Question

Let a differentiable function $f$ satisfy $f(x)+\underset{3}{\overset{x}{\int }}\frac{f(t)}{t}dt=\sqrt{x+1},x\ge 3$. Then $12f(8)$ is equal to :

• A. 19
• B. 17
• C. 1
• D. 34

Answer 1

Answer: (B)

Differentiate w.r.t. x
$f^{\prime }(x)+\frac{f(x)}{x}=\frac{1}{2\sqrt{x+1}}$
$\text{ I.F. }=e^{\int \frac{1}{x}dx}=e^{\ln x}=x$
$xf(x)=\int \frac{x}{2\sqrt{x+1}}dx$
$x+1=t^2$
$=\int \frac{t^2-1}{2t}2tdt$
$xf(x)=\frac{t^3}{3}-t+c$
$xf(x)=\frac{(x+1{)}^{3/2}}{3}-\sqrt{x+1}+c$
Also putting $x=3$ in given equation $f(3)+0=\sqrt{4}$
$f(3)=2$
$\Rightarrow C=8-\frac{8}{3}=\frac{16}{3}$
$f(x)=\frac{\frac{(x+1{)}^{3/2}}{3}-\sqrt{x+1}+\frac{16}{3}}{x}$
$f(8)=\frac{9-3+\frac{16}{3}}{8}=\frac{34}{24}$
$\Rightarrow 12f(8)=17$