Limit Continuity And Differentiability Ques 39
- Let $F(x)=f(x) g(x) h(x)$ for all real $x$, where $f(x), g(x)$ and $h(x)$ are differentiable functions. At same point $x_0, F^{\prime}\left(x_0\right)=21 F\left(x_0\right), f^{\prime}\left(x_0\right)=4 f\left(x_0\right), g^{\prime}\left(x_0\right)=-7 g\left(x_0\right)$ and $h^{\prime}\left(x_0\right)=k h\left(x_0\right)$, then $k=\ldots$.
(1997C, 2M)
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Answer:
Correct Answer: 39.$(24)$
Solution: Given, $F(x)=f(x) \cdot g(x) \cdot h(x)$
On differentiating at $x=x_0$, we get
$ \begin{aligned} F^v\left(x_0\right)=f^{\prime}\left(x_0\right) \cdot g\left(x_0\right) h\left(x_0\right)+f\left(x_0\right) \cdot g^{\prime}\left(x_0\right) h\left(x_0\right) +f\left(x_0\right) g\left(x_0\right) h^{\prime}\left(x_0\right) \end{aligned} $ $\quad$ ……..(i)
where, $\quad F^v\left(x_0\right)=21 F\left(x_0\right), f^{\prime}\left(x_0\right)=4 f\left(x_0\right)$
$ g^{\prime}\left(x_0\right)=-7 g\left(x_0\right) \text { and } h^{\prime}\left(x_0\right)=k h\left(x_0\right) $
On substituting in Eq. (i), we get
$21 F\left(x_0\right)=4 f\left(x_0\right) g\left(x_0\right) h\left(x_0\right)-7 f\left(x_0\right) g\left(x_0\right) h\left(x_0\right) $ $+k f(x) g\left(x_0\right) h\left(x_0\right) $
$\Rightarrow \quad 21=4-7+k,\left[\text { using } F\left(x_0\right)=f\left(x_0\right) g\left(x_0\right) h\left(x_0\right)\right] $
$\therefore \quad k=24$
BETA
