JEE MAIN - Mathematics (2022 - 29th June Evening Shift - No. 16)
Let f and g be twice differentiable even functions on ($$-$$2, 2) such that $$f\left( {{1 \over 4}} \right) = 0$$, $$f\left( {{1 \over 2}} \right) = 0$$, $$f(1) = 1$$ and $$g\left( {{3 \over 4}} \right) = 0$$, $$g(1) = 2$$. Then, the minimum number of solutions of $$f(x)g''(x) + f'(x)g'(x) = 0$$ in $$( - 2,2)$$ is equal to ________.
Answer
4
Explanation
Let $h(x)=f(x) \cdot g^{\prime}(x)$
As $f(x)$ is even $f\left(\frac{1}{2}\right)=\left(\frac{1}{4}\right)=0$
$\Rightarrow f\left(-\frac{1}{2}\right)=f\left(-\frac{1}{4}\right)=0$
and $g(x)$ is even $\Rightarrow g^{\prime}(x)$ is odd
and $g(1)=2$ ensures one root of $g^{\prime}(x)$ is 0 .
So, $h(x)=f(x) \cdot g^{\prime}(x)$ has minimum five zeroes
$\therefore h^{\prime}(x)=f^{\prime}(x) \cdot g^{\prime}(x)+f(x) \cdot g^{\prime \prime}(x)=0$,
has minimum 4 zeroes
As $f(x)$ is even $f\left(\frac{1}{2}\right)=\left(\frac{1}{4}\right)=0$
$\Rightarrow f\left(-\frac{1}{2}\right)=f\left(-\frac{1}{4}\right)=0$
and $g(x)$ is even $\Rightarrow g^{\prime}(x)$ is odd
and $g(1)=2$ ensures one root of $g^{\prime}(x)$ is 0 .
So, $h(x)=f(x) \cdot g^{\prime}(x)$ has minimum five zeroes
$\therefore h^{\prime}(x)=f^{\prime}(x) \cdot g^{\prime}(x)+f(x) \cdot g^{\prime \prime}(x)=0$,
has minimum 4 zeroes
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