Question

Let $f (x)$ and $g(x)$ be differentiable functions on (0, 2] such that $f"(x) - g"(x) = 0, f'(1) = 2g'(1) = 4, f(2) = 3g(2) = 9.$ Then $f (x)- g(x)$ at $x = 3/2$ is

• A. 0
• B. 2
• C. 10
• D. 5

Answer 1

Answer: (D)

5

Answer 2

$f ''(x) - g''(x) = 0$
Integrating, we get
?? $f'(x) - g'(x) = c$ ??? ....(i)
Put $x = 1$
??$f'(1) = g'(1) = c$
$\Rightarrow \:\:\: c = 2$?????$[ \therefore \:\: f'(1) = 2g' (1) = 4]$
Substituting the value of c in (i), we get
??$f'(x) - g'(x) = 2$ ??? ......(ii)
Integrating (ii), we get
??$f(x) - g(x) = 2x + c'$ .......(ii)
$\Rightarrow$ Put $x = 2$
?? $f(2) -g(2) = 4 + c'$
$\Rightarrow \:\: c' = 2$ ? ???$[ \therefore \:\: f(2) = 2g (2) = 9]$
Substituting $c'$ in (iii)
????$f(x)-g(x) = 2x + 2$
$\Rightarrow \:\: f\left(\frac{3}{2}\right) -g\left(\frac{3}{2}\right) =2 \times\frac{3}{2} + 2 = 5$