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Let \mathrm{f \text { and } g} be twice differentiable even functions on \mathrm{(-2,2)} such that \mathrm{f\left(\frac{1}{4}\right)=0, f\left(\frac{1}{2}\right)=0, f(1)=1 \text { and } g\left(\frac{3}{4}\right)=0, g(1)=2}. Then, the minimum number of solutions of \mathrm{f(x) g^{\prime \prime}(x)+f^{\prime}(x) g^{\prime}(x)=0 \text { in }(-2,2)} is equal to __________.

Option: 1

4


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

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

So \mathrm{h(x)} is continuous and differentiable.

Now as  \mathrm{f\left(\frac{1}{4}\right)=f\left(\frac{1}{2}\right)=0} and  \mathrm{f(x)} is even function, so \mathrm{f\left ( \frac{-1}{4}\right )=f\left ( \frac{-1}{2} \right )=0} which means \mathrm{h\left(-\frac{1}{2}\right)=h\left(-\frac{1}{4}\right)=h\left(\frac{1}{4}\right)=h\left(\frac{1}{2}\right)=0}

So  \mathrm{-\frac{1}{2},-\frac{1}{4}, \frac{1}{4}, \frac{1}{2}} are 4 roots of \mathrm{h(x)}

Also \mathrm{g'(0)=0}  ( \mathrm{g(x)} is even diffrential function \mathrm{\Rightarrow h(0)=0 }

From minimum roots of  \mathrm{h(x),g(x)}  may look something like this

\mathrm{\therefore h(x)} has minimum 5 roots

-\frac{1}{2},-\frac{1}{4}, 0, \frac{1}{4}, \frac{1}{2}

So, \mathrm{h'(x)=f'g'+fg''} has at least 4 roots

Hence answer is 4

Posted by

Shailly goel

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