Q92 / 150 · MHT CET 2026 (16 April Shift 1) · Mathematics

If $y^{\frac{1}{m}} + y^{\frac{-1}{m}} = 2x$ , then $(x^2 - 1)y_1^{\ 2} =$

  • A
    $my^2$
  • Correct answer
    B
    $m^2 y^2$
  • C
    $m^2 y$
  • D
    $my$

Solution

Given $y^{\dfrac{1}{m}} + y^{-\dfrac{1}{m}} = 2x$

Differentiating both sides with respect to $x$:

$\dfrac{1}{m} y^{\dfrac{1}{m} - 1} y_1 - \dfrac{1}{m} y^{-\dfrac{1}{m} - 1} y_1 = 2$

$\dfrac{y_1}{m y} \left( y^{\dfrac{1}{m}} - y^{-\dfrac{1}{m}} \right) = 2$

$y_1 \left( y^{\dfrac{1}{m}} - y^{-\dfrac{1}{m}} \right) = 2my$

Squaring both sides:

$y_1^2 \left( y^{\dfrac{1}{m}} - y^{-\dfrac{1}{m}} \right)^2 = 4m^2 y^2$

Using the identity $(a - b)^2 = (a + b)^2 - 4ab$:

$y_1^2 \left[ \left( y^{\dfrac{1}{m}} + y^{-\dfrac{1}{m}} \right)^2 - 4 \right] = 4m^2 y^2$

Substituting $y^{\dfrac{1}{m}} + y^{-\dfrac{1}{m}} = 2x$:

$y_1^2 \left[ (2x)^2 - 4 \right] = 4m^2 y^2$

$y_1^2 (4x^2 - 4) = 4m^2 y^2$

$4(x^2 - 1)y_1^2 = 4m^2 y^2$

$(x^2 - 1)y_1^2 = m^2 y^2$

Answer: $m^2 y^2$

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

Relevant Questions