Sol. The equations of the tangents to the curves y = f(x) and y = dt at arbitrary points on them are Y – f(x) = f ′ (x) (X –x) .. (i)

and, Y – dt = f(x) (X –x) ... (ii)

respectively.

It is given that (i) and (ii) intersect at the same point on X- axis. Therefore, putting y = 0 and equating X- coordinates obtained from (i) and (ii), we get

x – = x –

⇒ =

⇒ log = log f(x) + log C [On integration]

⇒ = Cf(x) ... (iii)

It is given that at x = 0, f(0) =1 and =

∴ = C × 1 ⇒ C =

Putting C = in (iii), we get

= f(x)

Differentiating both sides with respect to x, we get

f(x) = f ′ (x)

⇒ 2f(x) = f ′ (x)

⇒ = 2

⇒ log |f(x)| = 2x + log C 1

⇒ f(x) = C 1 e 2x At x = 0, we have f(0) = 1

∴ C 1 = 1 Hence, f(x) = e

2x