ﻟﻸﺳﺘﺎﺫ ﺍﻟﻘﺪﻳﺮ /ﻋﻠﻰ ﺍﻟﺪﻳﻦ ﻳﺤﻴﻰ ٠١١١٩٦٦٠٦١٦
ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ -ﺍﻟﺼﻒ ﺍﻟﺜﺎﻟﺚ ﺍﻟﺜﺎﻧﻮﻯ
ً )أوﻻ(
اﺸ
ا ﺠﺎور :
ﻋﻤﻠﻴﺔ ا ﻔﺎﺿﻞ ﺗ ﻮن
،ﻋﻤﻠﻴﺔ ا
ﻞ ﺗ ﻮن
+
ﻗﺎ q
ﻗﺘﺎ q
ً )أوﻻ( ﺗﻔﺎﺿﻞ إﺣﺪى ا وال ﺴﺎوى ﺣﺎﺻﻞ ﺑﺎﻹﺷﺎرة ا ﺒ ﻨﺔ داﺧﻞ ا ﺜﻠﺚ . ب أى دا
ﺑﺎﻹﺷﺎرة ا ﺒ ﻨﺔ داﺧﻞ ا ﺜﻠﺚ .
ﺑ ﻨﻤﺎ ﺹ /ﺗﻌ ﺹ /ﻗﺪ ﺗﻌ
¤Ù
¤Ù
ﺸﺘﻘﺔ ﺹ ﺑﺎﻟ ﺴﺒﺔ إ ﺱ
ﺸﺘﻘﺔ ﺹ ﺑﺎﻟ ﺴﺒﺔ إ ا ﺘﻐ ا ﺴﺘﻘﻞ ﺎ .أى أن
§Ùأ §Ù ،أ §Ù ،أ ...... ،وﻫﻜﺬا ¤Ù
Ùﻉ
Ùﻥ
ﻠﺤﻮﻇﺔ ) : (٢ﻋﻨﺪ إﺛﺒﺎت ﺻﺤﺔ اﻟﻌﻼﻗﺎت
) (١ﺸﺘﻖ اﻟﻄﺮﻓ ﺛﻢ ﺸﺘﻖ اﻟﻄﺮﻓ
ا ﺸﺘﻘﺎت :
ﺮة أﺧﺮى دون ﺗﻐﻴ
ا ﺪود
ﻗﺘﺎ q
با ا
ﻮن = §Ù : ¤Ù y¤
ﺹ /ﻣﻔﻬﻮم أوﺳﻊ ﻣﻦ §Ùﻷن §Ùﺗﻌ
-ﺟﺎ q
ﻇﺎ q
ً )ﺛﺎﻧﻴﺎ( ﺗ ﺎ ﻞ ﺣﺎﺻﻞ
ﻠﺤﻮﻇﺔ ): (١
ﻋﻜﺲ
ﻇﺘﺎ q
§y
ا ﻮﺳﻴﻂ أو ا ﺎراﻣ .و
-ﺟﺘﺎ q
ﺟﺘﺎ q
ا ﺎه ﺣﺮ ﺔ ﻋﻘﺎرب ا ﺴﺎﻋﺔ ً : ا ﺎ )ﺛﺎﻧﻴﺎ( ا ﺸ
ﻗﺎ q
ﺟﺎ q
ا ﺎه
ﺣﺮ ﺔ ﻋﻘﺎرب ا ﺴﺎﻋﺔ
إذا ﻧﺖ :ﺹ = د )ﻥ( ،ﺱ = د )ﻥ( ﻓﺈن ا ﺘﻐ )ﻥ( ﺴ
ﻮﺿﻊ
) (٢إذا أردﻧﺎ ا ﻌﻮ ﺾ ﻋﻦ ﻗﻴﻤﺔ ا ﺸﺘﻘﺔ اﻷو ﻳﻔﻀﻞ إ ﺎدﻫﺎ
اﻷﺧﺮﺗ
)(٣
ﺴﺎوى ا اﻟﺔ ا ﺎ ﺔ
ﻫﺎ ﺶ ا ﻞ
ﺗﻔﻜ ﻧﺎ اﻻﺳﺘﻌﺎﻧﺔ ﺑﺎﻟﻌﻼﻗﺔ ا ﻌﻄﺎة
ﻣﻌﻈﻢ ا ﻤﺎر ﻦ ﻧﻀﻊ
ﺗﺬﻛﺮ أن :
ﻗﻮاﻋﺪ إ ﺎد ا ﺸﺘﻘﺔ اﻷو :
ﻠﺤﻮﻇﺔ :
· ﺹ = ﺍ ) ﺣﻴﺚ ﺍ = ﺛﺎﺑﺖ ( ﲤﺲ ﺹ = ﺻﻔﺮ /
٢ / ﺹ = ﻇﺎ qﲤﺲ ﺹ = ﻗﺎ q
ﲤ ﺹ = /ﻥ × ﺱ · ﺹ = ﺱﻥ ﺲ
ﺑ ﻨﻤﺎ ﺹ = ﻇﺎ = p3ﻇﺎ = ٣] = ٥٦٠ﻣﻘﺪار ﺛﺎﺑﺖ ﲤﺲ ﺹ = /ﺻﻔﺮ
ﻥ –١
· ﺹ = ﺍ ﺱﻥ ﲤﺲ ﺹ = /ﺍ × ﻥ ﺱ
ﻛﺬ ﻚ þ :ﻇﺎ ﺱ .ﺱ = ﻗﺎ ٢ﺱ +ث ﺑ ﻨﻤﺎ
ﻥ –١
ﲤ ) Ùﺹ(ﻥ = ﻥ ﺹﻥ – §Ù × ١ · ﺹ = د )ﺱ( ﺲ ¤Ù ¤Ù
þﻇﺎ p3ﺱ = þﻇﺎ ٥٦٠ﺱ = ٣] þﺱ = ] ٣ﺱ +ث
أى : ُ ُ ُ ﺸﺘﻘﺔ داﻟﺔ ﺮﻓﻮﻋﺔ ﻷس = ﺸﺘﻘﺔ اﻷس × ﺸﺘﻘﺔ ﻣﺎ ﺖ اﻷس ُ ُ ﻛﺬ ﻚ :ﺸﺘﻘﺔ داﻟﺔ ﻣﺜﻠﺜﻴﺔ ﺮﻓﻮﻋﺔ ﻷس = ﺸﺘﻘﺔ اﻷس × ُ ﺸﺘﻘﺔ ا اﻟﺔ ا ﺜﻠﺜﻴﺔ ) ﺖ اﻷس ( × ﺸﺘﻘﺔ ﻣﺎﺑﺪاﺧﻞ ا اﻟﺔ
ﺗﺬﻛﺮ أن :
· ﺟﺎ ٢ﺱ +ﺟﺘﺎ ٢ﺱ = ١ﺉ – ١ﺟﺘﺎ ٢ﺱ = ﺟﺎ ٢ﺱ
– ١ ،ﺟﺎ ٢ﺱ = ﺟﺘﺎ ٢ﺱ
· + ١ﻇﺎ ٢ﺱ = ﻗﺎ ٢ﺱ ﺉ ﻗﺎ ٢ﺱ – ﻇﺎ ٢ﺱ = ١
ا ﺜﻠﺜﻴﺔ
،ﻇﺎ ٢ﺱ = ﻗﺎ ٢ﺱ – ١
· ﺸﺘﻘﺔ ﺣﺎﺻﻞ
ب دا
=
ﺸﺘﻘﺔ اﻷو × ا ﺎﻧﻴﺔ +ﺸﺘﻘﺔ ا ﺎﻧﻴﺔ × اﻷو
· + ١ﻇﺘﺎ ٢ﺱ = ﻗﺘﺎ ٢ﺱ ﺉ ﻗﺘﺎ ٢ﺱ – ﻇﺘﺎ ٢ﺱ = ١
· ﺸﺘﻘﺔ ﺧﺎرج ﻗﺴﻤﺔ دا
،ﻇﺘﺎ ٢ﺱ = ﻗﺘﺎ ٢ﺱ – ١
=
ﺸـــﺘﻘﺔ اﻟ ﺴـــﻂ× ا ﻘـــﺎم -ﺸــــــﺘﻘﺔا ﻘﺎم× اﻟ ﺴـــﻂ ـــﺮ ﻊ ا ﻘـــﺎم
· ﺟﺎ ٢ﺱ = ٢ﺟﺎ ﺱ ﺟﺘﺎ ﺱ
· ﻗﺎﻋﺪة ا ﺴﻠﺴﻠﺔ = §Ù :
· ﺟﺘﺎ ٢ﺱ = ﺟﺘﺎ ٢ﺱ – ﺟﺎ ٢ﺱ = ٢ﺟﺘﺎ ٢ﺱ – ٢ – ١ = ١ﺟﺎ ٢ﺱ
¤Ù
ﺫ¤ g · ﻇﺎ ٢ﺱ = g -1ﺫ ﺱ
§y
y¤
أوﺟﺪ §Ùإذا ﻧﺖ ﺹ ﺴﺎوى : ¤Ù
)) (١ﺍ( ٢ﺟﺎ س – ٣ﻇﺘﺎ س )ﺝ( ﻗﺎ س ﻇﺎ س
١
)ﺏ( ﺟﺘﺎ س ٤ +ﻗﺎ س
) (ﻗﺘﺎ س ﻇﺘﺎ س
ﻟﻸﺳﺘﺎﺫ ﺍﻟﻘﺪﻳﺮ /ﻋﻠﻰ ﺍﻟﺪﻳﻦ ﻳﺤﻴﻰ ٠١١١٩٦٦٠٦١٦
ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ -ﺍﻟﺼﻒ ﺍﻟﺜﺎﻟﺚ ﺍﻟﺜﺎﻧﻮﻯ iﺱ )ﻩ( i +1ﺱ )ﺍ( ﺹ ٢ = /ﺟﺘﺎ ﺱ ٣ +ﻗﺘﺎ ٢ﺱ
j -1ﺱ ) و( j +1ﺱ
ا ﻞ
)ﺏ( ﺑﺎﻻﺷﺘﻘﺎق ﺑﺎﻟ ﺴﺒﺔ إ
) § - /ﺫ( § + ¤ ﺇ ) ﺱ ٢ + ٢ﺱ ﺹ ( ﺹ ٢ – = /ﺱ ﺹ – ﺹ ٢ﺇ ﺹ = + ¤ ) ¤ﺫ§ ( ) (٤أوﺟﺪ §Ùإذا ن : ¤Ù
)ﺏ( ﺹ – = /ﺟﺎ ﺱ ٤ +ﻗﺎ ﺱ ﻇﺎ ﺱ
)ﺍ( ﺱ ﺟﺘﺎ ﺹ +ﺹ ﺟﺘﺎ ﺱ = ١
)ﺝ( ﺹ = /ﻗﺎ ﺱ × ) Ùﻇﺎﺱ( +ﻇﺎ ﺱ × ) Ùﻗﺎ ﺱ( ¤Ù
¤Ù
٢
)ﺏ( ٣ﺹ = ﺟﺎ ﺱ ﺟﺘﺎ ٢ﺹ
= ﻗﺎ ﺱ ﻗﺎ ٢ﺱ +ﻇﺎ ﺱ ﻗﺎ ﺱ ﻇﺎ ﺱ = ﻗﺎ ﺱ ) ﻗﺎ ﺱ +ﻇﺎ ﺱ (
٢
) (ﺹ – = /ﻗﺘﺎ ﺱ ﻇﺘﺎ ﺱ ﻇﺘﺎ ﺱ +ﻗﺘﺎ ﺱ )– ﻗﺘﺎ ٢ﺱ(
)ﺍ( ﺑﺎﻻﺷﺘﻘﺎق ﺑﺎﻟ ﺴﺒﺔ إ
= – ﻗﺘﺎ ﺱ ﻇﺘﺎ ٢ﺱ – ﻗﺘﺎ ٢ﺱ
=
i + ¤ g ¤ iﺫ i - ¤ g ¤ﺫ ¤ g ¤
¤g¤i
)ﺏ( ﺑﺎﻻﺷﺘﻘﺎق ﺑﺎﻟ ﺴﺒﺔ إ
ﺫ
j+ ¤ h ¤ jﺫj- ¤ h ¤ j+ ¤ h ¤ﺫ ¤ h ¤
)( ¤ j +1
ﺫ
) (٢أوﺟﺪ §Ùإذا ﻧﺖ ﺹ ﺴﺎوى : ¤Ù
)ﺍ( ﻇﺘﺎ ) ﺱ( ٣ + ٢
=
f ¤ fﺫ§ ﺇ ﺹ= / + 3ﺫ eﺫ§ ¤ e
) (٥أوﺟﺪ §Ùﻠﻤﻨﺤﻨﻴﺎت اﻵﺗﻴﺔ ﻋﻨﺪ اﻟﻘﻴﻢ ا ﻌﻄﺎة :
ﺫ¤ h ¤ j
¤Ù
)( ¤ j +1
ﺫ
)ﺍ( ﺱ = ) ﻥ ) ( ٧ +ﻥ – ، ( ٢ﺹ = ) ﻥ ) ( ١ + ٢ﻥ – ( ٢ ،
ا ﻞ
)ﺍ( ﺱ = ) ﻥ ) ( ٧ +ﻥ – = ( ٢ﻥ ٥ + ٢ﻥ – ١٤ﺇ ﺱ = ٢ﻥ ٥ + ،ﺹ = ) ﻥ ) ( ١ + ٢ﻥ – = ( ٢ﻥ ٢ – ٣ﻥ + ٢ﻥ – ٢ﺇ ﺹ ٣ = /ﻥ ٤ – ٢ﻥ ١ + °3 y§ §Ùﺫ 1 +°4 - ﺇ = = ﺫ5 +° y¤ ¤Ù
)ﺝ( ﺹ = /ﺟﺘﺎ )ﻗﺎ ٣ﺱ –) × (٢ﻗﺘﺎ ٣ﺱ ٢ﻇﺘﺎ ٣ﺱ ٦ × ٢ﺱ ( ٢
ﺫ5 +
،ﻋﻨﺪ ﻥ = ٢ﺉ
= ٦ﻗﺘﺎ ٢ ) ٣ﺱ +ﺑﺐ ( ﻇﺘﺎ ) ٢ﺱ +ﺑﺐ (
´ 3S4 = §Ùﺫ -ﺫ = 8 ´4S 3 ¤Ùﺫ 9 1+
) (٦أوﺟﺪ ﻗﻴﻤﺔ ا ﺎراﻣ
+ﺱ ٢ﻗﺎ 1ﻇﺎ 1 - × 1 ﺱﺫ ﺱ ﺱ – ﻗﺎ 1ﻇﺎ 1 ﺱ ﺱ
S - §Ùﺫ = 4 ¤Ù
ﺉ
ﺫ 3 ،ﺹ= / )ﺝ( ﺱ = / 1 +°4S ﺫ -°3Sﺫ
)ﻩ( ﺹ ٣ – = /ﻗﺘﺎ ٢ ) ٢ﺱ +ﺑﺐ ( × ] -ﻗﺘﺎ ) ٢ﺱ +ﺑﺐ ( ﻇﺘﺎ ) ٢ﺱ +ﺑﺐ ( [ × ٢
ﺉ
-°3S4 y§ §Ùﺫ = = 1+°4S 3 y¤ ¤Ù
ﻉ اﻟ ﻳ ﻮن ﻋﻨﺪﻫﺎ ﻠﻤﻨﺤ :
ﺱ = ٢ﻉ ٥ – ٣ﻉ ٤ – ٢ﻉ ، ١٢ +ﺹ = ٢ﻉ + ٢ﻉ – ٤
) (٣أوﺟﺪ §Ùإذا ن :
ﺎس أﻓ وآﺧﺮ رأ
. ا ﻞ
)ﺍ( ﺱ ٥ – ٣ﺱ ﺹ +ﺹ ٤ = ٣ﺱ
1+ ¬4 ﺱ ٦ = /ﻉ ١٠ – ٢ﻉ – ، ٤ﺹ ٤ = /ﻉ ١ +ﺉ = §Ù 6 ¤Ùﻉ ﺫ 4 - ¬10- إذا ن ا ﻤﺎس أﻓ :ﺇ ﻣﻴﻞ ا ﻤﺎس = ٠ﺇ ٠ = §Ùﺇ ٤ﻉ ٠ = ١ + ¤Ù ﺇ ﻉ = 1- 4
)ﺏ( ﺱ ٢ﺹ +ﺹ ٢ﺱ = ٢٥ )ﺍ( ﺑﺎﻻﺷﺘﻘﺎق ﺑﺎﻟ ﺴﺒﺔ إ
¤Ù
p3 ، 1 = y§ = §Ùﻋﻨﺪ = θ 4 y¤ ¤Ùﺫq i
= ٦ﻗﺎ ٢ﺱ ﻇﺎ ٢ﺱ – ٦ﻗﺘﺎ ٣ﺱ ﻇﺘﺎ ٣ﺱ
ﺱ:
،ﻋﻨﺪ ﻥ = ١ﺇ = 1+ 4 - 3 = §Ùﺻﻔﺮ
)ﺏ( ﺱ = ﻗﺎ ١ – θ ٢ﺇ ﺱ ٢ = /ﻗﺎ θﻗﺎ θﻇﺎ ، θﺹ = /ﻗﺎ θﻇﺎ θﺇ
) (ﺹ ٣ = /ﻗﺎ ٢ﺱ × ﻇﺘﺎ ٢ﺱ × –) ٢ + ٢ﻗﺘﺎ ٣ﺱ × ﻇﺘﺎ ٣ﺱ ( × ٣
¤Ù
،
ﻥ=٢
ا ﻞ
1 )ﺏ( ﺹ = /ﻗﺎ ]ﺱ /٢/ /–/ﻇﺎ ]ﺱ × /٢ /–/ ﺫ - ¤Sﺫ
)و(
،
/
)ﺍ( ﺹ – = /ﻗﺘﺎ ) ٢ﺱ ٢ × ( ٣ + ٢ﺱ = – ٢ﺱ ﻗﺘﺎ ) ٢ﺱ( ٣ + ٢
ﺹ ٢ = /ﺱ ﻗﺎ 1 ﺱ = ٢ﺱ ﻗﺎ 1 ﺱ
p3=θ 4
)ﺝ( ﺱ = ] ٣ﻥ ، /٢ /– /ﺹ = ] ٤ﻥ /١ /+/
)و( ﺱ ٢ﻗﺎ 1 ﺱ
= – ٦ﺱ ﺟﺘﺎ )ﻗﺘﺎ ٣ﺱ × ( ٢ﻗﺘﺎ ٣ﺱ ٢ﻇﺘﺎ ٣ﺱ
ﻥ=١
)ﺏ( ﺱ = ﻗﺎ ، ١ – θ ٢ﺹ = ﻇﺎ θ
) ٣ (ﻗﺎ ٢ﺱ ٢ +ﻗﺘﺎ ٣ﺱ
)ﻩ( – ﻗﺘﺎ ٢ ) ٣ﺱ +ﺑﺐ (
ﺱ ٣ :ﺹ = /ﺟﺘﺎ ﺱ ﺟﺘﺎ ٢ﺹ – ٢ﺟﺎ ٢ﺹ × ﺹ
/
× ﺟﺎ ﺱ ﺇ ﺹ ٢ + ٣ ) /ﺟﺎ ٢ﺹ ﺟﺎ ﺱ ( = ﺟﺘﺎ ﺱ ﺟﺘﺎ ٢ﺹ
)ﺏ( ﻗﺎ ]ﺱ / ٢ /–/
)ﺝ( ﺟﺎ ) ﻗﺘﺎ ٣ﺱ( ٢
ﺱ :ﺟﺘﺎ ﺹ +ﺱ )– ﺟﺎ ﺹ ( ﺹ + /ﺹ /ﺟﺘﺎ ﺱ +
/ﺹ§f -¤ e ﺇ ﺹ = §e¤-¤f
ﺫ )( ¤ i +1 )( ¤ i +1 )و( ﺹ( ¤ h ¤ j -0)( ¤ j -1) - ( ¤ h ¤ j +0)( ¤ j + 1) = / ﺫ )( ¤ j +1
=
ا ﻞ
ﺹ )– ﺟﺎ ﺱ( = ٠ﺇ ﺹ –) /ﺱ ﺟﺎ ﺹ +ﺟﺘﺎ ﺱ ( = – ﺟﺘﺎ ﺹ +ﺹ ﺟﺎ ﺱ
)ﻩ( ﺹ( ¤ g ¤ i +0) ¤ i - ¤ g ¤ i ( ¤ i + 1) = / ﺫ )( ¤ i +1 =
ﺱ ٢ :ﺱ ﺹ +ﺱ ٢ﺹ + /ﺹ ٢ + ٢ﺱ ﺹ ﺹ ٠ = /
ا ﻞ
،إذا ن ا ﻤﺎس رأ
ﺉ ٣ﺱ ٥ – ٢ﺹ – ٥ﺱ ﺹ ٣ + /ﺹ ٢ﺹ ٤ = /
¤ 3 - 4 /ﺫ §5 + ﺇ ) ٣ﺹ ٥ – ٢ﺱ ( ﺹ ٣ – ٤ = /ﺱ ٥ + ٢ﺹ ﺇ ﺹ = § 3ﺫ ¤ 5 -
ّ ﻣﻌﺮف ﺇ ا ﻘﺎم = ٠ :ﺇ ﻣﻴﻞ ا ﻤﺎس ﻏ
ﺇ ٦ﻉ ١٠ – ٢ﻉ – ٠ = ٤ﺉ ﻉ = ٢أ ،ﻉ = 1- 3
٢
ﻟﻸﺳﺘﺎﺫ ﺍﻟﻘﺪﻳﺮ /ﻋﻠﻰ ﺍﻟﺪﻳﻦ ﻳﺤﻴﻰ ٠١١١٩٦٦٠٦١٦
ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ -ﺍﻟﺼﻒ ﺍﻟﺜﺎﻟﺚ ﺍﻟﺜﺎﻧﻮﻯ
)) (١٠ﺍ( إذا ﻧﺖ ﺱ + ٢ﺹ ٩ = ٢أﺛﺒﺖ أن :
) (٧ﺑﺎﺳﺘﺨﺪام اﻻﺷﺘﻘﺎق ا ﺎراﻣ ى أوﺟﺪ :
Ùﺫ § ٠ = ١ + ٢( §Ù ) + ﺹ ﺫ
)ﺍ( ﺸﺘﻘﺔ ﺱ ١ + ٢ﺑﺎﻟ ﺴﺒﺔ إ üﺱﺫ 1 - ﺱ ﻋﻨﺪ ﺱ = ١ )ﺏ( ﺸﺘﻘﺔ + 8 üﺱﺫ ﺑﺎﻟ ﺴﺒﺔ إ ﺱ 1+ )ﺝ( ﺸﺘﻘﺔ ﺱ – ﺟﺎ ﺱ ﺑﺎﻟ ﺴﺒﺔ إ – ١ﺟﺘﺎ ﺱ ﻋﻨﺪ ﺱ = p 3 ا ﻞ
)ﺍ( ﺹ = ﺱ ١ + ٢ﺇ ﺹ ٢ = /ﺱ ¤
،ﻉ = ¤ üﺫ 1 -ﺇ ﻉ = /
)ﺏ( ﺹ = ü
¤ + 8ﺫ
ﺇ ﺹ= /
ﺉ = y§ = §Ù ¬Ù
¬y
) ﺱ (1+ﺫ
ü
)ﺏ( إذا ﻧﺖ ﺹ = ﻇﺎ ﺱ أﺛﺒﺖ أن : Ùﺫ § = ٢ﺹ ) + ١ﺹ( ٢ ﺫ
¤Ù
=
¤Ù Ùﺫ ﺫ § §Ù §Ù § ( ٠=١+ = ٠ﺉ ﺹ ﺫ )+ × ﺇ +١ﺹ+ Ù ¤Ù ¤Ùﺫ ¤Ù ¤Ù ¤Ù Ùﺫ §Ù ٢ § = ٢ﻗﺎ ﺱ × ﻗﺎ ﺱ ﻇﺎ ﺱ = ٢ﻗﺎ ٢ﺱ ﻇﺎ ﺱ (١) ..... )ﺏ( = ¤Ùﻗﺎ ﺱ ﺇ ¤Ùﺫ
1
4 = §Ù 3 ¬Ù
،ﰈ ٢ﺹ ) + ١ﺹ ٢ = ( ٢ﻇﺎ ﺱ ) + ١ﻇﺎ ٢ﺱ ( = ٢ﻇﺎ ﺱ × ﻗﺎ ٢ﺱ (٢) ........ ﻣﻦ ) (٢) ، (١ﺉ
)ﺝ( ﺑﻔﺮض ﺹ = ﺱ – ﺟﺎ ﺱ ﺇ ﺹ – ١ = /ﺟﺘﺎ ﺱ ،ﺑﻔﺮض ﻉ = – ١ﺟﺘﺎ ﺱ ﺇ ﻉ = /ﺟﺎ ﺱ ﺉ ،ﻋﻨﺪﻣﺎ ﺱ = ٦٠ = p 3
٥
ﻣﻦ :
) (٨أوﺟﺪ ا ﺸﺘﻘﺔ ا ﺎ ﺔ
)ﺍ( ﺹ = ﺱ ٢ – ٤ﺱ٥ + ٢
¤ f -1 y§ §Ù = = y¬ ¬Ù ¤e
Ùﺫ § ¤Ùﺫ
ا ﻞ
)ﺏ( ﻉ ٢ ) ٤ = /ﻥ – ٢ ) ٨ = ٢ × ٣( ١ﻥ – ٣( ١ﺉ ﻉ ٢ ) ٤٨ = //ﻥ – ( ١
٢
اﺸ
) ﺱ ¤ ´1- 1´ (1- ) (د ) /ﺱ( = ) ﺱ (1-ﺫ -0ﺫ) (1- ) ´1´ (1- ¤ ﺫ ﺉ د) //ﺱ( = = ) ﺱ 3 (1- ) ﺱ 4(1- ﺫ
