Question

Let $y=f(x)$ be drawn with $f(0)=2$ and for each real number ' $a$ ' the line tangent to $y=f(x)$ at $(a, f(a))$, has $x$-intercept $(a-2)$. If $f(x)$ is of the form of $k\,\, e^{p x}$, then $\left(\frac{k}{p}\right)$ has the value equal to

Answer 1

We have $f(0)=2$
Now $y-f(a)=f'(a)[x-a]$
For $x$-intercept $y=0$, so
$x=a-\frac{f(a)}{f'(a)}=a-2$
$ \Rightarrow \frac{f(a)}{f'(a)}=2$
$\Rightarrow \frac{f'(a)}{f(a)}=\frac{1}{2}$
$\therefore $ On integrating both sides w.r.t. $a$, we get
$\ln f(a)=\frac{a}{2}+C$
$f(a)=C e^{a / 2}$
$f(x)=C e^{x / 2}$
$f(0)=C$
$\Rightarrow C=2$
$\therefore f(x)=2 e^{x / 2}$
Hence $k=2, p=\frac{1}{2}$
$\Rightarrow \frac{k}{p}=4$