2010/2011, sem. zimowy
Funkcje cyklometryczne Zad.1. Wyznaczyć dziedzinę danej funkcji:
√
(1) f (x) =
− log cos x,
(11) f (x) = parccos(2x + 1), (20) f (x) = arccos( 3x−1 ), 2
√
(2) f (x) =
1−x2 ,
q x+1
(21) f (x) = arctg(12 −
2x
√
),
2x−4
(12) f (x) = arcsin
,
x2+1
√
x
(3) f (x) = log x· tg x , x2+3x+5
q
(22) f (x) = arccos(x2 − 4),
√
(13) f (x) =
ln 2x ,
x−1
log(2x2−8)
(4) f (x) =
,
(23) f (x) = arcsin x−3 , 3−3x
x−4
q 3x−4
2
(14) f (x) = ln
,
(5) f (x) = 10
4x+3
3 arcsin x ,
(24) f (x) = arccos 2−x , x−5
(6) f (x) = 2 − arccos 2x − 1, (15) f (x) = arcsin3(ln x + 5), q
√
(25) f (x) =
arcsin(x + 1 ),
3
(7) f (x) = ln
1 − sin 2x,
(16) f (x) =
1
,
arccos(1−ln x)
q
√
(26) f (x) =
1
,
(8) f (x) = ln
1 − cos 3x,
p
arccos x
2
2
(17) f (x) =
2x − 3,
√
√
(27) f (x) =
x+4
,
(9) f (x) = arcsin(2−x) , p
(18) f (x) =
5 −
5 − x,
arcsin x−2
x2−4x+4
6
√
(10) f (x) = arcsin
3x,
(19) f (x) = arcsin(x − 1), (28) f (x) =
2
√
+ arctg 1 .
3x+1
x
(29) f (x) = ln(4x − 3) + arcsin x+2 .
4
Zad.2. Obliczyć:
(1) 3 arcsin 1 − 2 arccos 0, (5) 2 arccos(− 1 ) − arctg1, 2
(2) 4arctg1 − arcctg(−1),
√
√
√
(6) arctg 3 − 2 arcsin 1 , (3) 1 arccos
3 + arctg(− 3),
3
2
2
2
√
√
√
(4) arcctg 3 − 3 arcsin
2 ,
(7) cos(3 arcsin
3 + arccos(− 1 )).
2
2
2
Zad.3. Naszkicować wykresy funkcji: (1) f (x) = arcsin(x − 3), (3) f (x) = arctgx + π ,
(5) f (x) = 2arcctgx,
2
(2) f (x) = arccos(x + 2), (4) f (x) = arcctgx − 1,
(6) f (x) = −arctgx.
1