CBSE Class 12 Mathematics sample paper 2023-24: question 28(b)

Hints

Open one hint at a time and try again before the next.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

1. Let $I={\int }_0^{\pi /4}\log (1+\tan x)dx$ ...(i). Using ${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$: $I={\int }_0^{\pi /4}\log \left(1+\tan \left(\frac{\pi }{4}-x\right)\right)dx$.

   1 mark

2. $1+\tan \left(\frac{\pi }{4}-x\right)=1+\frac{1-\tan x}{1+\tan x}=\frac{2}{1+\tan x}$, so $I={\int }_0^{\pi /4}\log 2dx-{\int }_0^{\pi /4}\log (1+\tan x)dx={\int }_0^{\pi /4}\log 2dx-I$ [by (i)].

   1 mark

3. $2I=\frac{\pi }{4}\log 2\Rightarrow I=\frac{\pi }{8}\log 2$.

   1 mark

Answer: $\frac{\pi }{8}\log 2$

Revise the chapter

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