Question

lf $f\left(x\right)=3e^{x^2}$, then $f^{\prime }\left(x\right)-2xf\left(x\right)+\frac{1}{3}f\left(0\right)-f^{\prime }\left(0\right)$ is equal to

• A. 0
• B. 1
• C. $\frac{7}{3}{e}^{x^2}$
• D. $e^{x^2}$

Answer 1

The correct option is B 1

$f\left(x\right)=3e^{x2}$
$\Rightarrow f^{\prime }\left(x\right)=3\left(2x\right)e^{x^2}=2xf\left(x\right)$
$f^{\prime }\left(x\right)-2xf\left(x\right)=0$
$\Rightarrow \frac{1}{3}f\left(0\right)=1$
$\Rightarrow f^{\prime }\left(0\right)=0$
$\Rightarrow f\left(x\right)-2xf\left(x\right)+\frac{1}{3}f\left(0\right)-f^{\prime }\left(0\right)=1$