የሒሳብ መግቢያ ፈተና፣ 2017 ዓ.ም. (2025)

Because \(x< 0\), the absolute value opens as \(|x| = -x\). The numerator becomes \(4x + 10(-x) = -6x\), so \(f(x) = \frac{-6x}{2x} = -3\). The \(x\) cancels safely because \(x\) is never zero in this problem. Option A is the trap: it comes from writing \(|x| = x\), which holds only when \(x\) is positive.

Write \(n = \lfloor x\rfloor\), the greatest integer that is not larger than \(x\). The equation becomes \(n^2 + 5n + 6 = 0\), which factors as \((n+2)(n+3) = 0\), so \(n = -2\) or \(n = -3\). Now read each case back into \(x\): \(\lfloor x\rfloor = -2\) means \(-2 \le x < -1\), and \(\lfloor x\rfloor = -3\) means \(-3 \le x < -2\). The two pieces sit side by side, so together they give \([-3,-1)\). Option A carries the same left endpoint and tempts for that reason, but \(x\) can never reach \(-1\), let alone go up to 0.

An inequality of the form \(|u| < 8\) means \(-8 < u < 8\). With \(u = 2-x\) this is \(-8 < 2-x < 8\). Subtract 2 from all three parts: \(-10 < -x < 6\). Now multiply by \(-1\), which turns both inequality signs around: \(-6 < x < 10\). Option A is exactly what you get if you forget to reverse the signs at that last step.

The two boundary curves meet where \(-x^2+4 = 2x-4\), that is \(x^2+2x-8 = 0\), so \(x = -4\) or \(x = 2\). The shaded region only exists between those two \(x\) values, so the domain of \(R\) is \([-4,2]\). An inverse relation swaps the coordinates of every point, so the range of \(R^{-1}\) is the domain of \(R\), which is \([-4,2]\) and not \([-4,4]\). That makes D the false statement. Option B looks like the weak one, but the \(y\) values really do run from \(-12\) at \(x=-4\) up to \(4\) at the top of the parabola, so B is true.

A square root sign always returns the non negative root, so \(\sqrt{u^2} = |u|\) for every real \(u\). Taking \(u = x-1\) gives \(f(x) = |x-1|\). Option B fails as soon as \(x < 1\). At \(x = 0\), for example, \(f(0) = \sqrt{(-1)^2} = 1\), while \(x-1 = -1\).

Read the power as a root: \(x^{\frac{2}{3}} = \left(\sqrt[3]{x}\right)^2\). A cube root accepts every real number, negatives included, since \(\sqrt[3]{-8} = -2\). So the domain is all of \(\mathbb{R}\). Squaring afterwards can never produce a negative result, and \(x = 0\) gives 0, so the outputs fill \([0,\infty)\). Option D is the common slip: it treats the fractional power like a plain square root and cuts away the negative inputs that a cube root handles without trouble.