A) \[k=e\left( 1-\frac{1}{a} \right)\]
B) \[k=e(1+a)\]
C) \[k=e(2-a)\]
D) The equality is not possible
Correct Answer: A
Solution :
Let \[f(x)=\log x\,\,\Rightarrow \,\,{f}'\,(x)=\frac{1}{x}\] Therefore, given function\[={f}'(a)+k{f}'(e)=1\] \[\Rightarrow \,\,\frac{1}{a}+\frac{k}{e}=1\,\,\Rightarrow \,\,k=e\,\left( \frac{a-1}{a} \right)\] Aliter : Apply L-Hospital?s rule to find both the limits.You need to login to perform this action.
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