LYS1 Matematik Denemesinin Çözümleri Mayıs 2012 ÖzdeBir

LYS 1

ÖZ-DE-BÝR YAYINLARI

ÜNÝVERSÝTE HAZIRLIK

A

MATEMATÝK DENEME SINAVI

1.

$

1 1 — – — 2 3 —––––––– 1 1 — + — 2 3

1 — 6 —–– 5 — 6

% $ % –1

=

6.

–1

A = {2, 6, 10, ..., 102} B = {0, 3, 6, 9, ..., 99}

=5

A∩B = {6, 18, ..., 90} Yanýt: D

s(A∩B) = 8 çýkar.

Yanýt: A

2.

2M3 + 3M3 = 5M3 =

M75 ll Yanýt: C

7.

4x + 1 ≡ 9 (mod 12) 4x ≡ 8 (mod 12) x ≡ 2 (mod 3) x = 2, 5, 8, 11 olmalý.

3.

Yanýt: D

(1 + 2 + 4 + 24 + 120 + ... + 9!)2

144424443 5 ile tam bölünür.

9 un 5 ile bölümünden kalan 4 tür.

Yanýt: E

8.

4.

1 — ≡ 2 ⇒ 2⋅4 ≡ 1 (mod 7) 4 1 — ≡ 5 ⇒ 3⋅5 ≡ 1 (mod 7) 3 2∗(2Δ5) 123

x⋅y = 9

3

Mll xy = 3

2∗3 = 5

Yanýt: B

Yanýt: B

5.

10 – N M = –––––– den N çift olmalý.

2

9.

N = –10, –8, –6, –4, –2, 0, 2, 4, 6, 8, 10

f(3) = 32012⋅22012 f(4) = 4

deðerlerini almalýki M tamsayýsý çýksýn.

2012 2012

f(3)⋅f(4) = 9

11 tane.

Yanýt: C

⋅3

2012 2012

⋅8

= f(9) Yanýt: D

A 10.

A

A

A 15.

g(f(x)) = 0 123

A

2k – 12 + 4 – 1020 = –4 k 2 = 1024

0

k = 10

f(x) = 0 ise, x = 2, –2 dir.

Yanýt: C

–2⋅2 = –4 Yanýt: A

11.

16.

y – x = 5xy

a4 + 2a3 + a2 = (a2 + a)2 = 1 Yanýt: A

5xy 3 + ––––– = 3 – 1 = 2 x–y Yanýt: E

12.

a = 2 + M2 ⇒ a2 – 4a + 2 = 0

17.

2 a+ — =4 a 4 2 a + ––– = 12 2 a

2x2 – 3x – 5 ≤ 0 –1

x2 –9 ≤ 0

5/2

–3

3

x = –1, 0, 1, 2 olur.

Yanýt: E Yanýt: C

13.

a + b = –3, a⋅b = –t 2 2 a + 2ab + b = 9

18.

y

15 – 2t = 9 6 = 2t

(2,4)

t=3 Yanýt: E

(a,a2) 0

x

Dik kenarlarýn eðimleri çarpýmý –1 olur.

14.

4 a2 — ⋅ — = –1 2 a

P(1) = a0 + a1 + a2 + a3 + a4 tür.

P(1) = 108 çýkar.

Yanýt: C

1 a=–— 2

1 b= — 4 Yanýt: A

2

A 19.

A

A

f(a + b) = a2 + 2ab + b2 + 2a + 2b+ 3

A 24.

A

n = 64 64

i + i2 + i3 + i4 + ... Σ ik = 1442443 k=1

2 2 f(a) + f(b) = a + b + 2a + 2b + 6

f(a) + f(b) = f(a) + f(b) den

0

2ab = 3 3 ab = — 2

olduðundan sonuç 0 olur.

Yanýt: A Yanýt: C

20.

25.

x = 1 deki teðetin eðimi 5 tir.

S/2

2⋅1 + a = 5 ⇒ a = 3 z

x=1⇒6=1+3+b

S/2

b=2 3– 2=1 Yanýt: C

21.

2

2

$ % + $ —S2 %

2, 3, 5, 7, 11, 13, 17, 19, 23, 29

S z2 = — 2

⇒ (2, 3), (2, 5), (2, 11), (2, 17), (2, 29)

P1 = 4S

5 5 1 —––––– = — = — C(10,2) 45 9

S P2 = 4 ––– M2

2

2

2

S S S S = ––– + ––– = ––– ⇒ z = ––– 4 4 2 M2

S P3 = 4 –––––– (M2)2 Yanýt: B ....................

S Pk = 4 –––––– (M2)2–1 ∞

22.

Σ Pk = 4S k=1

n = 1, 5, 25, 125, 625 olmalý.

T

Y

S S S 1 + ––– + ––––– + ––––– + ...

M2 (M2)2 (M2)3

f(1) + f(5) + f(25) + f(125) + f(625) ⎛ ⎜ 1 = 4S ⎜ ⎜ 1 ⎜⎜ 1 – 2 ⎝

= 0 + 1 + 2 + 3 + 4 = 10 Yanýt: D

23.

n = 3,

a3 = 2a2 + t ⇒ a3 = 10 + t

n = 4,

a4 = 2a3 + t ⇒ 20 + 3t

n=5

33 = 2(20 + 3t) + t

⎞ ⎟ ⎟ ⎟ ⎟⎟ ⎠

M2 S = ––– olduðundan 8

∞

M2 —––– = —––– = Σ Pk = 4 ––– 8 $ M2 – 1 % M2 – 1 k=1

t = –1

M2

1

M2 + 1 —––––––––––– (M2 – 1)(M2 + 1)

M2 + 1 = ––––––– = M2 + 1 olur.

