Qormaatilee darban

Qormaata galuumsa Herregaa, bara 2005 A.L.I (2013)

Gaaffilee dhugaa qormaata galuumsa yunivarsiitii kutaa 12 kan Herregaa, akkuma barattoonni bara 2005 A.L.I itti fudhatanitti. Tokkoon tokkoon gaaffii deebii sirrii fi ibsa qaba.

Bilisaan shaakaluu jalqabaa Waraqaa gaaffii

Damee lamaanuu
Eenyutu fudhata
2005 A.L.I (2013)
Bara qormaataa
Gaaffilee 60 deebii waliin
Kuusaa kana keessa

Gaaffilee qormaata kanaa yaalaa

Deebii banuun dura tokkoon tokkoon gaaffii ofii keessaniin deebisaa. Ibsi gara kutaa kitaaba barnootaa sirriitti waan qajeelchuuf, yeroo gaaffii dogoggortan fuula kam irra deebi'uu akka qabdan ni beektu.

Gaaffii 1

If \(\{a_n\}\) is a sequence such that \(a_1 = 2\), and \(a_{n+1} = a_n + 4\) for all \(n \geq 1\), then \(\sum_{n=1}^{35} a_n\) is equal to :

A.2460
B.2458
C.2450
D.2442

Deebii fi ibsa ilaalaa

Deebii: C 2450

Maaliif

The rule \(a_{n+1} = a_n + 4\) means every term is 4 more than the one before it, so this is an arithmetic sequence with \(a_1 = 2\) and \(d = 4\). From the 1st term to the 35th there are 34 steps, so \(a_{35} = 2 + 34(4) = 138\). The sum of an arithmetic sequence is \(S_n = \frac{n}{2}(a_1 + a_n)\), so \(S_{35} = \frac{35}{2}(2 + 138) = 35 \times 70 = 2450\). The common slip is to count 35 steps and write \(a_{35} = 142\), which gives a total of 2520. A total that is not on the list is your signal to recount the steps.

Gaaffii 2

If \(f(x) = x^2 + 2\ln x\), then \(\lim_{h \to 0} \dfrac{f(2+h)-f(2)}{h}\) ?

Deebii fi ibsa ilaalaa

Deebii: A 5

Maaliif

That limit is the definition of the derivative of \(f\) at \(x = 2\), so find \(f'\) and put 2 into it. For \(f(x) = x^2 + 2\ln x\), the derivative is \(f'(x) = 2x + \frac{2}{x}\). At \(x = 2\) this gives \(4 + 1 = 5\). Option B, 4, is what you get if you differentiate only the \(x^2\) part and forget that \(2\ln x\) contributes \(\frac{2}{x}\).

Gaaffii 3

If \(f(x) = e^{2x} + x - 3\cos x\), then what is \(f''(x)\)?

A.\(e^{2x} + 1 - 3\sin(x)\)
B.\(e^{2x} + 1 + 3\sin(x)\)
C.\(4e^{2x} - 3\cos(x)\)
D.\(4e^{2x} + 3\cos(x)\)

Deebii fi ibsa ilaalaa

Deebii: D \(4e^{2x} + 3\cos(x)\)

Maaliif

Differentiate twice. The first derivative is \(f'(x) = 2e^{2x} + 1 + 3\sin x\): the chain rule pulls a factor of 2 out of \(e^{2x}\), and the derivative of \(-3\cos x\) is \(+3\sin x\). Differentiating again, the constant 1 disappears and \(3\sin x\) becomes \(3\cos x\), so \(f''(x) = 4e^{2x} + 3\cos x\). Option C keeps a minus sign in front of the cosine. That minus sign was already used up at the first step, when \(-3\cos x\) turned into \(+3\sin x\).

Gaaffii 4

Which one of the following intervals does \(f(x) = x^4 + 4x\) increase?

A.\((-\infty, -1]\)
B.\((-\infty, 0]\)
C.\([-1, \infty)\)
D.\((-\infty, \infty)\)

Deebii fi ibsa ilaalaa

Deebii: C \([-1, \infty)\)

Maaliif

A function increases on the interval where its derivative is not negative. Here \(f'(x) = 4x^3 + 4 = 4(x^3 + 1)\), and \(x^3 + 1 \geq 0\) exactly when \(x^3 \geq -1\), which means \(x \geq -1\). So \(f\) increases on \([-1, \infty)\). Option A is the mirror image of the answer: on \((-\infty, -1]\) the derivative is negative, so the function is falling there.

Gaaffii 5

Which one of the following is the simplest form of \(\left|3+4i\right| - \dfrac{25i}{3+4i}\)

A.5 ? 5i
B.5 + 5i
C.1 + 3i
D.1 � 3i

Deebii fi ibsa ilaalaa

Deebii: D 1 � 3i

Maaliif

Handle the two pieces separately. The modulus is \(|3+4i| = \sqrt{9+16} = 5\). For the fraction, multiply top and bottom by the conjugate \(3-4i\): \(\frac{25i(3-4i)}{(3+4i)(3-4i)} = \frac{75i - 100i^2}{25} = \frac{100 + 75i}{25} = 4 + 3i\). So the expression is \(5 - (4+3i) = 1 - 3i\). Option C, \(1 + 3i\), comes from subtracting the real part but leaving the imaginary part with the sign it had before the subtraction.

Gaaffii 6

If \(Z = \cos(\pi/10) + i\sin(\pi/10)\), then what is the value of \(Z^5\)?

A.\(\pi/2 + \pi/2\ i\)
B.\(1/2 + 1/2\ i\)
C.\(i\)
D.\(1 + i\)

Deebii fi ibsa ilaalaa

Deebii: C \(i\)

Maaliif

The number is already in the form \(\cos\theta + i\sin\theta\) with \(\theta = \frac{\pi}{10}\), so use de Moivre's rule: raising to the power 5 multiplies the angle by 5. That gives \(Z^5 = \cos\frac{5\pi}{10} + i\sin\frac{5\pi}{10} = \cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\). Since \(\cos\frac{\pi}{2} = 0\) and \(\sin\frac{\pi}{2} = 1\), the value is \(i\). Option B comes from raising the two numbers \(\cos\frac{\pi}{10}\) and \(\sin\frac{\pi}{10}\) to the power 5 one by one, instead of multiplying the angle by 5.

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Kunneen gaaffilee dhugaa qormaata galuumsa Herregaa bara 2005 A.L.I ti?

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