AP DSC School Assistant Math Model Test 2

Concept:The given function $f(x)=\sqrt{\cos(\sin x)} + \sin^{-1}\left(\frac{1+x^{2}}{2x}\right)$ is defined only where both terms are defined. The square root requires $\cos(\sin x) \ge 0$, which is always true for all real $x$. The inverse sine term $\sin^{-1}(y)$ is defined only when $-1 \le y \le 1$. So we need to find $x$ satisfying $-1 \le \frac{1+x^{2}}{2x} \le 1$.Explanation:1. First term $\sqrt{\cos(\sin x)}$: For any real $x$, $\sin x$ always lies in $[-1,1]$. The cosine of any value in $[-1,1]$ is always non-negative (since $\cos \theta \ge 0$ for $|\theta| \le 1 < \pi/2$). Hence $\cos(\sin x) \ge 0$ for all $x \in \mathbb{R}$. So the first term is defined for all real numbers.2. Second term $\sin^{-1}\left(\frac{1+x^{2}}{2x}\right)$: The inverse sine function $\sin^{-1}(t)$ is defined only when $-1 \le t \le 1$. Here $t = \frac{1+x^{2}}{2x}$. So we require: $$-1 \le \frac{1+x^{2}}{2x} \le 1$$ We solve the two inequalities separately. Inequality (i): $\frac{1+x^{2}}{2x} \ge -1$ Multiply both sides by $2x$ carefully. Since $x$ appears in denominator, we consider cases. Alternatively, rewrite as: $$\frac{1+x^{2}}{2x} + 1 \ge 0 \quad \Rightarrow \quad \frac{1+x^{2}+2x}{2x} \ge 0 \quad \Rightarrow \quad \frac{(x+1)^{2}}{2x} \ge 0$$ The numerator $(x+1)^{2} \ge 0$ always, and equals $0$ only when $x=-1$. For the fraction to be $\ge 0$, the denominator $2x$ must be positive (since numerator is non-negative). So $x > 0$ gives a positive fraction; $x < 0$ gives a negative fraction (unless numerator is 0). The fraction equals $0$ only when $x=-1$ (which satisfies $\ge 0$). So inequality (i) holds for $x > 0$ or $x = -1$. But we need both inequalities simultaneously. It's easier to combine them into a single chain inequality. Start from: $$-1 \le \frac{1+x^{2}}{2x} \le 1$$ Multiply by $2x$ — but sign of $x$ matters. Instead, split into two cases: $x>0$ and $x<0$ (note $x=0$ is not allowed because denominator becomes zero). Case $x>0$: Multiply the double inequality by $2x$ (positive, no sign flip): $$-2x \le 1+x^{2} \le 2x$$ Left part: $-2x \le 1+x^{2} \Rightarrow 0 \le x^{2}+2x+1 = (x+1)^{2}$, always true (zero at $x=-1$, but $x>0$ so strict). Right part: $1+x^{2} \le 2x \Rightarrow x^{2}-2x+1 \le 0 \Rightarrow (x-1)^{2} \le 0$. This is only possible when $(x-1)^{2}=0$, i.e., $x=1$. So for $x>0$, the only solution is $x=1$. Case $x<0$: Multiply by $2x$ (negative flips inequality direction): $$-2x \ge 1+x^{2} \ge 2x$$ This is equivalent to two inequalities: $1+x^{2} \le -2x$ (from left) and $1+x^{2} \ge 2x$ (from right). The second inequality: $1+x^{2} \ge 2x \Rightarrow (x-1)^{2} \ge 0$ always true (zero at $x=1$, but $x<0$ so holds). The first inequality: $1+x^{2} \le -2x \Rightarrow x^{2}+2x+1 \le 0 \Rightarrow (x+1)^{2} \le 0$, which holds only when $(x+1)^{2}=0$, i.e., $x=-1$. So for $x<0$, the only solution is $x=-1$.3. Combining both cases, the second term is defined only for $x=-1$ and $x=1$. Thus the domain of $f(x)$ is the intersection of the domain of the first term (all real numbers) and the domain of the second term ($\{-1,1\}$). Hence the domain is $\{-1,1\}$.Answer:The function $f(x)$ is defined only for $x \in \{-1, 1\}$. Hence option A is correct.