ﺫ¬ ¬ = y§ = §Ù = y¤ ¤Ùﺫ ¬ -ﺫ ¬ 1 -
–٣
ا ﺠﺎور ﺗﻤﺜﻴﻼ
ً
//
د )س( ﻛﺜ ة ﺣﺪود
1-
) ﺱ (1 -ﺫ
،ﺣﺪد ﻣﻨﺤ
داﻟﺔ . ا ﻞ
6-
د )ﺱ( = ا ﻨﺤ )ﺏ( = داﻟﺔ ﻣﻦ ا رﺟﺔ ا ﺎ ﺔ ،د )ﺱ( = ا ﻨﺤ )ﺝ( = /
) ﺱ 4(1-
داﻟﺔ ﻣﻦ ا رﺟﺔ ا ﺎﻧﻴﺔ ،د) //ﺱ( = ا ﻨﺤ )ﺍ( = داﻟﺔ ﻣﻦ ا رﺟﺔ اﻷو
) (٩إذا ﻧﺖ ﺹ = ﺟﺎ ﺍ ﺱ اﺳﺘﻜﺸﻒ ﻧﻤﻂ اﻻﺷﺘﻘﺎق ا ﺘﺘﺎ ﺛﻢ أوﺟﺪ ﺹ). (٢٥
ا ﻞ
أﺟﺐ ﻋﻦ اﻷﺳﺌﻠﺔ اﻵﺗﻴﺔ :
ﺹ = ﺟﺎ ﺍ ﺱ ﺇ ﺹ = /ﺍ ﺟﺘﺎ ﺍ ﺱ = ﺍ ﺟﺎ ) ﺍ ﺱ ( p + ﺫ
ﺇ ﺹ – = //ﺍ ٢ﺟﺎ ﺍ ﺱ = ﺍ ٢ﺟﺎ ) ﺍ ﺱ +ﺑﺐ ( ﺇ ﺹ
p3 = ﺍ ٣ﺟﺎ ) ﺍ ﺱ + ﺫ
ا ﻞ
د )س( ،د )س( ﺣﻴﺚ /
ﺉ د) //ﺱ( = ٤ﺟﺘﺎ ٢ﺱ ﺉ د) ///ﺱ( = – ٨ﺟﺎ ٢ﺱ
´1´ (1- ¤ ) 3 -0ﺫ ﺉ د) ///ﺱ( = ) ﺱ 6(1 -
ﻋﻨﺪ ﻉ = ٢
) (١٢ﻳ ً ﺑﻴﺎﻧﻴﺎ ﻨﺤﻨﻴﺎت ا وال :د )س( ،
)ﺝ( د )ﺱ( = ﺟﺘﺎ ) ٢ﺱ +ﺑﺐ ( = – ﺟﺘﺎ ٢ﺱ ﺉ د ) /ﺱ( = ٢ﺟﺎ ٢ﺱ
=
،
§ 3Ù 3¤Ù
3Ù Ùﺫ – ¬Ù ٤ = ¤Ù § = )3ﻉ –× (١ § = -ﺫ، 1 ،ﻋﻨﺪﻣﺎ ﻉ = ٢ﺇ ¤Ùﺫ 3¤Ùﺫ ) 3ﻉ – ) 3 = 1 × ٤ – ( ١ﻉ – ، ٥ – ( ١ﻋﻨﺪﻣﺎ ﻉ = ٢ﺇ 3 = § 3Ù 4 3¤Ù ﺫ¬ -ﺫ 4 ﺫ
ﺉ ﻉ ٢ ) ١٩٢ = ///ﻥ – ( ١
=
=٢ﺹ)+١ﺹ (
Ùﺫ§ ) ¬ ¬Ù 1´ ¬ - 1´ (1 - ﺇ 1 - = ¤Ùﺫ × - = 1ﺫ ) 1ﻉ – ( ١ ﺫ= × ﺫ ﺫ¬ -ﺫ ¤Ù ) ¬ (1 - ) ¬ (1-
) (د )ﺱ( = ﺱ ﺱ 1-
)ﺍ( ﺹ ٤ = /ﺱ ٤ – ٣ﺱ ﺉ ﺹ ١٢ = //ﺱ ٤ – ٢ﺉ ﺹ ٢٤ = ///ﺱ
٢
ﺱ ٢ = /ﻉ – ، ٢ﺹ ٢ = /ﻉ ﺇ
) ﺏ( ﻉ = ) ٢ﻥ – ٤ ( ١
)ﺝ( د )ﺱ( = ﺟﺘﺎ ) ٢ﺱ +ﺑﺐ (
Ùﺫ§ ¤Ùﺫ
) (١١إذا ﻧﺖ ﺱ = ﻉ ٢ – ٢ﻉ ،ﺹ = ﻉ ٢أوﺟﺪ :
3S 60 f - 1 §Ù = = 3 ¬Ù 60 e
ﺇ
¤Ù
٢ §Ù
) ﺱ (1 +ﺫ
،ﻋﻨﺪ ﺱ = ١ﺇ
ﺱ :ﺇ ٢ﺱ ٢ +ﺹ ٠ = §Ù
إ ﺱ +ﺹ ٠ = §Ùﺑﺎﻻﺷﺘﻘﺎق ﺮة أﺧﺮى ﺑﺎﻟ ﺴﺒﺔ إ س :
ﺱ
ü
ا ﻞ
)ﺍ( ﺱ + ٢ﺹ ٩ = ٢ﺑﺎﻻﺷﺘﻘﺎق ﺑﺎﻟ ﺴﺒﺔ إ
¤ + 8ﺫ
¤ ´1- 1´ (1+ ¤ ) / ،ﻉ = ﺱ ﺇ ﻉ = ) ﺱ (1+ﺫ ﺱ 1+ ¤ + 8ﺫ
¤Ù
ﺉ y§ = §Ù = ¤ ü ٢ﺫ 1 - y¬ ¬Ù
¤ üﺫ 1 -
¤Ù
///
) (١إذا ن ﺱ + ٢ﺹ = ٢ﺱ – ﺹ أوﺟﺪ §Ùﻋﻨﺪﻣﺎ ﺱ = ١ ¤Ù
٣
= – ﺍ ﺟﺘﺎ ﺍ ﺱ
) (٢إذا ﻧﺖ ﺹ = ٢ﺱ – ١ ) ٢ﺱ ( ﻓﺄﺛﺒﺖ أن :
( ﺇ ﺹ) = (٤ﺍ ٤ﺟﺎ ﺍ ﺱ = ﺍ ٤ﺟﺎ ) ﺍ ﺱ ٢ +ﺑﺐ (
Ùﺫ § ٣ + ٢( §Ù ) +ﺱ = ١ ﺹ ﺫ
ﻥp ﺇ ﺹ)ﻥ( = ﺍﻥ ﺟﺎ ) ﺍ ﺱ + ﺫ 5ﺫ p ( = ﺍ ٢٥ﺟﺎ ) ﺍ ﺱ = ( p +ﺍ ٢٥ﺟﺘﺎ ﺍ ﺱ ﺇ ﺹ) = (٢٥ﺍ ٢٥ﺟﺎ ) ﺍ ﺱ + ﺫ ﺫ
¤Ù
(
¤Ù
) (٣إذا ن ﺱ ﺹ = ٧ﻓﺄﺛﺒﺖ أن :
٣
ﺱÙ ٣ﺫ § ¤Ùﺫ
= ١٤
ﻟﻸﺳﺘﺎﺫ ﺍﻟﻘﺪﻳﺮ /ﻋﻠﻰ ﺍﻟﺪﻳﻦ ﻳﺤﻴﻰ ٠١١١٩٦٦٠٦١٦
ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ -ﺍﻟﺼﻒ ﺍﻟﺜﺎﻟﺚ ﺍﻟﺜﺎﻧﻮﻯ
Ùﻉ = ٢ﺱ – = §Ù ، ٣ﺱ١ + ٢ ) (٤إذا ن
اﻷول = – ﻣﻘﻠﻮب ﻣﻴﻞ ا ﻤﺎس ﻠﻤﻨﺤ ا ﺎ ﻋﻨﺪ ﻧﻔﺲ
¤Ù
¤Ù Ùﺫ ﻉ ﻓﺄوﺟﺪ ﺫ ﻋﻨﺪ ﺱ = ١ §Ù
ا ﻘﻄﺔ .
) (١٠ﻣﻌﺎدﻟﺔ ا ﻤﺎس ﺑﻤﻌﻠﻮﻣﻴﺔ ﻣﻴﻠﻪ ﻡ وﻧﻘﻄﺔ ا ﻤﺎس)ﺱ،١ﺹ(١
) (٥إذا ن ٢ﺱ ﺹ ٥ = ٣ +ﺱ ٢ﻓﺄﺛﺒﺖ أن :
:ﺹ – ﺹ = ١ﻡ ) ﺱ – ﺱ( ١
Ùﺫ ﺱ § ٥ = ( §Ù ) ٢ + ﺫ
¤Ù
¤Ù
وﻣﻌﺎدﻟﺔ اﻟﻌﻤﻮدى
) (٦إذا ن ﺟﺎ ٣ﺱ – ﺟﺎ ٢ﺹ = ٠أﺛﺒﺖ أن : Ùﺫ ٢ﻇﺘﺎ ٢ﺹ § – ٩ = ٢( §Ù ) ٤ ﺫ
¤Ù
¤Ù
) (١أوﺟﺪ ﻣﻌﺎدﻟ ا ﻤﺎس واﻟﻌﻤﻮدى ﻠﻤﻨﺤ
) (٧إذا ن ﺹ = ﻗﺎ ﺱ ﻓﺄﺛﺒﺖ أن :
Ùﺫ ﺹ § = ٢( §Ù ) +ﺹ ٣ ) ٢ﺹ( ٢ – ٢ ﺫ
¤Ù
¤Ù
) (٨أوﺟﺪ ﻣﻴﻞ ا ﻤﺎس ﻠﻤﻨﺤ ﻋﻨﺪ p = q 4
ﻋﻨﺪ ا ﻘﻄﺔ اﻟ ﺗﻘﻊ
)ﺍ( ﺹ = ﻗﺘﺎ ٣ ) ٣ﺱ( ١ + ٢
¤Ù
)ﺏ( ﺹ = ﻇﺎ ) ﻇﺘﺎ ٣ﺱ (
واﻟﻌﻤﻮدى ﻠﻤﻨﺤ ٣ﺱ + ٢ﺹ ١٢ = ٢ﻋﻨﺪ ا ﻘﻄﺔ )– ( ٣ ، ١
) (١ﻹ ﺎد ﻣﻴﻞ ا ﻤﺎس ﻨﺤ ﻋﻨﺪ ﻧﻘﻄﺔ ا ﻤﺎس ﻧﻮﺟﺪ §Ù ¤Ù
ا ﻞ
¤ 3 - §Ù = ٣ﺱ + ٢ﺹ ١٢ = ٢ﺇ ٦ﺱ ٢ +ﺹ ٠ = §Ùﺇ § ¤Ù ¤Ù
ﺛﻢ ﻧﻌﻮض ﻓﻴﻬﺎ ﺑﻨﻘﻄﺔ ا ﻤﺎس ) (٢ﻹ ﺎد ﻣﻴﻞ ا ﻤﺎس إذا ُﻋﻠﻤﺖ ﻣﻌﺎدﻟﺔ ا ﻤﺎس ﻧﻮﺟﺪ §Ù ﻠﻤﺴﺘﻘﻴﻢ أو ﻧﻘﺴﻢ ) – ﻣﻌﺎ ﻞ ﺱ ( ُ ) (٣ﻹ ﺎد ﻣﻴﻞ ا ﻤﺎس إذا ﻋﻠﻤﺖ زاو ﺔ ﻣﻴﻠﻪ
ا ﻘﻄﺔ )– ( ٣ ، ١ﺇ ﻣﻴﻞ ا ﻤﺎس = ١ﺉ ﻣﻌﺎدﻟﺔ ا ﻤﺎس
¤Ù
ا ﺴ ﻨﺎت
) أى ﻧﻀﻊ ﻣﻘﺎم ا ﻜ
ﻮر ا ﺴ ﻨﺎت
ارﺗﻔﺎﻋﻪ = ٣وﺣﺪة ﻃﻮل ،وﻃﻮل ﻗﺎﻋﺪﺗﻪ = ٦ = ٤ + ٢وﺣﺪة ﻃﻮل ﺇ ﺴﺎﺣﺔ ا ﺜﻠﺚ = ٩ = ٦ × ٣ × 1وﺣﺪات ﺮ ﻌﺔ ﺫ
ﻮر ا ﺴ ﻨﺎت = ﻏ ﻣﻌﺮف
) (٣أوﺟﺪ ﻣﻌﺎدﻟ ا ﻤﺎس واﻟﻌﻤﻮدى ﻠﻤﻨﺤ ،ﺹ = ] + ٢ﺟﺎ θﻋﻨﺪ p = θ
ا ﺎﺗﺞ = ﺻﻔﺮ (
ا ﻞ
ﻣﻴﻞ ا ﻤﺎس = ﻣﻴﻞ ا ﺴﺘﻘﻴﻢ
¤Ù
ﻣﺘﻤﺎﺳﺎن ﻋﻨﺪ ﻧﻘﻄﺔ إذا ﻧﺖ ﻫﺬه ا ﻘﻄﺔ
ﻣﻴﻞ ا ﻤﺎس ﻠﻤﻨﺤ ا ﺎ ﻋﻨﺪ ﻫﺬه ا ﻘﻄﺔ .
ﻫﺬه ا ﻘﻄﺔ ﺗﻘﻊ
y¤
qe-
4
Sﺫ S 3ﺫ S 3ﺫ Sﺫ ﺇ ا ﻘﻄﺔ ) ﺫ ،ﺫ ( ،ﻣﻴﻞ اﻟﻌﻤﻮدى = ، ١ﺱ = ﺫ ،ﺹ = ﺫ Sﺫ S 3ﺫ ( أى أن : = –)×١ﺱ – ﺉ ﻣﻌﺎدﻟﺔ ا ﻤﺎس :ﺹ – ﺫ ﺫ Sﺫ S 3ﺫ ( =)×١ﺱ– ﺱ +ﺹ – ، ٠ = ٢] ٢ﻣﻌﺎدﻟﺔ اﻟﻌﻤﻮدى :ﺹ – ﺫ ﺫ
ﻣﻨﻬﻤﺎ و ﺬ ﻚ ﻣﻴﻞ ا ﻤﺎس ﻠﻤﻨﺤ اﻷول ﺴﺎوى
) (٩ﻳ ﻮن ا ﻨﺤﻨ
4
y qf = – ﻇﺘﺎ ، θﻋﻨﺪ p = θﺇ ﻣﻴﻞ ا ﻤﺎس = – ١ ﺇ = § = §Ù
ﻣﻴﻞ ا ﻤﺎس = – ﻣﻘﻠﻮب ﻣﻴﻞ ا ﺴﺘﻘﻴﻢ
ﻣﺘﻘﺎﻃﻌﺎن
ﺱ = ﺟﺘﺎ θ
ﺱ = ﺟﺘﺎ θﺇ ﺱ – = /ﺟﺎ ، θﺹ = ] + ٢ﺟﺎ θﺇ ﺹ = /ﺟﺘﺎ θ
) (٧إذا ن ا ﻤﺎس ﻊﻋ ﺴﺘﻘﻴﻢ ﻣﻌﻠﻮم ﻓﺈن
ﺗﻘﻊ
)– ، ( ٠ ، ٤ﰈ ﻣﻴﻞ اﻟﻌﻤﻮدى = – ١ﺉ ﻣﻌﺎدﻟﺔ اﻟﻌﻤﻮدى
:
ا ﺴ ﻨﺎت ) ( ٠ ، ٢ﺇ ا ﺜﻠﺚ ا ﺤﺪد ﺑﻤﺤﻮر ا ﺴ ﻨﺎت وا ﻤﺎس واﻟﻌﻤﻮدى
) (٦إذا ن ا ﻤﺎس ﺴﺘﻘﻴﻢ ﻣﻌﻠﻮم ﻓﺈن
) (٨ﻳ ﻮن ا ﻨﺤﻨ
:
ﺹ – – = ٣ﺱ – ، ١ﺑﻮﺿﻊ ﺹ = ٠ﺇ ﺱ = ٢ﺇ اﻟﻌﻤﻮدى ﻳﻘﻄﻊ ﻮر
) (٤ﻣﻴﻞ ا ﻤﺎس ا ى ﻳﻮازى ﻮر ا ﺴ ﻨﺎت = ﺻﻔﺮ
) (٥ﻣﻴﻞ ا ﻤﺎس اﻟﻌﻤﻮدى
،ﻋﻨﺪ
ﺹ – = ٣ﺱ ١ +و ﻮﺿﻊ ﺹ = ٠ﺇ ﺱ = – ٤ﺇ ا ﻤﺎس ﻳﻘﻄﻊ ﻮر
) ﻣﻌﺎ ﻞ ﺹ (
ا ﻮﺟﺐ ﻓﺈن ﻣﻴﻞ ا ﻤﺎس = ﻇﺎ ﻩ
ا ﻨﺤ و ﺣﺪاﺛﻴﻬﺎ ا ﺴ
) (٢أوﺟﺪ ﺴﺎﺣﺔ ا ﺜﻠﺚ ا ﺤﺪود ﺑﻤﺤﻮر ا ﺴ ﻨﺎت وا ﻤﺎس
ﺗﺬﻛﺮ أن :
ﺫp 3
ﺫp ﺹ = + ٣ﻗﺎ ﺱ ﺇ = §Ùﻗﺎ ﺱ ﻇﺎ ﺱ ،ﻋﻨﺪ ﺱ = 3 ¤Ù 1 ،ﺹ= ١ﺉ ﺇ ﻣﻴﻞ ا ﻤﺎس = ، ٣] ٢ﻣﻴﻞ اﻟﻌﻤﻮدى = ﺫ 3S ﺫp ﺫp ا ﻘﻄﺔ ) ( ١ ، 3ﺇ ﻣﻌﺎدﻟﺔ ا ﻤﺎس :ﺹ – ) ٣] ٢ = ١ﺱ – ( 3 أى أن ٣] ٢ :ﺱ – ﺹ ٣] 4 -ﺑﺐ ، ٠ = ١ +ﻣﻌﺎدﻟﺔ اﻟﻌﻤﻮدى : 3 ﺫp ﺫ ﺹ – ) 1 - = ١ﺱ – ( 3أى أن :ﺱ ٣] ٢ +ﺹ – 3 – ٣] ٢ﺑﺐ = ٠ ﺫ 3S
:ﺱ = ٢ﺟﺘﺎ ، q ٣ﺹ = ٥ﺟﺎq ٣
ﺎ ﻳﺄ :
ﺹ = + ٣ﻗﺎ ﺱ
ا ﻞ
Ùﺫ § =٢ ) (٩إذا ن ﺱ = ﻗﺎ ] ، qﺹ = ﻇﺎ qﻓﺄﺛﺒﺖ أن : ﺫ
) (١٠أوﺟﺪ ا ﺸﺘﻘﺔ اﻷو
:ﺹ – ﺹ ) × 1 – = ١ﺱ – ﺱ( ١ ﻡ
ا ﻌﺎﻣﺪ ﻋﻨﺪ ﻧﻘﻄﺔ إذا ﻧﺖ
أى أن :ﺱ – ﺹ ٠ = ٢] +
ﻣﻨﻬﻤﺎ و ﺬ ﻚ ﻣﻴﻞ ا ﻤﺎس ﻠﻤﻨﺤ
٤
ﻟﻸﺳﺘﺎﺫ ﺍﻟﻘﺪﻳﺮ /ﻋﻠﻰ ﺍﻟﺪﻳﻦ ﻳﺤﻴﻰ ٠١١١٩٦٦٠٦١٦
ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ -ﺍﻟﺼﻒ ﺍﻟﺜﺎﻟﺚ ﺍﻟﺜﺎﻧﻮﻯ ) (٤إذا ﻧﺖ ا ﻘﻄﺔ ) ( ٢ ، ١إﺣﺪى ﻧﻘﻂ ﺗﻘﺎﻃﻊ ا ﻨﺤﻨ ﺹ – ٢ﺱ ، ٣ = ٢ﺱ ﺹ = ٢ﻫﻞ ﻳﺘﻌﺎﻣﺪ ا ﻨﺤﻨ ا ﻘﻄﺔ ؟ ﻓ ّ إﺟﺎﺑﺘﻚ .
: ﻋﻨﺪ ﻫﺬه
اﻟﻌﻤﻮدﻳ
§Ùﺱ ﺹ ،ﺱﺹ =٢ = ﺹ – ٢ﺱ ٣ = ٢ﺇ ٢ﺹ ٢ – §Ùﺱ = ٠ﺇ ¤Ù ¤Ù 1 ﺇ ﺱ + §Ùﺹ = ٠ﺇ - = §Ùﺹ ،ﻋﻨﺪ ا ﻘﻄﺔ ) : ( ٢ ، ١ﻡ = ١ ¤Ù
¤Ù
ﺱ
١ – = (٢ –) × 1ﺉ ،ﻡ 1- = ٢ﺫ = – ٢ﺇ ﻡ × ١ﻡ = ٢ﺫ
ﻣﺘﻌﺎﻣﺪان أى أن :ا ﻨﺤﻨﻴﺎن ﻳﺘﻘﺎﻃﻌﺎن
ﺫ
،ﺹ = ﺱ ٣ – ٢ﺱ – ٢ﻳﺘﻘﺎﻃﻌﺎن
) (١٤إذا ن ا ﻨﺤﻨﻴﺎن ) :ﺱ – ﺍ ( + ٢ﺹ، ٨ = ٢ ) ﺱ +ﺍ ( + ٢ﺹ ٨ = ٢ﻣﺘﻘﺎﻃﻌ
ﻋﻨﺪ ﻧﻘﻄﺔ ﺗﻘﺎﻃﻌﻪ ﻣﻊ ﻮر ا ﺴ ﻨﺎت .
واﻟﻌﻤﻮدى ﻋﻠﻴﻪ ﻠﻤﻨﺤ
ﻮر ا ﺴ ﻨﺎت ﺛﻢ أوﺟﺪ
) (٣أوﺟﺪ ﻣﻌﺎدﻟﺔ ا ﻤﺎس ﻨﺤ ا اﻟﺔ ﺹ = ﺱ ﺟﺘﺎ ٢ﺱ +ﺟﺎ ﺱ
) (١ا ﻌﺪل ا ﺰﻣ ﻫﻮ إ ﺎد ا ﺸﺘﻘﺔ اﻷو ﺑﺎﻟ ﺴﺒﺔ ﻠﺰﻣﻦ و
ﻋﻨﺪ ﻧﻘﻄﺔ اﻷﺻﻞ .
ا ﻌﺪل ا ﺰﻣ ﺑﺎ ﻮﺟﺐ إذا ن ﻳ اﻳﺪ ) ﻳﺘﻤﺪد أ ،ﻳ ا ﻢ أ،
وذ ﻚ
ﻳ ﺘﻌﺪ أ ،ﻳﺼﺐ ( ،و ) (٢ا ﻜﻤﻴﺔ
) (٤إذا ن ﻣﻢ ﺍﺏ ﺝ ﰲ ﻣﻢ ﻩ و ﻓﺈن :
ﺹ = ٢ﺟﺎ ﺱ +ﺟﺘﺎ ﺱ
ﺍB Ùﻩ
ﺹ = ٣ﺱ + ٣ﺹ ٦ – ٢ﺱ
· ا ﺜﻠﺚ :ﺴﺎﺣﺔ ﺳﻄﺤﻪ = 1ﺫ × ﻃﻮل اﻟﻘﺎﻋﺪة × اﻻرﺗﻔﺎع = 1ﺫ ﺍ /ﺏ /ﺟﺎ ﺝ =
) (٨أوﺟﺪ ا ﻘﻂ ا ﻮاﻗﻌﺔ ا ﻨﺤ ﺱ – ٢ﺱ ﺹ +ﺹ٣ = ٢ ً واﻟ ﻳ ﻮن ﻋﻨﺪﻫﺎ ا ﻤﺎس ﻮاز ﺎ ﻮر ا ﺼﺎدات ﺛﻢ أوﺟﺪ ) (٩أوﺟﺪ ا ﻘﻂ ا ﻮاﻗﻌﺔ
=
ﺏÜ ﻩc
=
ﺍÜ cÙ
) (٥ﻣﻦ ا ﻔﻴﺪ ﺗﺬﻛﺮ اﻟﻌﻼﻗﺎت اﻵﺗﻴﺔ :
٩ +ﻋﻨﺪ ﻧﻘﻄﺔ ﺗﻘﺎﻃﻌﻪ ﻣﻊ ا ﺴﺘﻘﻴﻢ ﺹ = ﺱ
ﻣﻌﺎدﻟﺔ اﻟﻌﻤﻮدى ﻋﻨﺪ
أى ﻈﺔ = ا ﻜﻤﻴﺔ اﻻﺑﺘﺪاﺋﻴﺔ +ا ﻌﺪل × ا ﺰﻣﻦ
= ﺮ ﻊ ﻓﺮق ا ﺴ ﻨﺎت +ﺮ ﻊ ﻓﺮق ا ﺼﺎدات
ا ﺴ ﻨﺎت وذ ﻚ ﻋﻨﺪ ا ﻘﻄﺔ ) . ( ١ ، ١
) (٧أوﺟﺪ ﻣﻌﺎدﻟﺔ اﻟﻌﻤﻮدى
ب(.
) (٣ﺮ ﻊ ا ﻌﺪ ﺑ ﻧﻘﻄﺘ
ﺱ ٣ + ٢ﺱ ﺹ +ﺹ ٥ = ٢ﻣﻊ اﻻ ﺎه ا ﻮﺟﺐ ﺤﻮر
-B) BSﺍ( yÜ-B)( y B-B )( y
و ذا ن ﻣ ﺴﺎوى اﻷﺿﻼع ﻓﺈن ﺴﺎﺣﺘﻪ = 3Sل٢ 4 · ا ﺮ ﻊ :ﺴﺎﺣﺘﻪ = ل ، ٢ﻴﻄﻪ = ٤ل ،ﻃﻮل ﻗﻄﺮه = ل ]٢
ﻣﻨﻬﺎ .
