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 [30-Jan-2024 Shift 1]

Correct Answer is : f′(x)=0 for atleast two x in (0,2)

f^{'}(x)=0 for atleast two x in (0,2)

f^{' '}(x)=0 for exactly one x in (0,1)

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

JEE Main 30-Jan-2024 Shift 1 Solved Paper

Section: Mathematics

Question : 7 of 90

Marks: +1, -0

Let g:R→R be a non constant twice differentiable such that g′(‌

)=g′(‌

). If a real valued function f is defined as f(x)=‌

[g(x)+g(2−x)], then

[30-Jan-2024 Shift 1]

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