VITEEE 2014 Paper

It is clear that f (x) has a definite and unique value for each x ∊ [ 1,5].
Thus, for every point in the interval [1,5], the value of f (x) exists.
So, f(x) is continuous in the interval [1,5].
Also, f ' (x) =

−x

√25−x2

, , which clearly exists
for all x in an open interval (1,5).
So, f (x) is differentiable in (1,5).
So, there must be a value c ∊ [1,5] such that
f ' (c) =

f(5)−f(1)

5−1

=

−√24

4

=

0−2√6

4

= −

√6

2

But f ' (c) =

−c

√25−c2

⇒

−c

√25−c2

= -

√6

2

⇒ 4x2 = 6 (25 - x2)
⇒ 4c2 = 150 - 6c2 ⇒ 10c2 = 150
⇒ c2 = 15 ⇒ x = ± √15
∴ c = √15 ∊ [1 , 5]