وﻗﻄﺮاه ﻣﺘﻌﺎﻣﺪان وﻣ ﺴﺎو ﺎن .
ا ﻨﺤ
· ا ﺴﺘﻄﻴﻞ :ﺴﺎﺣﺘﻪ = اﻟﻄﻮل × اﻟﻌﺮض ،
ﺹ = ) ﺱ ٢ – ٢ﺱ ٢ – ٥ ) ( ١ +ﺱ ( واﻟ ﻳ ﻮن ا ﻤﺎس
ﻴﻄﻪ = ) اﻟﻄﻮل +اﻟﻌﺮض ( × ٢وﻗﻄﺮاه ﻣ ﺴﺎو ﺎن
ﻋﻨﺪﻫﺎ ﻮاز ﺎ ﻮر ا ﺴ ﻨﺎت ﺛﻢ أوﺟﺪ ﻣﻌﺎدﻟﺔ اﻟﻌﻤﻮدى ﻋﻨﺪ
· ا اﺋﺮة :ﺴﺎﺣﺘﻬﺎ = ﺑﺐ ﻗﻖ، ٢
ﻣﻨﻬﺎ .
٥
ﻮن
ﻮن ا ﻌﺪل ﺑﺎ ﺴﺎﻟﺐ إذا ن ﻳ ﻨﺎﻗﺺ
) ﻳﻨﻜﻤﺶ أ ،ﻳﻘ ب أ ،ﻳﻨﺼﻬﺮ أ ،ﻳ
) (٥أوﺟﺪ ﻗﻴﺎس ا ﺰاو ﺔ اﻟ ﻳﺼﻨﻌﻬﺎ ا ﻤﺎس ﻠﻤﻨﺤ
ا ﻨﺤ
٤ﺱ + ٢ﺹ ٢٠ = ٢ﻋﻨﺪ ا ﻘﻄﺔ
).(٤–،١
ﻣﻌﺎدﻟﺔ ا ﻤﺎس ﻠﻤﻨﺤ ﻋﻨﺪ ا ﻘﻄﺔ ﺱ = – ١
ﻋﻨﺪ ا ﻘﻄﺔ ﺱ = ٠
ا ﻌﺎﻣﺪ ﻓﺄوﺟﺪ ﻗﻴﻤﺔ ﺍ
) (١٥أوﺟﺪ ﺴﺎﺣﺔ ا ﺜﻠﺚ ا ﺤﺪود ﺑﻤﺤﻮر ا ﺼﺎدات وا ﻤﺎس
ﺹ = ﺱ + ٣ﺱ ٢ +ﻋﻨﺪ أى
) (٦أوﺟﺪ ﻣﻌﺎدﻟﺔ ا ﻤﺎس ﻠﻤﻨﺤ
ا ﻌﺎﻣﺪ ﻋﻨﺪ ا ﻘﻄﺔ
)(٤–،١
) (١أوﺟﺪ ﻣﻴﻞ ا ﻤﺎس ﻠﻤﻨﺤ ﺱ + ٢ﺹ ٢ – ٢ﺱ ٤ +ﺹ – ٠ = ٨
ﺫ¤ g ) (٤أوﺟﺪ ﻣﻌﺎدﻟﺔ ا ﻤﺎس ﻨﺤ ا اﻟﺔ ﺹ = g -1ﺫ ﺱ ﻋﻨﺪ ا ﻘﻄﺔ ﺱ = p 6
ﺹ = ٣ﺱ ٥ – ٢ﺱ – ٢
) (١٣أﺛﺒﺖ أن ا ﻨﺤﻨ
أﺟﺐ ﻋﻦ اﻷﺳﺌﻠﺔ اﻵﺗﻴﺔ :
ﻧﻘﻄﺔ ﻋﻠﻴﻪ ﻳﻤﻴﻞ ﺑﺰاو ﺔ ﺣﺎدة
ﺹ = ﺱ – ٢ﺱ ، ٢ +ﺹ = ٣ﺱ – ﺱ٢
ﻣﺘﻤﺎﺳﺎن ﻋﻨﺪ ا ﻘﻄﺔ ) ( ٢ ، ١وأوﺟﺪ ﻣﻌﺎدﻟﺔ ا ﻤﺎس ا ﺸ ك
) (٢ﺑﺮﻫﻦ
ﺱﺹ=ﺱ+ﺹ
ا ﺴﺘﻘﻴﻢ ٤ﺱ – ﺹ ٠ = ٣ +
) (١٢أﺛﺒﺖ أن ا ﻨﺤﻨ
ا ﻌﺎﻣﺪ ﻋﻨﺪ ا ﻘﻄﺔ ). ( ٢ ، ١
أن ا ﻤﺎس ﻠﻤﻨﺤ
) (١١أوﺟﺪ ﻣﻌﺎدﻟ ا ﻤﺎﺳ
ﻠﻤﻨﺤ
ا ان ﻳﻮاز ﺎن ا ﺴﺘﻘﻴﻢ ٢ﺱ +ﺹ ٠ = ٨ +
ا ﻞ
ﺎﺳﺎ ا ﻨﺤﻨ
) (١٠أوﺟﺪ ﻣﻌﺎدﻟ ا ﻤﺎﺳ
ﻨﺤ ا اﻟﺔ د ) ﺱ ( = ﺱ 3 + ﺱ
ﻗ ﻴﻄﻬﺎ = ٢ﺑﺐ ﻖ
ﻟﻸﺳﺘﺎﺫ ﺍﻟﻘﺪﻳﺮ /ﻋﻠﻰ ﺍﻟﺪﻳﻦ ﻳﺤﻴﻰ ٠١١١٩٦٦٠٦١٦
ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ -ﺍﻟﺼﻒ ﺍﻟﺜﺎﻟﺚ ﺍﻟﺜﺎﻧﻮﻯ
ً · ﺴﺎﺣﺔ أى ﻀﻠﻊ ﻣﻨﺘﻈﻢ ﻋﺪد أﺿﻼﻋﻪ ﻥ ﺿﻠﻌﺎ :
ﺇ = òﺱ × ٢ﻉ ﺇ
ﻡ = ﻥ ﺱ ٢ﻇﺘﺎ 180 ﻥ 4
ﺇ
BÙ ،ﺗﻨﻌﺪم ا ﺰ ﺎدة ﻋﻨﺪﻣﺎ Ùﻥ
· ا ﻜﻌﺐ :ﺣﺠﻤﻪ = ل ، ٣ﻃﻮل ﻗﻄﺮه = ل ]، ٣ ﻴﺔ = ٦ل٢
ﺳﻮر رأ
ً
ارﺗﻔﺎﻋﻪ
٣أﻣﺘﺎر .ﻓﺈذا اﻧﺰﻟﻖ اﻟﻄﺮف ﺍ ﻣﺒﺘﻌﺪا ﻋﻦ ا ﺴﻮر ﺑﻤﻌﺪل
ﺴﺎﺣﺘﻬﺎ ا ﺴﻄﺤﻴﺔ = ٤ﺑﺐ ﻗﻖ٢
5ﻡ /د ،أوﺟﺪ ﻣﻌﺪل ﻫﺒﻮط اﻟﻄﺮف ﺏ ﻋﻨﺪﻣﺎ ﺗﺼﻞ إ 4
ﺣﺎﻓﺔ ا ﺴﻮر .
د ،وﺗﺰداد ﺴﺎﺣﺔ ﺳﻄﺤﻪ
أوﺟﺪ ﻃﻮل ﺣﺮف ا ﻜﻌﺐ
kÙ
ﻈﺔ ﻣﺎ ﺑﻤﻌﺪل ٠٧٢ﺳﻢ / ٣د ،
ﺇ × ١٢ = ٠٧٢ﺱ × ٠٠٢ﺉ ﺱ = ٣ﺳﻢ
ﺇ Ùﻡ = ١٢ﺱ ¤Ù kÙ kÙ
Ùﻥ
اﻟﻌﻠﻮى ﻋﻨﺪﻣﺎ ﻳ ﻮن ﻗﻴﺎس ا ﺰاو ﺔ ﺑ ا ﺴﻠﻢ واﻷرض . p
kÙ ﺹ ﺇ ﺱ ﻇﺎ = ٦٠ﺹ ﺇ ﺹ = ] ٣ﺱ ،ﰈ ﻇﺎ = θ ﺱ ﺇ - = §Ùﺱ × ٣] ١٠ – = ٣٠ﺳﻢ /د Ùﻥ 3Sﺱ
ﺷ
ﺫ
ﺏ /ﺴﺎوى ١ﺳﻢ /د ﻓﺄوﺟﺪ ﻣﻌﺪل
ﻣﻦ ﺍ ، /ﻕ ) ﻻ ﺍ( ﻋﻨﺪ ا ﻠﺤﻈﺔ اﻟ ﻳ ﻮن ﻓﻴﻬﺎ ﺏ
،ﻋﻨﺪ ﺏ = ٨ﺳﻢ ﻓﺈن ﺍ = ٦ﺳﻢ /
ﺇ
ﺹ
/
Ùﺍy ×٨×1 +١ ×٦×1 ﺫ ﺫ Ùﻥ
ﺱ
ﻣﺘﻮازى ﺴﺘﻄﻴﻼت ﻗﺎﻋﺪﺗﻪ ﺮ ﻌﺔ
ﻳ ﻨﺎﻗﺺ ﺑﻤﻌﺪل ٢ﺳﻢ /د .أوﺟﺪ ﻣﻌﺪل ﺗﺰاﻳﺪ ﺣﺠﻤﻪ ﻋﻨﺪﻣﺎ
ﻳ ﻮن ﻃﻮل ﺿﻠﻊ ﻗﺎﻋﺪﺗﻪ ٥ﺳﻢ وارﺗﻔﺎﻋﻪ ٢٠ﺳﻢ ،ﺑﻌﺪ
دﻗﻴﻘﺔ ﻳﺘﻮﻗﻒ ﺗﻐ ﺣﺠﻢ ﻣﺘﻮازى ا ﺴﺘﻄﻴﻼت ﻋﻦ ا ﺰ ﺎدة .
/
= ٠ﺉ
Ùﺍy Ùﻥ
ﺏ
=٠
ﺝ ﺍ
/
ﺏ
ﺍ /
/
ﺝ
= 3 - = 6 -ﺳﻢ /ث 8
4
Ù ٢ /ﺍ Ùﺍy Ùﺏ y = ﻇﺎ ﺍ +ﺏ ﻗﺎ ﺍ ،ﻇﺎ ﺍ = ﺍ yﺇ ﺏ /ﻇﺎ ﺍ = ﺍ /ﺇ Ùﻥ Ùﻥ Ùﻥ ﺏy ﺍ Ù ﺍ Ù = – / ٠١٢ث = 3-ﺉ ﺇ × ٢( 5 ) ٨ + 3 × ١ 4 4 4 Ùﻥ Ùﻥ
،ﻃﻮل ﺿﻠﻌﻬﺎ ﻳ اﻳﺪ ﺑﻤﻌﺪل ١ﺳﻢ /د وارﺗﻔﺎﻋﻪ
ﺑﻔﺮض أﺑﻌﺎد ﻣﺘﻮازى ا ﺴﺘﻄﻴﻼت
ﺝ ،ﺴﺎﺣﺘﻪ ﺛﺎﺑﺘﺔ و ﺴﺎوى ٢٤
Ù / 1ﺏ Ù / 1 yﺍy +ﺏ ﻡ = 1ﺍ × /ﺏ ٢٤ = /ﺇ ﺍ ﺫ ﺫ Ùﻥ ﺫ Ùﻥ
ﺱ + ٢ﺹ = ٢ل ٢ﺣﻴﺚ ل = ﻃﻮل ا ﺴﻠﻢ Ùﻥ
1ﺫ
ا ﻞ
3
θ
ﺝ
ﺴﺎوى ٨ﺳﻢ .
ﺑﻤﻌﺪل ٣٠ﺳﻢ /ث .أوﺟﺪ ﻣﻌﺪل اﻧﺰﻻق اﻟﻄﺮف اﻟﻌﻠﻮى
ﺇ ٢ﺱ ٢ + ¤Ùﺹ ٠ = §Ùﺇ
ﻩ
× ¤Ù kÙ 3 §Ù ﻡ/د = ﺇ ﺱ = ٤ﺳﻢ ﺇ 5 Ùﻥ
ﺳﻢ ، ٢إذا ن ﻣﻌﺪل ﺗﻐ
ﺗﻐ
٣ﻡ ﺹ
ﺱ
+ 9Sﺱ
) (٥ﺍﺏ ﺝ ﻣﺜﻠﺚ ﻗﺎﺋﻢ ا ﺰاو ﺔ
أرض أﻓﻘﻴﺔ وﻃﺮﻓﻪ اﻟﻌﻠﻮى ) (٢ﻳﺮﺗ ﺰ ﺳﻠﻢ ﺑﻄﺮﻓﻪ ا ﺴﻔ ً ﺣﺎﺋﻂ رأ .إذا اﻧﺰﻟﻖ اﻟﻄﺮف ا ﺴﻔ ﻣﺒﺘﻌﺪا ﻋﻦ ا ﺎﺋﻂ
¤Ù × ¤ - = §Ù ﺹ kÙ Ùﻥ
ﺇ ﺹ=
15ﺫ = + ٩ ) ١٥ﺱ( ٢
٥ﻡ ﺍ
ﺏ
3 + ٩ ) 1 -ﺱ ( ٢ﺫ × ٢ﺱ
،ﻋﻨﺪﻣﺎ ﺗﻨﻄﺒﻖ ﺏ
kÙ
ا ﻞ
+ 9Sﺱ ﺫ = 3 ﺹ 5
ﺇ × ١٥ = §Ù
BÙ = ٣ﺱ ٠٥٤ = ٠٠٢ × ٩ × ٣ = ¤Ù ٢ﺳﻢ / ٣د ،ﰈ = òﺱ ٣ﺇ Ùﻥ
ﺝ ﺇ ﺍ=
+ 9Sﺱ ﺫ
،ﻣﻦ اﻟ ﺸﺎﺑﻪ :ﺍÜÙ = Ù ﺍ Bﺏ Ú
ﺣﺠﻤﻪ ﺣﻴ ﺌﺬ . ﺑﻔﺮض ﻃﻮل ﺣﺮف ا ﻜﻌﺐ = ﺱ ﺇ ﻡ = ٦ﺱ ٢ﺇ
4
ﻣﻢ ﺍﺝ ﻗﺎﺋﻢ ا ﺰاو ﺔ
ﻫﺬه ا ﻠﺤﻈﺔ وﻣﻌﺪل ا ﺰ ﺎدة
ا ﻞ
ا ﻞ
5 = ¤Ùﻡ /د
ﻜﻌﺐ ﻳﺘﻤﺪد ﺑﺎ ﺮارة ﻓ داد ﻃﻮل ﺣﺮﻓﻪ ﺑﻤﻌﺪل ٠٠٢ﺳﻢ /
اﺸ
أرض أﻓﻘﻴﺔ و ﺈﺣﺪى ﻧﻘﻄﻬﺎ
ﺍ
) (٣ﺟﺴﻢ ﻣﻌﺪ
= ٠ﺉ ﻥ = ٥دﻗﻴﻘﺔ
) (٤ﻣﺎﺳﻮرة ﻣﻴﺎه ﻃﺮﻓﺎﻫﺎ ﺍ ،ﺏ وﻃﻮ ﺎ ٥أﻣﺘﺎر ،ﺴ ﻨﺪ ﺑﻄﺮﻓﻬﺎ
ﻗﻉ · اﻻﺳﻄﻮاﻧﺔ :ﺣﺠﻤﻬﺎ = ﺑﺐ ﻗﻖ ٢ﻉ ،ﺴﺎﺣﺘﻬﺎ ا ﺎﻧ ﻴﺔ = ٢ﺑﺐ ﻖ
) (١
BÙ = + ٥ ) ٢ﻥ ( ) ٢ – ٢٠ﻥ ( – + ٥ ) ٢ﻥ ( Ùﻥ
٢
= + ٥ ) ٢ﻥ ( ) ٢ – ٢٠ﻥ – – ٥ﻥ ( = + ٥ ) ٢ﻥ ( ) ٣ – ١٥ﻥ (
ﺮ ﻊ ﻗﻄﺮه = ﺱ + ٢ﺹ + ٢ﻉ ٢
3
kÙ
+ ٥ ) :ﻥ ( + ٥ ) ،ﻥ ( ٢ – ٢٠ ) ،ﻥ ( ﺉ + ٥ ) = òﻥ ( ٢ – ٢٠ ) × ٢ﻥ (
ﺴﺎﺣﺘﻪ ا ﺎﻧ ﻴﺔ = ﻴﻂ ﻗﺎﻋﺪﺗﻪ × اﻻرﺗﻔﺎع ،
· ا ﻜﺮة :ﺣﺠﻤﻬﺎ = 4ﺑﺐ ﻗﻖ، ٣
kÙ
× ١٥٠ = ٢٠ × ١ × ٥ﺳﻢ / ٣د ،ﺑﻔﺮض اﺑﻌﺎد ﻣﺘﻮازى ا ﺴﺘﻄﻴﻼت ﻋﻨﺪ أى ﻈﺔ
· ﻣﺘﻮازى ا ﺴﺘﻄﻴﻼت :ﺣﺠﻤﻪ = ﺱ × ﺹ × ﻉ ،
ﺴﺎﺣﺘﻪ ا
BÙ Ùﻥ
= ﺱ ٢ + ¬Ù × ٢ﺱ × × ¤Ùﻉ = )٢ + (٢ –) × ٢(٥
أﺟﺐ ﻋﻦ اﻷﺳﺌﻠﺔ اﻵﺗﻴﺔ :
ﻢ
) (١ﺗﺘﺤﺮك ﻧﻘﻄﺔ
ا ﻨﺤ
ﺗﻐ إﺣﺪاﺛﻴﻬﺎ ا ﺴ
ا ﻞ
ﺹ=ﺱ– ﺱ ﺱ ﺫ 1+
ﻓﺈذا ن ﻣﻌﺪل
ﺑﺎﻟ ﺴﺒﺔ ﻠﺰﻣﻦ ﻋﻨﺪ ﺱ = ] ٢ﺴﺎوى
. ٩أوﺟﺪ ﻋﻨﺪ ﻧﻔﺲ ا ﻘﻄﺔ ﻣﻌﺪل ﺗﻐ إﺣﺪاﺛﻴﻬﺎ ا ﺼﺎدى
:ﺱ،ﺱ،ﻉ
٦
ﻟﻸﺳﺘﺎﺫ ﺍﻟﻘﺪﻳﺮ /ﻋﻠﻰ ﺍﻟﺪﻳﻦ ﻳﺤﻴﻰ ٠١١١٩٦٦٠٦١٦
ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ -ﺍﻟﺼﻒ ﺍﻟﺜﺎﻟﺚ ﺍﻟﺜﺎﻧﻮﻯ
ﺱ ﺱ · lﺍ = 1-ﻮ ﻩ ﺍ ﺉ lﻩ ١ = 1- ﺱ ﺱ
ﺑﺎﻟ ﺴﺒﺔ ﻠﺰﻣﻦ .
ﺱ¬ 0
) (٢أﺳﻄﻮاﻧﺔ ﺗﺘﻤﺪد ﺑﺎﻧﺘﻈﺎم ﻴﺚ ﺗﻈﻞ ﺘﻔﻈﺔ ﺸ ﻬﺎ ﻓﺈذا ن
· lﻮ ﻩ ﺱ = ﳘﺲ l ،ﻮ ﻩ ﺱ = -ﳘﺲ
ﻃﻮل ﻧﺼﻒ ﻗﻄﺮﻫﺎ ﻗﻖ ﻳﺰداد ﺑﻤﻌﺪل ٠٢ﺳﻢ /ث وارﺗﻔﺎﻋﻬﺎ
ﻉ ﻳﺰداد ﺑﻤﻌﺪل ٠١ﺳﻢ /ث .أوﺟﺪ ﻣﻌﺪل ا ﻐ
ﺱ¬ ¦
ﺣﺠﻢ
اﻷﺳﻄﻮاﻧﺔ ﻋﻨﺪﻣﺎ ﻗﻖ = ٢ﺳﻢ ،ﻉ = ٥ﺳﻢ .
ﺱ¬ 0
ﺱ¬ 0
· lﻩﺱ = l ، ١ﻩﺏ = ﻩﺏ ) ﺣﻴﺚ :ﺏ ﻱ ( ò ﺱ¬ ¦
ﺱ¬ 0
) (٣ﻣﺘﻮازى ﺴﺘﻄﻴﻼت أﺑﻌﺎده ١٢ ، ٤ ، ٣ﺳﻢ ،إذا ن ﻣﻌﺪل
ﺗﺰاﻳﺪ ﺑﻌﺪه اﻷول ٢ﺳﻢ /ث ،وﻣﻌﺪل ﺗﺰاﻳﺪ ﺑﻌﺪه ا ﺎ ١ﺳﻢ/ث
)) (١ﺍ( 1 + ١ ) l ﺱ
،وﻣﻌﺪل ﺗﻨﺎﻗﺺ ﺑﻌﺪه ا ﺎﻟﺚ ٣ﺳﻢ /ث ،ﻓﺄوﺟﺪ ﺣﺠﻢ ﻣﺘﻮازى أى ﻈﺔ ،وﻣﻌﺪل ﺗﻐ ﺣﺠﻤﻪ
ا ﺴﺘﻄﻴﻼت
ﻧﻬﺎﻳﺔ ٢ﺛﺎﻧﻴﺔ
،وﻣﺎ ﺴﺎﺣﺘﻪ ا ﺴﻄﺤﻴﺔ ﻋﻨﺪﺋﺬ ؟
) (٤ﻧﻘﻄﺔ ﺗﺘﺤﺮك ا ﻘﻄﺔ
اﺴ
ﺱ¬ ¦
ﺱ 1 ﺱ 1 ﺇ ا ﻘﺪار = = 5 [ ( 1 + ١ ) l ] = 5 [ ( 1 + ١ ) ] l
)ﺏ( ا ﻘﺪار = ٢( 1 + ١ ) lﺱ )+ ١
ﺑﺎﻟ ﺴﺒﺔ ﻠﺰﻣﻦ ﺴﺎوى ﻣﻌﺪل ﺗﻐ إﺣﺪاﺛﻴﻬﺎ ا ﺼﺎدى
وﻋﻤﻮدﻳﺔ
)) (٢ﺍ(
ﺴﺎﻓﺔ ٥أﻣﺘﺎر
ﺹ¬ ¦
) (٣اﺛﺒﺖ أن
¥
ﻥ¬¦
= l
ﻥ¬¦
ﻗﻮاﻋﺪ ﻫﺎﻣﺔ :
· l
ﺱ¬ 0
ﺍ
ﺱ
ﺍ
ﺱ¬ 0
ا ﻞ ﲤ٠ ﳘ ﺉ ﺹ ﺲ ﲤ ﺲ ﺇ ﺱ= ، 3ﺱ ﺲ 1 ﺹ
ﺹ
٣
[ = ٣ﻩ ﺹ
ﺹ
1 ٢ﺹ [ = ٢ -ﻩ
ﺹ
ﺹ
ﺹ¬ ¦
ﺹ
lﻥ] ﻮﻩ)ﻥ –(١+ﻮﻩﻥ[=١
ﻥ¬ ¦
ا ﻞ
ا ﻘﺪار = lﻥ ﻮ ﻥ l = 1 +ﻥ ﻮ ) ( 1 + ١
ﺍÙ ﺏ
ﺱ
ﺱ¬ ¦
ا ﻘﺪار = ) lﺹ ( 1 -ﺹ ( 1 - ١ ) l = ١ -ﺹ ×) = ١ -( 1 - ١ﻩ
· ﻠﺤﻮﻇﺔ ﻫﺎﻣﺔ :إذا ن ﺍ ،ﺏ ،ﺝ ، ،و أﻋﺪاد ﺴ ﻴﺔ ﻓﺈن :
+1) oﺱ (
ﺱ
ﳘ ﲤ ﺲ ﳘ ﺉ ﺹ ﺲ )ﺝ( ﻧﻔﺮض + ١ﺱ = ﺹ ﺇ ﺱ = ﺹ – ، ١ﺱ ﲤﺲ ﺲ
ﻥ =0
= ﻮ ﻩ ﺉ l
ﺱ¬ ¦
ﺹ ¬0
1 ﳘ=º ،ﻩ = ........... + 1 + 1 + 1 + ١ﺲ ﻥ 3 ﺫ 1
ﻩ
) ﺏ(
٢( 1 - ١ ) lﺱ
+1ﺱ
ا ﻘﺪار = ] + ١ ) lﺹ (
1 =+١) lﺱ(ﺱ
+1) oﺱ (
٢
) lﺱ (ﺱ
ﺱ
()ﻩ
B + ١ ) lﺱﺍ ( Ü ±ﺱ _ و = ﺱ¬ ¦ ُ ) ﺴﺘﺨﺪم ﻞ اﻷﺳﺌﻠﺔ ا ﻮﺿﻮﻋﻴﺔ ﻓﻘﻂ (
(
)ﺏ( ﻧﻔﺮض ﺹ = 1-ﺉ ﺱ = ( ١ - ) × 1 = 1-
´ ¤Ùﺍ ﺏ¤ ﻩ
ﺱ¬ ¦
ﺹ ¬0
ﺑﻌﺪ ٥أﻣﺘﺎر ﻣﻦ ﻧﻬﺎﻳﺔ
=ﻩ
ﺱ
ﺱ
ﺇ ا ﻘﺪار = ] + ١ ) lﺹ (
ا ﺴﺒﺎق وﻣﻌﺪل إﻗ اﺑﻪ ﻘﻄﺔ ا ﻬﺎﻳﺔ ١٠م /ث .
