Taylor series expansion of $f (x) = \int_{0}^{x} 3^{-} \frac{t^{2}}{2} d t 0$ around $x = 0$ has the form
$f (x) = a_{0} + a_{1} x + a_{2} x^{2} + . . .$
The coefficient $a_{2}$ (correct to two decimal places) in equal to
0

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Solution

The correct option is A 0
$f (x) = \int_{0}^{x} e^{(\frac{t^{2}}{2})} d t$
Then $f^{''} (x) = e^{\frac{x^{2}}{2}}$
and $f^{''} (x) = - x e^{\frac{x^{2}}{2}}$
Taylor series exp of f(x) about x=0 is
$f (x) = f (0) + x f^{'} (0) + x^{2} 2! f^{''} (0) + . . . . . .$
Hence on comparison ,
$a_{2} = \frac{f^{''} (0)}{2!} = \frac{0}{2!} = 0$

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