CBSE Class 12 Mathematics 2026: question 21(b)

Hints

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Step-by-step solution

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1. Differentiate both sides with respect to $x$:
   $\sqrt{3}\left(2x+2y\frac{dy}{dx}\right)=4\left(x\frac{dy}{dx}+y\right)$

   1 mark

2. Collect the $\frac{dy}{dx}$ terms: $\frac{dy}{dx}(2\sqrt{3}y-4x)=4y-2\sqrt{3}x$, so
   $\frac{dy}{dx}=\frac{2y-\sqrt{3}x}{\sqrt{3}y-2x}$

   ½ mark

3. At $\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$: numerator $=\sqrt{3}-\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}$, denominator $=\frac{3}{2}-1=\frac{1}{2}$.

   $\therefore \frac{dy}{dx}=\frac{\sqrt{3}/2}{1/2}=\sqrt{3}$

   ½ mark

Answer: $\frac{dy}{dx}=\sqrt{3}$

Revise the chapter

• Chapter: Continuity and Differentiability (notes, formulas, where marks are lost and every past question)
• Topic: Implicit Differentiation