Let $\mathrm{g}: \mathrm{R} \rightarrow \mathrm{R}$ be a non constant twice differentiable such that $g^{\prime}\left(\frac{1}{2}\right)=g^{\prime}\left(\frac{3}{2}\right)$. If a real valued function f is defined as $\mathrm{f}(\mathrm{x})=\frac{1}{2}[\mathrm{~g}(\mathrm{x})+\mathrm{g}(2-\mathrm{x})]$, then

(A) $\mathrm{f}^{\prime \prime}(\mathrm{x})=0$ for atleast two x in $(0,2)$

(B) $\mathrm{f}^{\prime \prime}(\mathrm{x})=0$ for exactly one x in $(0,1)$

(C) $\mathrm{f}^{\prime \prime}(\mathrm{x})=0$ for no x in $(0,1)$

(D) $\mathrm{f}^{\prime}\left(\frac{3}{2}\right)+\mathrm{f}^{\prime}\left(\frac{1}{2}\right)=1$

JEE Main · 2024 · 30 Jan 2024 · Shift 1 · Application of Derivatives

Let

g : R \to R

be a non constant twice differentiable such that

g^{'} (\frac{1}{2}) = g^{'} (\frac{3}{2})

. If a real valued function f is defined as

f (x) = \frac{1}{2} [g (x) + g (2 - x)]

, then

Answer: (A)

f^{'} (x) = \frac{1}{2} [g^{'} (x) - g^{'} (2 - x)]

, which vanishes at

x = \frac{1}{2}, 1, \frac{3}{2}

(at

x = 1

trivially, since

g^{'} (1) - g^{'} (1) = 0

; at

x = \frac{1}{2}

and

x = \frac{3}{2}

using the given

g^{'} (\frac{1}{2}) = g^{'} (\frac{3}{2})

).

Applying Rolle's theorem to

f^{'}

over

(\frac{1}{2}, 1)

and over

(1, \frac{3}{2})

shows

f^{''}

has a root in each interval — so

f^{''} (x) = 0

for at least two values of

x

(0, 2)

JEE Main · 2024 · 30 Jan 2024 · Shift 1 · Single correct option