Question

Let f and g be twice differentiable even functions on (–2, 2) such that
\(ƒ(\frac{1}{4})=0, ƒ(\frac{1}{2})=0, ƒ(1) =1\) and \(g(\frac{3}{4}) = 0 , g(1)=2\)
.Then, the minimum number of solutions of f(x)g′′(x) + f′(x)g′(x) = 0 in (–2, 2) is equal to_____.

Answer 1

Correct Answer: 4

The correct answer is 4
Suppose h(x) = f(x).g′(x)
As f(x) is even
\(ƒ(\frac{1}{2}) = (\frac{1}{4}) = 0\)
\(⇒ ƒ(-\frac{1}{2}) = ƒ(-\frac{1}{4}) = 0\)
and g(x) is even ⇒ g′(x) is odd
and g(1) = 2 ensures one root of g′(x) is 0.
So , h(x) = f(x).g′(x) has minimum fives zeroes
Therefore h′(x) = f′(x).g′(x) + f(x).g′(x)=0,
has minimum 4 zeroes