Verified Solution • 2025–26 Session moderate

QJEEM22Ma0e64: JEE Main 2022 Mathematics - Class 12 JEEM22Ma0e64 Solution: Let f and g be twice differentiable even functions on ( − 2, 2) …

Let f and g be twice differentiable even functions on (−2, 2) such that f(14)=0, f(12)=0, f(1)=1 and g(34)=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 ________.

Step-by-Step Verified Solution

Let h(x)=f(x)⋅g′(x)h(x)=f(x) \cdot g^{\prime}(x)

As f(x)f(x) is even f(12)=(14)=0f\left(\frac{1}{2}\right)=\left(\frac{1}{4}\right)=0

⇒f(−12)=f(−14)=0\Rightarrow f\left(-\frac{1}{2}\right)=f\left(-\frac{1}{4}\right)=0

and g(x)g(x) is even ⇒g′(x)\Rightarrow g^{\prime}(x) is odd

and g(1)=2g(1)=2 ensures one root of g′(x)g^{\prime}(x) is 0 .

So, h(x)=f(x)⋅g′(x)h(x)=f(x) \cdot g^{\prime}(x) has minimum five zeroes

∴h′(x)=f′(x)⋅g′(x)+f(x)⋅g′′(x)=0\therefore h^{\prime}(x)=f^{\prime}(x) \cdot g^{\prime}(x)+f(x) \cdot g^{\prime \prime}(x)=0 ,

has minimum 4 zeroes

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