CBSE Class 12 Mathematics 2023: question 24

Hints

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Step-by-step solution

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1. By the chain rule,

   $\frac{dy}{dx}=2\left(x+\sqrt{x^2-1}\right)\left(1+\frac{x}{\sqrt{x^2-1}}\right)$

   $=2\left(x+\sqrt{x^2-1}\right)\cdot \frac{\sqrt{x^2-1}+x}{\sqrt{x^2-1}}$

   $=\frac{2{\left(x+\sqrt{x^2-1}\right)}^2}{\sqrt{x^2-1}}$

   1½ marks

2. Since ${\left(x+\sqrt{x^2-1}\right)}^2=y$, we get $\sqrt{x^2-1}\frac{dy}{dx}=2y$.

   Squaring both sides: $(x^2-1){\left(\frac{dy}{dx}\right)}^2=4y^2$. Hence proved.

   ½ mark

Answer: $(x^2-1){\left(\frac{dy}{dx}\right)}^2=4y^2$ (proved)

Revise the chapter

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