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Differentiation question

2022 · 29 Jun · Shift 2 · Q39
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  5. /2022 · 29 Jun · Shift 2 · Q39

Differentiation question

2022 · 29 Jun · Shift 2 · Q39

JEE MainMathematicsDifferentiationNumerical+4 / −1
Let f and g be twice differentiable even functions on (−-− 2, 2) such that f(14)=0f\left( {{1 \over 4}} \right) = 0f(41​)=0, f(12)=0f\left( {{1 \over 2}} \right) = 0f(21​)=0, f(1)=1f(1) = 1f(1)=1 and g(34)=0g\left( {{3 \over 4}} \right) = 0g(43​)=0, g(1)=2g(1) = 2g(1)=2. Then, the minimum number of solutions of f(x)g′′(x)+f′(x)g′(x)=0f(x)g''(x) + f'(x)g'(x) = 0f(x)g′′(x)+f′(x)g′(x)=0 in (−2,2)( - 2,2)(−2,2) is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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