ﺱ¬ ¦
٥ 1
ﺱ
ﺱ¬ ¦
)ﺍ( ﻧﻔﺮض ﺹ = 3 ﺱ
.أوﺟﺪ ﻣﻌﺪل ﺗﻐ ا ﺰاو ﺔ اﻟ ﺗﺪور ﺑﻬﺎ ا ﻣ ا
ﺱ¬ 0
ﺱ¬ ¦
( 3 + ١ ) lﺱ
)ﺝ(
ﺴﺎر ﺴﺘﻘﻴﻢ ﺑﺎ ﺎه ﺧﻂ
ﻩ = 1 +١) l ﺱ (ﺱ ،ﻩ
ﺱ
ﺱ¬ ¦
ﻋﺔ ﺗﻐ ا ﺴﺎﻓﺔ ﺑ ﻨﻬﻤﺎ ﺑﻌﺪ ٢ﺛﺎﻧﻴﺔ .
ﺮﺻﺪ ﺣﺮ ﺔ ا ﻼﻋﺐ ﻋﻨﺪﻣﺎ ن
ﺱ
1 ﻩ5
٥ ﺱ = ] = ( 1 + ١) l × ٢[ ( 1 + ١) lﻩ = ١ × ٢ﻩ
ﺴﺎر ا ﺴﺒﺎق و ﻧﻔﺲ ا ﺴﺘﻮى اﻷﻓ
ﻠﻤ ﺴﺎﺑﻘ
ﺱ
ﺱ¬ ¦
) (٥ﻳﺮﺗﻔﻊ ﺑﺎ ﻮن ﺑﻤﻌﺪل ﺛﺎﺑﺖ ١٠ﻣ /ث وﻋﻨﺪﻣﺎ ن ارﺗﻔﺎع ً ﻋﺔ ﻣﻨﺘﻈﻤﺔ ا ﺎ ﻮن ٢٢٠ﻣ ا .ﺮت ﻣﻦ ﺘﻪ ﺳﻴﺎرة ﺴ
ا ﻬﺎﻳﺔ ،و ﻧﺖ إﺣﺪى ﻣ ات ﺧﻂ ا ﻬﺎﻳﺔ
ﺱ
ﺱ¬ ¦
ﺑﺎﻟ ﺴﺒﺔ ﻠﺰﻣﻦ .
) (٦
ا ﻞ
ﺱ
ا ﻠﺤﻈﺔ اﻟ ﺑ ﻮن ﻓﻴﻬﺎ ﻣﻌﺪل ﺗﻐ إﺣﺪاﺛﻴﻬﺎ
ﺳﺒﺎق ١٠٠ﻣ ،ﺮى ﻻﻋﺐ
ﺱ¬ ¦
ﲤ٠ ﳘ ﻓﺈن ﺹ ﺲ ﲤ ﺲ )ﺍ( ﺑﻔﺮض ﺹ = ، 1ﻋﻨﺪﻣﺎ ﺱ ﺲ
ا ﻨﺤ ﺱ ﺹ = ﺱ +ﺹ – ٥أوﺟﺪ ﻮﻗﻊ
٥٠ﻣ /ث .أوﺟﺪ
¤1 (5
)ﺏ( 1 + ١ ) l ﺱ
( ﺫ5+ ¤
=١
٧
ﻩ ﻥ
ئﻩ )( 1 +1 ﻥ =l 1 ﻥ
ﺱ ¬0
ﻥ¬¦
ئﻩ ) +1ﺱ (
ﺱ
ﻩ
ﻥ
= ١ﺣﻴﺚ ﺱ = 1 ﻥ
١-
ﻟﻸﺳﺘﺎﺫ ﺍﻟﻘﺪﻳﺮ /ﻋﻠﻰ ﺍﻟﺪﻳﻦ ﻳﺤﻴﻰ ٠١١١٩٦٦٠٦١٦
ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ -ﺍﻟﺼﻒ ﺍﻟﺜﺎﻟﺚ ﺍﻟﺜﺎﻧﻮﻯ
· ﺹ = ﻩﺏ ) ﺣﻴﺚ :ﺏ ﻱ ( òﺉ = §Ùﺻﻔﺮ ¤Ù ُ · ﺣﺎﻟﺔ وﺟﻮد داﻟﺔ ﺮﻓﻮﻋﺔ ﻷس ﻋﺒﺎرة ﻋﻦ داﻟﺔ أﺧﺮى ﻧﺄﺧﺬ
أﺟﺐ ﻋﻦ اﻷﺳﺌﻠﺔ اﻵﺗﻴﺔ :
ﺫ5 + ¤ )) (١ﺍ( أ ﻤﻞ ) l : ﺫ 1+ ¤ ﺱ¬ ¦
ﺫ5 + ¤ )ﺏ( أوﺟﺪ ) l : ﺫ 1+ ¤ ﺱ¬ ¦
) (٢أوﺟﺪ l :
ﺱ¬ 0
) (٣أوﺟﺪ l :
ﺱ¬ 0
) (٤أوﺟﺪ l :
ﻩ4ﺱ
) (٥أوﺟﺪ l :
ﺧﻮاص ا ﻠﻮ ر ﺘﻤﺎت ﺛﻢ ﺮى ﻋﻤﻠﻴﺔ اﻻﺷﺘﻘﺎق .
( ٣ﺱ . ٢ +
1- ¤ e 35ﺱ
) (١أوﺟﺪ
¤g
) (٣ﺹ =
oﻩ )5 +1ﺱ( 3 ﺫ 3¤
) (٥ﺹ =
ﺱ
+ﻩ
٦ﺱ
– ٢ﺱ ٣
(
ﻗﺎ ٢ﺱ
) (١٠ﺹ = ٣ – ٥ﻮ ﻩ ﺱ
-1ﺫئﻩ ¤ ) (١٢ﺹ = ئﻩ ¤ ٢
٢
) (٣ﺹ = /
٢ﺱ
) (١٦ﺹ = ﺱ
ﺱ
) (١٨ﺹ = ٣ﺱ × ٢
ﺱ
ا ﻞ
¤Ù 1 × د )ﺱ( · ] Ùﻮ ﻩد )ﺱ( [ = ﺩ)(¤ ¤Ù
i - ¤ g ¤Úﺫ ¤Ú ´ ¤
gﺫﺱ
٦ﺱ
) (٤ﺹ ٦ + ٢ = /ﻩ
) (٥ﺹ ٢ – × 1 = /ﺱ × ﻩ ﺫ
· ] Ùﺍد )ﺱ( [ = ﺍد )ﺱ( × د) /ﺱ( × ﻮ ﻩﺍ
) (٦ﺹ ) ٣ = /ﻩ
¤Ù · ] Ùﻮ ﺍد )ﺱ( [ = × 1د )ﺱ( × ﻮ ﻩ ﺩ)(¤ ¤Ù ﺍ
٢ﺱ
=)٦ﻩ
/
٢ﺱ
+ﻩ
= –ﺱﻩ
–٢ﺱ ٢
+ﻩ
( ×)٢ﻩ
–٢ﺱ
) (٧ﺹ ٢) = /ﺱ ٥ ( ٢ +
:
– ٧ﺱ٢
()ﻩ
ﺱ ٢ + ٢ﺱ
٤ﺱ
) (٢ﻮ ﻩﻩ = ١
) (٩ﺹ = /ﻩ
) (٤ﻮ ﻩﺱﻥ = ﻥ ﻮ ﻩﺱ
٢ﺱ
=ﻩ
) (٦ﻮ ﻩ ﺍ = ﻮ ﻩﺍ -ﻮ ﻩﺏ ﺏ
٢ﺱ
=٢ﻩ
oﻩ ﺍ ) (٨ﻮ ﺏﺍ = oﻩ ﺏ
)(١٠
٨
Ù × ) ¤Ùﺍ
×٢ﺱﺍ
٢ﺱ
×ﺍ
ﺱ٥ – ٢
ﺱ٥ – ٢
ﺱ٥ – ٢
3- 1 ﺹ × ٣ – = /ﺱ = ﺱ
٢ﺱ
– ﻩ
– ٧ﺱ٢
– ٢ﻩ
–٢ﺱ
–٤ﺱ
(
(
ﻮ )٢=٥ﺱ ٥(١+
) (٨ﺹ ٢ = /ﻗﺎ ﺱ × ﻗﺎ ﺱ ﻇﺎ ﺱ × ٣
ﻩ
ﻗﺎ ٢ﺱ
(+ﺍ
٣
) (١٤ﺹ = ٢ﺱ ﻮ ﻩﺱ
) (٢ﺹ = /ﻩﺱ × ﺟﺘﺎ ﺱ +ﺟﺎ ﺱ × ﻩﺱ = ﻩﺱ ) ﺟﺘﺎ ﺱ +ﺟﺎ ﺱ (
/
) (٧ﻮ ﻩﺱ × ﻮ ﺱﻩ = ١
ﺱ٥ – ٢
) (١ﺹ ٢ = /ﻩﺱ – ٢ﺟﺎ ٢ﺱ
· ] Ùﻩد )ﺱ( [ = ﻩد )ﺱ( × د) /ﺱ(
) (٥ﻮ ﻩﺍﺏ = ﻮ ﻩﺍ +ﻮ ﻩﺏ
) (٨ﺹ = ٢
) (١٧ﺹ = ) ﺟﺎ ﺱ (
ﻼﺣﻈﺎت ﻫﺎﻣﺔ :
) (٣ﻩ
ﺫ
) (٦ﺹ = ) ﻩ
ﻩ
) (١٥ﺹ = ﺱ ئﻩ ¤
¤Ù ﺱ · ﺹ = ﺍ ﺉ = §Ùﺍﺱ ﻮ ﻩﺍ ¤Ù · ﺹ = ﻮ ﺱ ﺉ 1 = §Ù ﻩ ¤Ùﺱ 1 1 §Ù = ﻮ ﻩ = ·ﺹ= ﻮ ﺱ ﺉ o ¤ ¤Ùﻩ ﺍ ﺱ ﺍ ﺍ
ﻩ
– ٧ﺱ٢ 1
٢ﺱ
) (١٣ﺹ = ﻮ ﻩ) ٧ﺱ – ( ٣
ﻮ ﺱ
gﺱ
) (١١ﺹ = ﺱ ٢ﻮ ﻩﺱ
=ﺱ
) (٤ﺹ = ٢ﺱ +ﻩ
) (٩ﺹ = ﻩ ٢ﺱ × ﺍ
) (٦أوﺟﺪ l :ﻩ 3ﺱ 1- ﺱ¬ f -1S 0ﺱ
) (١ﻮ ﻩ = ١ﺻﻔﺮ
ﻩﺱ
) (٧ﺹ = ٥
3ﺱ 1-
· ﺑﻌﺾ ﺧﻮاص ا ﻠﻮ ر ﺘﻢ اﻟﻄﺒﻴ
) (٢ﺹ = ﻩﺱ ﺟﺎ ﺱ
ﺱ ٢ + ٢ﺱ
5 + 1 ) oﺱ (
ﺱ · ﺹ = ﻩ ﺉ = §Ùﻩ
ﺎ ﻳﺄ :
) (١ﺹ = ٢ﻩﺱ +ﺟﺘﺎ ٢ﺱ
3ﺱ -ﺫ ¤
ﺱ¬ 0
ﺱ¬ 0
ا ﻠﻮ ر ﺘﻢ اﻟﻄﺒﻴ
( ٣ﺱ . .......... = ٢ +
ﻠﻄﺮﻓ وﻧ ﺴﻂ اﻟﻌﻼﻗﺔ ا ﻌﻄﺎة ﺑﺎﺳﺘﺨﺪام
×ﻮ ٣ ﻩ
ﺱ٥ – ٢
× ﻮ ﺍ+ﺍ ﻩ
Ù ) ¤Ùﻩ ×
ﺱ٥ – ٢
)س ﻮ ﺍ (١+ ﻩ
ﺱ ٢ + ٢ﺱ
٢ﺱ
×٢ﻩ
(
٢ﺱ
ﻮ ٥ ﻩ
ﻟﻸﺳﺘﺎﺫ ﺍﻟﻘﺪﻳﺮ /ﻋﻠﻰ ﺍﻟﺪﻳﻦ ﻳﺤﻴﻰ ٠١١١٩٦٦٠٦١٦
ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ -ﺍﻟﺼﻒ ﺍﻟﺜﺎﻟﺚ ﺍﻟﺜﺎﻧﻮﻯ ) (١١ﺹ ٢ = /ﺱ ﻮ ﺱ × 1 +ﺱ = ٢ﺱ ) ٢ﻮ ﺱ ( ١ +
ﺹ 1 = /ﺇ ﻣﻴﻞ ا ﻤﺎس = ١ﺉ ﻣﻴﻞ اﻟﻌﻤﻮدى = – ١ ﺱ
ﺱ ﻩ 1 - /ﺱﺫ ´ ئﻩﺱ - 1) ¤ -ﺫئﻩ( ¤ 1) (١٢ﺹ = = ﺫ ﺫ ﺱ )ئﻩ ( ¤ )ئﻩﺱ ( 14 ) (١٣ﺹ = / 1ﺫ × ٧)٢س–= ٧×(٣ 7ﺱ 3 - )( 3 - ¤ 7 3 ٣ ﺱ × ٢ﺱ ٢ = ٢ﺱ ) ٢ﻮ ﺱ( ٣ + ٣ ) (١٤ﺹ ٤ = /ﺱ × ﻮ ﺱ + ﻩ ﻩ 1 ´1 /ئﻩﺱ ¤ ´ ¤ -ئﻩﺱ 1 - ) (١٥ﺹ = = ﺫ ﺫ )ئﻩﺱ ( )ئﻩﺱ ( ﻩ
) (١٦ﺑﺄﺧﺬ ﻮ ر ﺘﻢ اﻟﻄﺮﻓ ﺇ
ﻩ
ﺹ
ﻩ
ﻩ
) (١٧ﺑﺄﺧﺬ ﻮ ر ﺘﻢ اﻟﻄﺮﻓ ﺇ
٢ﺱ
ﻩ ﺫ
ﻩ
ﻩ
+ﻮ ﺟﺎ ﺱ × = ١ﺱ ﻇﺘﺎ ﺱ +ﻮ ﺟﺎ ﺱ
) (١٨ﺑﺄﺧﺬ ﻮ ر ﺘﻢ اﻟﻄﺮﻓ
ﻩ
ﻩ
ﺹ
ﻩ
ﻸﺳﺎس ﻩ :
ﺫ§y = ﻮ + ٣ﺹ /ﻮ ٢ ﺹ ﻩ ﻩ
ﺇ ٢ﻮ ﺹ=ﺱ ﻮ +٣ﺹ ﻮ ٢ﺇ ﻩ
ا ﻞ
ﺏ ﺍBﺹ = /ﺍ ﻩ ﺱ × -ﺫﺏ = ﺱﺫ ﺱ ﺏ ﺏ ﺏ ﺏ ﺫ ﺇ ﺹ = //ﺫ ﺍ Bﻩ ﺱ - +ﺍ × Bﻩ ﺱ × -ﺏ = ﺫ ﺍ Bﻩ ﺱ +ﺍ Bﻩ ﺱ ﺱ4 ﺱ3 ﺱ3 ﺱﺫ ﺱﺫ ﺏ ﺏ ﺏ ﺏ ﺫ ﺱ -ﺍB ﺍB ﺫ ﺍB ﺇ اﻷﻳﻤﻦ = ﺱ )ﺍ ﻩ ﺱ ( × ) 3ﻩ ﺱ 4 +ﻩ ﺱ ( × ٢ +ﺍ ﻩ × ﺱﺫ ﺱ ﺱ
ﻩ
ﻸﺳﺎس ﻩ :ﻮ ﺹ = ﺱ ﻮ ﺟﺎ ﺱ
ﺱ
ﻩ
أﺛﺒﺖ أن :
ﺱ ﺹ ﺹ ٢ + //ﺹ ﺹ – /ﺱ ﺹ٠ = ٢/
ﺇ ﺹ ) = /ﺟﺎ ﺱ( ) ﺱ ﻇﺘﺎ ﺱ +ﻮ ﺟﺎ ﺱ (
ﻩ
ﺏ ﺱ
) (٤إذا ﻧﺖ ﺹ = ﺍ ﻩ
) +١ﻮ ﺱ(
ﻩ
ﻩ
§
ﻩ
ئﻩ 3 ﺇ ﺹ= / ﺫ -ئﻩ ﺫ ﺹ
ﺏ ﺱ
ﺏ ﺱ
ﺏ ﺫ ﺫB ﺍB– ﺱ ) ﺫ × ﻩ ﺱ ( = ٢ﺍ ﺫ ﻩ ﺱ ﺱ
ﺫB ﺍﺫ Bﺫ Úﺱ ﺱ3
ﺫ§y ﺇ – ﺹ /ﻮ = ٢ﻮ ٣ﺇ ﺹ ) /ﺫ – ﻮ = ( ٢ﻮ ٣ ﻩ
ﻣﻦ ا ﻨﺤﻨﻴﺎت ا ﺎ ﺔ ﻋﻨﺪ ﻗﻴﻢ ﺱ ا ﻌﻄﺎة ) (١أوﺟﺪ ا ﺸﺘﻘﺔ اﻷو
)ﺍ( ﺹ = ﻮ ٥ﺱ
،
) ﺏ( ﺹ = ٤ﻮ ) ٣ﺱ ( ١ +
،
ﺱ=١
،
ﺱ=١
،
ﺱ=٣
٢
)ﺝ( ﺹ = ﻮ ) ٢ﺱ( ٣ – ٢ ٢
٤
٢
) ( ﺹ = ) ٣ﻮ ﺱ (
)ﺏ( ﺹ = ٣ﻩﻗﺎ ﺱ – ﻩ )ﺝ( ﺹ = ٧
٣+ﻩ
ﻗﺎ
) (٢أوﺟﺪ ا ﺸﺘﻘﺔ اﻷو
p 3
ﺱ
+ﻩ ﺟﺎ ﺱ
ﺎ ﻳﺄ :
)ﺍ( ﺹ = – ٥ﻇﺎ ﺱ ﻮ ﻩ ) ٢ﺱ( ٩ + ٣
ﺇ ﻣﻴﻞ ا ﻤﺎس = – ١٦ﻮ ﻩ ٢
) (ﺹ ٢ × ٣ = /ﻮ ﺱ × ﻮ ﻩ = 6ﻮ ﺱ ﻮ ﻩ ﺇ ﻣﻴﻞ ا ﻤﺎس = ٢ﻮ ٣ﻮ ﻩ
ﺱﺫ
) ﺏ( ﺹ =
ﻩ
)ﺝ( ﺹ =
¤o
ﻮ ) ﺱﺫ ﻩ 3ﺱ
) (ﺹ = ) ﺟﺎ ﺱ (
ﺱ
ﺍﺏ /ﻷﻗﺮب ﺛﻼﺛﺔ أرﻗﺎم ﻋ
]ﺱ /
) ٣ﺱ ٥ – ٢ﺱ ( ٥ + ﻮ ﻩﺱ
) (ﺹ = ﻩ
ا ﻞ
ﺹ = ﻮ ﻩ ٢ﺱ ﻋﻨﺪ ا ﻘﻄﺔ
ﺍ ) ، ١ﻮ ﻩ ( ٢ﻳﻘﻄﻊ ﻮر ا ﺴ ﻨﺎت
ﺎ ﻳﺄ :
٢ )ﺍ( ﺹ = ﻩﺱ – ﺱ +ﻗﺎ ٢ﺱ
)ﺍ( ﺹ × ٥ × 1 = /ﻮ ﻩ ﺇ ﻣﻴﻞ ا ﻤﺎس = 1ﻮ ﻩ ¤5 ﺫ ٢ ٢ ﺫ1 3 ﻮﻩ = ٣ﻮﻩ × ﻮ ﻩ ﺇ ﻣﻴﻞ ا ﻤﺎس = )ﺏ( ﺹ × ٤ = / 1 + 1´ 3 1+ ¤ 3 16ﺱ ﻮ ﻩ )ﺝ( ﺹ × ٤ = /ﺫ ٤ × 1ﺱ ﻮ ﻩ = ﺫ ¤ﺫ ٢ 3 - ٢ ﺫ3 - ¤
) (٣إذا ن اﻟﻌﻤﻮدى ﻠﻤﻨﺤ
+
ﻩ
ﺱ=٢
ﺫB ﺱ
ﺫB ﺍﺫ Bﺫ Úﺱ ﺱ3
= ﺻﻔﺮ = اﻷ
أﺟﺐ ﻋﻦ اﻷﺳﺌﻠﺔ اﻵﺗﻴﺔ :
) (٢أوﺟﺪ ﻣﻴﻞ ا ﻤﺎس
) +ئﻩ ﺫ (0-
ﺫ
= ) üئﻩ ﺫ(ﺫ ) +ئﻩ ﺫ(ﺫ = ] ٢ﻮ ٠٩٨ = ٢وﺣﺪة ﻃﻮل . ﻩ
ﻩ
ﻩ
ﻩﺫ (
ﺫ
- 1ئﻩ
ﻩ
ﺇ ﺹ ٢ = /ﺹ ) + ١ﻮ ﺱ ( = ٢ﺱ ¤f ﺹ = yﺱ × ﺹ ¤e
)ü
ﺏ ) ﻮ ﻩ ( ٠ ،ﺇ ﺍﺏ =
ﻸﺳﺎس ﻩ :ﻮ ﺹ = ٢ﺱ ﻮ ﺱ
ﺹ ٢ = yﺱ × + 1ﻮ ﺱ × + ١ ) ٢ = ٢ﻮ ﺱ ( ﺱ
ﺇ ﻣﻌﺎدﻟﺔ اﻟﻌﻤﻮدى :ﺹ -ئﻩ ﺫ = ١أى :ﺱ – ﺹ = – ١ﻮ = ٢ﻮ ﻩ ﻩ ﺫ ﺱ 1- ﻩ ﻩ ﺇ ا ﺴﺘﻘﻴﻢ ﻳﻘﻄﻊ ﻮر ا ﺴ ﻨﺎت ،ﺑﻮﺿﻊ ﺹ = ٠ﺉ ﺱ = ﻮ ﻩ ﺫ
( ﻇﺎ ﺱ
ﺱﻩ
)ﻩ( ﺹ = ﻩ
) (٣أوﺟﺪ Ùﺫ § ¤Ùﺫ
ا ﻘﻄﺔ ﺏ ،أوﺟﺪ ﻃﻮل
ﺔ.
ﺎ ﻳﺄ :
)ﺍ( ﺹ = ﺱ ﻮ س
ا ﻞ
٩
ﺫ ﺫB – ﺍﺫ ﻩ ﺱ
ﺫB ﺱ
–
ﻟﻸﺳﺘﺎﺫ ﺍﻟﻘﺪﻳﺮ /ﻋﻠﻰ ﺍﻟﺪﻳﻦ ﻳﺤﻴﻰ ٠١١١٩٦٦٠٦١٦
ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ -ﺍﻟﺼﻒ ﺍﻟﺜﺎﻟﺚ ﺍﻟﺜﺎﻧﻮﻯ ٤ﻥ
)ﺏ( ﺱ = ﻩ ٣ﻥ ،ﺹ = ﻩ
ا ﻞ 1ﻩ )ﺍ( = + ١ ) þﻩ – ٢ﺱ ( ﺱ = ﺱ – ﺫ )ﺏ( = 31ﺱ ٢ + ٣ﻩﺱ +ث ¤ﺫ1+ Ú )ﺝ( = 1 +ﻩ ٣ﺱ +ث ﺫ3 1 + Ú
) (٤إذا ن :ﻩﺱ ﺹ = ﺱ + ٢ﺹ ﻓﺄﺛﺒﺖ أن : ) §Ùﺱ ﻩﺱ ﺹ – ٢ = ( ١ﺱ – ﺹ ﻩ
ﺱﺹ
¤Ù
) (٥إذا ﻧﺖ :ﺱ ٢ﺹ = ﺍﺏ ﻮ ﻩ ﺱ ﻓﺄﺛﺒﺖ أن : ¤Ù
) ﺏ( ) þﺱ – ( ٣ﻩ
ﺱ ٦ – ٢ﺱ ٥ +
)ﺍ( = ﻩﺟﺎ ﺱ +ﺱ + ٣ث
· þﻩﺱ .ﺱ = ﻩﺱ +ث
)ﺏ( = ٢) þ 1ﺱ –(٦ﻩ ﺫ
· þﻩﺍ ﺱ +ﺏ .ﺱ = 1ﻩﺍ ﺱ +ﺏ +ث ) ﺣﻴﺚ ﺍ ﻵ ﺻﻔﺮ (
ئﻩ 3 )) (٤ﺍ( þ ﺱ
د )ﺱ(
+ث · þﻩد )ﺱ( × د) /ﺱ( .ﺱ = ﻩ ُ أى أن :ﺗ ﺎ ﻞ ا اﻟﺔ اﻷﺳﻴﺔ × ﺸﺘﻘﺔ اﻷس = ا اﻟﺔ اﻷﺳﻴﺔ +ث ﺱ ) ( · þﺩ ¤ yﺱ = ﻮ ﻩ | د )ﺱ( | +ث ﺩ)(¤
أوﺟﺪ
ً
+ث
)ﺍ(
)ﺏ( = 51ﻩ
ﺱ
) ¤ﺫ +ﺫ(¤Ù )ﺝ( þ 1+ ¤ 6 + 3¤
)ﺍ(
)ء( = – ٢] – þ ٣ﻩ
)) (٢ﺍ( þ
Úﺱ ¤ -Ú + Úﺱ
ﺱ
ﺫ ﺱ 4 -ﺱ ) ﺏ( þﺫ
+ ¤ﺫ¤
¤f þﻇﺘﺎ ﺱ ﺱ = þ ¤e
ا ﻞ
ﺱ = ﻮ ﻩ | ﺟﺎ ﺱ | +ث
) - ¤ﺫ() + ¤ﺫ( +¤ﺫ ﺱ=þ )ﺏ( = þ ﺱ + ¤ ) ¤ﺫ(
)ﺝ( = – 1 × ٦ﻩ ٠٢ص +ث = – ٣٠ﻩ ٠٢ﺹ +ث ﻥ = – ٣ﻩ
ﺫ
)) (٦ﺍ( þﻇﺘﺎ ﺱ .ﺱ
+ث
ﺫ0.
