Question

If $f(x)$, $g(x)$ be twice differential functions on $[0,2]$ satisfying $f^{"}(x) = g^{"}(x), f^{'}(1) = 2g^{'}(1) = 4$ and $f(2) = 3g(2) = 9$, then $f(x) - g(x)$ at $x = 4$ equals.

• A. 0
• B. 10
• C. 8
• D. 2

Answer 1

Answer: (B)

10

Answer 2

We have, $f^{"}(x) = g^{"}(x).$ On integration,

we get $f^{'}(x) = g^{'}(x) + C$ ……. (i)

Putting x=1, we get

$f^{'}(1) = g(1) + C$

⇒$4 = 2 + C$

⇒ $C = 2$

$\therefore f^{'}(x) = g^{'}(x) + 2$

Integrating w.r.t.x, we get f(x) =g(x) +2x+$c_{1}$ ……. (ii)

Putting x = 2, we get.

$f(2) = g(2) + 4 + c_{1}$

⇒ $9 = 3 + 4 + c_{1}$

⇒ $c_{1} = 2$

$\therefore f(x) = g(x) + 2x + 2$

Putting x=4, we get f(4) -g(4) =10.