Q69 / 150 · MHT CET 2022 (10 Aug Shift 2) · Mathematics

If $x=t+\frac{1}{t}$ and $y=t-\frac{1}{t}$, the $\frac{\mathrm{d} y}{\mathrm{~d} x}=$

  • A
    $\frac{1-t^2}{1+t^2}$
  • Correct answer
    B
    $\frac{t^2+1}{t^2-1}$
  • C
    $\frac{1+t^2}{1-t^2}$
  • D
    $\frac{t^2-1}{t^2+1}$

Solution

$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\frac{\mathrm{d} y}{\mathrm{~d} t}}{\frac{\mathrm{d} x}{\mathrm{~d} t}}=\frac{1+\frac{1}{t^2}}{1-\frac{1}{t^2}}=\frac{t^2+1}{t^2-1}$

Hence, the correct answer is option: 'B'.

Relevant Questions