Let and be twice differentiable even functions on such that , and . Then, the minimum number of solutions of in is equal to ________.

4 Explanation: Solution 1. Identify the derivative: The expression is the derivative of . Let . 2. Use Symmetry: Since and are even, and are odd. Thus, and . This implies . 3. Find roots of : . . By Rolle's theorem on between and (and their even counterparts), must have at least one root between and . Let's call it , so . 4. Count the roots of : Using the roots of at , and , there are at least 3 intervals in , giving 3 roots for . By symmetry (even/odd properties), there will be a similar count on the negative side. Result: 4

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JEE 2022 — Mathematics PYQ

JEE | Mathematics | 2022

Let $f$ and $g$ be twice differentiable even functions on $(-2, 2)$ such that $f(\frac{1}{4}) = 0, f(\frac{1}{2}) = 0, f(1) = 1$ , and $g(\frac{3}{4}) = 0, g(1) = 2$ . Then, the minimum number of solutions of $f(x)g''(x) + f'(x)g'(x) = 0$ in $(-2, 2)$ is equal to ________.

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