Q62 / 180 · COMEDK 2023 (Memory Based) · Mathematics

The slope of the tangent to the curve, $y=x^2-x y$ at $\left(1, \frac{1}{2}\right)$ is

  • A
    $\frac{4}{3}$
  • B
    $\frac{2}{3}$
  • Correct answer
    C
    $\frac{3}{4}$
  • D
    $\frac{3}{2}$

Solution

Given curve, $y=x^2-x y$. On differentiating the equation, $y=x^2-x y$ w.r.t. $x$, we $\begin{aligned} & \text { get } \frac{d y}{d x}=2 x-\left(x \frac{d y}{d x}+y\right) \\ & \Rightarrow \quad(1+x) \frac{d y}{d x}=2 x-y \Rightarrow \frac{d y}{d x}=\frac{2 x-y}{1+x} \\ & \Rightarrow\left(\frac{d y}{d x}\right)_{(1,1 / 2)}=\frac{2(1)-\left(\frac{1}{2}\right)}{1+(1)}=\frac{3 / 2}{2}=\frac{3}{4} \end{aligned}$

Hence, the correct answer is option: 'C'.

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