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If the three function $f(x), g(x)$ and $h(x)$ are such that
$h(x)=f(x) \cdot g(x)$ and $f^{\prime}(x) \cdot g^{\prime}(x)=c$
where $c$ is constant, then
$\frac{f^{\prime \prime}(x)}{f(x)}+\frac{g^{\prime \prime}(x)}{g(x)}+\frac{2 c}{f(x) \cdot g(x)}$
is equal to
Options:
$h(x)=f(x) \cdot g(x)$ and $f^{\prime}(x) \cdot g^{\prime}(x)=c$
where $c$ is constant, then
$\frac{f^{\prime \prime}(x)}{f(x)}+\frac{g^{\prime \prime}(x)}{g(x)}+\frac{2 c}{f(x) \cdot g(x)}$
is equal to
Solution:
2311 Upvotes
Verified Answer
The correct answer is:
$\frac{h^{\prime \prime}(x)}{h(x)}$
Given, $h(x)=f(x) \cdot g(x)$ and $f^{\prime}(x) \cdot g^{\prime}(x)=c$
Now, $h^{\prime}(x)=f^{\prime}(x) \cdot g(x)+f(x) \cdot g^{\prime}(x)$
$h^{\prime \prime}(x)=f^{\prime \prime}(x) \cdot g(x)+f^{\prime}(x) \cdot g^{\prime}(x)$
$+f^{\prime}(x) \cdot g^{\prime}(x)+f(x) \cdot g^{\prime \prime}(x)$
$h^{\prime \prime}(x)=f^{\prime \prime}(x) \cdot g(x)+f(x) \cdot g^{\prime \prime}(x)$
$+2 f^{\prime}(x) \cdot g^{\prime}(x)$
$h^{\prime \prime}(x)=f^{\prime \prime}(x) \cdot g(x)+f(x) \cdot g^{\prime \prime}(x)+2 c \quad \ldots$ (i)
Now, we find
$$
\frac{f^{\prime \prime}(x)}{f(x)}+\frac{g^{\prime \prime}(x)}{g(x)}+\frac{2 c}{f(x) \cdot g(x)}
$$
$$
\begin{aligned}
&=\frac{f^{\prime \prime}(x) \cdot g(x)+g^{\prime \prime}(x) \cdot f(x)+2 c}{f(x) \cdot g(x)} \\
&=\frac{h^{\prime \prime}(x)}{h(x)} \quad[\text { from Eq. (i)] }
\end{aligned}
$$
Now, $h^{\prime}(x)=f^{\prime}(x) \cdot g(x)+f(x) \cdot g^{\prime}(x)$
$h^{\prime \prime}(x)=f^{\prime \prime}(x) \cdot g(x)+f^{\prime}(x) \cdot g^{\prime}(x)$
$+f^{\prime}(x) \cdot g^{\prime}(x)+f(x) \cdot g^{\prime \prime}(x)$
$h^{\prime \prime}(x)=f^{\prime \prime}(x) \cdot g(x)+f(x) \cdot g^{\prime \prime}(x)$
$+2 f^{\prime}(x) \cdot g^{\prime}(x)$
$h^{\prime \prime}(x)=f^{\prime \prime}(x) \cdot g(x)+f(x) \cdot g^{\prime \prime}(x)+2 c \quad \ldots$ (i)
Now, we find
$$
\frac{f^{\prime \prime}(x)}{f(x)}+\frac{g^{\prime \prime}(x)}{g(x)}+\frac{2 c}{f(x) \cdot g(x)}
$$
$$
\begin{aligned}
&=\frac{f^{\prime \prime}(x) \cdot g(x)+g^{\prime \prime}(x) \cdot f(x)+2 c}{f(x) \cdot g(x)} \\
&=\frac{h^{\prime \prime}(x)}{h(x)} \quad[\text { from Eq. (i)] }
\end{aligned}
$$
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