Continuity and DifferentiabilityMedium
Question
Let $f(x)$ and $g(x)$ be twice differentiable functions satisfying $f^{''}(x) = g^{''}(x)$ for all $x \in R$, $f^{'}(1) = 2{\text{ }g}^{'}(1) = 4$ and $g(2) = 3f(2) = 9$. Then $f(25) - g(25)$ is equal to :
Options
A.20
B.40
C.-20
D.-40
Solution
Sol. [Key Concept: Integrate $f'' = g''$ twice; the given values fix the two constants, so $f(25) - g(25) = 2(25) + C_{2}$.]
$$f^{''}(x) = g^{''}(x) $$Integrate
$$f^{'}(x) = g^{'}(x) + C_{1} $$Put $x = 1$
$${f^{'}(1) = g^{'}(1) + C_{1} }{4 = 2 + C_{1} }{C_{1} = 2 }{\Rightarrow f^{'}(x) = g^{'}(x) + 2 }{f(x) = g(x) + 2x + C_{2} }$$Put $x = 2$
$${f(2) = g(2) + 4 + C_{2} }{3 = 9 + 4 + C_{2} }{C_{2} = - 10 }{f(x) = g(x) + 2x - 10 }$$Put $x = 25$
$$f(x) - g(25) = 40$$
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