If f(x) = 3 e^ x^2 , then f'(x) - 2xf(x) + 1 3 f(0) - f'(0) = - Vidyadip

MCQ

If $f(x) = 3{e^{{x^2}}}$, then $f'(x) - 2xf(x) + {1 \over 3}f(0) - f'(0) = $

A
$0$
✓
$1$
C
${7 \over 3}{e^{{x^2}}}$
D
None of these

✓

Answer

Correct option: B.

$1$

b
(b) We have $f(x) = 3{e^{{x^2}}}.$

Differentiating w.r.t. $x,$ we get $f'(x) = 6x{e^{{x^2}}}$;

$\therefore f(0) = 3$ and $f'(0) = 0$

==>$f'(x) - 2xf(x) + \frac{1}{3}f(0) - f'(0)$

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

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