BITSAT 2013 Solved Question Paper

The equation of the given curve is
y =

1

x−3

, x ≠ 3
The slope of the tangent to the given curve at any point (x, y) is given by

dy

dx

=

−1

(x−3)2

For tangent having slope 2, we must have
2 =

−1

(x−3)2

⇒ 2 (x−3)2 = - 1 ⇒ (x−3)2 = -

1

2

which is not possible as square of a real number cannot be negative.
Hence, there is no tangent to the given curve having slope 2.