A) \[0\]
B) \[2ab\]
C) \[ab(a+b)\]
D) \[ab\]
Correct Answer: C
Solution :
Given, \[f(x)=b{{e}^{ax}}+a{{e}^{bx}}\] On differentiating w.r.t. x, we get \[f'(x)=ab\,{{e}^{ax}}+ab\,{{e}^{bx}}\] Again, differentiating, we get \[f''(x)={{a}^{2}}b{{e}^{ax}}+a{{b}^{2}}\,{{e}^{bx}}\] \[\Rightarrow \] \[f''(0)={{a}^{2}}b+a{{b}^{2}}\] \[=ab(a+b)\]
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