١‫ﭘﺎﺳﺦ ﻧﺎﻣﮫ‬
‫ردﯾﻒ‬
3
8x6 − 7x3 − 1 = 0 x
= t 8t2 − 7t − 1 = 0 ⇒ ∆ = 49 + 32 = 81
 t =1 ⇒ x3 =1 ⇒ x =1
7±9 
=
⇒  −1
t
−1
−1
16
⇒ x3 =
⇒x=
t =
8
2
 8
− Sx + p =
( S =−1/5,P =−7) x
0x
2
2
(‫اﻟﻒ‬
+ 1/ 5x − 7 =0
3
2
x=
− 7 0x22
− 14 0
 x + 3x =
2
∆ = b2 − 4ac = 9 − 4( 2 )( −14 ) = 9 + 112 = 121
⇒ x2 +
 , −3 + 11
=
= 2
−b ± ∆ −3 ± 121 −3 ± 11  x
4
x
⇒
=
=
=
⇒
3
−
− 11 −7
2a
4 =
2(2 )
x,, =
4
2

x
١
(‫ب‬
2
x −2
=
( x − 1)( x + 1) x + 1 x ( x − 1)
nj R°µ] ³IµU Joò
→ x.x − 2x ( x − 1) = ( x − 2 )( x + 1)
x ( x − 1)( x + 1)
−
⇒ x 2 − 2x2 + 2x= x 2 − x − 2 ⇒ 2x2 − 3x − 2= 0
∆ = b2 − 4ac = ( −3 ) − 4( 2 )( −2 ) = 9 + 16 = 25
2
٢
 , 3+5
=
= 2
−b ± ∆ 3 ± 25  x
4
=
⇒x
=
⇒
3
−
5
1
2a
2(2 )
x,, =
=−

4
2

Eq] ¾M Eq] u²IU
→
ABC: AB||ME 
MD ED
=
MA EB
x + 4 10
= ⇒ 10x = 5x + 20 ⇒ x + 4
5
x

BN BE
u²IU
+
BDC:DE ||EN →
Eq] ¾M Eq] NC DE
5 3
⇒ + ⇒ 5y = 30 ⇒ y = 6
10 y
⇒
٣
‫ﻣﯽداﻧﯿﻢ ﮐﮫ ﺳﻬﻤﯽﻫﺎ روی داﻣﻨﮥﺷﺎن ﯾﻌﻨﯽ ‪ ‬ﯾﮏﺑﮫﯾﮏ ﻧﯿﺴﺘﻨﺪ‪ .‬ﺑﺮای اﺛﺒﺎت ا�ﻦ‬
‫ﻣﻮﺿﻮع ﻣﯽﺗﻮاﻧﯿﻢ ﻧﻤﻮدار ﺳﻬﻤﯽ را رﺳﻢ ﮐﻨﯿﻢ‪.‬‬
‫‪٤‬‬
‫(‬
‫اﮔﺮ داﻣﻨﮫ را ) ∞‪ 4,+‬ﯾﺎ ‪ −∞,4‬ﻓﺮض ﮐﻨﯿﻢ‪ ،‬آن ﮔﺎه ‪ f‬ﯾﮏﺑﮫﯾﮏ ﺧﻮاﻫﺪ ﺷﺪ‪.‬‬
‫‪y=x-  x  0 ≤ x⟨3‬‬
‫‪0 ≤ x⟨1 ⇒  x  = 0 ⇒ y = x − 0 = x‬‬
‫‪٥‬‬
‫‪1 ≤ x⟨2 ⇒  x  =1 ⇒ y = x − 1‬‬
‫‪2 ≤ x⟨3 ⇒  x  = 2 ⇒ y = x − 2‬‬
sin76° + cos194°
cos104° − sin284°
=
=
(
) (
)
cos ( 90 + 14 ) − cos ( 270 + 14 )
sin 90° − 14° + 6 cos 180° + 14°
°
°
°
°
cos14 − 6 cos14 −5cos14 5
5
5 1
5
cos
cot
x
=
=
=
=
14°
14° =
°
°
°
2
2
2 m 2m
− sin14 − sin14
12sin14
°
°
°
٦
٧
1
=
6log
23
1
log2 2=
log2 2(0=
/ 3 ) 0/6
6x =
3
log72 = log 32 x22 = log32 + log23 = 2log3 + 3log2
=
6log
2
3
(
)
= 2 ( 0 / 4 ) + 3 ( 0 / 3 ) = 0 / 8 + 0 / 9 = 1/ 7
٨
⇒ 6log3 2 + log72 = 0/6 + 1/ 7 = 2 / 3
٩
1
2
2
3x + 1) log( 3 ) + log( 3 + x )
log2 + log(=
 


