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