If A = {x in R : sin^{-1}(sqrt{x^2+x+1}) in [ | vedclass

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(C) For $A$,the domain of $\sin^{-1}(u)$ is $u \in [-1, 1]$. Since $\sqrt{x^2+x+1} \ge 0$,we require $0 \le x^2+x+1 \le 1$. Solving $x^2+x+1 \ge 0$: T

Vedclass · Accepted Answer

$A^{C} \cap B = [\frac{\pi}{3}, \frac{\pi}{2}]$

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(C) For $A$,the domain of $\sin^{-1}(u)$ is $u \in [-1, 1]$. Since $\sqrt{x^2+x+1} \ge 0$,we require $0 \le x^2+x+1 \le 1$. Solving $x^2+x+1 \ge 0$: The discriminant $D = 1-4 = -3  0$ for all $x \in R$. Solving $x^2+x+1 \le 1$: $x^2+x \le 0 \implies x(x+1) \le 0$,so $x \in [-1, 0]$. Thus,$A = [-1, 0]$. For $B$,we find the range of $y = \sin^{-1}(\sqrt{x^2+x+1})$ for $x \in [-1, 0]$. Let $f(x) = x^2+x+1 = (x+\frac{1}{2})^2 + \frac{3}{4}$. For $x \in [-1, 0]$,the minimum value of $f(x)$ is $\frac{3}{4}$ (at $x = -1/2$) and the maximum value is $1$ (at $x = -1$ or $x = 0$). Thus,$\sqrt{x^2+x+1} \in [\sqrt{3}/2, 1]$. Then $y = \sin^{-1}(\sqrt{x^2+x+1}) \in [\sin^{-1}(\sqrt{3}/2), \sin^{-1}(1)] = [\frac{\pi}{3}, \frac{\pi}{2}]$. So $B = [\frac{\pi}{3}, \frac{\pi}{2}]$. Checking options: $A = [-1, 0]$ and $B = [\frac{\pi}{3}, \frac{\pi}{2}]$. $A \cap B = \phi$ (empty set). $A \cap B^C = A \setminus B = [-1, 0] \setminus [\frac{\pi}{3}, \frac{\pi}{2}] = [-1, 0]$. $A^C \cap B = B \setminus A = [\frac{\pi}{3}, \frac{\pi}{2}] \setminus [-1, 0] = [\frac{\pi}{3}, \frac{\pi}{2}]$. Therefore,option $C$ is correct.

If $A = \{x \in R : \sin^{-1}(\sqrt{x^2+x+1}) \in [-\frac{\pi}{2}, \frac{\pi}{2}]\}$ and $B = \{y \in R : y = \sin^{-1}(\sqrt{x^2+x+1}), x \in A\}$,then which of the following is true?

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