Question

If $y=f\left(x\right)$ be a function satisfying the relation $y^2-x^{2}y=x$, then which of the following may hold good for $y=f\left(x\right)$ ?

• A. $f^{\prime }\left(x\right)=\frac{1+2xf\left(x\right)}{2f\left(x\right)-x^2}$
• B. $f^{\prime }\left(x\right)=\frac{f\left(x\right)+2x{f}^2\left(x\right)}{f^2\left(x\right)+x}$
• C. $f^{\prime }\left(1\right)=1+\frac{2}{\sqrt{5}}$
• D. $f^{\prime }\left(1\right)=1-\frac{2}{\sqrt{5}}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $f^{\prime }\left(x\right)=\frac{f\left(x\right)+2x{f}^2\left(x\right)}{f^2\left(x\right)+x}$
B $f^{\prime }\left(x\right)=\frac{1+2xf\left(x\right)}{2f\left(x\right)-x^2}$
C $f^{\prime }\left(1\right)=1+\frac{2}{\sqrt{5}}$
D $f^{\prime }\left(1\right)=1-\frac{2}{\sqrt{5}}$

$y^2-x^{2}y=x$ ....(1)
$2y{y}^{\prime }-x^2{y}^{\prime }-2xy=1$
$y^{\prime }(2y-x^2)=1+2xy$
$\Rightarrow y^{\prime }=\frac{1+2xy}{2y-x^2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{1+2xf\left(x\right)}{2f\left(x\right)-x^2}$ .....(2)
Hence, option A is correct.
Multiplying an dividing numerator and denominator (2) by $f\left(x\right)$
$\Rightarrow f^{\prime }\left(x\right)=\frac{f\left(x\right)+2x{f}^2\left(x\right)}{2f^2\left(x\right)-x^{2}f\left(x\right)}$
Hence, option B is correct.
Now, $y^2-x^{2}y-x=0$
$y=\frac{x^2\pm \sqrt{x^4+4x}}{2}$
$y^{\prime }=\frac{1}{2}(2x\pm \frac{4x^3+4}{2\sqrt{x^4+4x}})$
$y^{\prime }=x\pm \frac{x^3+1}{\sqrt{x^4+4x}}$
$\Rightarrow y^{\prime }=1\pm \frac{2}{\sqrt{5}}$
Hence, option C and D are correct.