हिंदी

Let f and g be twice differentiable even functions on (–2, 2) such that f(14)=0,f(12)=0,f(1)=1, and g(34) = 0, g(1) = 2. Then, the minimum number of solutions of f(x)g''(x) + f'(x)g'(x)

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प्रश्न

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 ______.

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MCQ
रिक्त स्थान भरें
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उत्तर

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.

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Differentiation of the Sum, Difference, Product, and Quotient of Two Functions
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