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Let f(x) and g(x) be twice differentiable functions satisfying f''(x)=g''(x) for all x R, f'(1)=2g'(1)=4 and g(2)=3f(2)=9. Then f(25)-g(25) is equal to :
(2) 40
f''(x)=g''(x) f'(x)-g'(x)=c1
f'(1)=4, g'(1)=2 c1=2
f(x)-g(x)=2x+c2
g(2)=9, f(2)=3 3-9=4+c2 c2=-10
f(25)-g(25)=50-10=40
Answer: (2)
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