For the equation cos ^{-1} |x| + cos ^{-1} |2 | vedclass

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(C) Given equation: $\cos ^{-1} |x| + \cos ^{-1} |2x| = \pi$.Since the range of $\cos ^{-1} y$ is $[0, \pi]$,for the sum to be $\pi$,we analyze the do

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$1$

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(C) Given equation: $\cos ^{-1} |x| + \cos ^{-1} |2x| = \pi$.Since the range of $\cos ^{-1} y$ is $[0, \pi]$,for the sum to be $\pi$,we analyze the domain: $|x| \leq 1$ and $|2x| \leq 1$,which implies $|x| \leq 1/2$.Case $1$: $x = 0$.Substituting $x = 0$: $\cos ^{-1}(0) + \cos ^{-1}(0) = \pi/2 + \pi/2 = \pi$. Thus,$x = 0$ is a solution.Case $2$: $x 
eq 0$.Let $\cos ^{-1} |x| = \alpha$ and $\cos ^{-1} |2x| = \beta$. Then $\alpha + \beta = \pi$,which implies $\beta = \pi - \alpha$.Taking cosine on both sides: $\cos(\beta) = \cos(\pi - \alpha) = -\cos(\alpha)$.Substituting back: $|2x| = -|x|$.Since $|2x| \geq 0$ and $-|x| \leq 0$,this equality holds only if $|2x| = 0$ and $|x| = 0$,which leads back to $x = 0$.Therefore,the only real solution is $x = 0$. The number of real solutions is $1$.

For the equation $\cos ^{-1} |x| + \cos ^{-1} |2x| = \pi$,the number of real solution$(s)$ is

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