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Mathematics · 2026 · 3 marks
CBSE 2026 · Region 1 · Set 1 · Q28
If I_1=∫_-π / 4^π / 4 (d x)/(1+ cos 2 x) and I_2=∫_-1 / 2^1 / 2|x| d x, then show that I_1-4 I_2=0.
If $\displaystyle \mathrm{I}_{1}=\int_{-\pi / 4}^{\pi / 4} \frac{\mathrm{~d} x}{1+\cos 2 x}$ and $\displaystyle \mathrm{I}_{2}=\int_{-1 / 2}^{1 / 2}|x| \mathrm{d} x$, then show that $\displaystyle \mathrm{I}_{1}-4 \mathrm{I}_{2}=0$.
Official answer
The statement is proved by evaluating I₁ = $\displaystyle 1$ and I₂ = $\displaystyle 1$/$\displaystyle 4$, giving I₁ - 4I₂ = $\displaystyle 1$ - $\displaystyle 4$($\displaystyle 1$/$\displaystyle 4$) = $\displaystyle 0$
Marking-scheme solution
$\displaystyle \mathrm{I}_1 = 2\int_0^{\pi/4}\dfrac{dx}{1+\cos 2x} \quad \left(\dfrac{1}{1+\cos 2x}\ \text{is an even function}\right)$
$\displaystyle = 2\int_0^{\pi/4}\dfrac{dx}{2\cos^2 x} = \int_0^{\pi/4}\sec^2 x\,dx$
$\displaystyle = \left[\tan x\right]_0^{\pi/4} = 1$
$\displaystyle \mathrm{I}_2 = 2\int_0^{1/2}|x|\,dx \quad \left(|x|\ \text{is an even function}\right)$
$\displaystyle = 2\int_0^{1/2} x\,dx$
$\displaystyle = \left[x^2\right]_0^{1/2} = \dfrac14$
$\displaystyle \mathrm{I}_1 - 4\mathrm{I}_2 = 1 - 4\times\dfrac14 = 0$
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CBSE Class 12 Mathematics past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.