ﺫ ﺱ (ﺱ ﺱ=+١) þ
)ﺝ( = ) þﺱ – ( 3 + ٦ﺱ = 1ﺱ ٦ – ٢ﺱ ٣ +ﻮ ﻩ | ﺱ | +ث
ﺱ
– ] ٢ﻥ
1 5 5ﻮﻩ|ﺱ|+ث ﺱ ( ﺱ = ﺱ– ٢ ٢) þﺱ – ×3 3
= ﺱ ٢+ﻮﻩ|ﺱ|+ث
) ٢] ٣ þ (ﻩ– ] ٢ﻥ ﻥ
– ] ٢ﻥ
ﺫ ﺱ 4 -ﺱ ) ﺏ( þ ¤ﺫ -ﺫ ¤
ا ﻞ
) - ¤ﺫ() + ¤ﺫ( +¤ﺫ ﺱ=þ )ﺏ( = þ ﺱ - ¤ ) ¤ﺫ(
)ﺍ( = ﺑﺐ þﻩ ﺱ = ﺑﺐ ﻩ +ث –٥ﻉ
ا ﻞ
Sﺱ
ا ﻞ ﺱ
¤ﺫ
)ﺝ( ] ) þﺱ – ٢( 3ﺱ
) ﺏ( – þﻩ – ٥ﻉ ﻉ
)ﺝ( ٦ – þﻩ ٠٢ﺹ ﺹ
ﺱ
6ﺱﺫ 5 - ﺱ )) (٥ﺍ( þ 3ﺱ
ﻩ
)) (١ﺍ( þﺑﺐ ﻩﺱ ﺱ
4 ) ﺏ( þ ¤ 3ئﻩ 5
ﺱ
ﺫئﻩ ¤ 1 ﺫ ﺱ ﺱ = ﺫ 3ﻮ ﻩ | ﺱ | +ث ﺱ= þ )ﺝ( = þ 3 ¤ 3ئﻩ ¤
ﻣﻦ ا
+ث
1 4 ﺱ × 1ﺱ = 43ﻮ | ٥ﺱ | +ث )ﺏ( = þ 3 ئﻩ 5
) 1ﺍ( B + ¤ · ) þﺍ ﺱ +ﺏ (ﻥ ﺱ = × ﻥ 1+ ﺍ ﻥ1+ ]ﺩ ) [( ¤ +ث · ] þد )ﺱ( [ ﻥ د) /ﺱ( ﺱ = ﻥ 1+ ﺍ ﻩ = ﻮ ﺏﺍ · = 1ﻮ ﺏﻩ ، ﺏ ﻩﺏ
ﻼت اﻵﺗﻴﺔ :
ﺱ=ﻩ
ﺱ ٦ – ٢ﺱ ٥ +
1 ﺱ ﺱ = ﻮ ﻩ ٣ﻮ ﻩ | ﺱ | +ث )ﺍ( = þﻮ ﻩ× ٣
ﻥ1+
· ﻮ ﺏﺍ × ﻮ ﺍﺝ = ﻮ ﺏﺝ
ﺱ ٦ – ٢ﺱ ٥ +
ئ )ﺝ( þﻩ 3ﺱ ¤ئﻩ ¤
· 1 þﺱ= ﻮﻩ|ﺱ|+ث ﺗﺬﻛﺮ أن :
ﺱ
ا ﻞ
ﺍ
+ث
)) (٣ﺍ( ) þﺟﺘﺎ ﺱ ﻩﺟﺎ س ٣ +ﺱ ( ٢ﺱ
Ù ٢ﺫ § ٥ +ﺱ ٤ + §Ùﺹ = ٠ ﺱ ﺫ
¤Ù
–٢ﺱ
+ث
ﺫ ﺱ (ﺱ ﺱ=+١)þ
= ﺱ – ٢ﻮﻩ|ﺱ|+ث
¤3ﺫ 6 + 1 )ﺝ( = þ 1+ ¤ 6 + 3¤ 3
)ﺏ( ) þﺱ ٢ + ٢ﻩﺱ( ﺱ
)ﺝ( ) þﺱ ٢ﻩ +ﻩ ٣ﺱ( ﺱ
١٠
ﺱ = 31ﻮ ﻩ | ﺱ ٦ + ٢ﺱ + | ١ +ث
ﻟﻸﺳﺘﺎﺫ ﺍﻟﻘﺪﻳﺮ /ﻋﻠﻰ ﺍﻟﺪﻳﻦ ﻳﺤﻴﻰ ٠١١١٩٦٦٠٦١٦
ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ -ﺍﻟﺼﻒ ﺍﻟﺜﺎﻟﺚ ﺍﻟﺜﺎﻧﻮﻯ ) (٧ﻣﻴﻞ ا ﻤﺎس ﻨﺤ ا اﻟﺔ د ﻋﻨﺪ أى ﻧﻘﻄﺔ ﻋﻠﻴﻪ ) ﺱ ،ﺹ ( ﺴﺎوى 1 ﺫÚ- ¤
ﺴﺘﺨﺪم ﻚ ا ﺸﺘﻘﺔ اﻷو :
،و ن د )ﻩ( = ، 1أوﺟﺪ د ) ٢ﻩ( ا ﻞ
ﺫ
) (١
1 1 ﺇ ﺹ= 1 þﺱ= þ 1ﺫ ﺱ= ﰈ = §Ù ﺫ ¤Ù ﺫ ﺫÚ - ¤ ﺫÚ - ¤ ﺫÚ - ¤ 1 1 1 ﻮ ﻩ | ٢ﺱ – ﻩ | +ث ،ﰈ د )ﻩ( = ﺫ ﺇ ﺫ = ﺫ ﻮ ﻩ | ٢ﻩ – ﻩ | +ث ﺇ ث = ﺻﻔﺮ ﺇ د )ﺱ( = 1ﻮ ﻩ | ٢ﺱ – ﻩ | ﺇ د ) ٢ﻩ( = 1ﻮ ﻩ | ٤ﻩ – ﻩ | ﺫ ﺫ 1 = ) 1ﻮ ﻩ + ٣ﻮ ﻩ ﻩ ( = ﺫ ) ﻮ ﻩ( ١ + ٣ 1ﻮ ﻩ ٣ﻩ = ﺫ ﺫ
) (٢ﻧﻮﺟﺪ ا ﺸﺘﻘﺔ اﻷو )(٣
ﺴﺐ ﻗﻴﻢ ﺱ .
اﻟ ﺗﻴﺐ ٣٠٠ ، ٢٠٠وﺣﺪة .أوﺟﺪ ﻣﺒﻴﻌﺎت ا ﻞ ﻣﺒﻴﻌﺎت ا ﺼﻨﻊ ﺇ
) (١ﺣﺪد ﻓ ات اﻟ اﻳﺪ وﻓ ات ا ﻨﺎﻗﺺ
ﻥ
)ﺏ( ﺭ )ﺱ( = ﺱ ﺱ ﺫ 1+
ﺑﻌﺪ اﺳﺒﻮﻋ ﻡ = ٢٠٠ﺇ = ٢٠٠ﺍ ﻮ ﻩ + ٢ث (١) ........
،د ) /ﺱ( = ٠ﻋﻨﺪﻣﺎ ﺱ ٦ – ٢ﺱ ٠ = ٥ +ﺇ ) ﺱ – ) ( ٥ﺱ – ٠ = ( ١
ﺇ = ١٠٠ﺍ ) ﻮ ﻩ – ٤ﻮ ﻩ = ( ٢ﺍ ﻮ ﻩ ٢ﺉ ﺍ = 100ﺑﺎ ﻌﻮ ﺾ )(٢ ئﻩ ﺫ
ﺇ ﺱ = ٥أ ،ﺱ = ١
ئﻩ ﺫ
ﺇ ا اﻟﺔ ﻣ اﻳﺪة
--ﻣﺘﻨﺎﻗﺼﺔ
[]٥،١
اﻟﺔ د ﺣﻴﺚ :
ا ﻞ
) þ ،ﺱ ٢ﻩ +ﻩ ٣ﺱ +ﺑﺐ ( ٤ﺱ
د )ﺱ( = ﺱ – ٢ﺟﺘﺎ ﺱ ﺇ د ) /ﺱ( = ٢ + ١ﺟﺎ ﺱ ،ﻋﻨﺪﻣﺎ ٢ + ١ﺟﺎ ﺱ = ٠
¤Ù þ ، ¤S
p7 ٥ ﺇ ﺟﺎ ﺱ = 1 -ﺉ ﺱ = = ٢١٠ 6 ﺫ
ﻩ ¤S
ﺇ ا اﻟﺔ ﻣ اﻳﺪة
¤6 + 9 ) þ (٦ ¤ﺫ ¤ 3 +
ﺱ
eﺫ + ¤ﺫ¤ f þ ، eﺫ + ¤ﺫ 1- ¤ e
3 )þ (٧ ﺫ ¤ﻩ ¤
ﺱ
þ ،ﻇﺘﺎ ٣ﺱ ﺱ
،ا اﻟﺔ ﻣﺘﻨﺎﻗﺼﺔ
) (٣ﻳﻮﺿﺢ ا ﺸ
ﺱ
٠
٧ﺑﺐ ٦
---
ﻣ اﻳﺪة
+ﺱ ﻮ ﻩ ( ٣ﺱ
p11 ٥ أ ،ﺱ = = ٣٣٠ 6
١١ﺑﺐ ٦
٢ﺑﺐ
þ ،ﻗﺘﺎ ﺱ .ﺱ
ﻩﺱ ﺫÚ +ﺱ (ﺱ )þ ، )) þ (٥ Ú ﺫ ﺱ ¤ﻩ ¤S
ﺳﻠﻮك د )ﺱ(
اﻟﻔ ﺗ – [ :ﳘﺲ ، ١ [ ، ] ١ ،ﳘﺲ ] ،وﻣﺘﻨﺎﻗﺼﺔ
+++
) þ (٤ﻗﺎ ﺱ .ﺱ
ﻣ اﻳﺪة
د ) /ﺱ(
د )ﺱ( = ﺱ – ٢ﺟﺘﺎ ﺱ > ٠ ،ﺱ > ٢ﺑﺐ
أﺟﺐ ﻋﻦ اﻷﺳﺌﻠﺔ اﻵﺗﻴﺔ :
) þ (٣ﻇﺎ ﺱ .ﺱ
+++
) (٢ﺣﺪد ﻓ ات اﻟ اﻳﺪ وا ﻨﺎﻗﺺ
ﺱ
١
ﻣ اﻳﺪة
ﺇ ﺑﻌﺪ ٨أﺳﺎﺑﻴﻊ ﻓﺈن ﻡ = + ١ ) ١٠٠ﻮ ٤٠٠ = ( ٨ ٢
) ) ٩ þ (٢ﻩﺱ × ٨( ١ +ﻩﺱ ﺱ
٥
+++
100ئ ﻥ ﻩ ١٠٠ = ١٠٠ +ﻮ ٢ﻥ + ١ ) ١٠٠ = ١٠٠ +ﻮ ٢ﻥ ( ﺇ ﻡ= ئﻩ ﺫ
3
ا ﻞ
د )ﺱ( = ﺱ ٩ – ٣ﺱ ١٥ + ٢ﺱ ﺇ د ) /ﺱ( = ٣ﺱ ١٨ – ٢ﺱ ١٥ +
،ﺑﻌﺪ ٤أﺳﺎﺑﻴﻊ ﻡ = ٣٠٠ﺇ = ٣٠٠ﺍ ﻮ ﻩ + ٤ث (٢) ..........ﺑﺎﻟﻄﺮح )(١) – (٢
) þ (١ﺫ ﻩ ٣ﺱ – ٢ﺱ
ﻫﺬه اﻟﻔ ة
ﺎ ﻳﺄ :
)ﺍ( د )ﺱ( = ﺱ ٩ – ٣ﺱ ١٥ + ٢ﺱ
ﲨ(٠ ﺇ ﻡ = ﺍ ﻮ ﻩ | ﻥ | +ث = ﺍ ﻮ ﻩ ﻥ +ث ) ﻷن ﻥ ﺲ
ﺇ ٢ × 100 = ٣٠٠ﻮ ﻩ + ٢ث ﺉ ث = ١٠٠
ﻫﺬه اﻟﻔ ة .
Ùﻡ = ﺍ ﺣﻴﺚ ﺍ = ﺛﺎﺑﺖ
Ùﻥ
ﻗﻴﻤﺔ ﻣﻦ ﻗﻴﻢ ﺱ ا ﺎ ﺔ
و ذا ﻧﺖ د) /ﺱ( > ٠ﻓﺈن ا اﻟﺔ ﺗ ﻮن ﻣﺘﻨﺎﻗﺼﺔ
ا ﺼﻨﻊ ﺑﻌﺪ ٨اﺳﺎﺑﻴﻊ . ﻥ
ﻣﻘﺎﻣﻬﺎ ﺑﺎ ﺼﻔﺮ ) ﻻﺣﺘﻤﺎل أن ﺗ ﻮن ﻏ ﻣﻌﺮﻓﺔ (
ﻓﺈذا ﻧﺖ د) /ﺱ( < ٠ﻓﺈن ا اﻟﺔ ﺗ ﻮن ﻣ اﻳﺪة
ا ﺰﻣﻦ ﺑﺎﻷﺳﺎﺑﻴﻊ ،و ﻧﺖ ﻣﺒﻴﻌﺎت ا ﺼﻨﻊ ﺑﻌﺪ أﺳﺒﻮﻋ و ٤
ﰈ ﻡ ¨ 1ﺣﻴﺚ ﻡ
اﻟﺔ ﺛﻢ ﺴﺎو ﻬﺎ ﺑﺎ ﺼﻔﺮ أو ﺴﺎوى
) (٤ﻧﺒﺤﺚ إﺷﺎرة د) /ﺱ( ﻗﺒﻞ و ﻌﺪ
ً ) (٨إذا ن ﻣﻌﺪل ﺗﻐ ﻣﺒﻴﻌﺎت أﺣﺪ ا ﺼﺎﻧﻊ ﻳ ﻨﺎﺳﺐ ﻋﻜﺴﻴﺎ ﻣﻊ أﺳﺎﺑﻴﻊ
ﺪد ﺎل ا اﻟﺔ
+++ ﻣ اﻳﺪة
ﻣﺘﻨﺎﻗﺼﺔ
p11 p7 اﻟﻔ ﺗ ٢ ، 6 [ ، ] 6 ، ٠ [ :ﺑﺐ ] p11 p7 ] ، [ 6 6
ا ﺠﺎور ﻣﻨﺤ
د) /ﺱ(
اﻟﺔ د ﺣﻴﺚ د )ﺱ( ﻛﺜ ة ﺣﺪود . ّ )ﺍ( ﻋ ﻓ ات اﻟ اﻳﺪ وﻓ ات ا ﻨﺎﻗﺺ
اﻟﺔ د
)ﺏ( أوﺟﺪ ﻤﻮﻋﺔ ﺣﻞ ا ﺘﺒﺎﻳﻨﺔ د )ﺱ( < ٠ //
١١
ﺱ
د )ﺱ( /
ﺳﻠﻮك د )ﺱ(
ﻟﻸﺳﺘﺎﺫ ﺍﻟﻘﺪﻳﺮ /ﻋﻠﻰ ﺍﻟﺪﻳﻦ ﻳﺤﻴﻰ ٠١١١٩٦٦٠٦١٦
ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ -ﺍﻟﺼﻒ ﺍﻟﺜﺎﻟﺚ ﺍﻟﺜﺎﻧﻮﻯ ا ﻞ
ا اﻟﺔ ﻣﺘﻨﺎﻗﺼﺔ
اﻟﻔ ﺗ
،ا اﻟﺔ ﻣ اﻳﺪة
اﻟﻔ ة [ – ] ١ ، ٣ﻷن ا ﻨﺤ ﻳﻘﻊ ا ﺰء ا ﻮﺟﺐ ﻣﻦ ﺻﺺ
ﻷن ﻣﻨﺤ د ) /ﺱ(
– [ :ﳘﺲ ، ١ [ ، ] ٣ - ،ﳘﺲ ]
ﻫﺎﺗ اﻟﻔ ﺗ ﻳﻘﻪ
ﺴﺘﺨﺪم ﻚ ا ﺸﺘﻘﺔ اﻷو :
ا ﺰء ا ﺴﺎﻟﺐ ﻣﻦ ﻮر ا ﺼﺎدات
) (١
) (٢ﻧﻮﺟﺪ ا ﺸﺘﻘﺔ اﻷو
) (٤ﺣﺪد ﻓ ات ﺗﺰاﻳﺪ وﺗﻨﺎﻗﺺ ا اﻟﺔ :د )ﺱ( = ﺱ – ﻩﺱ .أوﺟﺪ ا ﻘﻂ ا ﺮﺟﺔ ﺛﻢ ﺣﺪد ﻓ ات اﻟ اﻳﺪ وﻓ ات ا ﻨﺎﻗﺺ
د )ﺱ( = ﺱ – ﻩ
ﺱ
ا ﻞ
ﺱ
اﻟﺔ د
)(٣
ﺱ
ﺱ = ، ٠ﻋﻨﺪﻣﺎ ﺱ > ٠ﺇ د ) /ﺱ( < ٠ﺉ ا اﻟﺔ ﻣ اﻳﺪة
أو ﻓ ات اﻟ اﻳﺪ وا ﻨﺎﻗﺺ
ا وال اﻵﺗﻴﺔ :
) (١
ﻣﻦ
ﻠﻴﺔ
ﻮﺟﺒﺔ ﻧﺖ ا ﻘﻄﺔ
ﻮﻗﻊ ﻗﻴﻤﺔ ﺻﻐﺮى ﻠﻴﺔ
د) //ﺍ( < ٠ﺉ ﻋﻨﺪ ﺱ = ﺍ ﺗﻮﺟﺪ ﻗﻴﻤﺔ ﺻﻐﺮى ﻠﻴﺔ
د )ﺍ( > ٠ﺉ ﻋﻨﺪ ﺱ = ﺍ ﺗﻮﺟﺪ ﻗﻴﻤﺔ ﻋﻈ
ﻼﺣﻈﺎت ﻫﺎﻣﺔ :
ﻠﻴﺔ
) (١إذا ﻧﺖ ) ﺍ ،ﺏ ( ﻧﻘﻄﺔ ﺗﻤﺎس ﻓﺈن د )ﺍ( = ﺏ
) (٢إذا ﻧﺖ ) ﺍ ،ﺏ ( ﻧﻘﻄﺔ ﺣﺮﺟﺔ
) (٢د )ﺱ( = ) ﺱ ٣( ٤ +
· ﺣﺪد ﻓ ات اﻟ اﻳﺪ وﻓ ات ا ﻨﺎﻗﺺ 3 - ¤ ) Sﺫ( ﺫ
د )ﺍ( = ﺏ ،د) /ﺍ( = ﺻﻔﺮ
ﺎ ﻳﺄ :
) (٣إذا ﻢ ﺗﺘﻐ
) (٤د )ﺱ( = ٦ﺱ – ٣ﻮ ﻩ ﺱ
اﻟﺔ د ﻓﺈن :
إﺷﺎرة د) /ﺱ( ﺣﻮل ﺍ ﻓﺈﻧﻪ ﻋﻨﺪ ﺱ = ﺍ ﺗﻮﺟﺪ
ﻧﻘﻄﺔ ﺣﺮﺟﺔ ﻟ ﺴﺖ ﻋﻈ
٢
ﻠﻴﺔ أو ﺻﻐﺮى
ﻠﻴﺔ
ﺑﻔﺮض اﻟﻔ ة ا ﻐﻠﻘﺔ
) (٥د )ﺱ( = ٢ﻩ ٢ﺱ – ٤ﺱ
-¤ﺫ ) (٦د )ﺱ( = +¤ﺫ
]ﺍ،ﺏ[:
) (١ﻧﻮﺟﺪ ا ﻘﻂ ا ﺮﺟﺔ
) (٨أوﺟﺪ ﻗﻴﻢ ﺍ ،ﺏ ،ﺝ ،
ا ﻘﻄﺘ
ﻴﺚ ﻘﻖ ا ﻨﺤ :
د )ﺱ( = ﺍ ﺱ + ٣ﺏ ﺱ + ٢ﺝ ﺱ +ا
) (١ﻳﻤﺮ ﺑﻨﻘﻄﺔ اﻷﺻﻞ .
ﺍ ،ﺏ:
ً
أ ﻫﺬه اﻟﻘﻴﻢ ﻴﻌﺎ = اﻟﻘﻴﻤﺔ اﻟﻌﻈ ً أﺻﻐﺮ ﻫﺬه اﻟﻘﻴﻢ ﻴﻌﺎ = اﻟﻘﻴﻤﺔ ا ﺼﻐﺮى ا ﻄﻠﻘﺔ
ً وط ا ﺎ ﺔ ﻣﻌﺎ :
ﻧﻘﻄﺔ ﺣﺮﺟﺔ ﻋﻨﺪ ﺱ = ١
ا ﻄﻠﻘﺔ ،
) (٣ﻣﻌﺎدﻟﺔ ا ﻤﺎس ﻠﻤﻨﺤ ﻋﻨﺪ ا ﻘﻄﺔ ) ، ٢د ) ( (٢ﻋﻠﻴﻪ
) (١إذا ن د )ﺱ( = 1ﺱ ٨ – ٣ﺱ ٣ +ﻓﺄوﺟﺪ اﻟﻘﻴﻢ اﻟﻌﻈ
٩ :ﺱ +ﺹ = ٢٠
3
وا ﺼﻐﺮى ا ﺤﻠﻴﺔ
اﻟﺔ د وذ ﻚ ﺑﺄن ﺴﺎوى ا ﺸﺘﻘﺔ اﻷو
ﺑﺎ ﺼﻔﺮ و ﺴﺐ ﻗﻴﻢ ﺱ ّ ﻧﻌﻮض ا اﻟﺔ اﻷﺻﻠﻴﺔ د )ﺱ( ﺑ ﻞ ا ﻘﻂ ا ﺮﺟﺔ و ﺬ ﻚ )(٢
) (٧د )ﺱ( = ﺟﺎ ﺱ +ﺟﺘﺎ ﺱ > ٠ ،ﺱ > ٢ﺑﺐ
)(٢
ﻧﻘﻄﺔ ﺣﺮﺟﺔ
//
د )ﺱ( = ٩ﺱ – ﺱ٣
) (٣د )ﺱ( = – ٣
ﺴﺐ ﻗﻴﻢ ﺱ .
· إذا ﺗﻐ ت إﺷﺎرة د) /ﺱ( ﻣﻦ ﺳﺎ ﺔ إ
· أوﺟﺪ ا ﻘﻂ ا ﺮﺟﺔ ﺛﻢ ﻋ
ﻣﻘﺎﻣﻬﺎ ﺑﺎ ﺼﻔﺮ ) ﻻﺣﺘﻤﺎل أن ﺗ ﻮن ﻏ ﻣﻌﺮﻓﺔ (
ﻮﻗﻊ ﻗﻴﻤﺔ ﻋﻈ
،ﻋﻨﺪﻣﺎ ﺱ < ٠ﺇ د ) /ﺱ( > ٠ﺉ ا اﻟﺔ ﻣ اﻳﺪة
ّ
اﻟﺔ ﺛﻢ ﺴﺎو ﻬﺎ ﺑﺎ ﺼﻔﺮ أو ﺴﺎوى
) (٤ﻧﺒﺤﺚ إِﺷﺎرة د) /ﺱ( ﻗﺒﻞ و ﻌﺪ · إذا ﺗﻐ ت إﺷﺎرة د) /ﺱ( ﻣﻦ ﻮﺟﺒﺔ إ ﺳﺎ ﻪ ﻧﺖ ا ﻘﻄﺔ
ﺇ د ) /ﺱ( = – ١ﻩﺱ – ١ ،ﻩ = ٠ﻋﻨﺪﻣﺎ ﻩ = ١ﺉ
أﺟﺐ ﻋﻦ اﻷﺳﺌﻠﺔ اﻵﺗﻴﺔ :
ﺪد ﺎل ا اﻟﺔ
اﻟﺔ د . ا ﻞ
٣
٢
د )ﺱ( = 1ﺱ – ٩ﺱ ٣ +ﺇ د) /ﺱ( = ﺱ – ، ٩ﺱ ٠ = ٩ – ٢ﻋﻨﺪﻣﺎ ) 3 ﺱ – ) ( ٣ﺱ ٠ = ( ٣ +ﺉ ﺱ = _ ٣
ﻋﻨﺪ ﺱ > – ٣ﺇ د ) /ﺱ( < ، ٠ﻋﻨﺪ ﺱ ﻱ [ – ] ٣ ، ٣ﺇ د ) /ﺱ( > ، ٠
ﻋﻨﺪ ﺱ < ٣ﺇ د ) /ﺱ( < ٠ﺇ ﻋﻨﺪ ﺱ = – ٣ﺗﻮﺟﺪ ﻗﻴﻤﺔ ﻋﻈ = د )– ، ٢١ = ٣ + (٣ –) ٩ – ٣(٣ –) 1 = (٣ 3
١٢
ﻠﻴﺔ
ﻟﻸﺳﺘﺎﺫ ﺍﻟﻘﺪﻳﺮ /ﻋﻠﻰ ﺍﻟﺪﻳﻦ ﻳﺤﻴﻰ ٠١١١٩٦٦٠٦١٦
ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ -ﺍﻟﺼﻒ ﺍﻟﺜﺎﻟﺚ ﺍﻟﺜﺎﻧﻮﻯ ﻋﻨﺪ ﺱ = ٣ﺗﻮﺟﺪ ﻗﻴﻤﺔ ﺻﻐﺮى ﻠﻴﺔ = د )١٥ – = ٣ + (٣) ٩ – ٣(٣) 1 = (٣
3
اﻟﺔ :د )ﺱ( = ü 3ﺱ ﺫ ﻗﻴﻤﺔ ﺻﻐﺮى ﻠﻴﺔ
) (٢أﺛﺒﺖ أن
ﺫ 3
د )ﺱ( = ü 3ﺱﺫ = ﺱ
أﺟﺐ ﻋﻦ اﻷﺳﺌﻠﺔ اﻵﺗﻴﺔ :
ا ﻞ
ﺇ د ) /ﺱ( = ﺫ ﺱ 3
1=3
) (١إذا ﻋﻠﻢ أن :د )ﺱ( = ﺍ ﺱ + ٢ﺏ ﺱ ٢ +
ﺫ S3 3ﺱ
ﻋﻨﺪ ) ( ٤ ، ١أوﺟﺪ ﻗﻴﻤﺔ
/ / ،د )ﺱ( ﻏ ﻣﻌﺮﻓﺔ ﻋﻨﺪ ﺱ = ، ٠ﻋﻨﺪ ﺱ > ٠ﺇ د )ﺱ( > ، ٠ﻋﻨﺪ ﺱ < · ٠أوﺟﺪ اﻟﻘﻴﻢ اﻟﻌﻈ
ﺇ د ) /ﺱ( < ٠ﺇ ﻋﻨﺪ ﺱ = ٠ﺗﻮﺟﺪ ﻗﻴﻤﺔ ﺻﻐﺮى ﻠﻴﺔ = د)٠ = (٠
) (٣ﻫﻞ اﻟﺔ :د )ﺱ( = ﺱ ٣ + ٣ﺱ – ٤ﻗﻴﻢ ﻋﻈ ﻠﻴﺔ ؟ ﻓ ّ إﺟﺎﺑﺘﻚ .
