CBSE Class 12 Mathematics sample paper 2026-27: question 33

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Step-by-step solution

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1. Multiply the numerator and denominator inside the root by $(a-x)$: $I={\int }_{-a}^a\sqrt{\frac{(a-x{)}^2}{a^2-x^2}}dx={\int }_{-a}^a\frac{a-x}{\sqrt{a^2-x^2}}dx$ [since $a-x\ge 0$ on $[-a,a]$].

   ½ mark

2. Split the integral: $I={\int }_{-a}^a\frac{a}{\sqrt{a^2-x^2}}dx-{\int }_{-a}^a\frac{x}{\sqrt{a^2-x^2}}dx=I_1-I_2$

   1 mark

3. $\frac{a}{\sqrt{a^2-x^2}}$ is an even function, so $I_1=2a{\int }_0^a\frac{dx}{\sqrt{a^2-x^2}}$ [using ${\int }_{-a}^{a}f(x)dx=2{\int }_0^{a}f(x)dx$ when $f(-x)=f(x)$].

   1 mark

4. $I_1=2a{\left[{\sin }^{-1}\frac{x}{a}\right]}_0^a=2a\cdot \frac{\pi }{2}=\pi a$

   1 mark

5. $\frac{x}{\sqrt{a^2-x^2}}$ is an odd function, so $I_2=0$ [using ${\int }_{-a}^{a}f(x)dx=0$ when $f(-x)=-f(x)$].

   1 mark

6. $I=I_1-I_2=\pi a$

   ½ mark

Answer: $\pi a$

Revise the chapter

• Chapter: Integrals (notes, formulas, where marks are lost and every past question)
• Topic: Properties of Definite Integrals