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

• A. 4
• B. 3
• C. 2
• D. 1

Answer 1

Answer: (A)

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 )= Ce ^{ a / 2} $
$f ( x )= Ce ^{ x / 2}$
$f (0)= C$ $\Rightarrow C =2$
$\therefore f ( x )=2 e ^{ x / 2}$
Hence$ k =2, p =\frac{1}{2}$