For a in R - {0},if a cos x + a sin x + a = 2 | vedclass

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(A) Given the equation $a \cos x + a \sin x + a = 2K + 1$. We can rewrite the expression as $a(\cos x + \sin x) + a = 2K + 1$. Using the identity $\co

Vedclass · Accepted Answer

$\left[\frac{a - 1 - a\sqrt{2}}{2}, \frac{a - 1 + a\sqrt{2}}{2}\right]$

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(A) Given the equation $a \cos x + a \sin x + a = 2K + 1$. We can rewrite the expression as $a(\cos x + \sin x) + a = 2K + 1$. Using the identity $\cos x + \sin x = \sqrt{2} \cos(x - \frac{\pi}{4})$,we get $a[\sqrt{2} \cos(x - \frac{\pi}{4}) + 1] = 2K + 1$. Since $-1 \leq \cos(x - \frac{\pi}{4}) \leq 1$,the range of $\sqrt{2} \cos(x - \frac{\pi}{4}) + 1$ is $[1 - \sqrt{2}, 1 + \sqrt{2}]$. Multiplying by $a$ (assuming $a > 0$),we have $a(1 - \sqrt{2}) \leq 2K + 1 \leq a(1 + \sqrt{2})$. Solving for $K$: $a - a\sqrt{2} - 1 \leq 2K \leq a + a\sqrt{2} - 1$. Thus,$K \in \left[\frac{a - a\sqrt{2} - 1}{2}, \frac{a + a\sqrt{2} - 1}{2}\right]$. This matches option $A$.

For $a \in R - \{0\}$,if $a \cos x + a \sin x + a = 2K + 1$ has a solution,then $K$ lies in the interval

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