Question

If \(x = 4t^3+3,y = 3t^4+4\) and \(\frac{\frac{d^2x}{dy^2}}{(\frac{dx}{dy})^n}\) is constant then the value of \(n\) is

• A. \(1\)
• B. \(2\)
• C. \(6\)
• D. \(5\)

Hint:

Find dx/dy as a function of t, then differentiate again with respect to y.

Answer 1

The Correct Option is D

Step 1: Approach
Write the ratio with unknown $n$ and require the power of $t$ to vanish.

Step 2: Quantities
$\dfrac{dx}{dy}=\dfrac{12t^2}{12t^3}=t^{-1}$. Differentiating $t^{-1}$ with respect to $y$: $-t^{-2}\cdot\dfrac{dt}{dy}$ and $\dfrac{dt}{dy}=\dfrac1{12t^3}$.

Step 3: Result
$\dfrac{d^2x}{dy^2}=-\dfrac{1}{12}t^{-5}$. Dividing by $t^{-n}$ gives $-\dfrac1{12}t^{n-5}$.

Step 4: Constant
We need $n-5=0$. So $n=5$, option (D).

Final Answer:
The ratio equals -t^(n-5)/12, which is constant only for n = 5, option (D). \[ \boxed{5} \]