Question

Let $f\left(x\right)=e^x\sin x$ be a continuous function in $R.$ If $f\left(x\right)=0$ has atleast one real solution, then interval of $x$ can be

• A. $(-1,1)$
• B. $(0,1)$
• C. $(1,2)$
• D. $(2,3)$

Answer 1

The correct option is A $(-1,1)$

Given : $f\left(x\right)=e^x\sin x$ is continuous in $R$.
So, $I.V.T.$ is applicable in any interval.
Now, $e^x>0\forall x\in R$
and $\sin x>0,x\in (0,\pi )$
From $I.V.T.:$ there is no solution for $f\left(x\right)=0$ in $(0,3)$
but $f(-1)=e^{-1}\sin (-1)<0,f\left(1\right)=e\cdot \sin 1>0$
$\Rightarrow f\left(x\right)=0$ has atleast one solution in $(-1,1).$