JEE Advanced 2015 · Paper 2 · Mathematics Q54

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MathematicsMultiple correct+4 / -2Question 54 of 60

Let $f,g:[-1,2]\to\mathbb R$ 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)

f'(x)-3g'(x)=0

has exactly three solutions in

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

f'(x)-3g'(x)=0

has exactly one solution in

(-1,0)

f'(x)-3g'(x)=0

has exactly one solution in

(0,2)

f'(x)-3g'(x)=0

has exactly two solutions in

(-1,0)

and exactly two solutions in

(0,2)

Answer and worked solution

Official answer: B, C

The auxiliary function. Let $h=f-3g$ . Then $h(-1)=3$ , $h(0)=3$ and $h(2)=3$ .

Rolle's theorem. $h'$ has at least one zero in $(-1,0)$ and one in $(0,2)$ .

At most one. $h''$ never vanishes on either interval, so $h'$ is strictly monotonic there and each zero is unique.

(B), (C)

Topics:Rolle's and mean value theorems

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JEE Advanced 2015 · Paper 2, Mathematics question 54