Question

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 `g(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 ______.

पर्याय

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Answer 1

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 `g(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 4.

Explanation:

Given `f(1/4) = 0, f(1/2) = 0, f(1) = 1`, and `g(3/4)` = 0, `g(1)` = 2

Let p(x) = f(x)g'(x)

⇒ p'(x) = f'(x)g'(x) + f(x)g"(x)

∵ f(x) is on even function

∴ `f(1/4) = f(-1/4) = f(1/2) = f(-1/2)` = 0

So, f(x) = 0 has minimum 4 roots

Also, given g(x) is an even function

⇒ `g(3/4) = g(-3/4)` = 0

∴ g(x) = 0 has minimum 2 roots.

⇒ g'(x) = has minimum one root.

So, p(x) = 0 has minimum 5 roots.

⇒ p'(x) = 0 has minimum 4 roots.