If $f (x) = b e^{a x} + a e^{b x}$ , then $f^{''} (0) =$

0

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2 a b

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a b (a + b)

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a b

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Solution

The correct option is D $a b (a + b)$
Given $f = b e^{a x} + a e^{b x}$
$\frac{d f}{d x} = b e^{a x} a + a e^{b x} b = a b (e^{a x} + e^{b x})$

$\frac{d^{2} f}{d x^{2}} = a b (e^{a x} a + e^{b x} b)$
So at $x = 0,$ $\frac{d^{2} f}{d x^{2}} = a b (a + b)$

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