ﺇ ﻻ ﻳﻮﺟﺪ ﻗﻴﻢ ﻋﻈ
-1ﺱ
اﻟﺔ
د )ﺱ( = =
ﺱﺫ
-1ﺱ
ﺫ¤ - ¤ﺫ
) -1ﺱ
ا ﻞ
،ﺱ ﻵ ١ﺇ د ) /ﺱ( =
اﻟﺔ :
ﺫ ¤ ´ (1 - ) - ( ¤ - 1) ¤ﺫ ) - 1ﺱ ( ﺫ
)ﺍ( د )ﺱ( = ١٠ﺱ ﻩ
اﻟﺔ د ﺣﻴﺚ :
،د ) /ﺱ( = ١٠ﻩ – ﺱ – ١٠ﺱ ﻩ
)ﺏ( د )– ، ٢ – = (١د ) ، ١٢ = (٣د =
¤4 - 4ﺫ
) ﺱﺫ (1 +
ﺫ
و ذا ﻧﺖ د) //ﺱ( > ٠ﻓﺈن ا ﺤﺪب ﻷ
) (٣إذا ﺗﻐ ت إﺷﺎرة د) //ﺱ( ﻼﺣﻈﺎت ﻣﺔ :
/
) (١إذا ﻧﺖ ) ﺍ ،ﺏ ( ﻧﻘﻄﺔ إﻧﻘﻼب
ا ﻄﻠﻘﺔ = ٣٦٨ﻋﻨﺪ ﺱ = ، ١
)ﺱ( = )4
¤ﺫ 1+
)
(1+
) (٢ﻧﻘﻄﺔ اﻻﻧﻘﻼب ﺗﻘﻊ
ﺫ
اﻟﻌﻈ
ﺇ د ) /ﺱ( = ٠ﻋﻨﺪﻣﺎ ﺱ = _ ١ﺉ د )٢ = (١
ﺇ ﺗﻮﺟﺪ ﻗﻴﻤﺔ ﻋﻈ
اﻟﺔ د ) أو ﻧﻘﻄﺔ ﺣﺮﺟﺔ
اﻟﺔ د) /ﺱ( ( ﻓﺈن :د )ﺍ( = ﺏ ،د) //ﺍ( = ﺻﻔﺮ
( -ﺫ ¤4´ ¤
ﺱﺫ
ﺟﺎﻧ ا ﻘﻄﺔ ﻓﺈن ا ﻘﻄﺔ
ﻧﻘﻄﺔ إﻧﻘﻼب
= ٠ﻋﻨﺪﻣﺎ ﺱ = ١
اﻟﻘﻴﻤﺔ ا ﺼﻐﺮى ا ﻄﻠﻘﺔ = ٠ﻋﻨﺪ ﺱ = ٠
اﻟﺔ د) /ﺱ(
ﻓﺈذا ﻧﺖ د) //ﺱ( < ٠ﻓﺈن ا ﺤﺪب ﻷﺳﻔﻞ
ا ﻞ
ﺉ د ) ١٠ = (١ﻩ – ٣٦٨ = ١ﺇ اﻟﻘﻴﻤﺔ اﻟﻌﻈ
ط
) (٢ﻧﺒﺤﺚ إﺷﺎرة د) //ﺱ( ﻗﺒﻞ و ﻌﺪ ا ﻘﻂ ا ﺮﺟﺔ
ﺉ د ) ، ٠ = (٠د ) ٤٠ = (٤ﻩ – ٠٧٣٣ = ٤ –ﺱ
@ 0< ¤
ّ ﻣﻌﺮﻓﺔ ﻌﻞ د) //ﺱ( = ٠أ ،ﻏ ) (١ﻧﻮﺟﺪ ﻗﻴﻢ ﺱ اﻟ ً ّ ﻣﻌﺮﻓﺔ أﻳﻀﺎ ) أى ا ﻤﺎس رأ ( . أن ﺗ ﻮن د) /ﺱ( ﻏ
،ﺱ ﻱ ]– [ ٣ ، ١
و
اﻟﻔ ة ] – [ ٣ ، ٣
ﺴﺘﺨﺪم ﻚ ا ﺸﺘﻘﺔ ا ﺎﻧﻴﺔ :
)ﺍ( د )ﺱ( = ١٠ﺱ ﻩ– ﺱ ،ﺱ ﻱ ] [ ٤ ، ٠
–ﺱ
@ 0³ ¤
ﻋﻨﺪ ﺱ = ٢ﺗﻮﺟﺪ ﻗﻴﻤﺔ ﺻﻐﺮى ﻠﻴﺔ = د )٤ – = (٢
4ﺱ )ﺏ( د )ﺱ( = ﺱ ﺫ 1+
) (٧اﺣﺴﺐ اﻟﻘﻴﻢ اﻟﻘﺼﻮى
اﻟﺔ د ﺣﻴﺚ :
ﻠﻴﺔ = د ) ، ٠ = (٠ﻋﻨﺪ ﺱ < ٢ﺇ د ) /ﺱ( > ٠ﺉ
) (٥أوﺟﺪ اﻟﻘﻴﻢ اﻟﻘﺼﻮى ا ﻄﻠﻘﺔ
) (٦د )ﺱ( = ﺱ – ﻮ ﻩﺱ
،ﺱ<٠
ï د )ﺱ( = ý ¤ ïþﺫ -ﺫ ¤
ﻋﻨﺪ ﺱ > ٠ﺇ د ) /ﺱ( > ، ٠ﻋﻨﺪ ﺱ ﻱ [ ] ٢ ، ٠ﺇ د ) /ﺱ( < ٠ﺉ ﻋﻨﺪ
ﺱ = ٠ﺗﻮﺟﺪ ﻗﻴﻤﺔ ﻋﻈ
) (٥د )ﺱ( = ﺱ ﻩﺱ
،ﺱﻱ]]٢،٠
¤ 3 - 3¤ üﺫ
،د ) /ﺱ( = ٠ﻋﻨﺪﻣﺎ ﺱ ) – ٢ﺱ ( = ٠ﺇ ﺱ = ٠أ ،ﺱ = ٢
(ﺫ
واﻟﻘﻴﻢ ا ﺼﻐﺮى ا ﺤﻠﻴﺔ
) (٤د )ﺱ( = ﺟﺎ ﺱ +ﺟﺘﺎ ﺱ ،ﺱ ﻱ ] ٢ ، ٠ﺑﺐ [
ً
) (٤أوﺟﺪ اﻟﻘﻴﻢ اﻟﻌﻈ وا ﺼﻐﺮى ا ﺤﻠﻴﺔ ﺫ ً د )ﺱ( = ﺱ ﻣﺒ ﻨﺎ ﻧﻮﻋﻬﺎ .
اﻟﺔ د ﺣﻴﺚ :
) (٣د )ﺱ( = - 3ﺱ ،ﺱ ﻱ{٢،١–}– ò ¤ﺫ - ¤ -ﺫ
د )ﺱ( = ﺱ ٣ + ٣ﺱ – ٤ﺇ د ) /ﺱ( = ٣ﺱ ٣ + ٢ﻵ ) ٠ﻮﺟﺒﺔ داﺋﻤﺎ ( أو ﺻﻐﺮى ﻠﻴﺔ
ﻣﻦ ﺍ ،ﺏ وﺣﺪد ﻧﻮع ا ﻘﻄﺔ .
) (٢د )ﺱ( = ﺱ – ٤ﺱ٣
وﺻﻐﺮى
ا ﻞ
ﺎ ﻧﻘﻄﺔ ﺣﺮﺟﺔ
ا ﺼﻐﺮى ا ﻠﺘﺎن
ﺑﻌﺪﻳﻦ ﻣ ﺴﺎو
) (٣إذا ﻢ ﺗﺘﻐ إﺷﺎرة د )ﺱ(
اﻫﺎ .
ﺟﺎﻧ ا ﻘﻄﺔ ا ﺮﺟﺔ ﻓﺈن
//
ﻣﻄﻠﻘﺔ = ١٢ﻋﻨﺪ ﺱ = ٣
ﻣﻦ اﻟﻘﻴﻤﺘ
ا ﻘﻄﺔ ﻟ ﺴﺖ ﻧﻘﻄﺔ اﻧﻘﻼب و ﻜﻨﻬﺎ ﺣﺮﺟﺔ ﻓﻘﻂ .
وﺗﻮﺟﺪ ﻗﻴﻤﺔ ﺻﻐﺮى ﻣﻄﻠﻘﺔ = – ٢ﻋﻨﺪ ﺱ = – ١
) (١ﺣﺪد ﻓ ات ا ﺤﺪب ﻷ ا ﻨﺤﻨﻴﺎت اﻵﺗﻴﺔ :
وا ﺤﺪب ﻷﺳﻔﻞ
)ﺍ( د )ﺱ( = ﺱ ٤ – ٢ﺱ ٢ +
١٣
ﻣﻦ
)ﺏ( ﺭ )ﺱ( = ﺱ ٤ – ٤ﺱ٣
ﻟﻸﺳﺘﺎﺫ ﺍﻟﻘﺪﻳﺮ /ﻋﻠﻰ ﺍﻟﺪﻳﻦ ﻳﺤﻴﻰ ٠١١١٩٦٦٠٦١٦
ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ -ﺍﻟﺼﻒ ﺍﻟﺜﺎﻟﺚ ﺍﻟﺜﺎﻧﻮﻯ
ﺇ ﻣﻌﺎدﻟﺔ ا ﻤﺎس :ﺹ – ) ٣ = ٢ﺱ – ( ١أى ٣ :ﺱ – ﺹ – ٠ = ١
ا ﻞ
)ﺍ( د )ﺱ( = ﺱ ٤ – ٢ﺱ ٢ +ﺇ د ) /ﺱ( = ٢ﺱ – ٤ ً ﺪب ﻷﺳﻔﻞ . ﺇ د) //ﺱ( = ) ٢ﻮﺟﺒﺔ داﺋﻤﺎ ( ﺉ ا ﻨﺤ
)ﺏ( ﺭ )ﺱ( = ﺱ ٤ – ٤ﺱ ٣ﺇ ﺭ ) /ﺱ( = ٤ﺱ ١٢ – ٣ﺱ
) (٤ﻳﻤﺜﻞ ا ﺸ
٢
ﺪب ﻷ
ا ﻨﺤ
و ﺪب ﻷﺳﻔﻞ
[]٢،٠
ﺪب ﻷﺳﻔﻞ ﺪب ﻷ
وا ﺤﺪب ﻷﺳﻔﻞ
د )ﺱ( = S3ﺱ ،ﺭ )ﺱ( = ﺱ
ﺫ 3
اﺧﺘﺒﺎر ا ﺸﺘﻘﺔ ا ﺎﻧﻴﺔ . د )ﺱ( = = ¤S3ﺱ
ا ﻞ 1 3
ﺪب ﻷﺳﻔﻞ
ﺇ د ) /ﺱ( = 1ﺱ 3
ﺫ 3
ﺇ ا ﻤﺎس رأ
ﺪب ﻷﺳﻔﻞ 3
53
13
//
[ – ﳘﺲ ، ٠ [ ، ٠ ،ﳘﺲ ]
) (٣إذا ﻧﺖ د )ﺱ( = ý ïþ
ا ﺤﺪب ﻷ
ﻏ ﻣﻌﺮﻓﺔ ﻋﻨﺪ ﺱ = ٠
43
ﺪب ﻷ ﺫ
¤ 3ﺫ 3¤ -
[ ، ٠ﳘﺲ ]
ا ﻨﺤ
،د ) ) ٠ < (١ﻮﺟﺐ( ﺇ ﻳﻮﺟﺪ ﻗﻴﻤﺔ ﺻﻐﺮى ﻠﻴﺔ = د )١٢ – = (١
) (٦ارﺳﻢ ا ﺸ
د )ﺱ( = ١٢ﺱ – ﺱ ٣ﺇ د ) /ﺱ( = ٣ – ١٢ﺱ ، ٢د) //ﺱ( = – ٦ﺱ ﺑﻮﺿﻊ د )ﺱ( = ٠ﺉ ﺱ = _ ، ٢ﺑﻮﺿﻊ د )ﺱ( = ٠ﺉ ﺱ = ٠ /
@ 1- £ ¤
ﺫ) (3 + ¤
ا ﻨﺤ
ﺪب ﻷﺳﻔﻞ
ﻧﻘﻄﺔ اﻻﻧﻘﻼب
//
٢
٢-
ﺱ - - - - + + + + - - - -ﺩ) /ﺱ(
ﺭ )ﺱ(
ﺪب ﻷ
اﻟﻌﺎم ﻨﺤ
ا اﻟﺔ د )ﺱ( = ١٢ﺱ – ﺱ٣
ا ﻞ
ﻏ ﻣﻌﺮﻓﺔ ﻋﻨﺪ ﺱ = ٠
@ 1- > ¤
ﻣﺘﻨﺎﻗﺼﺔ ﻣ اﻳﺪة
ﻣﺘﻨﺎﻗﺼﺔ
د )ﺱ(
---ﺪب ﻷ
٠
ﺱ + + + +د) //ﺱ( ﺪب ﻷﺳﻔﻞ
ﻧﻮﺟﺪ ﻧﻘﻂ ا ﻘﺎﻃﻊ ﻣﻊ ﻮر ا ﺴ ﻨﺎت :
ﺣﺪد ﻓ ات
د )ﺱ(
ص
ﺑﻮﺿﻊ د )ﺱ( = ٠ﺉ ﺱ = ٠أ ،ﺱ = _ ٣] ٢ ﻧﻮﺟﺪ ﻧﻘﻂ ا ﻘﺎﻃﻊ ﻣﻊ ﻮر ا ﺼﺎدات :
س
١ ٢ ٣ ٤ ٥
١٥ ١٠ ٥
ﺑﻮﺿﻊ ﺱ = ٠ﺉ ﺹ = ٠
) (٧ارﺳﻢ ﺷ
@ 1->¤
ﺪب ﻷﺳﻔﻞ ﺪب ﻷﺳﻔﻞ
ً ً ﻣﺎ ﻨﺤ ا اﻟﺔ ﺹ = د )ﺱ( إذا ﻋﻠﻤﺖ أن :
(١د ﻣﺘﺼﻠﺔ ﺎ ﺎ ] ، ٠ﳘﺲ ] ،د ) ، ٣ = (٤د )١ = (٠
(٢د) /ﺱ( < ٠ﻋﻨﺪﻣﺎ ﺱ < ٠
(٣د) //ﺱ( < ٠ﻋﻨﺪﻣﺎ ﺱ > ، ٤د، ٠ = (٤) // د) //ﺱ( > ٠ﻋﻨﺪﻣﺎ ﺱ < ٤
١٠ ﺱ + + + + + + + + - - - -ﺭ) //ﺱ(
[ – ﳘﺲ ] ١ ، ١ – [ ، ] ١ – ،
ﻠﻴﺔ = د )– ٢٠ = (٣
//
،د) //ﺱ( = ٠ﻋﻨﺪﻣﺎ – ١ ) ٦ﺱ ( = ٠ﺉ ﺱ = ١ ﺪب ﻷ
//
ﺇ د )– ) ٠ > (٣ﺳﺎﻟﺐ( ﺇ ﻳﻮﺟﺪ ﻗﻴﻤﺔ ﻋﻈ
ü ﺫ ïï 1- > ¤ @ ( 3 + ¤ ) üï / @ 1- < ¤ ،د )ﺱ( = ý د )ﺱ( = ý ¤ 3 ïþﺫ 1- £ ¤ @ 3 ¤ - ï ïﻏ ـ ﻮﺟــﻮدة @ 1 - = ¤ þ üﺫ @ 1- > ¤ // ﺇ ﻻ ﻳﻮﺟﺪ ﺎس ﻋﻨﺪ ﺱ = – ، ١د )ﺱ( = ý 1 - < ¤ @ ¤ 6 - 6þ
[ ، ١ﳘﺲ ]
ا ﻞ
//
¤ 3 - ¤6ﺫ
ﺪب ﻷ
وا ﺼﻐﺮى
ﺱ ٢ +ﺱ – ٠ = ٣ﺉ ﺱ = – ٣أ ،ﺱ = ، ١د )ﺱ( = ٦ﺱ – ٦
اﻻﻧﻘﻼب وﻣﻌﺎدﻟﺔ ﺎس ا ﻨﺤ ﻋﻨﺪﻫﺎ .
ﻣﻦ ا ﺮﺳﻢ :
اﻟﺔ د ﺣﻴﺚ د )ﺱ( = ﺱ ٣ –٣ﺱ ٩ – ٢ﺱ
٢
وا ﺤﺪب ﻷﺳﻔﻞ ﻨﺤ ا اﻟﺔ د ،وأوﺟﺪ ﻧﻘﻂ ا ﻞ
ﻧﻘﻄﺔ اﻧﻘﻼب ﻋﻨﺪ ﺱ = ١ﻷن د )ﺱ( ﺗﻐ ت إﺷﺎرﺗﻬﺎ ﻗﺒﻞ
د )ﺱ ( = ﺱ ٣ + ٣ﺱ ٩ – ٢ﺱ ﺇ د ) /ﺱ( = ٣ﺱ ٦ + ٢ﺱ – = ٩ﺻﻔﺮ ﻋﻨﺪﻣﺎ
٠ ﺱ - - -ﺭ) //ﺱ(----
( 3 + ¤ ) üï
[ – ، ] ١ ، ٢و ﺪب ﻷﺳﻔﻞ //
ا ﺤﻠﻴﺔ
ﻏ ﻣﻌﺮﻓﺔ ﻋﻨﺪ ﺱ = ٠ 9
اﻟﻔ ﺗ :
[]٥،١
و ﻌﺪ ﺱ = ١ﻱ ﺎل ا اﻟﺔ د .
د )ﺱ(
ﻋﻨﺪ ﺱ = ، ٠ﺭ )ﺱ( = -ﺫ ﺱ
ﺪب ﻷ
ﻣﻦ ا ا
ا ﻞ
ﺪب ﻷ
)ﺏ( ا ﻨﺤ
،ﰈ د (٠) /ﻏ ﻣﻌﺮﻓﺔ ﺇ
٠ ﺱ + + + +د) //ﺱ( ----
/
)ﺍ( ا ﻨﺤ
وﺣﻘﻖ إﺟﺎﺑﺘﻚ ﺑﺎﺳﺘﺨﺪام
[ – ﳘﺲ ، ] ٠ ،و ﺪب ﻷ
ﺇ ﺭ )ﺱ( = ﺫ ﺱ
ﻣﻦ ا ﺮﺳﻢ :
ا ﻨﺤ
ﺫ3
9
ﺪب ﻷ
،ﺭ )ﺱ( = ﺱ
ﺭ )ﺱ(
د
) (٥ﺑﺎﺳﺘﺨﺪام اﺧﺘﺒﺎر ا ﺸﺘﻘﺔ ا ﺎﻧﻴﺔ أوﺟﺪ اﻟﻘﻴﻢ اﻟﻌﻈ
ﻋﻨﺪ ﺱ = ، ٠د) //ﺱ( = -ﺫ ﺱ
،ا ﻨﺤ
ﺪب ﻷﺳﻔﻞ
)ﺏ( ﻫﻞ ﺗﻮﺟﺪ ﻧﻘﻂ اﻧﻘﻼب ﻨﺤ ﻫﺬه اﻟﻔ ة ؟ ﻓ ّ إﺟﺎﺑﺘﻚ .
[ – ﳘﺲ ، ٢ [ ، ] ٠ ،ﳘﺲ ]
) (٢ﺣﺪد ﻓ ات ا ﺤﺪب ﻷ
ا ﻤﺎس رأ
وا ﺤﺪب ﻷﺳﻔﻞ ﻨﺤ ا اﻟﺔ د
٠ ٢ ﺱ + + + + - - - - + + + +ﺭ) //ﺱ(
وﻣﻦ ا ﺮﺳﻢ :
ا ﺠﺎور ﻣﻨﺤ د )ﺱ(
اﻟﺔ ا ﺘﺼﻠﺔ د . ّ )ﺍ( وﺿﺢ ﻓ ات ا ﺤﺪب ﻷ
،ﺭ) //ﺱ( = ١٢ﺱ ٢٤ – ٢ﺱ ﺑﻮﺿﻊ ﺭ) //ﺱ( = ٠ﺉ ١٢ﺱ ) ﺱ – ٠ = ( ٢ ﺉ ﺱ = ٠أ ،ﺱ = ٢
اﻟﻔ ة [ – ] ٥ ، ٢
//
) ( ١ ، ٠ ) ، ( ٣ ، ٤ﻧﻘﺎط
ﺭ )ﺱ(
ا ﻨﺤ
ا اﻟﺔ ﻣ اﻳﺪة ﻋﻨﺪ ﺱ < ، ٠وا ﻨﺤ
ﻷﺳﻔﻞ ﻋﻨﺪ س > ، ٤و ﺪب ﻷ
٢
) ( ٢ ، ١ﺇ ﻣﻴﻞ ا ﻤﺎس ﻋﻨﺪﻫﺎ = ٣ = (١) ٣ – (١) ٦
) ( ٣ ، ٤ﻧﻘﻄﺔ اﻧﻘﻼب
١٤
ا ﻞ
ص
ﺪب
ﻋﻨﺪ ﺱ < ٤ س
ﻟﻸﺳﺘﺎﺫ ﺍﻟﻘﺪﻳﺮ /ﻋﻠﻰ ﺍﻟﺪﻳﻦ ﻳﺤﻴﻰ ٠١١١٩٦٦٠٦١٦
ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ -ﺍﻟﺼﻒ ﺍﻟﺜﺎﻟﺚ ﺍﻟﺜﺎﻧﻮﻯ ﺹ = ﺱ + ٣ﺍ ﺱ + ٢ﺏ ﺱ
) (٨ا ﻨﺤ
ً ) أو ﺻﻐﺮى ( ﺗﺒﻌﺎ ﻟﻄﻠﺐ ا ﺴﺄﻟﺔ . ّ ) (٥ﺗﺬﻛﺮ ﻴﻊ اﻟﻌﻼﻗﺎت ا ﻨﺪﺳﻴﺔ اﻟ ﺳﺒﻖ وأن ذﻛﺮﻧﺎﻫﺎ
ﻧﻘﻄﺔ اﻧﻘﻼب ﻋﻨﺪ
) ( ٢ ، ٢أوﺟﺪ : ً أوﻻ :ﻗﻴﻤﺔ ﻣﻦ ﺍ ،ﺏ ً ﺛﺎﻧﻴﺎ :ﻮﻗﻊ اﻟﻘﻴﻢ اﻟﻌﻈ وا ﺼﻐﺮى ا ﺤﻠﻴﺔ .
درس ا ﻌﺪﻻت ا ﺰﻣﻨﻴﺔ ا ﺮﺗﺒﻄﺔ .
ا ﻞ
) (١أوﺟﺪ أ
د )ﺱ( = ﺱ + ٣ﺍ ﺱ + ٢ﺏ ﺱ ﺇ د) /ﺱ( = ٣ﺱ ٢ + ٢ﺍ ﺱ +ﺏ ،د) //ﺱ( = ٦ﺱ ٢ +ﺍ
ﺩ) //ﺱ( = ٠ﻋﻨﺪ ﺱ = ٢ﺉ ﺍ = – ( ٢ ، ٢ ) ، ٦ﻱ ا ﻨﺤ
ﺇ د )ﺱ( = ﺱ ٦ – ٣ﺱ ٩ + ٢ﺱ ،د ) /ﺱ( = ٣ﺱ ١٢ – ٢ﺱ ٩ +
،د) //ﺱ( = ٦ﺱ – ، ١٢ﻧﻀﻊ د ) /ﺱ( = ٠ﻓﻨﺤﺼﻞ و
داﺧﻞ داﺋﺮة ﻃﻮل ﻧﺼﻒ ﻗﻄﺮﻫﺎ ١٢ﺳﻢ .
