QJEEAdv15Mad914: JEE Advanced 2015 Mathematics - Class 12 JEEAdv15Mad914 Solution: Let be continuous functions which are twice differentiable on…

Question

Verified Solution • 2025–26 Session moderate

Let

f, g :

\left[ { - 1,2} \right] \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 $(f-3g)''$ never vanishes. Then the correct statement(s) is (are)

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

Accepted Answer

Let $F(x) = f(x) - 3g(x)$

$\therefore$ $F( - 1) = 3$ , $F(0) = 3$ and $F(2) = 3$

So, $F'(x)$ will vanish at least twice in

$( - 1,0) \cup (0,2)$ .

$\because$ $F''(x) > 0$ or $< 0$ , $\forall x \in ( - 1,0) \cup (0,2)$

Hence, $f'(x) - 3g'(x) = 0$ has exactly one solution in $( - 1,0)$ and one solution in $(0,2)$ .

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