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Let $$f$$ be a differentiable function such that $${x^2}f(x) - x = 4\int\limits_0^x {tf(t)dt} $$, $$f(1) = {2 \over 3}$$. Then $$18f(3)$$ is equal to :

JEE · Math · previous-year question

  1. A.160correct
  2. B.210
  3. C.150
  4. D.180

Answer

A. 160

Explanation

Given that $$ x^2 f(x)-x=4 \int_0^x t f(t) d t $$ On differentiating both sides with respect to $x$, we get $$ \begin{array}{rlrl} & 2 x f(x)+x^2 f^{\prime}(x)-1 =4 x f(x) \\\\ &\Rightarrow x^2 f^{\prime}(x)-2 x f(x)-1 =0 \\\\ &\Rightarrow x^2 \frac{d y}{d x}-2 x y =1 ~~~~~~~(Let, y=f(x) ]\\\\ &\frac{d y}{d x}-\frac{2}{x} y =\frac{1}{x^2} \end{array} $$ On comparing above equation with $\frac{d y}{d x}+P y=Q$, where $P=-\frac{2}{x}, Q=\frac{1}{x^2}$ Now, IF $=e^{\int(-2 / x) d x}=e^{-2 \log x}=1 / x^2$ Solution is $y\left(\frac{1}{x^2}\right)=\int\left(\frac{1}{x^2}\right) \frac{1}{x^2} d x$ $$ \begin{array}{ll} &\Rightarrow \frac{y}{x^2}=\int \frac{1}{x^4} d x \Rightarrow \frac{y}{x^2}=-\frac{1}{3 x^3}+C \\\\ &\Rightarrow y =-\frac{1}{3 x}+C x^2 \end{array} $$ Given, $f(1)=\frac{2}{3}$. So, $\frac{2}{3}=-\frac{1}{3}+C \Rightarrow C=1$ Thus, $y=f(x)=-\frac{1}{3 x}+x^2$ $$ \therefore 18 f(3)=18\left\{9-\frac{1}{9}\right\}=18 \times \frac{80}{9}=160 $$

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