ﺉ ﺏ=٩
ا ﻘﻂ ا ﺮﺟﺔ
ﻋﻨﺪ ﺱ = ٥و ﺎﺧﺘﺒﺎرﻫﺎ ﺪ ﻋﻨﺪﻫﺎ ﻗﻴﻤﺔ ﺻﻐﺮى ﻠﻴﺔ
،ﺱ = ١وﻋﻨﺪﻫﺎ ﻗﻴﻤﺔ ﻋﻈ
ا ﻞ
ﺑﻔﺮض ﻡ = ﺱ ﺳﻢ
ﰈ ﻣﻢ ﺍﺏ ﺝ ﻣ ﺴﺎوى ا ﺴﺎﻗ
اﻟﺔ
ﻠﻴﺔ
ﺇ ﻡ = -144 üﺱﺫ × ) + ١٢ﺱ ( ﺇ
Ùﻡ Ùﺱ
= -144 üﺱﺫ + ١٢ ) -ﺱ ( ×
¤ -144ﺫ -ﺫ¤ - ¤1ﺫ Ùﻡ ، = ¤ -144Sﺫ Ùﺱ
ﻣﻦ ا وال اﻵﺗﻴﺔ وأوﺟﺪ ﻧﻘﻂ اﻻﻧﻘﻼب
) (١ﺹ = ﺱ ٩ – ٣ﺱ ٢٤ + ٢ﺱ – ١٠ ) (٢ﺹ = ﺱ ٢٤ – ٤ﺱ١٠ + ٢
ﺇ ﺴﺎﺣﺔ ا ﺜﻠﺚ ﺗ ﻮن أ
ﺇ أ
اﻟﻌﻈ
ﺹ = ﺱ ) ﺱ – ٢( ٣
ﻮﺿﺤﺎ ﻋﻠﻴﻪ ﻮاﻗﻊ اﻟﻘﻴﻢ
ا ﺼﻨﺪوق اﻟ
وا ﺼﻐﺮى ا ﺤﻠﻴﺔ وﻧﻘﻂ اﻻﻧﻘﻼب إن وﺟﺪت .
) (٦ارﺳﻢ ا ﺸ
ﺫ5ﺫ ﺇ ﺹ = ﺱﺫ
~٢د = (١) /د١ = (١ -) /
ﻌﻞ ا
ﻔﺔ أﻗﻞ ﻣﺎ ﻳﻤ ﻦ .
٢
ﺉ ﺣﺠﻤﻪ = ﺱ ﺹ = ٢٥٢
،ﺗ ﻠﻔﺔ اﻟﻄﻼء = ت = ٥٠ﺱ ٢٠ + ٢ﺱ ٤ × ٣٠ + ٢ﺱ ﺹ
ﺫ5ﺫ 40+ 3¤ 70ﺫ30 ﺇ ت = ٧٠ﺱ ١٢٠ + ٢ﺱ ﺹ = ٧٠ﺱ ١٢٠ + ٢ﺱ × ﺫ = ﺱ ﺱ ﺫ 3 10) ¤ /ﺫ 40+ ¤ 70) - ( ¤ﺫ 40 - 3¤140 1´ (30ﺫ30 = ﺇ ت = ﺱﺫ ﺱﺫ
ﺱ<٠
) (١ﺗ ﻮ ﻦ ﻗﺎﻋﺪة ا اﻟﺔ ﻣﻦ ﺧﻼل ا ﻄﻠﻮب
ً
ﺻﻨﺪوق ﺳﻌﺘﻪ ٢٥٢ﻣ ا ﻜﻌﺒﺎ ،وﻗﺎﻋﺪﺗﻪ
ﺑﻔﺮض أﺑﻌﺎد ا ﺰان :ﺱ ،ﺱ ،ﺹ ﻣ
~١د ) ٢ = (١ -د ) ، ٤ = (٠د )٠ = (١
~٣د) //ﺱ( > ٠
] ٣ﺳﻢ٢
ً
ا ﻞ
اﻟﻌﺎم ﻨﺤ ا اﻟﺔ ا ﺘﺼﻠﺔ د إذا ن :
ﺱ > ، ٠د) /ﺱ( < ٠
ﻣﺎﻳﻤ ﻦ
--- +++ -- ١٢٦
ﻡ/
/
) (٢ﺧﺰان ُ ﺮ ﻌﺔ .ﻳﺮاد ﻃﻼؤه ﻣﻦ ا اﺧﻞ ﺑﻤﺎدة زﻟﺔ ،ﻳﺘ ﻒ اﻟﻘﺎع ٥٠ ً ً ﻣ ﺮ ﻊ ،و ﺘ ﻒ اﻟﻐﻄﺎء ٢٠ﺟﻨﻴﻬﺎ ﻣ ﺮ ﻊ ﺟﻨﻴﻬﺎ ً ﻣ ﺮ ﻊ ،أوﺟﺪ أﺑﻌﺎد ،ﻛﻤﺎ ﻳﺘ ﻒ ا ﻮاﻧﺐ ٣٠ﺟﻨﻴﻬﺎ
) (٤د )ﺱ( = ﺱ | ﺱ – | ٤ ﺛﻢ ارﺳﻢ ا ﺸ
ﺝ
= ٠ﻋﻨﺪﻣﺎ ٢ﺱ ١٢ + ٢ﺱ – ٠ = ١٤٤
ﺴﺎﺣﺔ = ]٥٤ = ( ٦ + ١٢ ) × /٣٦/ –/ ١٤٤
ﺷ
اﻟﻌﺎم ﻠﻤﻨﺤ
ﺫ¤
ﺫ ¤ -144Sﺫ
وﻣﻦ ا ﺮﺳﻢ ﺪ أن ﻋﻨﺪ ﺱ = ٦ﺗﻮﺟﺪ ﻗﻴﻤﺔ ﻋﻈ
) (٣د )ﺱ( = ﺱ٧ – ٢ ) (٥ﻋ
ﺏ
ﻡ ١٢ﺱ
ﺇ ﺱ ٦ + ٢ﺱ – ٠ = ٧٢ﺇ ) ﺱ – ) ( ٦ﺱ ٠ = ( ١٢ +ﺉ ﺱ = ٦أ، ﺱ = ) ١٢ -ﺮﻓﻮض (
ً
١٢
،ﺴﺎﺣﺔ ﻣﻢ ﺍﺏ ﺝ = ﻡ = 1ﺏ ﺝ × ﺍ = ﺏ × ﺍ ﺫ
أﺟﺐ ﻋﻦ اﻷﺳﺌﻠﺔ اﻵﺗﻴﺔ :
ﻓ ات اﻟ اﻳﺪ وا ﻨﺎﻗﺺ ﻠﻤﻨﺤ
ﺇ ﺍ / ﻊﻋ ﺏ ﺝ /و ﻨﺼﻔﻪ
ﺍ
ﺇ ﺍ + ١٢ ) = ﺱ ( ﺳﻢ ﺇ ﺏ -144 ü = ﺱﺫ ﺳﻢ
· أدرس ﺪب ﻣﻨﺤ ) إن ُوﺟﺪت ( :
ﺴﺎﺣﺔ ﺜﻠﺚ ﻣ ﺴﺎوى ا ﺴﺎﻗ ﻳﻤ ﻦ رﺳﻤﻪ
،ت ٠ = /ﺉ ١٤٠ﺱ ٠ = ٣٠٢٤٠ – ٣ﺇ ﺱ ٠ = ٢١٦ – ٣ﺉ ﺱ = ٦
ا ﺴﺄﻟﺔ
،ت ٠ < /أﻗﻞ ﻣﺎﻳﻤ ﻦ ﻋﻨﺪ ﺱ = ٦ﺉ أﺑﻌﺎد ا ﺰان ٧ ، ٦ ، ٦ :ﻣ
) (٢ﻣﻦ اﻟﻌﻼﻗﺔ ا ﺬﻛﻮرة ﺑﺎ ﺴﺄﻟﺔ ﻧﻮﺟﺪ ﻣﺘﻐ ﺑﺪﻻﻟﺔ اﻵﺧﺮ ّ وﻧﻌﻮض ﺑﻪ ﻗﺎﻋﺪة ا اﻟﺔ
)(٣
ﺴﺘﻮى إﺣﺪا ﻣﺘﻌﺎﻣﺪ ُرﺳﻢ ﺍﺏ ﰐ ﻳﻤﺮ ﺑﺎ ﻘﻄﺔ ﺝ ) ( ٢ ، ٣
و ﻘﻄﻊ ﻮر اﻹﺣﺪاﺛﻴﺎت
) (٣ﺸﺘﻖ ﻗﺎﻋﺪة ا اﻟﺔ و ﺴﺎوى ا ﺸﺘﻘﺔ ﺑﺎ ﺼﻔﺮ ﻠﻮﺻﻮل ﻟﻘﻴﻤﺔ
ا ﻘﻄﺘ
ﺍ ،ﺏ ،أﺛﺒﺖ أن أﺻﻐﺮ
ﺴﺎﺣﺔ ﻠﻤﺜﻠﺚ ﺍو ﺏ ﺴﺎوى ١٢وﺣﺪة ﺮ ﻌﺔ ﺣﻴﺚ و
ا ﺘﻐ .
ﻧﻘﻄﺔ اﻷﺻﻞ ) . ( ٠ ، ٠
) (٤ﻧ ﺒﺖ أن ا اﻟﺔ ﻋﻨﺪ ﻫﺬة اﻟﻘﻴﻤﺔ ﺗ ﻮن ﻗﻴﻤﺔ ﻋﻈ
ا ﻞ
١٥
ﻟﻸﺳﺘﺎﺫ ﺍﻟﻘﺪﻳﺮ /ﻋﻠﻰ ﺍﻟﺪﻳﻦ ﻳﺤﻴﻰ ٠١١١٩٦٦٠٦١٦
ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ -ﺍﻟﺼﻒ ﺍﻟﺜﺎﻟﺚ ﺍﻟﺜﺎﻧﻮﻯ ﺱ= ﺫ ﺉ ﺹ= 6 ﺱ 3ﺹ
ﻣﻢ ﺍ ﺝ ﰲ ﻣﻢ ﺝ ﻩ ﺏ ﺇ
،ﺴﺎﺣﺔ ﻣﻢ ﺍو ﺏ = ﻡ = ) 1ﺱ ) ( ٣ +ﺹ ( ٢ + ﺫ ١ﺇ ﻡ=) 1ﺱ +٦=(٢+ 6 )(٣+ﺱ ٩+ﺱ ﺱ ﺫ ،ﻡ ٩ – ١ = /ﺱ
٢-
،ﻡ ١٨ = //ﺱ
ﺏ
) ، ٣ﺝ( ٢ ﺍ
٣-
ﺱ ٣
ﻧﻘﻄﺔ ﻩ
ا اﺋﺮة ورﺳﻢ ﻣﻨﻬﺎ ﺎس آﺧﺮ
ا ﺴﺎﺑﻘ
ﺹ ﻩ ٢ و
، ﺝ
،ﻡ ٠ < (٣) //اﻗﻞ ﻣﺎﻳﻤ ﻦ ﺉ ﻡ = ١٢ = 9 + ٣ + ٦وﺣﺪة ﺮ ﻌﺔ
) þ (١ﺍ .ﺱ = ﺍﺱ +ث ) ﺣﻴﺚ ﺍ ﻱ ( ò
3
) (٤إذا ن ﻴﻂ ﻗﻄﺎع داﺋﺮى = ١٢ﺳﻢ ،أوﺟﺪ ﻗﻴﺎس زاو ﺔ
) þ (٢ﺱﻥ .ﺱ =
ﻣﺎ ﻳﻤ ﻦ .
ﻗ +ل = ١٢ﺉ ل = ٢ – ١٢ﻗﻖ ﻴﻂ اﻟﻘﻄﺎع = ١٢ﺇ ٢ﻖ
ﻗ = ٦ﻗﻖ – ﻗﻖ ،ﺴﺎﺣﺔ اﻟﻘﻄﺎع = 1ل ﻗﻖ = ٢ – ١٢ ) 1ﻗﻖ ( × ﻖ ﺫ
ﺫ
وﻟ
Ð ﻦﻩ = ®
) þ (٦ﻗﺎ ٢ﺱ .ﺱ = ﻇﺎ ﺱ +ث
=٢
) þ (٧ﻗﺎ ﺱ ﻇﺎ ﺱ .ﺱ = ﻗﺎ ﺱ +ث
) þ (٨ﻗﺘﺎ ﺱ ﻇﺘﺎ ﺱ .ﺱ = -ﻗﺘﺎ ﺱ +ث ﻧﺘﺎﺋﺞ :
أﺟﺐ ﻋﻦ اﻷﺳﺌﻠﺔ اﻵﺗﻴﺔ :
1 ﺍ~ þﺟﺎ ) ﺍﺱ +ﺏ ( .ﺱ = – ﺟﺘﺎ )ﺍﺱ +ﺏ( +ث ﺍ 1 ﺏ~ þﺟﺘﺎ ) ﺍﺱ +ﺏ ( .ﺱ = ﺟﺎ ) ﺍﺱ +ﺏ ( +ث ﺍ 1 ﺝ~ þﻗﺎ ) ٢ﺍﺱ +ﺏ ( .ﺱ = ﻇﺎ ) ﺍﺱ +ﺏ ( +ث ﺍ 1 þ ~ﻗﺎ ) ﺍﺱ +ﺏ ( ﻇﺎ ) ﺍﺱ +ﺏ ( ﺱ = ﻗﺎ ) ﺍﺱ +ﺏ ( +ث ﺍ 1 و~ þﻗﺘﺎ ) ﺍﺱ +ﺏ ( ﻇﺘﺎ ) ﺍﺱ +ﺏ ( ﺱ = -ﻗﺘﺎ ) ﺍﺱ +ﺏ (+ث ﺍ
) (١ﻗﺴﻤﺖ ﻗﻄﻌﺔ ﻣﻦ ا ﺴﻠﻚ ﻃﻮ ﺎ ٣٤ﺳﻢ إ ﺟﺰأﻳﻦ وﺛ ا ﺰء اﻷول
ﺷ
ﺴﺘﻄﻴﻞ ﻋﺮﺿﻪ ﺱ ﺳﻢ ،ﻃﻮ ﺿﻌﻒ ﻋﺮﺿﻪ
وﺛ ا ﺰء ا ﺎ
ﺮ ﻊ .أوﺟﺪ ﻗﻴﻤﺔ ﺱ
ﺷ
ﻤﻮع ﺴﺎﺣ ا ﺴﺘﻄﻴﻞ وا ﺮ ﻊ أﺻﻐﺮ ﻣﺎ ﻳﻤ ﻦ . اﺸ
)(٢
ﻳ ﻮن
ا ﻘﺎﺑﻞ : ﺍ ﺝ
ﺍﺏ ﻳﻘﻄﻊ ﻮر ا ﺼﺎدات
) þ (٩ﻩﺱ ﺱ = ﻩﺱ +ث
ﻥ
ﺍ ،و ﻘﻄﻊ ﻮر ا ﺴ ﻨﺎت
) 1 þ (١٠ﺱ = ﻮ ﻩ | ﺱ | +ث ﺣﻴﺚ ﺱ ﻵ ٠ ﺱ
ﺏ
ﻼﺣﻈﺎت ﻫﺎﻣﺔ :
و ﻤﺮ ﺑﺎ ﻘﻄﺔ ﺝ ) . ( ٣ ، ٤أوﺟﺪ أﺻﻐﺮ ﺴﺎﺣﺔ ﺴﻄﺢ
) (١إذا ﻧﺖ ا ﺰاو ﺔ ا ﻮﺟﻮدة داﺧﻞ ا اﻟﺔ ا ﺜﻠﺜﻴﺔ ﻋﺪدﻳﺔ ﻓﺈﻧﻨﺎ ً ﻧﻌﺎ ﻠﻬﺎ ﻣﻌﺎ ﻠﺔ ا ﺎﺑﺖ .ﻓﻤﺜﻼ :
ﻣﻢ ﺍوﺏ ﺣﻴﺚ )و( ﻧﻘﻄﺔ اﻷﺻﻞ . ) (٣ﻳﺮاد ﻋﻤﻞ ﺧﺰان أﺳﻄﻮا ا ﺸ
و ﺪون ﻏﻄﺎء وذ ﻚ ﺑﺎﺳﺘﺨﺪام
٧٥ﺑﺐ ﻡ ٢ﻣﻦ ا ﺼﺎج .أوﺟﺪ أﺑﻌﺎد ا ﺰان أ
þﺟﺎ . ٥ ٦٠ﺱ = ﺱ ﺟﺎ + ٥ ٦٠ث þ ،ﻗﺎ . p ٢ﺱ = ﺱ ﻗﺎ + p ٢ث
ﻳ ﻮن ﺣﺠﻤﻪ
6
ﻣﺎ ﻳﻤ ﻦ .
) (٤أوﺟﺪ ﻧﻘﻄﺔ
ا ﺴﺎﻓﺔ ﺑ ﻨﻬﺎ و
ا ﻨﺤ
) (٥ﻠﻌﺐ
ﺷ
ﺹ = ] ٤ﺱ/٣/+ /
ﻴﺚ ﺗ ﻮن
ﺴﺘﻄﻴﻞ ﻳ ﺘ
ﺑﻨﺼ داﺋﺮﺗ
ﻴﻄﻪ ٤٢٠ﻣ ا ﻓﺄوﺟﺪ اﺑﻌﺎد ا ﻠﻌﺐ اﻟ
ﻣﺎﻳﻤ ﻦ . ) (٦ﺍﺏ /ﻗﻄﺮ
) (٢ﻣﻦ ا ﻬﻢ ﺗﺬﻛﺮ اﻟﻘﻮاﻧ
داﺋﺮة .رﺳﻢ ﺎﺳﺎن
6
اﻵﺗﻴﺔ :
ﺟﺎ ٢ﺱ = !٢؛ – !٢؛ ﺟﺘﺎ ٢ﺱ
ا ﻘﻄﺔ ) ( ٠ ، ٢أﻗﻞ ﻣﺎ ﻳﻤ ﻦ .
ً
+ث
) þ (٥ﺟﺘﺎ ﺱ .ﺱ = ﺟﺎ ﺱ +ث
ﻣﺎﻳﻤ ﻦ ،
//
ﺫ - 1ﺫ´ 3 ﺇ ﻩ= 3
ﻥ 1+
ﻥ 1+
) þ (٤ﺟﺎ ﺱ .ﺱ = – ﺟﺘﺎ ﺱ +ث
٢
ﺇ ﻡ ٢ – ٦ = /ﻗﻖ ،ﺑﻮﺿﻊ ﻡ = ٠ﺉ ﻗﻖ = ٣ﺳﻢ ،ﻡ > ٠أ /
ﺱ ﻥ 1+
+ث
) 1ﺍ( B + ¤ ) ) þ (٣ﺍﺱ +ﺏ (ﻥ ﺱ = × ﻥ 1+ ﺍ
ا ﻞ
ﰈ
اﻟ ﺗﻴﺐ .أﺛﺒﺖ أن أﺻﻐﺮ ﺴﺎﺣﺔ
ﺸﺒﻪ ا ﻨﺤﺮف ﺍﺏ ﺝ ﺴﺎوى ٢ﻗﻖ ٢وﺣﺪة ﺮ ﻌﺔ .
،ﻡ ٠ = /ﻋﻨﺪﻣﺎ ٩ – ١ﺱ ٠ = ٢ -ﺉ ﺱ ٩ = ٢ﺇ ﺱ = ٣
اﻟﻘﻄﺎع ا ى ﻌﻞ ﺴﺎﺣﺘﻪ أ
اﺋﺮة ﻗﻄﻊ ا ﻤﺎﺳ
ﺟﺘﺎ ٢ﺱ = ٢؛! ٢! +؛ ﺟﺘﺎ ٢ﺱ
،إذا ن
ﻇﺎ ٢ﺱ = ١ +ﻗﺎ ٢ﺱ
ﻌﻞ ﺴﺎﺣﺘﺔ أ
) þ (٣ﻩ ﻙ ﺱ +ﺍ .ﺱ = 1ﻩ ﻙ
ﻙﺱ+ﺍ
+ث ﺣﻴﺚ ﻙ ﻵ ٠
) þ (٤ﻩد )ﺱ( × د) /ﺱ( .ﺱ = ﻩد )ﺱ( +ث
اﺋﺮة ﻣﻦ ﺍ ،ﺏ ،أﺧﺬت
١٦
ﻟﻸﺳﺘﺎﺫ ﺍﻟﻘﺪﻳﺮ /ﻋﻠﻰ ﺍﻟﺪﻳﻦ ﻳﺤﻴﻰ ٠١١١٩٦٦٠٦١٦
ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ -ﺍﻟﺼﻒ ﺍﻟﺜﺎﻟﺚ ﺍﻟﺜﺎﻧﻮﻯ
) ( ) þ (٥ﺩ (¤ )yﺱ = ﻮ ﻩ | د )ﺱ( | +ث ﺩ ¤
ا ﻞ
]ﺩ ) [( ¤ ) ] þ (٦د )ﺱ( [ ﻥ د) /ﺱ( ﺱ = ﻥ 1+ ) ( ) þ (٧ﺩ ¤ yﺱ = S ٢ﺩ ) + ( ¤ث Sﺩ ) ( ¤
ﻥ1+
)ﺍ( ﺹ = ٢ ) ٤ﺱ ٢ × ٣( ٥ +ﺱ = ٢ ) ٨ﺱ ٣( ٥ +ﺱ
+ث
)ﺏ( ﺹ = ٢ﻩ ٢ﺱ – ٣ﺱ
ÐÙ0¬ - ¬Ù0Ð )ﺝ( ﺹ = Ðﺫ ﺗﻔﻜ ﻧﺎﻗﺪ ٢ :ﺱ ٢ +ﺹ ٠ = §Ùﺉ ﺹ = -ﺱ ﺱ ﺹ ¤Ù
ﺱ
) (٢أوﺟﺪ :
ُ ) (١إذا ﻋﻠﻢ ﻣﻴﻞ ا ﻤﺎس د) /ﺱ( ﻓﺈن ﻣﻌﺎدﻟﺔ ا ﻨﺤ
)ﺍ( ٣ þﺱ ) ﺱ ٤( ٣ + ٢ﺱ
:
ﺹ = þد) /ﺱ( .ﺱ ُ ) (٢إذا ﻋﻠﻢ ﻣﻌﺪل ﺗﻐ ﻣﻴﻞ ا ﻤﺎس د) //ﺱ( ﻓﺈن :
ا ﻞ
ﺍ~ إذا
ا ﻘﺪار = ﺫ ) þ 3ﺱ ٢ × ٤( ٣ + ٢ﺱ ﺱ = ﺫ þ 3ﻉ ٤ﻉ = ﺫ 51 × 3ﻉ + ٥ث 3 = ) 10ﺱ + ٥( ٣ + ٢ث
) ( = ﺩ ¤ﻓﺈن :
نﺩ
ﺩ)§ (
)ﺏ( ﺑﻔﺮض ﻉ = ﺱ ٤ – ٣ﺇ ﻉ = ٣ﺱ ٢ﺱ 1 ﺇ ا ﻘﺪار = 3
þد )ﺹ( .ﺹ = þد )ﺱ( .ﺱ ﺏ~ إذا
نﺩ
) ( = ﺩ )§( ﻓﺈن :
ﺩ ¤
) (٣أوﺟﺪ ا
ا
ﻞ ﺑﺎ ﻌﻮ ﺾ :
·
þد ] ﺭ )ﺱ( [ × ﺭ) /ﺱ( ﺱ :ﻧﻀﻊ ﺭ )ﺱ( = ﻉ
ﻞ ﻥSﺩ ) ( ¤ﻧﻀﻊ د )ﺱ( = ﻉ أ ،د )ﺱ( = ﻉ
ب دا
· ) þﺣﺎﺻﻞ
ﻥ
ا ﻘﺪار = ) þﻉ × ٢( 1 -ﻉ
Ð
ﺗﻔﻜ ﻧﺎﻗﺪ :
1 ﺫ
1 × 1ﻉ =)þ 1ﻉ٢– ٢ﻉ 1 ×(١+ﻉ ﺫﻉ
3 9 3 3 7 1 3 5 ﺫ ﺫ 5ﺫ ﺫ 1 = ) þ 1ﻉ ﺫ – ٢ﻉ ﺫ +ﻉ ﺫ( ﻉ = ) ﻉ ﺫ – × ٢ﻉ +ﻉ 3 5 7ﺫ 7 7ﺫ ﺫ 4 ﺫ 3 5 7 +ث = 3 ) ü 189ﺱ 3 ) ü 135 – (1 +ﺱ 3 ) ü 81 + (1 +ﺱ + (1 +ث 3 ﺫ(
) (٤أوﺟﺪ : )ﺍ( þ
ü
ا ﻘﺪار = ٣ – ١ ) þ 16-ﺱ( ٢ ٢ﺱ –٣
) ﺏ( ﺹ = ﻩ ا ﺘﻐ
ﺱ
¤ 3 - 1ﺫ
ﺱ )ﺏ( – ٣ ) þﺱ ( ﻩ ٦ﺱ – ﺱ ٢ﺱ ا ﻞ
)ﺍ( ﺑﻔﺮض ﻉ = ٣ – ١ﺱ ٢ﺇ ﻉ = – ٦ﺱ ﺱ
)ﺝ( ﺹ = ﻉ ﺣﻴﺚ ﻉ ،ل دوال
0ﺫ
)ﺏ( ﺑﻔﺮض ﻉ = ٣ﺱ ١ +ﺇ ﺱ = ﻉ 1 -ﺇ ﺱ = 1ﻉ 3 3
ﻛﺜ ة ﺣﺪود ،ﻡ = داﻟﺔ ﻣﺜﻠﺜﻴﺔ ،أ = داﻟﺔ أﺳﻴﺔ .
