Q109 / 160 · TS EAMCET 2020 (10 Sep Shift 2) · Mathematics
Let $f:[2,5] \rightarrow \mathbf{R}$ be a differentiatiable function and $\frac{f(5)}{f(2)}=1$. If there is a $c \in(2,5)$ such that $c f^{\prime}(c)=2 f(c)-2 c^3$, then $f(x)=$
- Correct answerA$-2 x^3+\frac{78}{7} x^2$
- B$x^3-8 x^2+17 x-10$
- C$x^3-6 x^2+3 x+10$
- D$x^3-7 x^2+10 x$
Solution
Given,
$c f^{\prime}(c)=2 f(c)-2 c^3=x f^{\prime}(x)=2 f(x)-2 x^3$
Put, $\quad f(x)=y \Rightarrow f^{\prime}(x)=\frac{d y}{d x}$
$\because \quad x \frac{d y}{d x}-2 y=-2 x^3$
$\frac{d y}{d x}-\frac{2 y}{x}=-2 x^2$
Here IF $=e^{\int \frac{-2}{x} d x}=e^{-2 \log x}=\frac{1}{x^2}$
$\therefore$ Solution of differential equation is
$\frac{y}{x^2}=\int \frac{-2 x^2}{x^2} d x+c$
$y=-2 x^3+c x^2$
$\because \quad f(x)=-2 x^3+c x^2$
Given, $\quad f(5)=f(2)$
$\because \quad-2(5)^3+c(5)^2=-2\left(2^3\right)+c(2)^2$
$-250+25 c=-16+4 c$
$21 c=234$
$c=\frac{78}{7}$
$\because \quad f(x)=-2 x^3+\frac{78}{7} x^2$
Hence, the correct answer is option: 'A'.