JEE Main 2024 Mathematics question
Let g : ℝ → ℝ be a non constant twice differentiable function such that g′(1/2) = g′(3/2). If a real valued function f is defined as , then
Answer: (C)
Solution
f′(x) = ½[g′(x) − g′(2 − x)]. Then f′(1/2) = ½[g′(1/2) − g′(3/2)] = 0, f′(3/2) = ½[g′(3/2) − g′(1/2)] = 0
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