+7\left(2\sin x\cos x\right)}{2\cos 5x-3\cos 3x+7\cos x}dx=\int \frac{2\sin x(2\cos 5x-3\cos 3x+7\cos x)}{2\cos 5x-3\cos 3x+7\cos x}dx=\int 2\sin xdx=-2\cos x+C$
$\Rightarrow f\left(x\right)=-2\cos x+C$
As, $f\left(0\right)=1,C=3$
$\Rightarrow f\left(x\right)=3-2\cos x$
and $f^{\prime }\left(x\right)=2\sin x,$
$f^{\prime \prime }\left(x\right)=2\cos x\Rightarrow f\left(x\right)+f^{\prime \prime }\left(x\right)=3$
Also, $f_{max}=5,f_{min}=1$