If y¹/m}+y^{-1/m}=2 x, then ≤ft(x²-1) y₁{ }²=

Correct Answer is : m2y2m² y²m2y2

Correct Answer is : m2y2m² y²m2y2

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MHTCET 16 Apr 2026 Shift 1 Paper

Question : 122 of 150

Marks: +1, -0

y^{\frac{1}{m}}+y^{\frac{-1}{m}}=2 x

\left(x^{2}-1\right) y_{1}{ }^{2}=

Solution:

Concept:

Differentiate the given equation and use algebraic identities to eliminate the term

\left(y^{1/m}-y^{-1/m}\right)^2

Explanation:

Given:

y^{1/m}+y^{-1/m}=2x

Differentiate both sides with respect to

x

\frac{1}{m}y^{1/m-1}\,y_1 - \frac{1}{m}y^{-1/m-1}\,y_1 = 2.

Factor out

\frac{y_1}{m y}

\frac{y_1}{m y}\left(y^{1/m}-y^{-1/m}\right)=2.

Rearranging and squaring:

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

Now evaluate the squared difference.

Since

y^{1/m}\cdot y^{-1/m}=1

, we have:

\left(y^{1/m}-y^{-1/m}\right)^2 = \left(y^{1/m}+y^{-1/m}\right)^2 - 4.

Using

y^{1/m}+y^{-1/m}=2x

\left(y^{1/m}-y^{-1/m}\right)^2 = 4x^2 - 4 = 4(x^2-1).

Substitute into the squared equation:

y_1^2 \cdot 4(x^2-1) = 4m^2y^2.

Divide by

4

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

Answer:

Option B:

m^2y^2

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