Let f and g be twice differentiable even functions on (-2, 2) such that f ( 1 4 ) = 0 , f ( 1 2 ) = 0 , f ( 1 ) = 1 and f ( 3 4 ) = 0 , g ( 1 ) = 2 Then, the minimum number of solutions of f ( x ) g " ( x ) + f ' ( x ) g ' ( x ) = 0 i n ( − 2 , 2 ) is equal to……….

Let f and g be twice differentiable even functions on (-2, 2) such that $f (\frac{1}{4}) = 0, f (\frac{1}{2}) = 0, f (1) = 1$ and $f (\frac{3}{4}) = 0, g (1) = 2$

Then, the minimum number of solutions of $f (x) g " (x) + f' (x) g' (x) = 0 i n (- 2, 2)$ is equal to……….

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Vishal Baghel | Contributor-Level 10

4 months ago

f (x) is an even function

$f (- \frac{1}{4}) = f (- \frac{1}{2}) = f (\frac{1}{2}) = f (\frac{1}{4}) = 0$

So, f (x) has at least four roots in (-2, 2)

$g (\frac{- 3}{4}) = g (\frac{3}{4}) = 0$

So, g (x) has at least two roots in (-2, 2)

now number of roots of f (x) $\cdot g " (x) = f' (x) \cdot g' (x) = 0$

It is same as number of roots of $\frac{d}{d x} (f (x) \cdot g' (x)) = 0$ will have atleast 4 roots in (-2, 2)

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