Question

If $f$ is strictly increasing and positive function, such that $x\underset{0}{\overset{x}{\int }}(1-t)\sin \left(f\right(t\left)\right)dt=2\underset{0}{\overset{x}{\int }}t\sin \left(f\right(t\left)\right)dt,$ where $x>0.$ Then the value of $f^{\prime }\left(x\right)\cot f\left(x\right)+\frac{3}{1+x}$ in the domain of $f\left(x\right)$ is

• A. 0
• B. 0.00
• C. 0.0

Answer 1

Answer: (A), (B), (C)

Given expression is
$x\underset{0}{\overset{x}{\int }}(1-t)\sin \left(f\right(t\left)\right)dt=2\underset{0}{\overset{x}{\int }}t\sin \left(f\right(t\left)\right)dt$
Differentiating w.r.t. $x$ using Leibnitz theorem, we get
$\underset{0}{\overset{x}{\int }}(1-t)\sin \left[f\right(t\left)\right]dt+x(1-x)\sin \left[f\right(x\left)\right]=2x\sin \left[f\right(x\left)\right]$
$\Rightarrow \underset{0}{\overset{x}{\int }}(1-t)\sin \left[f\right(t\left)\right]dt=x\sin \left[f\right(x\left)\right](1+x)$
Again differentiating w.r.t. $x$, we get
$(1-x)\sin \left[f\right(x\left)\right]=(x+x^2)\cos \left[f\right(x\left)\right]f^{\prime }\left(x\right)+(2x+1)\sin \left[f\right(x\left)\right]$
$\Rightarrow -3x\sin \left[f\right(x\left)\right]=(x+x^2)\cos \left[f\right(x\left)\right]f^{\prime }\left(x\right)$
$\Rightarrow \frac{-3x}{x(1+x)}=\cot \left[f\right(x\left)\right]f^{\prime }\left(x\right)$
$\Rightarrow f^{\prime }\left(x\right)\cot f\left(x\right)+\frac{3}{1+x}=0$