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JEE Advanced 2016Paper 2

Exam: JEE Advanced
Year: 2016
Paper: Paper 2
Subject: Mathematics
Chapter: Limits

JEE Advanced 2016 · Paper 2 · Mathematics · Limits

Let f: ℝ → (0, ∞) and g: ℝ → ℝ be twice differentiable functions...

Let

f: \mathbb{R} \to \left(0, \infty\right)

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 ) \ne 0

and

{g}^{′}( 2 ) \ne 0.\lim_{x \to 2} \dfrac{f( x )g( x )}{{f}^{′}( x ){g}^{′}( x )} = 1,

then

Options

A $f$ has a local minimum at $x = 2$
B $f$ has a local maximum at $x = 2$
C ${f}^{″}\left(2\right) > f\left(2\right)$
D $f\left(x\right) - {f}^{″}\left(x\right) = 0$ for at least one $x \in \mathbb{R}$

Correct answer

A. $f$ has a local minimum at $x = 2$

Step-by-step solution

Using L'Hospital Rule

( asitis\dfrac{0}{0} form )

\lim_{x \to 2} \dfrac{{f}^{′} ( x )g( x ) + f( x ){g}^{′} ( x )}{{f}^{″} ( x ){g}^{′} ( x ) + {f}^{′} ( x ){g}^{″} ( x )} = 1

⇒ \dfrac{f( 2 ){g}^{′} ( 2 )}{{f}^{″} ( 2 ){g}^{′} ( 2 )} = 1 ⇒ {f}^{″} ( 2 ) = f( 2 ) > 0

Also{f}^{′} ( 2 ) = 0\text{and}{f}^{″} ( 2 ) > 0

<br />

∴ x = 2 is local minima.

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