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# If f(x) = b${{\text{e}}^{{\text{ax}}}}$+ a${{\text{e}}^{{\text{bx}}}}$, then ${{\text{f}}^{''}}$(0) =${\text{A}}{\text{. 0}} \\ {\text{B}}{\text{. 2ab}} \\ {\text{C}}{\text{. ab(a + b)}} \\ {\text{D}}{\text{. ab}} \\$  Verified
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Hint - To solve this question we differentiate the equation f(x) twice. Then we substitute 0 in place of x to determine the answer.

Given f(x) = b${{\text{e}}^{{\text{ax}}}}$+ a${{\text{e}}^{{\text{bx}}}}$
⟹f’(x) = $\dfrac{{{\text{df}}}}{{{\text{dx}}}}$ = b${{\text{e}}^{{\text{ax}}}}$$\dfrac{{\text{d}}}{{{\text{dx}}}}$(ax) +a${{\text{e}}^{{\text{bx}}}}$$\dfrac{{\text{d}}}{{{\text{dx}}}}$(bx) = b${{\text{e}}^{{\text{ax}}}}$a +a${{\text{e}}^{{\text{bx}}}}$b = ab (${{\text{e}}^{{\text{ax}}}}$+ ${{\text{e}}^{{\text{bx}}}}$)
($\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{{\text{e}}^{\text{x}}}} \right) = {{\text{e}}^{\text{x}}}$)
⟹f “(x) = $\dfrac{{{{\text{d}}^2}{\text{f}}}}{{{\text{d}}{{\text{x}}^2}}}$ = ab ( a${{\text{e}}^{{\text{ax}}}}$ + ${{\text{e}}^{{\text{bx}}}}$b)
⟹f “(0) = ab(${{\text{e}}^0}$a+${{\text{e}}^0}$b) (${{\text{e}}^0}$=1)