Let a differentiable function f satisfy [f(x)+_3^x (f(t))/(t) d t=x+1, x 3].

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 :

Accepted Answer

Differentiating both sides we get,

$$
\begin{aligned}
& f^{\prime}(x)+\frac{f(x)}{x}=\frac{1}{2 \sqrt{x+1}} \\
& \Rightarrow \frac{d y}{d x}+\frac{y}{x}=\frac{1}{2 \sqrt{x+1}} \\
& \Rightarrow \mathrm{IF}=x \\
& \Rightarrow y x=\frac{1}{2} \int \frac{x}{\sqrt{x+1}} d x+c \\
& \Rightarrow y x=\frac{1}{2}\left(\frac{(x+1)^{\frac{3}{2}}}{\frac{3}{2}}-2(x+1)^{\frac{1}{2}}\right)+c \\
& x y=\frac{1}{3}(x+1)^{\frac{3}{2}}-(x+1)^{\frac{1}{2}}+c \\
& f(3)=2
\end{aligned}
$$

So, $x=3, y=2$

$\Rightarrow c=\frac{16}{3}$

Now, $x=8$

$8 f(8)=\frac{27}{3}-3+\frac{16}{3}=\frac{34}{3}$

$12 f(8)=\frac{34}{3} \times \frac{12}{8}=17$

Solutions · 1