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) e 0 and g' (2) e 0 . If \mathop \lim \limits_x \to 2 f( x )g( x ) \over f'( x )g'( x ) = 1, then

f has a local minimum at x=2, f(x)-f''(x)=0 for at least one x \in R

f(x)-f''(x)=0 for at least one x \in R

Application of Derivatives PYQ — JEE Advanced 2016 · Shift 2 · Q28 with Solution

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Correct answer: A, D

We have twice differentiable functions $f:\mathbb R\to (0,\infty),\qquad g:\mathbb R\to \mathbb R$ with continuous $f''$ and $g''$ on $\mathbb R$ .

Also, $f'(2)=0,\qquad g(2)=0,\qquad f''(2)\neq 0,\qquad g'(2)\neq 0$ and $\lim_{x\to 2}\frac{f(x)g(x)}{f'(x)g'(x)}=1.$

We must determine which options are correct.

Since $f'(2)=0$ and $f''$ is continuous, $f(x)=f(2)+\frac{f''(2)}{2}(x-2)^2+o\big((x-2)^2\big).$ Also, $f'(x)=f''(2)(x-2)+o(x-2).$

Since $g(2)=0$ and $g'(2)\neq 0$ , $g(x)=g'(2)(x-2)+o(x-2),$ $g'(x)=g'(2)+o(1).$

Now compute:

\left(g'(2)(x-2)+o(x-2)\right).$$ So the leading term is $$f(x)g(x)=f(2)g'(2)(x-2)+o(x-2).$$ Similarly, $$f'(x)g'(x)=\left(f''(2)(x-2)+o(x-2)\right)\left(g'(2)+o(1)\right) =f''(2)g'(2)(x-2)+o(x-2).$$ Therefore, $$\lim_{x\to 2}\frac{f(x)g(x)}{f'(x)g'(x)}=\frac{f(2)g'(2)}{f''(2)g'(2)}=\frac{f(2)}{f''(2)}.$$ But this limit is given to be $1$. Hence, $$\frac{f(2)}{f''(2)}=1\quad\Rightarrow\quad f''(2)=f(2).$$ Since $f:\mathbb R\to (0,\infty)$, we have $$f(2)>0.$$ Thus, $$f''(2)=f(2)>0.$$ --- 3. **Check option A: $f$ has a local minimum at $x=2$** We know $$f'(2)=0,\qquad f''(2)>0.$$ By the second derivative test, $f$ has a local minimum at $x=2$. So **A is correct**. --- 4. **Check option B: $f$ has a local maximum at $x=2$** A local maximum would require $f''(2)<0$ (under second derivative test), but here $$f''(2)>0.$$ So $x=2$ is not a local maximum. So **B is false**. --- 5. **Check option C: $f''(2)>f(2)$** From above, $$f''(2)=f(2).$$ Hence strict inequality is false. So **C is false**. --- 6. **Check option D: $f(x)-f''(x)=0$ for at least one $x\in\mathbb R$** Define $$h(x)=f(x)-f''(x).$$ Since $f''$ is continuous and $f$ is twice differentiable, $h$ is continuous on $\mathbb R$. At $x=2$, $$h(2)=f(2)-f''(2)=0.$$ Therefore there exists at least one real number $x$ (namely $x=2$) such that $$f(x)-f''(x)=0.$$ So **D is correct**. --- 7. **Final conclusion** Correct options are: $$\boxed{A, D}$$ This matches the stored correct answer.