Q170 / 240 · KCET 2026 · Mathematics
If $\sqrt{x} \sqrt[3]{y} = (x + y)^n$ and $x\dfrac{dy}{dx} - y = 0$, then $n = $
- A$1$
- B$\dfrac{6}{5}$
- Correct answerC$\dfrac{5}{6}$
- D$\dfrac{4}{9}$
Solution
Given equation: $\sqrt{x} \sqrt[3]{y} = (x + y)^n$
Taking natural logarithm on both sides:
$\dfrac{1}{2} \ln x + \dfrac{1}{3} \ln y = n \ln(x + y)$
Differentiating with respect to $x$:
$\dfrac{1}{2x} + \dfrac{1}{3y} \dfrac{dy}{dx} = \dfrac{n}{x + y} \left(1 + \dfrac{dy}{dx}\right)$
It is given that $x \dfrac{dy}{dx} - y = 0 \Rightarrow \dfrac{dy}{dx} = \dfrac{y}{x}$
Substituting $\dfrac{dy}{dx} = \dfrac{y}{x}$ in the differentiated equation:
$\dfrac{1}{2x} + \dfrac{1}{3y} \left(\dfrac{y}{x}\right) = \dfrac{n}{x + y} \left(1 + \dfrac{y}{x}\right)$
$\dfrac{1}{2x} + \dfrac{1}{3x} = \dfrac{n}{x + y} \left(\dfrac{x + y}{x}\right)$
$\dfrac{5}{6x} = \dfrac{n}{x}$
$n = \dfrac{5}{6}$
Answer: $\dfrac{5}{6}$
Taking natural logarithm on both sides:
$\dfrac{1}{2} \ln x + \dfrac{1}{3} \ln y = n \ln(x + y)$
Differentiating with respect to $x$:
$\dfrac{1}{2x} + \dfrac{1}{3y} \dfrac{dy}{dx} = \dfrac{n}{x + y} \left(1 + \dfrac{dy}{dx}\right)$
It is given that $x \dfrac{dy}{dx} - y = 0 \Rightarrow \dfrac{dy}{dx} = \dfrac{y}{x}$
Substituting $\dfrac{dy}{dx} = \dfrac{y}{x}$ in the differentiated equation:
$\dfrac{1}{2x} + \dfrac{1}{3y} \left(\dfrac{y}{x}\right) = \dfrac{n}{x + y} \left(1 + \dfrac{y}{x}\right)$
$\dfrac{1}{2x} + \dfrac{1}{3x} = \dfrac{n}{x + y} \left(\dfrac{x + y}{x}\right)$
$\dfrac{5}{6x} = \dfrac{n}{x}$
$n = \dfrac{5}{6}$
Answer: $\dfrac{5}{6}$
Hence, the correct answer is option: 'C'.