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  • Let a differentiable function  satisfy  <img alt="f(x)+\int_3^x \frac{f(t)}{t} d t=\sqrt{x+1}, x \geq 3" s

Let a differentiable function f satisfy  f(x)+\int_3^x \frac{f(t)}{t} d t=\sqrt{x+1}, x \geq 3    Then 12 f  (8) is equal to : 

 

Option: 1

34


Option: 2

1 


Option: 3

17


Option: 4

19


Answers (1)

best_answer

f(x)+\int_3^1 \frac{f(t)}{t} d t=\sqrt{x+1}, x \geq 3

Differentiate both side w.r.t. x

f^{1}(x)+\frac{f(x)}{x} = \frac{1}{\sqrt{x+1},} 

Above eqn. is linear differential equation

\text { I.f. }=e^{\int \frac{1}{x} d x}=e^{\ln x}=x

Solution is

\begin{aligned} & f(x) \cdot x=\int \frac{x}{2 \sqrt{x+1}} d x+C \\ & f(x) \cdot x=\frac{1}{2} \int\left(\frac{x+1}{\sqrt{x+1}}-\frac{1}{\sqrt{x+1}}\right) d x+C \\ & f(x) \cdot x=\frac{1}{2} \int\left(\sqrt{x+1}-\frac{1}{\sqrt{x+1}}\right) d x+C \\ & f(x) \cdot x=\frac{1}{2}\left[\frac{2}{3}(x+1)^{3 / 2}-2 \sqrt{x+1}\right]+C \\ \end{aligned}

\begin{aligned} \ & \because f(3)=2 \end{aligned}

than

\begin{aligned} & 2.3=\frac{1}{2}\left[\frac{2}{3} \times 8-2 \times 2\right]+C \\ & 6=\frac{1}{2}\left[\frac{16}{3}-4\right]+C \\ & 6=\frac{2}{3}+C \\ & C=\frac{16}{3} \end{aligned}

f(x) \cdot x=\frac{1}{2}\left[\frac{2}{3}(x+1)^{3 / 2}-2 \sqrt{x+1}\right]+\frac{16}{3}\\

Put x = 8

\mathrm{f}(8) \cdot 8=\frac{1}{2}\left[\frac{2}{3} \times 27-2 \times 3\right]+\frac{16}{3}

f(8) \cdot 8=\frac{1}{2}[12]+\frac{16}{3}\\

f(8) \cdot 8=6+\frac{16}{3}=\frac{34}{3}

12 f(8)=17

 

Posted by

Pankaj Sanodiya

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