´TÄnI«² ¦ Ä ¾M ®ÄkLU
´TÄnI«² ¦ Ä ¾M ®ÄkLU
⇒ log 2( 3x + 1) = log 3( 3 + x ) ⇒ 6x + 2 = 9 + 3x
7
⇒ 6x − 3x = 9 − 2 ⇒ 3x=7 ⇒ x= ( ¡ ¡ )
3
(
(
)
١٠
lim f ( x ) + g( x ) =lim f ( x ) + lim g( x ) =0 + ( −5 ) =−5 (‫اﻟﻒ‬
x→4
)
x→4
x→4
lim 3f ( x ) − 5g( g) =
3 lim f ( x ) − 5 limg( x ) =
3 ( 3 / 5 ) − 5 (6 ) =
−19/ 5 (‫ب‬
x→0
(
x→0
)
4
lim f ( x ) + x2 =
+ x −2
x→3
4
2
(
x→0
)
4
lim f ( x ) + lim x2 + lim x − 2 (‫پ‬
x→3
x→3
x→3
= 5 + 3 + 3 − 2= 625 + 9 + 1= 365

 


 

lim f x .g x=  limf x  x  limg x=
x6 18 (‫ت‬
 3=
+
x→( −1)
 x→ −1 +   x→ −1 + 
 ( )   ( ) 
 f x  lim+ f x
0
 x→4
lim 
=
= = 0 (‫ث‬

−5
x→4+ 
 g x  lim+ g x
( ( ) ( ))
( )
( )
( )
lim g( x )=
x→4
x→4
lim g( x )=
x→4
( )
( )
( )
−5
½kzº þÄo ÷ U (‫ج‬
١١
‫‪0‬‬
‫اﻟﻒ( ﺑﺎ ﺟﺎﮔﺬاری ﻋﺪد‪ ٢‬ﺑﮫ ﺟﺎی ‪x‬ﻫﺎ ﺣﺎﺻﻞ ﺣﺪ ﺑﺮاﺑﺮ‬
‫‪0‬‬
‫ﺑﺮ )‪ (x-٢‬ﺗﻘﺴﯿﻢ ﻣﯽﮐﻨﯿﻢ‪ .‬اﻟﺒﺘﮫ در ا�ﻦ ﺳﺆال‪ ،‬راﺣﺖﺗﺮ اﺳﺖ ﮐﮫ ﺻﻮرت و ﻣﺨﺮج را‬
‫اﺳﺖ ﻟﺬا ﺻﻮرت و ﻣﺨﺮج را‬
‫ﺗﺠﺰﯾﮫ ﮐﻨﯿﻢ‪:‬‬
‫‪١٢‬‬
‫‪12‬‬
‫اﻟﻒ(‬
‫‪5‬‬
‫) ‪( x − 2 ) ( x 2 + 2x + 4‬‬
‫=‬
‫‪lim‬‬
‫‪x→2‬‬
‫) ‪( x + 3)( x − 2‬‬
‫‪ke‬‬
‫=‬
‫| ‪x  + | x‬‬
‫| ‪ −0/ 1 + | x‬‬
‫‪−1 − x −1 − 0 −1 − 0‬‬
‫‪1‬‬
‫‪lim   = lim ‬‬
‫ب( ‪= lim = = = −‬‬
‫‪x +2‬‬
‫‪0+2 0+2‬‬
‫‪2‬‬
‫‪x→0− x + 2‬‬
‫‪x→0−‬‬
‫‪x→0− x + 2‬‬
‫‪‬‬
‫‪x | −1 −2‬‬
‫‪ 2x + 4 x⟨−1‬‬
‫‪y |2 0‬‬
‫‪‬‬
‫‪‬‬
‫‪x | −1 0 2‬‬
‫‪ 2‬‬
‫‪y x − 1 −1 ≤ x⟨2‬‬
‫‪y | 0 −1 3‬‬
‫‪‬‬
‫‪‬‬
‫‪x |2 5‬‬
‫‪ −x + 5 2⟨ x⟨5‬‬
‫‪y |3 0‬‬
‫‪‬‬
‫‪‬‬
‫‪١٣‬‬
‫)‬
‫(‬
‫)‬
‫ﺗﺎ�ﻊ ‪ f‬در ﺑﺎزهﻫﺎی ‪ 2,5‬و ‪ −∞,−1‬ﭘیﻮﺳﺘﮫ اﺳﺖ‪ ،‬وﻟﯽ در ﺑﺎزهﻫﺎی ‪  −2,2‬و‬
‫‪ −∞,−1‬ﻧﺎﭘیﻮﺳﺘﮫ اﺳﺖ‪.‬‬
‫(‬
‫‪١٤‬‬
‫‪20‬‬
‫) ‪n( S‬‬
‫=⇒ }‪{1,2,3,...,20‬‬
‫=‬
‫‪S‬‬
‫اﻟﻒ( ‪n( A ) 1‬‬
‫‪3‬‬
‫‪1‬‬
‫⇒= ‪A‬‬
‫‪n‬‬
‫‪A‬‬
‫=‬
‫⇒‬
‫‪P‬‬
‫‪A‬‬
‫=‬
‫) ( }{‬
‫= ‪( ) nS‬‬
‫‪( ) 20‬‬
B :jkø ·j¼M µ¤n ¦ Ä k¶IzÃQ ⇒ B =
{1,2,3,4,5,6,7,8,9}
9
9 ⇒ P (B ) =
⇒ n( B ) =
20
A=
{3} ⇒ A  B ={3} ⇒ P( A  B ) =201
(‫ب‬
1
P( A  B ) 20 1
P( A |B=
) PB= =
( ) 9 9
20
:‫ ﻟﺬا‬،‫ ﻫﻢ ﻣﺴﺘﻘﻞاﻧﺪ‬B′ ‫ و‬A′ ،‫ ﻣﺴﺘﻘﻞ ﺑﺎﺷﻨﺪ‬B ‫ و‬A ‫ﻣﯽداﻧﯿﻢ اﮔﺮ‬
P( A′  B′ ) = P( A′ ).P(B′ )
1 2
1 P( A ) =−
1
=
P( A′ ) =−
3 3
2 3
1 P(B ) =−
1
=
P(B′ ) =−
5 5
2 3 2 3 13
P( A′  B′ ) = P( A′ ) + P(B′ ) − P( A′  B′ ) = + − x =

