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Qormaata galuumsa Herregaa, bara 2017 A.L.I (2025)

Gaaffilee dhugaa qormaata galuumsa yunivarsiitii kutaa 12 kan Herregaa, akkuma barattoonni bara 2017 A.L.I itti fudhatanitti. Tokkoon tokkoon gaaffii deebii sirrii fi ibsa qaba.

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  • 2017 A.L.I (2025)

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Gaaffii 1

If \(x< 0\) , then, the simplest form of \(f(x) = \frac{4x + 10|x|}{2x}\) is equal to:

  1. A.3
  2. B.7
  3. C.-3
  4. D.-7
Deebii fi ibsa ilaalaa

Deebii: C -3

Maaliif

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.

Gaaffii 2

For the greatest integer function \(f(x) = |x|\) , if \(|x|^2 +5|x| + 6 = 0\) , what is the value of \(x?\)

  1. A.[-3,0)
  2. B.[0,2)
  3. C.[-3,-1)
  4. D.[-1,0)
Deebii fi ibsa ilaalaa

Deebii: C [-3,-1)

Maaliif

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.

Gaaffii 3

Which of the following intervals is the solution set of the inequality \(|2 - x|< 8?\)

  1. A.(-10,6)
  2. B.(-6,10)
  3. C.(-8,8)
  4. D.(-6,12)
Deebii fi ibsa ilaalaa

Deebii: B (-6,10)

Maaliif

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.

Gaaffii 4

Which one of the following is false about the relation \(R = \{(x,y)|x,y\in \Re ,y\leq -x^2 +4\) and \(y\geq 2x - 4?\)

  1. A.Domain of R is [-4,2]
  2. B.Range of R is [-12,4]
  3. C.p(2,1) is a point on the graph of R^{-1}
  4. D.Range of R^{-1}[-4,4]
Deebii fi ibsa ilaalaa

Deebii: D Range of R^{-1}[-4,4]

Maaliif

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.

Gaaffii 5

Which of the following is equal to \(f(x) = \sqrt{(x - 1)^2}\) , for every \(x\in \mathbb{R}\) ?

  1. A.g(x) = x + 1
  2. B.g(x) = x - 1
  3. C.g(x) = |x - 1|
  4. D.g(x) = |x| + 1
Deebii fi ibsa ilaalaa

Deebii: C g(x) = |x - 1|

Maaliif

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 \(x-1 = -1\).

Gaaffii 6

The domain and range of the function \(f(x) = 2x^{\frac{2}{3}}\) are respectively

  1. A.ℝ and ℝ
  2. B.[0,∞) and ℝ
  3. C.ℝ and [0,∞)
  4. D.[0,∞) and [0,∞)
Deebii fi ibsa ilaalaa

Deebii: C ℝ and [0,∞)

Maaliif

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.

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