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Qormaata galuumsa kutaa 12ffaa Herregaa, bara 2007 A.L.I (2015)

Gaaffilee dhugaa qormaata galuumsa yunivarsiitii kutaa 12 kan Herregaa, akkuma barattoonni bara 2007 A.L.I itti fudhatanitti. Tokkoon tokkoon gaaffii deebii sirrii fi ibsa qaba.

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  • Damee lamaanuu

    Eenyutu fudhata

  • 2007 A.L.I (2015)

    Bara qormaataa

  • Gaaffilee 62 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

Which one of the following is the equation of an ellipse with centre at (1, 4), vertices at (10, 4) and (1, 2) is :

  1. A.4(x−1)2+81(y−4)2=3244(x-1)^2 + 81(y - 4)^2 = 324
  2. B.(x−1)2+9(y−4)2=4(x-1)^2 + 9(y - 4)^2 = 4
  3. C.9(x−1)2+4(y−4)2=19(x - 1)^2 + 4(y - 4)^2 = 1
  4. D.2(x−1)2+9(y−4)2=42(x - 1)^2 + 9(y - 4)^2 = 4
Deebii fi ibsa ilaalaa

Deebii: A 4(x−1)2+81(y−4)2=3244(x-1)^2 + 81(y - 4)^2 = 324

Maaliif

The horizontal semi-axis is a=∣10−1∣=9a = |10 - 1| = 9 and the vertical semi-axis is b=∣4−2∣=2b = |4 - 2| = 2. The standard equation is (x−1)281+(y−4)24=1\frac{(x-1)^2}{81} + \frac{(y-4)^2}{4} = 1, which simplifies to 4(x−1)2+81(y−4)2=3244(x-1)^2 + 81(y-4)^2 = 324.

Gaaffii 3

Which of the following is convergent sequence?

  1. A.1, 1/2, 1, 1/3, 1, 1/4, …
  2. B.{(−1)n}n=1∞\{(-1)^n\}_{n=1}^{\infty}
  3. C.{10109−1100n}n=1∞\left\{10^{109} - \frac{1}{100^n}\right\}_{n=1}^{\infty}
  4. D.{n+sin⁡1n}n=1∞\left\{n + \sin\frac{1}{n}\right\}_{n=1}^{\infty}
Deebii fi ibsa ilaalaa

Deebii: C {10109−1100n}n=1∞\left\{10^{109} - \frac{1}{100^n}\right\}_{n=1}^{\infty}

Maaliif

As n→∞n \to \infty, 1100n→0\frac{1}{100^n} \to 0, so the sequence converges to the constant 1010910^{109}. The other sequences diverge by oscillation or tending to infinity.

Gaaffii 4

Let f(x)=2x(x2+1)4f(x) = 2x(x^2 + 1)^4, then which of the following is an anti- derivative of f(x)?

  1. A.2x5(x2+1)5+c\frac{2x}{5}(x^2 + 1)^5 + c
  2. B.25(x2+1)5+c\frac{2}{5}(x^2 + 1)^5 + c
  3. C.x5(x2+1)5+1\frac{x}{5}(x^2 + 1)^5 + 1
  4. D.15(x2+1)5−1\frac{1}{5}(x^2 + 1)^5 - 1
Deebii fi ibsa ilaalaa

Deebii: D 15(x2+1)5−1\frac{1}{5}(x^2 + 1)^5 - 1

Maaliif

Using the substitution u=x2+1u = x^2 + 1 with du=2x dxdu = 2x\,dx, we find ∫2x(x2+1)4 dx=15(x2+1)5+C\int 2x(x^2 + 1)^4\,dx = \frac{1}{5}(x^2 + 1)^5 + C. Setting the constant of integration C=−1C = -1 gives the anti-derivative in option D.

Gaaffii 5

Which one of the following is an arithmetic sequence?

