Continuity and DifferentiabilityHard
Question
f′(x) = g(x) and g′(x) = - f(x) for all real x and f(5) = 2 = f′(5) then f2(10) + g2(10) is -
Options
A.2
B.4
C.8
D.none of these
Solution
f′(x) = g(x) and g′(x) = - f(x)
Now
[f2(x) + g2(x)] = 2f(x) f′(x) + 2g(x) g′(x)
= 2f(x)g(x) - 2g(x)f(x) = 0
∴ f2(x) + g2(x) = constant
f2(5) + g2(5) = 4 + 4 = 8
∴ f2(10) + g2(10) = 8
Now
[f2(x) + g2(x)] = 2f(x) f′(x) + 2g(x) g′(x)= 2f(x)g(x) - 2g(x)f(x) = 0
∴ f2(x) + g2(x) = constant
f2(5) + g2(5) = 4 + 4 = 8
∴ f2(10) + g2(10) = 8
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