JEE Advanced · Mathematics · 16. Limits

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

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

Verified Solution

Answer & Solution

Correct Answer

(D) $f (x) - f^{″} (x) = 0$ for at least one $x \in R$

Step-by-step Solution

Detailed explanation

Using L'Hospital Rule

(a s i t i s \frac{0}{0} f o r m)

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

\Rightarrow \frac{f (2) g^{'} (2)}{f^{″} (2) g^{'} (2)} = 1 \Rightarrow f^{″} (2) = f (2) > 0

A l s o f^{'} (2) = 0 and f^{″} (2) > 0

∴

x = 2

is local minima.

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Let $f : R \to R$ be defined by $f (x) = \frac{x^{2} - 3 x - 6}{x^{2} + 2 x + 4}$ Then which of the following statements is(are) TRUE?

JEE Advanced 2021 Easy

Match the following.

	List – I		List – II
(A)	Let $y (x) = \cos (3 \cos^{- 1} x), x ϵ [- 1, 1], x \neq \pm \frac{\sqrt{3}}{2} .$ Then $\frac{1}{y (x)} \{(x^{2} - 1) \frac{d^{2} y (x)}{d x^{2}} + x \frac{d y (x)}{d x}\}$ equals	(P)	1
(B)	Let $A_{1}, A_{2}, \dots, A_{n} (n > 2)$ be the vertices of a regular polygon of n sides with its centre at the origin. Let $\vec{a_{k}}$ be the position vector of the point $A_{k}, k = 1, 2, \dots n .$ If $\|\sum_{k = 1}^{n - 1} (\vec{a_{k}} \times \vec{a_{k + 1}})\| = \|\sum_{k = 1}^{n - 1} (\vec{a_{k}} \cdot \vec{a_{k + 1}})\|$ , then the minimum value of n is	(Q)	2
(C)	If the normal from the point $P (h, 1)$ on the ellipse $\frac{x^{2}}{6} + \frac{y^{2}}{3} = 1$ is perpendicular to the line $x + y = 8,$ then the value of h is	(R)	8
(D)	Number of positive solutions satisfying the equation $\tan^{- 1} (\frac{1}{2 x + 1}) + \tan^{- 1} (\frac{1}{4 x + 1}) = \tan^{- 1} (\frac{2}{x^{2}})$ is	(S)	9

JEE Advanced 2014 Hard

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Let f :R→0, ∞ and g :R→R be twice differentiable functions such that f ″ and g ″ are continuous functions on R . Suppose f ′ ( 2 )=g( 2 )=0, f ″( 2 )≠0 and g ′ ( 2 ) ≠0. lim x→2 f( x ) g( x ) f ′ ( x ) g ′ ( x ) =1, then

Step-by-step Solution

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