Q43 / 69 · JEE Advanced 2011 (Paper 1) · Mathematics
Let $f:[1, \infty) \rightarrow[2, \infty)$ be a differentiable function such that $f(1)=2$. If $6 \int_1^x f(t) d t=3 x f(x)-x^3$ for all $x \geq 1$, then the value of $f(2)$ is
Correct answer: 2.667
Solution
Given, $f(1)=\frac{1}{3}$ and $6 \int_1^x f(t) d t$ $=3 x f(x)-x^3$, for all $x \geq 1$
Using (Newton-Leibnitz formula),
On Differentiating both sides,
$$
\begin{aligned}
& 6 f(x) \cdot 1-0-3 f(x)+3 x f^{\prime}(x)-3 x^2 \\
\Rightarrow & 3 x f^{\prime}(x)-3 f(x)=3 x^2 \\
\Rightarrow & f^{\prime}(x)-\frac{1}{x} f(x)=x
\end{aligned}
$$
$$
\Rightarrow \frac{x f^{\prime}(x)-f(x)}{x^2}=1 \Rightarrow \frac{d}{d x}\left\{\frac{f(x)}{x}\right\}=1
$$
On integrating both sides,
$$
\begin{array}{ll}
& \frac{f(x)}{x}=x+C \quad\left[\because f(1)=\frac{1}{3}\right] \\
\Rightarrow \quad & \frac{1}{3}=1+C \Rightarrow \quad C=-\frac{2}{3} \\
& \text { Now, } \quad f(x)=x^2-\frac{2}{3} x \\
\Rightarrow \quad & f(2)=4-\frac{4}{3}=\frac{8}{3}
\end{array}
$$
Note Here, $f(1)=2$ does not satisfy given function.
$$
\therefore \quad f(1)=\frac{1}{3}
$$
For that, $f(x)=x^2-\frac{2}{3} x$ and $\quad f(2)=4-\frac{4}{3}=\frac{8}{3}$
Hence, the correct answer is 2.667.