Question

The integral $I=\int \frac{e^{\sqrt{x}} cos \left(e^{\sqrt{x}}\right)}{\sqrt{x}}dx$ $=f\left(x\right)+c$ (where, $c$ is the constant of integration) and $f\left(l n \left(\frac{\pi }{4}\right)\right)^{2}=\sqrt{2}$ . Then, the number of solutions of $f\left(x\right)=2e\left(\forall x \in R - \left\{0\right\}\right)$ is equal to

Answer 1

Given integral is $I=\int \frac{e^{\sqrt{x}} cos \left(e^{\sqrt{x}}\right)}{\sqrt{x}}dx$
Let $e^{\sqrt{x}}=t$
$\Rightarrow \frac{e^{\sqrt{x}}}{2 \sqrt{x}}dx=dt$
$\therefore I=2\int costdt$
$=2sin t+c$
$=2sin \left(e^{\sqrt{x}}\right)+c$
$\therefore f\left(x\right)=2sin\left(e^{\sqrt{x}}\right)$
which has range $=\left[- 2,2\right]$
Hence, $f\left(x\right)=2e$ has no solution