Q84 / 150 · MHT CET 2026 (18 April Shift 2) · Mathematics
Let $y = \sqrt[p]{x^3 y}$. If $\dfrac{dy}{dx} = \dfrac{3}{2}$ when $y = 1$, then the value of $p$ is equal to $\ldots$
- A$1$
- B$2$
- Correct answerC$3$
- D$4$
Solution
Given $y = \sqrt[p]{x^3 y}$
Raising both sides to the power $p$:
$y^p = x^3 y$
$y^{p-1} = x^3$
Differentiating both sides with respect to $x$:
$(p-1) y^{p-2} \dfrac{dy}{dx} = 3x^2$
When $y = 1$, from $y^{p-1} = x^3$, we get $x^3 = 1 \Rightarrow x = 1$.
Substituting $x = 1$, $y = 1$, and $\dfrac{dy}{dx} = \dfrac{3}{2}$ into the differentiated equation:
$(p-1) (1)^{p-2} \left(\dfrac{3}{2}\right) = 3(1)^2$
$(p-1) \dfrac{3}{2} = 3$
$p - 1 = 2$
$p = 3$
Answer: $3$
Raising both sides to the power $p$:
$y^p = x^3 y$
$y^{p-1} = x^3$
Differentiating both sides with respect to $x$:
$(p-1) y^{p-2} \dfrac{dy}{dx} = 3x^2$
When $y = 1$, from $y^{p-1} = x^3$, we get $x^3 = 1 \Rightarrow x = 1$.
Substituting $x = 1$, $y = 1$, and $\dfrac{dy}{dx} = \dfrac{3}{2}$ into the differentiated equation:
$(p-1) (1)^{p-2} \left(\dfrac{3}{2}\right) = 3(1)^2$
$(p-1) \dfrac{3}{2} = 3$
$p - 1 = 2$
$p = 3$
Answer: $3$
Hence, the correct answer is option: 'C'.