Given two curves: `y=f(x)` passing through the point `(0,1)` and `g(x)=int_(-oo)^xf(t)dt` passing through the point `(0,1/n)dot` The tangents drawn to both the curves at the points with equal abscissas intersect on the x-axis. Find the curve `y=f(x)dot`

Explanation

Equation of tangent to the curve `y=f(x)` is
`Y-y=f^(')(x)(X-y)`
Equation of tangents to the curve
`g(x) = y_(1)=int_(-infty)^(x) f(t)dt`
is `Y-y_(1) = f(x)(X-x)` `((dy_(1))/(dx) = g^(')(x) = f(x))`
Since, the tangents with equal abscissa intersect on the x-axis,
`x-y/(f^(')(x)) = x-y_(1)/(f(x))`
`(f(x))/(g^(')(x)) = y_(1)/(g^(')(x))`
`(g^(')(x))/(g(x)) = (g^('')(x))/(g^(')(x))`
Integrating both sides,
`"ln "g(x) = "ln "cg^(')(x)`
`g(x) = cg^(')(x)`
or `(g^(')(x))/(g(x)) = c`
Integrating both sides,
`g(x) = ke^(cx)`
`f(x)=g^(')(x) = kce^(cx)`
The curve `y=f(x)` passes through `(0,1)`. Thus, kc=1.
The curve `y=g(x)` passes through `(0,1/n)`. Thus, `k=1/n` or `c=n` or `f(x) = e^(nx)`