2–1

Yanýt: B

Yanýt: B 3

A 26.

A 15m = 3m⋅5m

A

den 5m kullanýlmalý.

A 30.

n = 2, 7, 12 alýnýrsa.

4 5⋅15⋅25 ⇒ 3⋅5 olur.

A

π tan(arctanx + arctank) = tan — 4 x+k 1–k —–––– = 1 x = —–––– 1 – kx 1+k Yanýt: E

m en çok 4 olur.

Yanýt: C

31.

27.

P(x) = x3 + ax2 + bx + c Kökler: 3, 1 + i, 1 – i

tanθ + cotθ = 4 2 2 tan θ + 2 + cot θ = 16

Kökler toplamý ⇒ –a = 5

2 2 Mtan lllllll θ + cot θ = M14 ll

a = –5 Yanýt: A

Yanýt: B

28.

|F| > 1

32.

argF –1 = –argF

| | 1 –– F

2 2i (1 + i) ––––––– = k ⇒ —– = k 2 –2i (1 – i)

k = –1

1 = ––– < 1 |F|

Yanýt: B

Buda C olabilir.

Yanýt: D

33.

29.

$ %

$

2 3 4 2011 S = log2 — ⋅ — ⋅ — ⋅...⋅ –––– 3 4 5 2012

(sint – cost)(sin2t + sint⋅cost + cos2t) –––––––––––––––––––––––––––––– sint – cost

$

1 S = log2 –––– 1006

sin2t sin t + cos t + —––– 2 2

$ %

2 3 S = log2 — + log2 — + ... + log2 3 4

2

2011 –––– % $ 2012

%

%

1 2S = –––– 1006

1 7 1+ — = — 6 6

Yanýt: B Yanýt: E 4

A 34.

A

A

log22 1 logab(2) = –––––––– = ––––––––––– log2(ab) log2a +log2b

A 38.

A

xsinx 0 lim —––––––––– = — 2 0 x→0 M2x lll +3 – M3 2+3 + M3) x⋅sinx⋅(M2xlll lim —––––––––––––––––– x→0 2x2

xy 1 = —–––––– = ––––– x+y 1 1 — + — x y

lim x ⋅ sinx x→0 2 ⋅ x ⋅ x

Yanýt: D

⋅2M3 = M3

1

Yanýt: E

35.

$ %

x = 16 ⇒ f

1 — 16

+ 2f(32) = log264

$ %

1 1 x = — ⇒ f(32) + 2f — 32 16

39.

1 = log2 — 8

f’(w) – f’(2) 0 lim —–––––––– = — le hospital w–2 0 w→2 lim f’’(x) = f’’(2) w→2

denklemlerinden f

2 f’(x) = 3x – 2

$ % 1 — 16

f’’(x) = 6x

f’’(2) = 12

yok edilirse Yanýt: D

f(32) = 5 bulunur.

Yanýt: E

40.

df(2sinx) ––––––– = 2cosx⋅f’(2sinx) dx π π π x = — ⇒ 2⋅cos —⋅f’ 2sin — 6 6 6

$

36.

y 1

%

M3 = 2⋅ —– ⋅f’(1) 2

123

0 + 2⋅1 = 2 (max)

5 –— 2 –1

0

–1

1

5 = M3⋅– — 2

x

5M3 = – –––– 2

0 – 2⋅1 = –2 (min)

Yanýt: A Yanýt: B

41.

37.

x = e olur.

lim f(x) = 0, lim g(x) = 0 x→2

1 – lnx f’(x) = –––––– = 0 x

x→2

1 f(e) = — e

2 – f(x) 2 lim —–––––– = —– = 2 1 + g(x) 1 x→2 Yanýt: E

Yanýt: A

5

A 42.

A

A

x2 + 2x + 4 (x – 2)(x2 + 2x + 4) f(x) = ––––––––––––––––– = ––––––––––– (x – 2)(x + 2) x+2

A 47.

x = –2 D.A

cos2x = 1 – 2sin2x 1 — 2

4 y = x + ––––––– x+2

A

π/3

π/3

0

0

∫

1 sinxdx = – — cosx 2

∫=

1 — 4 Yanýt: A

y = x eðik asimptot –2 + 1 = –1 Yanýt: B

48.

43.

y

D þýkký yanlýþtýr.

2x+1

x2

x ∈ (0, b) de f’ azalan f’’ < 0 x ∈ (b, 2) de f’ artan f’’ > 0

1

Yanýt: D –1

0

0

44.

lim tan x→0

$ % 6π —– 9

⇒ tan

$ % 2π —– 3

}

∫ x2dx = —31

–1

= –M3

1

∫ (2x + 1)dx = 2

Yanýt: A

0

x

1

1 7 2 +— = — 3 3

Yanýt: C

45.

h 0 lim —–––––––––– = — le hospital uygulanýrsa, 0 h→0 f(x + h) – f(x) 1 1 —–– = x ⇒ f’(x) = — f’(x) x

49.

Deðiþmez çünkü 1. satýr 3. satýra eklenmiþtir.

f(x) = ln(3x) olabilir.

Yanýt: D Yanýt: D

2

46.

∫ f’(x)dx = 1 – 1 – 6 = –6

50.

–2

2

∫ f’’(x)dx = f’(x)| –2 = f’(2) –f’(–2) = –2 – 2 = –4 2

–2

–6 – 4 = –10

E

1

1

2

x

R E ⋅

y

1

3

4

R E =

y+3=5

2 + 4x = 14

y=2

x=3

5

5

13

14

R

x+y=5 Yanýt: E

Yanýt: C 6