UPSEE Solved Model Paper 5

The equation of given is y = e2x = x1 ... (i)
putting x = 0 in eq (i) we get y = 1
∴ The given point is P (0,1)
on differentiating eq (i) wrt we get

dy

dx

= 2e2x+2x
∴ (

dy

dx

)(0,1) = 2
Equation of tangent at P (0 , 1) to eq (1) is y - 1 = 2 (x - 0)
⇒ 2x - y + 1 = 0 (ii)
∴ Required distance = length of perpendicular from (1,1) to the line 2x - y + 1 = 0
= 2−1+1√4+1 =

2

√5