Question

Let a differentiable function $f$ satisfy $f(x)+\int\limits_3^x \frac{f(t)}{t} d t=\sqrt{x+1}, x \geq 3$. Then $12 f(8)$ 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} d x}=e^{\ln x}=x $
$ x f(x)=\int \frac{x}{2 \sqrt{x+1}} d x$
$ x+1=t^2$
$ =\int \frac{t^2-1}{2 t} 2 t d t $
$ x f(x)=\frac{t^3}{3}-t+c$
$ x f(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 12 f (8)=17$