CBSE Class 12 Mathematics 2024: question 28(b)

Hints

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Step-by-step solution

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1. On $[1,3]$: $|x-1|=x-1$, $|x-3|=3-x$; $|x-2|=2-x$ on $[1,2]$ and $x-2$ on $[2,3]$. So $I={\int }_1^3(x-1)dx+{\int }_1^2(2-x)dx+{\int }_2^3(x-2)dx+{\int }_1^3(3-x)dx={\int }_1^{3}2dx+{\int }_1^2(2-x)dx+{\int }_2^3(x-2)dx$

   1½ marks

2. $=[2x{]}_1^3+{\left[\frac{(2-x{)}^2}{-2}\right]}_1^2+{\left[\frac{(x-2{)}^2}{2}\right]}_2^3=4+\frac{1}{2}+\frac{1}{2}=5$

   1½ marks

Answer: 5

Revise the chapter

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