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MHTCET 20 Apr 2026 Shift 2 Paper

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Question : 123 of 150

Marks: +1, -0

If

x=4 t^{3}+3, y=3 t^{4}+4

\frac{\frac{d^{2} x}{d y^{2}}}{\left(\frac{d x}{d y}\right)^{n}}

is constant then the value of

n

is

1
2
6
5

Solution:

Concept:

For parametric equations, use

\frac{dx}{dy} = \frac{dx/dt}{dy/dt}

and then differentiate again with respect to

y

.

Explanation:

Given

x = 4t^3 + 3

and

y = 3t^4 + 4

.

First, differentiate both with respect to

t

:

\frac{dx}{dt} = 12t^2

and

\frac{dy}{dt} = 12t^3

.

Therefore,

\frac{dx}{dy} = \frac{dx/dt}{dy/dt} = \frac{12t^2}{12t^3} = \frac{1}{t}

.

Now find

\frac{d^2x}{dy^2}

:

\frac{d^2x}{dy^2} = \frac{d}{dy}\left(\frac{dx}{dy}\right) = \frac{\frac{d}{dt}\left(\frac{1}{t}\right)}{\frac{dy}{dt}}

.

Since

\frac{d}{dt}(t^{-1}) = -t^{-2}

,

\frac{d^2x}{dy^2} = \frac{-t^{-2}}{12t^3} = -\frac{1}{12t^5}

.

Substitute into the given expression:

\frac{\frac{d^2x}{dy^2}}{\left(\frac{dx}{dy}\right)^n} = \frac{-\frac{1}{12t^5}}{\left(\frac{1}{t}\right)^n} = -\frac{1}{12}t^{n-5}

.

For this to be constant, the power of

t

must be zero:

n - 5 = 0

.

Thus,

n = 5

.

Answer:

n = 5

, which is option D.

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