JEE Advanced · 2015 · 24 May 2015 · Paper 2 · Application of Derivatives

Let

f, g : [- 1, 2] \to R

be continuous functions which are twice differentiable on the interval

(- 1, 2)

. Let the values of

f

and

g

at the points

- 1, 0

and 2 be as given in the following table:

	$x = - 1$	$x = 0$	$x = 2$
$f (x)$	3	6	0
$g (x)$	0	1	-1

In each of the intervals

(- 1, 0)

and

(0, 2)

the function

(f - 3 g)

'' never vanishes. Then the correct statement(s) is(are)

(A) $f^{'} (x) - 3 g^{'} (x) = 0$ has exactly three solutions in $(- 1, 0) \cup (0, 2)$
(B) $f^{'} (x) - 3 g^{'} (x) = 0$ has exactly one solution in $(- 1, 0)$ ✓ Correct
(C) $f^{'} (x) - 3 g^{'} (x) = 0$ has exactly one solution in $(0, 2)$ ✓ Correct
(D) $f^{'} (x) - 3 g^{'} (x) = 0$ has exactly two solutions in $(- 1, 0)$ and exactly two solutions in $(0, 2)$

Answer: (B), (C)

Solution

Define

h (x) = f (x) - 3 g (x)

. From the table:

h (- 1) = 3 - 0 = 3

h (0) = 6 - 3 = 3

h (2) = 0 - (- 3) = 3

. Since

h (- 1) = h (0) = h (2) = 3

, by Rolle's theorem,

h^{'} (x) = 0

has at least one root in

(- 1, 0)

and at least one in

(0, 2)

. Since

h^{''} (x) = (f - 3 g)^{''}

never vanishes in each interval,

h^{'}

is strictly monotonic in each interval, so

h^{'} (x) = 0

has exactly one solution in each interval. Thus (B) and (C) are correct.

JEE Advanced · 2015 · 24 May 2015 · Paper 2 · One or more correct options