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 (integer)
4
Solution
Let $h(x)=f(x) \cdot g^{\prime}(x)$
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As $f(x)$ is even $f\left(\frac{1}{2}\right)=\left(\frac{1}{4}\right)=0$
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$\Rightarrow f\left(-\frac{1}{2}\right)=f\left(-\frac{1}{4}\right)=0$
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and $g(x)$ is even $\Rightarrow g^{\prime}(x)$ is odd <br/><br/>
and $g(1)=2$ ensures one root of $g^{\prime}(x)$ is 0 .
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So, $h(x)=f(x) \cdot g^{\prime}(x)$ has minimum five zeroes
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$\therefore h^{\prime}(x)=f^{\prime}(x) \cdot g^{\prime}(x)+f(x) \cdot g^{\prime \prime}(x)=0$,
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has minimum 4 zeroes
About this question
Subject: Mathematics · Chapter: Differentiation · Topic: Derivatives of Standard Functions
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