Question

Let $f\left(x\right)=e^x-3x^2$, then which of the following options is/are correct

• A. $I.V.T.$ is applicable in $[1,2]$
• B. $I.V.T.$ is not applicable in $[1,2]$
• C. $f\left(x\right)=0$ has no real solution in $(0,1)$
• D. $f\left(x\right)=0$ has atleast one real solution in $(0,1)$

Answer 1

Answer: (A), (D)

$f\left(x\right)=0$ has atleast one real solution in $(0,1)$

Given : $f\left(x\right)=e^x-3x^2$ which is continuous in $R$
So, $I.V.T.$ is applicable in any interval.
Now, $f\left(0\right)=1-0=1>0$ and $f\left(1\right)=e-3<0$,
$f\left(2\right)=e^2-12<0$
From $I.V.T.$
$f\left(x\right)=0$ has atleast one solution in $(0,1)$