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  • Let f be a differentiable function such that   <img alt="x^2 f(x)-x=4 \int_0^x t f(t) d t, f(1)=\frac{2}{3}" src="https://entrancecorner.oncodecogs.com/gif.latex?x%5E2%20f%28x%29-x%3D4%20

Let f be a differentiable function such that   x^2 f(x)-x=4 \int_0^x t f(t) d t, f(1)=\frac{2}{3}    Then 18 f (3) is equal to : 

Option: 1

180


Option: 2

150


Option: 3

210


Option: 4

160


Answers (1)

best_answer

x^2 f(x)-x=4 \int_0^x t f(t) d t,

Differentiate w.r.t. x

\begin{aligned} & x^2 f^{\prime}(x)+2 x f(x)-1=4 x f(x) \\ & \text { Let } y=f(x) \\ & \Rightarrow x^2 \frac{d y}{d x}-2 x y-1=0 \\ & \frac{d y}{d x}-\frac{2}{x} y=\frac{1}{x^2} \\ & \text { I.F. }=e^{\int \frac{-2}{x} d x}=\frac{1}{x^2} \end{aligned}

Its solution is

\begin{aligned} & \frac{y}{x^2}=\int \frac{1}{x^4} d x+C \\ & \frac{y}{x^2}=\frac{-1}{3 x^3}+C \\ & \because f(1)=\frac{2}{3} \Rightarrow y(1)=\frac{2}{3} \end{aligned}

\begin{aligned} & \Rightarrow \frac{2}{3}=-\frac{1}{3}+C \\ & \Rightarrow C=1 \\ & \because y=-\frac{1}{3 x}+x^2 \\ & f(x)=-\frac{1}{3 x}+x^2 \\ & f(3)=-\frac{1}{9}+9=\frac{80}{9} \Rightarrow 18 f(3)=160 \end{aligned}

 

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rishi.raj

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