If the domain of the function f(x) = frac{cos | vedclass

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(B) For the function $f(x)$ to be defined,the following conditions must be satisfied:$1$. The argument of $\cos^{-1}$ must be in $[0, 1]$: $0 \leq \sq

Vedclass · Accepted Answer

$\frac{3}{2}$

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(B) For the function $f(x)$ to be defined,the following conditions must be satisfied:$1$. The argument of $\cos^{-1}$ must be in $[0, 1]$: $0 \leq \sqrt{x^2-x+1} \leq 1$.Squaring gives $0 \leq x^2-x+1 \leq 1$.$x^2-x+1 \leq 1 \Rightarrow x^2-x \leq 0 \Rightarrow x(x-1) \leq 0 \Rightarrow x \in [0, 1]$.$2$. The denominator must be non-zero and the argument of the square root must be positive: $\sin^{-1}(\frac{2x-1}{2}) > 0$.Since the range of $\sin^{-1}$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$,we also require $\frac{2x-1}{2} \leq 1$.So,$0 $0 $1 $\frac{1}{2} Taking the intersection of $x \in [0, 1]$ and $x \in (\frac{1}{2}, \frac{3}{2}]$,we get $x \in (\frac{1}{2}, 1]$.Thus,$\alpha = \frac{1}{2}$ and $\beta = 1$.Therefore,$\alpha + \beta = \frac{1}{2} + 1 = \frac{3}{2}$.

If the domain of the function $f(x) = \frac{\cos^{-1} \sqrt{x^2-x+1}}{\sqrt{\sin^{-1}(\frac{2x-1}{2})}}$ is the interval $(\alpha, \beta]$,then $\alpha + \beta$ is equal to:

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