JEE Main 2022 Mathematics question
Let f and g be twice differentiable even functions on (−2, 2) such that f(1/4) = 0, f(1/2) = 0, f(1) = 1 and g(3/4) = 0, g(1) = 2. Then, the minimum number of solutions of f(x)g″(x) + in (−2, 2) is equal to .
Answer: 4
Solution
Let , so + f(x)g″(x), and the required equation is . Since f is even, , giving 4 zeros of f. Since g is even,
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