UPSEE 2017 Solved Paper

Given,
y=−√x+2 ......(i)
And bisector of first quadrant is
y=x .......(ii)
On solving Eqs. (i) and (ii), we get
x=1,4
∴ From Eq. (i)
Points are (1,1) and (4,0)
But (4,0) not satisfy Eq. (ii)
∴ Point (1,1) is only point of intersection of curve (i) and line (ii).
Now, slope of tangent of curve (i) is

dy

dx

=

−1

2√x

∴ Slope of tangent at point (1,1) is

dy

dx

|(1,1)=

−1

2

∴ Slope of normal at point (1,1)
=

−1

dy dx |(1,1)

=

−1

− 1 2

=2
∴ Equation of the normal to the curve at point (1,1) will be
y−1=2(x−1)⇒2x−y−1=0