UPSC NDA Math Model Paper 5

Given: f(2) = 2 and f’(2) = 2.

To find:

lim

x→2

xf(2)−2f(x)

x−2

=?
Check the form by putting x = 2

lim

x→2

xf(2)−2f(x)

x−2

=

2(2)−2(2)

2−2

=

0

0

Apply L ' hospital's rule,

lim

x→2

xf(2)−2f(x)

x−2

=

lim

x→2

d dx (xf(2)−2f(x))

d dx (x−2)

=

lim

x→2

f(2)−2f′(x)

1

Differentiate the numerator and differentiate the denominator, we get
Now, take limit at x=2

lim

x→2

=f(2)−2f′(2)
=2−2(2)=−2
Option (3) is correct.