JEE Mains · Maths · STD 12 - 5. continuity and differentiation
Let \(g: R \rightarrow R\) 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 \(f^{\prime}(x)=0\) for atleast two \(x\) in \((0,2)\)
- B \(f^{\prime \prime}(x)=0\) for exactly one \(x\) in \((0,1)\)
- C \(\mathrm{f}^{\prime}(\mathrm{x})=0\) for no \(\mathrm{x}\) in \((0,1)\)
- D \(\mathrm{f}^{\prime}\left(\frac{3}{2}\right)+\mathrm{f}^{\prime}\left(\frac{1}{2}\right)=1\)
Answer & Solution
Correct Answer
(A) \(f^{\prime}(x)=0\) for atleast two \(x\) in \((0,2)\)
Step-by-step Solution
Detailed explanation
\(f^{\prime}(x)=\frac{g^{\prime}(x)-g^{\prime}(2-x)}{2}, f^{\prime}\left(\frac{3}{2}\right)=\frac{g^{\prime}\left(\frac{3}{2}\right)-g^{\prime}\left(\frac{1}{2}\right)}{2}=0\) Also…
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