8/20/2019 2015 Zadaci Sa Resenjima Sa ETF Iz Matematike Sa
Prijemnog Ispita
1/2
−10%
−1
k ∈ Z 0, 0010101 · 10k > 1001
k −6 5 −5
6 0
√ 2x + y = 1 2x +
√ 2y = 3
√ 2
2 √
2 − 1 0 23
√ 3
√ 6
6
x1 x2 x2 + x + 1 = 0
y1 = ax1 + x2 y2 =
x1 + ax2, (a ∈R)
y2 + (a + 1) y − a2 + a + 1 = 0
y2 + a2 + 1 y + 1 = 0
y
2
+ (a + 1) y + a2
− a + 1 = 0
y
2
+
a
2
+ 1
y + a2
− a + 1 = 0
k ∈ R, i2 = −1
1 + i
1 − i2015
+ −1 + 5ki
3i − 1 k
3
5 0
1
3 −1
2 3
d1 =
2 −√ 3 d2 15◦ 30◦ 45◦ 60◦
22, 5◦
3 : 2 7 : 5 4 : 3
8 : 5 7 : 4
a = 0, 10,1 b = 0, 20,2 c =
0, 30,3
b < c < a a < b < c
b < a < c c < b < a c < a
< b
cos
x − 3π
2
= −4
5
π
2 < x < π, sin
x
2 cos
5x
2
− 38125
82125
4125
1 −1
f (x)
+ f (f (x)) = x
f (x) = |x| + a a > 0,
1 0 2 3
4
A = 1
6
(log2 3)
3 − (log2 6)3 − (log2 12)3 + (log2 24)3
, 2A
1 36 72 144
64
8/20/2019 2015 Zadaci Sa Resenjima Sa ETF Iz Matematike Sa
Prijemnog Ispita
2/2
(x, y) x2 − 2x− y < 1
2 y + |x − 1| < 2
0 2 1 4
3
14
9
3√
x + 1√
x
n(n ∈N, x > 0) ,
√ x
1 48 84 5
21
N N
[105, 2 · 105) [2 · 105, 3 · 105) [3
· 105, 4 · 105) [4 · 105, 5 · 105) [5 · 105, 6
· 105)
a1, a2, . . . d = 1
a1+a2+
· · ·+a98 = 137
a2 + a4 + a6 + · · · + a98
88 93 103 127
141
x 43x − 24x+2 · 3x+1 + 20 · 12x · 3x ≥ 8 ·
6x 8x−1 + 6x
a, b c d
−∞ < a < b < c < d −1) A
B O OAB
1
e
2
e
3
e e 2e
logcos x sin x = 4 logsinx cos
x
0,
π
6
π6
, π
4
π4
, π
3
π3
, π
2
5π
6 , π
x +√
3√
x +
x +√
3+
x −√ 3√
x −
x −√ 3=√
x
[√
3, 2√
3) (2√
3, 3√
3) [3√
3, 6) [6, 8) ∅
![Zbirka Zadataka Sa Resenjima[1]](https://img.pdfslide.tips/doc/110x75/577d2b641a28ab4e1eaaa7de/zbirka-zadataka-sa-resenjima1.jpg)
