Question

Let $F\left(x\right)=f\left(x\right)g\left(x\right)h\left(x\right)$ for all real $x$, where $f\left(x\right),g\left(x\right),h\left(x\right)$ are differentiable functions. At some point $x_0$, if $F^{\prime }\left(x_0\right)=21F\left(x_0\right),f^{\prime }\left(x_0\right)=4f\left(x_0\right),g^{\prime }\left(x_0\right)=-7g\left(x_0\right)$ and
$h^{\prime }\left(x_0\right)=\lambda h\left(x_0\right)$, then $\lambda$ =

• A. $12$
• B. $-12$
• C. $24$
• D. $-24$

Answer 1

The correct option is C $24$

We have,
$F\left(x\right)=f\left(x\right)g\left(x\right)h\left(x\right)$
taking log
$\Rightarrow \log F\left(x\right)=\log f\left(x\right)+\log g\left(x\right)+\log h\left(x\right)$
Differentiating both sides w.r.t x, we get
$\frac{F^{\prime }\left(x\right)}{F\left(x\right)}=\frac{f^{\prime }\left(x\right)}{f\left(x\right)}+\frac{g^{\prime }\left(x\right)}{g\left(x\right)}+\frac{h^{\prime }\left(x\right)}{h\left(x\right)}$
$\Rightarrow \frac{F^{\prime }\left(x_0\right)}{F\left(x_0\right)}=\frac{f^{\prime }\left(x_0\right)}{f\left(x_0\right)}+\frac{g^{\prime }\left(x_0\right)}{g\left(x_0\right)}+\frac{h^{\prime }\left(x_0\right)}{h\left(x_0\right)}$
$\Rightarrow 21=4-7+\lambda$
$\Rightarrow \lambda =24$