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Differential Equation f″(x)=−f(x) Problem | JEE

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Question
If f′′(x)=−f(x)f''(x) = -f(x), then [f(2)]2+[f′(2)]2[f(4)]2+[f′(4)]2\dfrac{[f(2)]^2 + [f'(2)]^2}{[f(4)]^2 + [f'(4)]^2} equals:
A22
B12\dfrac{1}{2}
C11correct
DNone of these
Solution
Let g(x)=[f(x)]2+[f′(x)]2g(x) = [f(x)]^2 + [f'(x)]^2. Differentiate:
g′(x)=2f(x)f′(x)+2f′(x)f′′(x)=2f′(x)[f(x)+f′′(x)]=2f′(x)[f(x)−f(x)]=0.g'(x) = 2f(x)f'(x) + 2f'(x)f''(x) = 2f'(x)\big[f(x) + f''(x)\big] = 2f'(x)\big[f(x) - f(x)\big] = 0.
 So g(x)g(x) is constant. Hence
[f(2)]2+[f′(2)]2[f(4)]2+[f′(4)]2=g(2)g(4)=1.\frac{[f(2)]^2 + [f'(2)]^2}{[f(4)]^2 + [f'(4)]^2} = \frac{g(2)}{g(4)} = 1.
 Correct answer: (3)
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