If |cos x + sin x| + |cos x - sin x| = 2 sin | vedclass

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(B) Given the equation: $|\cos x + \sin x| + |\cos x - \sin x| = 2 \sin x$.For the right side $2 \sin x$ to be non-negative,we must have $\sin x \ge 0

Vedclass · Accepted Answer

$2$

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(B) Given the equation: $|\cos x + \sin x| + |\cos x - \sin x| = 2 \sin x$.For the right side $2 \sin x$ to be non-negative,we must have $\sin x \ge 0$,which implies $x \in [0, \pi]$.Case $1$: $0 \le x \le \frac{\pi}{4}$. Here $\cos x \ge \sin x$,so $(\cos x + \sin x) + (\cos x - \sin x) = 2 \cos x = 2 \sin x$ $\Rightarrow 	an x = 1$ $\Rightarrow x = \frac{\pi}{4}$.Case $2$: $\frac{\pi}{4}  0$ and $\cos x - \sin x Case $3$: $\frac{3\pi}{4} The solution set is $x \in [\frac{\pi}{4}, \frac{3\pi}{4}]$.Since $\frac{\pi}{4} \approx 0.785$ and $\frac{3\pi}{4} \approx 2.356$,the integral values of $x$ in this interval are $1$ and $2$.The maximum integral value is $2$.

If $|\cos x + \sin x| + |\cos x - \sin x| = 2 \sin x$ for $x \in [0, 2\pi]$,then the maximum integral value of $x$ is:

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