JEE MAIN - Mathematics (2024 - 30th January Morning Shift - No. 1)
Let $$g: \mathbf{R} \rightarrow \mathbf{R}$$ be a non constant twice differentiable function such that $$\mathrm{g}^{\prime}\left(\frac{1}{2}\right)=\mathrm{g}^{\prime}\left(\frac{3}{2}\right)$$. If a real valued function $$f$$ is defined as $$f(x)=\frac{1}{2}[g(x)+g(2-x)]$$, then
$$f^{\prime \prime}(x)=0$$ for atleast two $$x$$ in $$(0,2)$$
$$f^{\prime}\left(\frac{3}{2}\right)+f^{\prime}\left(\frac{1}{2}\right)=1$$
$$f^{\prime \prime}(x)=0$$ for no $$x$$ in $$(0,1)$$
$$f^{\prime \prime}(x)=0$$ for exactly one $$x$$ in $$(0,1)$$
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 $$\mathrm{f}^{\prime}\left(\frac{1}{2}\right)=\frac{\mathrm{g}^{\prime}\left(\frac{1}{2}\right)-\mathrm{g}^{\prime}\left(\frac{3}{2}\right)}{2}=0, \mathrm{f}^{\prime}\left(\frac{1}{2}\right)=0$$
$$\Rightarrow \mathrm{f}^{\prime}\left(\frac{3}{2}\right)=\mathrm{f}^{\prime}\left(\frac{1}{2}\right)=0$$
$$\Rightarrow \operatorname{rootsin}\left(\frac{1}{2}, 1\right)$$ and $$\left(1, \frac{3}{2}\right)$$
$$\Rightarrow \mathrm{f}^{\prime \prime}(\mathrm{x})$$ is zero at least twice in $$\left(\frac{1}{2}, \frac{3}{2}\right)$$
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