  1. A.3, 5, 7, 9, 11, …
  2. B.3, 6, 12, 24, 28, …
  3. C.-3, 6, -9, 12, -15, …
  4. D.1, 3, 6, 10, 15, 21, …
Deebii fi ibsa ilaalaa

Deebii: A 3, 5, 7, 9, 11, …

Maaliif

An arithmetic sequence has a constant difference between consecutive terms; for 3, 5, 7, 9, 11, the common difference is d=2d = 2.

Gaaffii 6

Suppose that p represents the statement “He missed the tournament”, q represents the statement “He got the gold medal.” And r represents the statement “he took a trip abroad.”. then which of the following symbolic expressions represents the statement: “If He took a trip abroad and he does not miss the tournament, then he get the gold medal.”?

  1. A.(r⇒q) ¬p(r \Rightarrow q)\ \neg p
  2. B.r∧(p⇒q)r \wedge (p \Rightarrow q)
  3. C.(r∧¬p⇒q)(r \wedge \neg p \Rightarrow q)
  4. D.¬(r∨p)∨q\neg(r \vee p) \vee q
Deebii fi ibsa ilaalaa

Deebii: C (r∧¬p⇒q)(r \wedge \neg p \Rightarrow q)

Maaliif

'He took a trip abroad' is rr, 'he does not miss the tournament' is ¬p\neg p, and their conjunction implies 'he got the gold medal' (qq), which is symbolically written as (r∧¬p⇒q)(r \wedge \neg p \Rightarrow q).

Gaaffii 8

The Ozone level (in ppb- parts per billion) on a sunny day in metropolitan area is given by formula p(t)= 80 + 12t � t2t^2 where t is time in hours and t = 0 corresponds to 9A.M. what is the rate of increase of Ozone level after 3-hrs(i.e. at 12 A.M)?

  1. A.6 ppb
  2. B.12 ppb
  3. C.107 ppb
  4. D.113 ppb
Deebii fi ibsa ilaalaa

Deebii: A 6 ppb

Maaliif

The rate of increase is given by the derivative p′(t)=12−2tp'(t) = 12 - 2t. At t=3t = 3 hours, p′(3)=12−2(3)=6p'(3) = 12 - 2(3) = 6 ppb.

Qormaata guutuu PDF dhaan buufadhaa

Gaaffilee 62 kuusaa kana keessa jiran hunda akka waraqaa qormaataa maxxanfameetti, furtuu deebii ibsa 57 waliin immoo akka faayilii lammaffaatti. Maxxansaa yookiin daataa malee dubbisaa.

Bilisa; herregas hin gaafatu.

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Qormaata kana taa'uu dura Herregaa qayyabadhu

Yaadannoon gabaabaan Herregaa kutaa 12 kaardii 33 qaba. Tokkoon tokkoon boqonnaa boqonnaadhaan kitaaba biyyaalessaa irraa qophaa'ee seericha, mallattoowwan isaa maal bakka akka bu'an fi dogoggora qabxii nama dhabsiisu ni ibsa.

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Kunneen gaaffilee dhugaa qormaata galuumsa Herregaa bara 2007 A.L.I ti?

Eeyyee. Qormaata galuumsa yunivarsiitii biyyaalessaa kan Herregaa barattoonni bara 2007 A.L.I (2015) fudhatan irraa fudhataman. Kuusaan keenya gaaffilee isaa 62 qaba; tokkoon tokkoon isaanii deebii fi ibsa gara kitaaba barnootaatti qajeelchu qabu.

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Lakkoofsa bilbilaatiin herrega bilisaa banadhaa. Qormaatichi guutuun qormaata fakkeenyaa sa'aatiin daangeffame ta'ee qabxii battalumaa kenna; fooyya'iinsi gargaaraa AI ji'atti qarshii 199 kan ta'e waan isin jalaa darbe hunda isinii ibsa.

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Eeyyee; bilisas. Waraqaan gaaffii faayilii tokko, furtuun deebii ibsa isaa waliin immoo faayilii biraa ti; kanaaf qormaata dursitanii yaaluu dandeessu. Maxxansaa, yookiin yeroo daataan hin jirre irra deebiif bilbila keessan irratti kaa'adhaa.

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