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 $k{e}^{px}$, then $\left(\frac{k}{p}\right)$ has the value equal to

Answer 1

Answer: 4

We have $f(0)=2$
Now $y-f(a)=f^{\prime }(a)[x-a]$
For $x$-intercept $y=0$, so
$x=a-\frac{f(a)}{f^{\prime }(a)}=a-2$
$\Rightarrow \frac{f(a)}{f^{\prime }(a)}=2$
$\Rightarrow \frac{f^{\prime }(a)}{f(a)}=\frac{1}{2}$
$\therefore$ On integrating both sides w.r.t. $a$, we get
$\ln f(a)=\frac{a}{2}+C$
$f(a)=C{e}^{a/2}$
$f(x)=C{e}^{x/2}$
$f(0)=C$
$\Rightarrow C=2$
$\therefore f(x)=2e^{x/2}$
Hence $k=2,p=\frac{1}{2}$
$\Rightarrow \frac{k}{p}=4$