JEE Advanced 2016 · Paper 2 · Mathematics Q45

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MathematicsMultiple correct+4 / -2Question 45 of 54

Let $f:\mathbb R\to(0,\infty)$ and $g:\mathbb R\to\mathbb R$ be twice differentiable functions such that $f''$ and $g''$ are continuous functions on $\mathbb R$ . Suppose $f'(2)=g(2)=0$ , $f''(2)\ne0$ and $g'(2)\ne0$ . If $\displaystyle\lim_{x\to2}\frac{f(x)g(x)}{f'(x)g'(x)}=1$ , then

f

has a local minimum at

x=2

f

has a local maximum at

x=2

f''(2)>f(2)

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

for at least one

x\in\mathbb R

Answer and worked solution

Official answer: A, D

L'Hôpital. Since $f'(2)=g(2)=0$ , the limit is $0/0$ :
$\lim_{x\to2}\frac{f'g+fg'}{f''g'+f'g''}=\frac{f(2)g'(2)}{f''(2)g'(2)}=\frac{f(2)}{f''(2)}=1.$

So $f''(2)=f(2)>0$ , because $f$ takes positive values.

(A) $f'(2)=0$ and $f''(2)>0$ : local minimum. True.
(B) False.
(C) $f''(2)=f(2)$ , not greater. False.
(D) $x=2$ satisfies $f(x)=f''(x)$ . True.

(A), (D)

Topics:L'Hospital's rule Maxima and minima

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JEE Advanced 2016 · Paper 2, Mathematics question 45