CUET UG 2025 — Mathematics Algebra

Since $\sin|{-x}| = \sin|x|$sin∣−x∣=sin∣x∣ and $\cos|{-x}| = \cos|x|$cos∣−x∣=cos∣x∣, the integrand $(\sin|x| + \cos|x|)$(sin∣x∣+cos∣x∣) is an even function.

For any even function integrated over a symmetric interval $[-a, a]$[−a,a]:

$$\int_{-a}^{a} f(x)\,dx = 2\int_{0}^{a} f(x)\,dx$$∫−aa​f(x)dx=2∫0a​f(x)dx

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$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\sin|x| + \cos|x|)\,dx = 2\int_{0}^{\frac{\pi}{2}} (\sin|x| + \cos|x|)\,dx$$∫−2π​2π​​(sin∣x∣+cos∣x∣)dx=2∫02π​​(sin∣x∣+cos∣x∣)dx

For $x \in [0,\, \pi/2]$x∈[0,π/2], $x$x is non-negative, so $|x| = x$∣x∣=x:

$$= 2\int_{0}^{\frac{\pi}{2}} (\sin x + \cos x)\,dx$$=2∫02π​​(sinx+cosx)dx

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$$= 2\Big[-\cos x + \sin x\Big]_{0}^{\frac{\pi}{2}}$$=2[−cosx+sinx]02π​​

At $x = \dfrac{\pi}{2}$x=2π​:

$$-\cos\frac{\pi}{2} + \sin\frac{\pi}{2} = -(0) + (1) = 1$$−cos2π​+sin2π​=−(0)+(1)=1

At $x = 0$x=0:

$$-\cos 0 + \sin 0 = -(1) + (0) = -1$$−cos0+sin0=−(1)+(0)=−1

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$$= 2\Big[(1) - (-1)\Big]$$=2[(1)−(−1)]

$$= 2 \times 2$$=2×2

$$= \boxed{4}$$=4​