KCET · Maths · Differentiation
If \(f(x)=b e^{a x}+a e^{b x}\), then \(f^{\prime \prime}(0)\) is equal to
- A 0
- B \(2 \mathrm{ab}\)
- C \(a b(a+b)\)
- D ab
Answer & Solution
Correct Answer
(C) \(a b(a+b)\)
Step-by-step Solution
Detailed explanation
Given,
\[
f(x)=b e^{a x}+a e^{b x}
\]
On differentiating w.r.t. \(x\), we get
\[
f^{\prime}(x)=a b e^{a x}+a b e^{b x}
\]
Again, differentiating, we get
\[
\]
\begin{aligned}
\mathrm{f}^{\prime \prime}(\mathrm{x}) &=\mathrm{a}^{2} b \mathrm{e}^{\mathrm{ax}}+\mathrm{ab}^{2} \mathrm{e}^{\mathrm{bx}} \\
\Rightarrow \quad \mathrm{f}^{\prime \prime}(0) &=\mathrm{a}^{2} \mathrm{~b}+\mathrm{ab}^{2} \\
&=\mathrm{ab}(\mathrm{a}+\mathrm{b})
\end{aligned}
$$
\[
f(x)=b e^{a x}+a e^{b x}
\]
On differentiating w.r.t. \(x\), we get
\[
f^{\prime}(x)=a b e^{a x}+a b e^{b x}
\]
Again, differentiating, we get
\[
\]
\begin{aligned}
\mathrm{f}^{\prime \prime}(\mathrm{x}) &=\mathrm{a}^{2} b \mathrm{e}^{\mathrm{ax}}+\mathrm{ab}^{2} \mathrm{e}^{\mathrm{bx}} \\
\Rightarrow \quad \mathrm{f}^{\prime \prime}(0) &=\mathrm{a}^{2} \mathrm{~b}+\mathrm{ab}^{2} \\
&=\mathrm{ab}(\mathrm{a}+\mathrm{b})
\end{aligned}
$$
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