We are given the curves $y=\int_{-\infty}^x f(t) d t$ through the point $\left(0, \frac{1}{2}\right)

Question

We are given the curves $y=\int_{-\infty}^x f(t) d t$ through the point $\left(0, \frac{1}{2}\right)$ and $y=f(x)$ where $f(x)>0$ and $f(x)$ is differentiable, $\forall x \in R$ passes through $(0,1)$. Tangents drawn to both the curves at the points with equal abscissae intersect on the same point on the X -axis. Then:

Professor PC Thomas Classes · Answer

We have the equations of the tangents to the curve $y=\int_{-\infty}^x f(t) d t$ and $y=f(x)$ at arbitrary points on them are 
$Y-\int_{-\infty}^x f(t) d t=f(x)(X-x)$  ...(i)
 And $Y-f(x)=f^{\prime}(x)(X-x)$ ...(ii) \ As Eqs. (i) and (ii) intersect at the same point on the X -axis Putting $Y=0$ and equating $x$ -coordinates, we have 
$x-\frac{f(x)}{f^{\prime}(x)}=x-\frac{\int_{-\infty}^x f(t) d t}{f(x)} \Rightarrow \frac{f(x)}{\int_{-\infty}^x f(t) d t}=\frac{f^{\prime}(x)}{f(x)} \Rightarrow \int_{-\infty}^x f(t) d t=c f(x)$ ....(iii)

As $f(0)=1 \Rightarrow \int_{-\infty}^0 f(t) d t=c 	imes 1 \Rightarrow c=\frac{1}{2}$ 
$\Rightarrow \int_{-\infty}^x f(t) d t=\frac{1}{2} f(x)$; differentiating both the sides and integrating and using boundary conditions, we get $f(x)=e^{2 x} ; y=2 e x$ is tangent to $y=e^{2 x}$. 
$	herefore \quad$ Number of solutions $=1$.
 Clearly, $f(x)$ is increasing for all $x$.
 $ 	herefore \lim _{x ightarrow \infty}\left(e^{2 x}ight)^{e^{-2 x}}=1\left(\infty^0 	ext { form }ight) $