Question

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

Given,

$f\left(x\right)=3e^{x^2}$

$\Rightarrow f^{\prime }\left(x\right)=6e^{x^2}x$

Now,

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

$=6e^{x^2}x-2x\left(3e^{x^2}\right)+\frac{1}{3}\left(3e^{0^2}\right)-6e^{0^2}\left(0\right)$

$=6e^{x^2}x-6e^{x^2}x+1-0$

$=0+1-0$

$=1$