KCET 2026 Mathematics Solved Paper

Concept:Use logarithmic differentiation and the given condition to find the value of $n$.Explanation:Given: $\sqrt{x} \sqrt[3]{y}=(x+y)^{n}$ and $x \frac{dy}{dx}-y=0$.From the second equation, $x \frac{dy}{dx}=y$, so $\frac{dy}{dx}=\frac{y}{x}$.Take logarithms on both sides of $\sqrt{x} \sqrt[3]{y}=(x+y)^{n}$.Write it as $\log\left(x^{1/2} y^{1/3}\right)=\log\left((x+y)^{n}\right)$.Using logarithm laws: $\frac{1}{2}\log x+\frac{1}{3}\log y=n\log(x+y)$.Differentiate both sides with respect to $x$: $\frac{1}{2x}+\frac{1}{3y}\frac{dy}{dx}=n\frac{1+\frac{dy}{dx}}{x+y}$.Substitute $\frac{dy}{dx}=\frac{y}{x}$.Left side becomes $\frac{1}{2x}+\frac{1}{3y}\cdot\frac{y}{x}=\frac{1}{2x}+\frac{1}{3x}=\frac{5}{6x}$.Right side becomes $n\cdot\frac{1+\frac{y}{x}}{x+y}=n\cdot\frac{\frac{x+y}{x}}{x+y}=\frac{n}{x}$.Equate the two sides: $\frac{5}{6x}=\frac{n}{x}$.Hence, $n=\frac{5}{6}$.Answer:$n=\frac{5}{6}$, so the correct option is C.