Question

If the equation $e^x-\sin x-k=0$ has atleast one real solution in $(0,\frac{\pi }{2})$, then the sum of possible integral value(s) of $k$ is
(Use $e^{\frac{\pi }{2}}=4.8$)

Answer 1

Let $f\left(x\right)=e^x-\sin x-k=0$ has atleast one solution in $(0,\frac{\pi }{2})$.
From $I.V.T.,$ we have
$f\left(0\right)\cdot f\left(\frac{\pi }{2}\right)<0$
$\Rightarrow (1-k)(e^{\frac{\pi }{2}}-1-k)<0$
$\Rightarrow k\in (1,e^{\frac{\pi }{2}}-1)$
$\Rightarrow k\in (1,3.8)$
So, possible integral values of $k$ are $2$ and $3.$