Question

If $f\left(x\right)$ satisfies the relation $f\left(x\right)=\cos x-\underset{0}{\overset{x}{\int }}{f}^{\prime }\left(t\right)(2\cos t+{\cos }^{2}t)dt$, then the range of $f\left(x\right)$ is

• A. $(-\infty ,-1)$
• B. $(-\infty ,1)$
• C. $(-\infty ,1]$
• D. $(-\infty ,-1]$

Answer 1

The correct option is C $(-\infty ,1]$

$f\left(x\right)=\cos x-\underset{0}{\overset{x}{\int }}{f}^{\prime }\left(t\right)(2\cos t+{\cos }^{2}t)dt$
Differentiating w.r.t. $x$, we get
$\Rightarrow f^{\prime }\left(x\right)=-\sin x-f^{\prime }\left(x\right)(2\cos x+{\cos }^{2}x)\Rightarrow f^{\prime }\left(x\right)(1+\cos x{)}^2=-\sin x\Rightarrow f^{\prime }(x)=-\frac{\sin x}{(1+\cos x{)}^2}$
Integrating w.r.t. $x$. we get
$\Rightarrow \int f^{\prime }\left(x\right)dx=-\int \frac{\sin x}{(1+\cos x{)}^2}dx$
Assuming $1+\cos x=t\Rightarrow -\sin xdx=dt$,
$\Rightarrow f\left(x\right)=\int \frac{1}{t^2}dx\Rightarrow f\left(x\right)=-\frac{1}{t}+Cf\left(x\right)=-\frac{1}{1+\cos x}+C$

Now, we know that $f\left(0\right)=1$, so
$C=\frac{3}{2}\Rightarrow f\left(x\right)=\frac{3}{2}-\frac{1}{1+\cos x}$
We know that,
$\cos x\in [-1,1]\Rightarrow 1+\cos x\in [0,2]\Rightarrow \frac{1}{1+\cos x}\in [\frac{1}{2},\infty )\therefore f\left(x\right)\in (-\infty ,1]$