 3 5 3 5 15
١٥
P( A ′ )XP(B′ )
 A : ½oÀp ²¼L¤ k¶IzÃQ
B : ½oÀp Sw»j ²¼L¤ k¶IzÃQ

=
P( A ) 2P(B=
/625,P( A ) ?
),P( A  B ) 0=
P ( A  B ) = P ( A ) + P (B ) − P
A  B)
(
½oÀp ²¼L¤ ·¼a kºH ®£Tv¶B‫و‬A
.jnHkº yTw»j ²¼L¤ nj ÁoÃYDU
= 2P(B ) + P(B ) − P( A ).P(B )

2P(B )
P( A ) = 2x ‫ ﻟﺬا ﺧﻮاﻫﯿﻢ داﺷﺖ‬،‫ ﻓﺮض ﮐﻨﯿﻢ‬X ‫ را‬P(B) ‫ �ﻬﺘﺮ اﺳﺖ‬،‫ﺑﺮای رﺣﺘﯽ ﮐﺎر‬
⇒ 0/625 = 2x + x − 2x2 ⇒ 2x2 − x + 0/625 = 0
١٦
:‫ﺧﻮاﻫﺪ �ﻮد‬
∆ = b2 − 4ac = −94( 2 )(0/625 ) = 9 − 5 = 4

5
x
=
⟩1( ¡ ¡ ù )

−b ± ∆ 3 ± 2
4
x
=
=
⇒
1
1 1
2a
4
x =
¡ ¡ ) ⇒ P( A ) =
2x =
2x =
(

4
4 2
R = max − min= 250 − 12= 328 (‫اﻟﻒ‬
١٧
‫‪,81,94,95,100‬‬
‫ب( ‪ ,102,130,180,250‬‬
‫‪Q3‬‬
‫‪80‬‬
‫‪,81‬‬
‫‪‬‬
‫‪80+81‬‬
‫‪80/5‬‬
‫‪2‬‬
‫‪12,31,35,42,43‬‬
‫‪ ,45,52,56,‬‬
‫= ‪Q2‬‬
‫‪Q1‬‬
‫ً‬
‫ﺗﻘﺮﯾﺒﺎ ‪ ٪٢٥‬دادهﻫﺎ ﻗﺒﻞ از ‪ Q1‬ﯾﻌﻨﯽ ‪ ٤٣‬ﻫﺴﺘﻨﺪ‪.‬‬
‫پ(‬
‫ً‬
‫ﺗﻘﺮﯾﺒﺎ ‪ ٪٥٠‬دادهﻫﺎ ﺑﻌﺪ از ﻣﯿﺎﻧﮫ ﯾﻌﻨﯽ ‪ ٨٠/٥‬ﻫﺴﺘﻨﺪ‪.‬‬
‫ت(‬
‫ً‬
‫ﺗﻘﺮﺑﺒﺎ دادهﻫﺎ ﻗﺒﻞ از ‪ Q3‬ﯾﻌﻨﯽ ‪ ١٠٠‬ﻫﺴﺘﻨﺪ‪.‬‬
‫ث(‬