)ﺍ( ﺹ = ) ٢ﺱ ( ٥ +
)ﺏ( þﺱ S3 ٢ﺫ 1 + ¤ﺱ
4ﺫ
ﺣﻴﺚ ﺣﺮوﻓﻬﺎ ﺗﻌ اﻵ :ل = داﻟﺔ ﻮ ر ﺘﻤﻴﺔ ،ﻙ = داﻟﺔ
٤
ﻼت اﻵﺗﻴﺔ :
3ﻉ + ( ٥ث = ٢ ) 1ﺱ – ٢ ) 3 + ٦( ٣ﺱ – + ٥( ٣ث = 61 ) 41ﻉ + ٦ 5
ا اﻟﺔ اﻷو × ﺗﻜﺎﻣﻞ ا ﺎﻧﻴﺔ – ) þﺗﻜﺎﻣﻞ ا ﺎﻧﻴﺔ × ﺗﻔﺎﺿﻞ اﻷو ( ً · ﺘﺎر ا اﻟﺔ ا ﺮاد ﺗﻔﺎﺿﻠﻬﺎ ﺗﺒﻌﺎ ﻟ ﺗﻴﺐ ﺣﺮوف ﻤﺔ ) ﻜﻤﺎ (
) (١أوﺟﺪ ﺗﻔﺎﺿ
+ث
ا ﻘﺪار = þﻉ ﺫ × 3 +ﻉ × ٤ 1ﻉ = ) þ 41ﻉ ٣ + ٥ﻉ ( ٤ﻉ ﺫ
(=
ﻣﻦ :
3
4
)ﺍ( ﺑﻔﺮض ﻉ = ٢ﺱ – ٣ﺇ ﺱ = ﻉ ﺫ 3 +ﺇ ﺱ = 1ﻉ ﺫ
أى أن ﺗﻔﺎﺿ ﺹ = ا ﺸﺘﻘﺔ اﻷو ﺑﺎﻟ ﺴﺒﺔ إ ﺱ .ﺗﻔﺎﺿ ﺱ
ا
4
3
ا ﻞ
ﺗﻔﺎﺿ ا اﻟﺔ ﺹ = د )ﺱ( ﻫﻮ :ﺹ = د) /ﺱ( ﺱ
ﻞ ﺑﺎ ﺠﺰئ :
ﺱ3
(4 -
5
ﺱ = þ 1ﻉ – ٥ﻉ = 1- × 1ﻉ – + ٤ث
)ﺍ( þﺱ ) ٢ﺱ – ٤( ٣ﺱ
· إذا ُوﺟﺪ
þ
)
3ﺱﺫ
1= 4 ﺫ( 3 ¤ )1
( 1) þﺹ = ( 1) þﺱ ﺩ ¤ ﺩ §
ا
) ﺱ(4 - 3
5
ﺱ
)ﺍ( ﺑﻔﺮض ﻉ = ﺱ ٣ + ٢ﺇ ﻉ = ٢ﺱ ﺱ
د) /ﺱ( = þد) //ﺱ( .ﺱ ﺛﻢ ﺹ = þد) /ﺱ( .ﺱ ﻼﺣﻈﺎت ﻫﺎﻣﺔ :
)ﺏ( þ
ﺱﺫ
ﺱ.
1ﺫ
1× – ٦ﺱ ﺱ = þ 16-ﻉ ﺫ ﻉ
1 = ٢ × 16-ﻉ ﺫ +ث = – 3 - 1 ü 31ﺱﺫ +ث
)ب( ﺑﻔﺮض ﻉ = ٦ﺱ – ﺱ ٢ﺇ ﻉ = ) ٢ – ٦ﺱ ( ﺱ = – ٣ ) ٢ﺱ ( ﺱ
٢ ا ﻘﺪار = þ 1ﻩ ٦ﺱ – ﺱ × – ٣ ) ٢ﺱ ( ﺱ = þ 1ﻩﻉ ﻉ = 1ﻩﻉ +ث ﺫ ﺫ ﺫ
=
إذا ن ﺱ + ٢ﺹ ٢٥ = ٢أوﺟﺪ ﺹ ﺑﺪﻻﻟﺔ ﺱ ،ﺹ ،ﺱ
١٧
٦ 1ﺱ – ﺱ٢
ﺫ
ﻩ
+ث
ﻟﻸﺳﺘﺎﺫ ﺍﻟﻘﺪﻳﺮ /ﻋﻠﻰ ﺍﻟﺪﻳﻦ ﻳﺤﻴﻰ ٠١١١٩٦٦٠٦١٦
ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ -ﺍﻟﺼﻒ ﺍﻟﺜﺎﻟﺚ ﺍﻟﺜﺎﻧﻮﻯ ) (٥أوﺟﺪ : )ﺍ( þ
ا ﻘﺪار = ﺱ ﻮ ﻩ ) ﺱ þ – ( ١ +
Úﺫ¤
Úﺫ3 + ¤
ﺗﻔﻜ ﻧﺎﻗﺪ :
ﺑﺎﺳﺘﺨﺪام ا
1 ) ﺏ( þ ) ¤ئﻩ ( ¤
ﺱ
ﺫ
ﺱ
ﻞ ﺑﺎ ﻌﻮ ﺾ أﺛﺒﺖ ﺻﺤﺔ اﻟﻘﻮاﻋﺪ ا ﺎ ﺔ :
)ﺍ( ] þد )ﺱ( [ ﻥ د) /ﺱ( ﺱ = ]ﺩ ) [ ( ¤ ﻥ 1+
ﻥ 1+
،ﻥ ﻵ–١
+ث
) ( )ﺏ( þﺩ (¤ )yﺱ = ﻮ ﻩ | د )ﺱ( | +ث ،د )ﺱ( ﻵ ٠ ﺩ ¤
ا ﻞ
þﺱ ﺱ = þﺱ 1 -1 +ﺱ = ( 1 - ١ ) þﺱ ﺱ 1+ ﺱ 1+ ﺱ 1+
= ﺱ – ﻮ ﻩ | ﺱ + | ١ +ث ﺑﺎ ﻌﻮ ﺾ
)ﺏ( ﺑﻔﺮض ﺹ = ﻮ ﻩ ﺱ ﺇ ﺹ = 1ﺱ ﺱ þ1 1 ،ﻉ = ﺱ ﺫ ﺱ ﺇ ﻉ =1ﺱ ﺫ 1-
ﺗﻔﻜ ﻧﺎﻗﺪ ) :ﺍ( ﺑﻔﺮض ﻉ = د )ﺱ( ﺇ ﻉ = د ) /ﺱ( ﺱ
اﻷﻳﻤﻦ = þﻉ ﻥ ﻉ = ﻉ ﻥ + 1+ث = ]ﺩ ) [( ¤ ﻥ 1+ ﻥ 1+ Ùﻉ = ﻮ ﻩ | ﻉ | +ث = ﻮ ﻩ | د )ﺱ( | +ث )ﺏ( اﻷﻳﻤﻦ = þ ﻉ
= ]ﺱ ) ﻮ ﻩ ]ﺱ – + ( ١ث
ا ﻌﻮ ﺾ ؟ ﻓ ّ إﺟﺎﺑﺘﻚ
Sﺫ 1 + ¤
3
ا ﻞ ﺇ ﺹ =٣ﺱ þ ٢ – 1ﺱﺇ ﻉ = ﻩ
ﻉ = ﻩ– ٢ﺱ ﺱ
= 1- ﺫ = 1- ﺫ
)ﺏ( ﺑﻔﺮض
ﺫ
ﺫ
ﺫ 3 –٢ﺱ +ث ﻩ × + )٣ﺱ (٥+ﻩ ﺫ ﺫ ) ٣ﺱ ( ٥ +ﻩ– ٢ﺱ – 3ﻩ– ٢ﺱ +ث 4 ﻉ 3- 1 ﻉ =٢ﺱ ، ٣+ﺱ = ﺇ ﺱ = ﻉ ﺫ ﺫ ٢ – 1-ﺱ
1-
ا ﻘﺪار = ) 1 þﻉ – × ( ٣ﻉ ﺫ × 1ﻉ = 1 4 ﺫ ﺫ
ا ﻘﺪار = -ﺫ 1ﺱ ﻩ – ٢ﺱ þ 1 +ﻩ – ٢ﺱ ﺱ = -ﺫ 1ﺱ ﻩ – ٢ﺱ – 1ﻩ – ٢ﺱ +ث 4
+(1ث = -ﺫ 1ﻩ – ٢ﺱ ) ﺱ + ﺫ
ﺱ ﺑﻄﺮ ﻘﺔ
ﺇ ا ﻘﺪار = ٣ ) 1-ﺱ ( ٥ +ﻩ – ٢ﺱ – ) þ ( 3 -ﻩ – ٢ﺱ ﺱ
)ﺏ( þﺱ ٢ﻩ ﺱ ٣ +ﺱ
ﺫ
ﺱ ) ﺏ( þ Sﺫ 3 + ¤ ¤4
ﺱ
)ﺍ( ٣ ) þﺱ ( ٥ +ﻩ – ٢ﺱ ﺱ : ،
ﺇ ﺹ=ﺱ )ﺍ( ﺑﻔﺮض ﺹ = ﺱ þ– ٢ﺱ ،ﻉ = ﻩ ﺱ ﺇ ﻉ = ﺫ ﻩ
)ﺏ( ﺑﻔﺮض ﺹ = ﺱ ٢ﺇ ﺹ = ٢ﺱ ﺱ þﺱ ٣+ ،ﻉ = ﻩﺱ ٣ +ﺱ ﺇ ﻉ = ﻩ
= )1ﺫ ﻉ 4
3
3 ﺫ
ﺗﻔﻜ ﻧﺎﻗﺪ :
=)þﻉ
ﺫ 3
–ﻉ
ل=ﺱ ﺇ ل=ﺱ ﺑﻔﺮض þﺱ ٣+ ،ﻡ = ﻩﺱ ٣ +ﺱ ﺇ ﻡ = ﻩ
– ٣ﻉ ﺫ
6
ﺑﻔﺮض ﻉ = ٢ﺱ ١ +ﺇ ﺱ = ﻉ 1 -ﺇ ﺱ = 1ﻉ ﺫ ﺫ
13
= ٢)3ﺱ (١+ 5
1 )þﻉ ﺫ
1ﺫ (ﻉ
1 – ٢ × ٣ﻉ ﺫ ( +ث = ) ü 1ﺫ S 3 – 3 (3 + ¤ﺫ + 3 + ¤ث
ا ﻘﺪار = ٢ ) þﻉ – × ( ٢ﻉ
ا ﻘﺪار = ﺱ ٢ﻩﺱ þ ٢ – ٣ +ﺱ ﻩﺱ ٣ +ﺱ (١) ...........
13
(ﻉ = 3ﻉ
5 3
5
× 1ﻉ =)þﻉ – ×(١ﻉ ﺫ
5 3
– 3ﻉ ﺫ
– ٢)3ﺱ (١+ ﺫ
ﺫ 3
ﺫ 3
13
ﻉ
+ث
+ث
) (٩أوﺟﺪ ﻣﻌﺎدﻟﺔ ا ﻨﺤ ا ﺎر ﺑﺎ ﻘﻄﺔ ) ( ١ ، ٠وا ى ﻣﻴﻞ ا ﻤﺎس
ﺇ þﺱ ﻩﺱ ٣ +ﺱ = ﺱ ﻩﺱ þ – ٣ +ﻩﺱ ٣ +ﺱ = ﺱ ﻩﺱ – ٣ +ﻩ ) : (١ا ﻘﺪار = ﺱ ٢ﻩﺱ ٢ – ٣ +ﺱ ﻩﺱ ٢ + ٣ +ﻩ
)ﺍ( þﻮ ﻩ) ﺱ ( ١ +ﺱ
ﺱ ٣+
)ﺏ( ) þﻮ ﻩ ﺱ ÷ ]ﺱ ( ﺱ
ا ﻞ
)ﺍ( ﺑﻔﺮض ﺹ = ﻮ ) ﺱ ( ١ +ﺇ ﺹ = 1 ﻩ ﺱ 1+ þ -
س
ﻋﻨﺪ أى ﻧﻘﻄﺔ ) ﺱ ،ﺹ ( واﻗﻌﺔ ﻋﻠﻴﻪ ﺴﺎوى ﺱ ¤ üﺫ 1+
ﺱ ٣+
= ﻩ ﺱ ) ٣ +ﺱ ٢ – ٢ﺱ ( ٢ +
،ﻉ =ﺱ
ﺗﻔﻜ ﻧﺎﻗﺪ :ﻫﻞ ﻳﻤﻜﻨﻚ إ ﺎد þ
+ث
٢ – 1-ﺱ
) (٧أوﺟﺪ :
ﺱ
ﻧﻔﺮض ﺹ = ٣ﺱ ٥ +
ا ﻞ
ﺑﺎ ﻌﻮ ﺾ
ﺫ
ا ﻘﺪار = ] 1ﺱ ﻮ ﻩ ﺱ – þ 1ﺱ ﺫ ﺱ = ] 1ﺱ ﻮ ﻩ ﺱ – ] ٢ × 1ﺱ +ث ﺫ ﺫ ﺫ ﺫ
3ﺱ 5+ )ﺍ( þ Úﺫ¤
ﺫ 1 )ﺏ( ﺑﻔﺮض ﻉ = ﻮ ﻩ ﺱ ﺇ ﻉ = ﺱ ﺱ ا ﻘﺪار = 1 þﺫ ﻉ = þﻉ – ٢ﻉ = – ﻉ – + ١ث = + 1 -ث = + 1-ث ئﻩ ﺱ ﻉ ﻉ
)ﺍ( þﺱ ﻩ– ٢ﺱ ﺱ
)(١
ﺇ ا ﻘﺪار = ﺱ ﻮ ﻩ ) ﺱ – ( ١ +ﺱ +ﻮ ﻩ | ﺱ + | ١ +ث
ﺫ Úﺫ ¤ 1 1 1 1ﻮﻩ|ﻉ |+ث = ﻉ ﺱ= þ ا ﻘﺪار = þ ﺫ ﺫ ﻉ ﺫ Úﺫ 3 + ¤ = 1ﻮ ﻩ | ﻩ ٢ﺱ + | ٣ +ث
) (٦أوﺟﺪ :
ﺱ 1+
) (٨أوﺟﺪ :
)ﺍ( ﺑﻔﺮض ﻉ = ﻩ ٢ﺱ ٣ +ﺇ ﻉ = ٢ﻩ ٢ﺱ ﺱ
ﻥ 1+
ﺱ ﺱ (١) ...........
ا ﻞ
§Ù ﻣﻴﻞ ا ﻤﺎس = = ¤Ùﺱ ¤ üﺫ 1 + ﺇ ﺹ = þﺱ ¤ üﺫ 1+ﺱ = ٢ ) þ 1ﺱ ( ¤ üﺫ 1+ﺱ ﺫ
= × 1ﺫ ) ﺱ( ١ + ٢ ﺫ
3
3 ﺫ
+ث ،ﰈ ) ( ١ ، ٠ﻱ ﻠﻤﻨﺤ
ﺇ (١+٠)1 =١
+ث ﺉ ث = ﺫ ﺇ ﻣﻌﺎدﻟﺔ ا ﻨﺤ :ﺹ = ) 1ﺱ( ١ + ٢ 3
ﺇ ﻉ =ﺱ
١٨
3
3 ﺫ
3
+ﺫ 3
3 ﺫ
ﻟﻸﺳﺘﺎﺫ ﺍﻟﻘﺪﻳﺮ /ﻋﻠﻰ ﺍﻟﺪﻳﻦ ﻳﺤﻴﻰ ٠١١١٩٦٦٠٦١٦
ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ -ﺍﻟﺼﻒ ﺍﻟﺜﺎﻟﺚ ﺍﻟﺜﺎﻧﻮﻯ
)ﺏ( þﻗﺎ ﺱ ) ﺟﺘﺎ ٢ﺱ +ﻇﺎ ﺱ ( ﺱ
.
)ﺝ( þﻗﺘﺎ ﺱ ) ﻇﺘﺎ ﺱ – ﻗﺘﺎ ﺱ ( ﺱ
أﺟﺐ ﻋﻦ اﻷﺳﺌﻠﺔ اﻵﺗﻴﺔ : ﺎ ﻳﺄ :
· أوﺟﺪ
)þ (
) ) þ (١ﺱ S ( ١ – ٢ﺱ 1+ﺱ )þ (٢
ﺱﺫ
ﺱ þ ،ﺱ ) ٣ﺱ ٥( ١ – ٢ﺱ
Sﺫ + ¤ﺫ
5 )S þ (٣ﺫ ¤S -ﺱ þ ،
Ú ¤ﺫ¤
) ﺫ(1+ ¤
ﺫ
fﺫ ﺱ
e -1ﺱ
ﺱ ا ﻞ
)ﺍ( ا ﻘﺪار = – ﺟﺘﺎ ﺱ +ﻇﺎ ﺱ +ث
)ﺏ( ا ﻘﺪار = ) þﻗﺎ ﺱ ﺟﺘﺎ ٢ﺱ +ﻗﺎ ﺱ ﻇﺎ ﺱ ( ﺱ = × 1 ) þﺟﺘﺎ ٢ﺱ +ﻗﺎ ﺱ ﻇﺎ ﺱ ( ﺱ
ﺱ
fﺱ
= ) þﺟﺘﺎ ﺱ +ﻗﺎ ﺱ ﻇﺎ ﺱ ( ﺱ = ﺟﺎ ﺱ +ﻗﺎ ﺱ +ث
) þ (٤ﻮ ﻩ ﺱ .ﺱ þ ،ﺱ ٣ﻮ ﻩ ﺱ .ﺱ
٢
)ﺝ( ا ﻘﺪار = ) þﻗﺘﺎ ﺱ ﻇﺘﺎ ﺱ – ﻗﺘﺎ ﺱ ( ﺱ = – ﻗﺘﺎ ﺱ +ﻇﺘﺎ ﺱ +ث
) e - 1ﺱ ()( ¤ e + 1
٢ ٢ ) ) þ (٥ﻮ ﻩ ﺱ ( ﺱ þ ،ﻗﺘﺎ ) ﺱ ( 3 +ﺱ
)(
) ٦ þ (٦ﻗﺎ ٢ﺱ ) ﻇﺎ ٢ﺱ +ﺟﺘﺎ ٢ ٢ﺱ ( ﺱ þ ،ﺱ ﺟﺎ ﺱ .ﺱ
= + ١ ) þﺟﺎ ﺱ ( ﺱ = ﺱ – ﺟﺘﺎ ﺱ +ث
ﺫ
eﺫ) ﺫ (3 - ¤ )þ (٧ ) f -1ﺫ (3 - ¤
)ﺍ( ) þﺟﺎ ٣س – ﻗﺎ ٢ ٢س ( س
) þ (٨ﺱ ٢ﺟﺘﺎ ) ﺱ ( ٥ + ٣ﺱ þ ،ﻇﺘﺎ ٣ﺱ .ﺱ
)ﺏ( ٦ þﻗﺎ ٢س ) ﻇﺎ ٢س +ﺟﺘﺎ ٢ ٢س ( س
) (٩أوﺟﺪ ﻣﻌﺎدﻟﺔ ا ﻨﺤ ا ى ﻣﻴﻞ ا ﻤﺎس ﻋﻨﺪ أى ﻧﻘﻄﺔ
)ﺝ( + ١ ] þﻇﺘﺎ ٣ ) ٢س – [ ( ١س
ﻋﻠﻴﻪ ﻳﻌﻄﻰ ﺑﺎﻟﻌﻼﻗﺔ = §Ùﺱ § f +1 ¤Ù
) ( ٢ ، ٠ا ﻮاﻗﻌﺔ ﻋﻠﻴﻪ
:ﺹ=٢ﺱ. ٢+
· ﺗﻢ ﺗﻠﺨﻴﺺ ﻗﻮاﻋﺪ ﺗ ﺎ ﻞ ا وال ا ﺜﻠﺜﻴﺔ ﺑﺎ ﺼﻔﺤﺔ اﻷو - ﻳﺮ
ﺮاﺟﻌﺘﻬﺎ .
· ﻣﻦ ا ﻔﻴﺪ ﺗﺬﻛﺮ اﻟﻌﻼﻗﺎت ا ﺜﻠﺜﻴﺔ اﻵﺗﻴﺔ : âﺟﺎ ٢ﺱ +ﺟﺘﺎ ٢ﺱ = ١
)þ (
ﺴﺎوى
– ٤ﺟﺎ ﺱ ﺟﺘﺎ ﺱ ،و ﻧﺖ ﻣﻌﺎدﻟﺔ ا ﻤﺎس ﻋﻨﺪ ا ﻘﻄﺔ
e -1ﺱ
) (٢أوﺟﺪ :
ﺱ ) þ ،ﻇﺎ ٢ﺱ ٢ +ﺟﺎ ٢ﺱ ( ﺱ
) (١٠أوﺟﺪ ﻣﻌﺎدﻟﺔ ا ﻨﺤ ا ى ﻣﻌﺪل ﺗﻐ ﻣﻴﻞ ا ﻤﺎس
e - 1ﺫ ﺱ ا ﻘﺪار = þ e -1ﺱ
ﺱ=þ
ﺱ
eﺫ ) ﺫ( 3 - ¤
) f - 1ﺫ( 3 - ¤
س ا ﻞ
1ﻇﺎ ٢ﺱ +ث )ﺍ( ا ﻘﺪار = – 31ﺟﺘﺎ ٣ﺱ – ﺫ
)ﺏ( ا ﻘﺪار = ٦ ) þﻗﺎ ٢ﺱ ﻇﺎ ٢ﺱ ٦ +ﻗﺎ ٢ﺱ ﺟﺘﺎ ٢ ٢ﺱ ( ﺱ = ٦ ) þﻗﺎ ٢ﺱ ﻇﺎ ٢ﺱ ٦ +ﺟﺘﺎ ٢ﺱ ( ﺱ =×٦ 1ﺟﺎ ٢ﺱ +ث = ٣ﻗﺎ ٢ﺱ ٣ +ﺟﺎ ٢ﺱ +ث 1ﻗﺎ ٢ﺱ × ٦ + ﺫ ﺫ )ﺝ( ا ﻘﺪار = þﻗﺘﺎ ٣ ) ٢ﺱ – ( ١ﺱ = – 31ﻇﺘﺎ ) ٣ﺱ – + ( ١ث f - 1ﺫ ) ﺫ( 3 - ¤ ﺱ ) (ا ﻘﺪار = þ ) f - 1ﺫ( 3 - ¤ ) f - 1ûùﺫ ) f + 1ûù ëé( 3 - ¤ﺫ é ( 3 - ¤ =þ ëﺱ
) f - 1ﺫ( 3 - ¤
+ ١ âﻇﺎ ٢ﺱ = ﻗﺎ ٢ﺱ
= + ١ ] þﺟﺘﺎ ) ٢ﺱ – [ ( ٣ﺱ = ﺱ +ﺟﺎ ) ٢ﺱ – + ( ٣ث
+ ١ âﻇﺘﺎ ٢ﺱ = ﻗﺘﺎ ٢ﺱ
) (٣أوﺟﺪ ﻣﻌﺎدﻟﺔ ا ﻨﺤ ا ى ﻳﻤﺮ ﺑﺎ ﻘﻄﺔ ) ( ٢ – ، ١وﻣﻴﻞ
âﺟﺎ ٢ﺱ = ٢ﺟﺎ ﺱ ﺟﺘﺎ ﺱ
ا ﻤﺎس ﻋﻨﺪ أى ﻧﻘﻄﺔ ﻋﻠﻴﻪ ) ﺱ ،ﺹ ( ﻫﻮ :
âﺟﺘﺎ ٢ﺱ = ٢ﺟﺘﺎ ٢ﺱ – ١
= ٣ﺑﺐ ﺟﺘﺎ ﺑﺐ ﺱ – ٢ﺑﺐ ﺟﺎ ﺑﺐ ﺱ
âﺟﺘﺎ ٢ﺱ = ٢ – ١ﺟﺎ ٢ﺱ
ا ﻞ
ﰈ ٣ = §Ùﺑﺐ ﺟﺘﺎ ﺑﺐ ﺱ – ٢ﺑﺐ ﺟﺎ ﺑﺐ ﺱ ¤Ù
ﺇ ﺹ = ٣ ) þﺑﺐ ﺟﺘﺎ ﺑﺐ ﺱ – ٢ﺑﺐ ﺟﺎ ﺑﺐ ﺱ ( ﺱ
) (١أوﺟﺪ :
= ٣ﺟﺎ ﺑﺐ ﺱ ٢ +ﺟﺘﺎ ﺑﺐ ﺱ +ث ،ﰈ ) ( ٢ – ، ١ﻱ ا ﻨﺤ
ﺇ – ٣ = ٢ﺟﺎ ﺑﺐ ٢ +ﺟﺘﺎ ﺑﺐ +ث ﺉ ث = ﺻﻔﺮ
)ﺍ( ) þﺟﺎ ﺱ +ﻗﺎ ٢ﺱ ( ﺱ
ﺇ ﻣﻌﺎدﻟﺔ ا ﻨﺤ
١٩
:ﺹ = ٣ﺟﺎ ﺑﺐ ﺱ ٢ +ﺟﺘﺎ ﺑﺐ ﺱ
ﺇ
ﻘﻖ ﻣﻌﺎد ﻪ
ﻟﻸﺳﺘﺎﺫ ﺍﻟﻘﺪﻳﺮ /ﻋﻠﻰ ﺍﻟﺪﻳﻦ ﻳﺤﻴﻰ ٠١١١٩٦٦٠٦١٦
ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ -ﺍﻟﺼﻒ ﺍﻟﺜﺎﻟﺚ ﺍﻟﺜﺎﻧﻮﻯ ﺍ
) (١إذا ﻧﺖ د )ﺱ( ﻓﺮدﻳﺔ وﻣﺘﺼﻠﺔ
)ﺏ( þﺟﺎ ) ٢ﺱ – ( ٥ﺱ
-ﺍ
،
þد )ﺱ( ﺱ = ﺻﻔﺮ ﺏ
ﺍ
þد )ﺱ( ﺱ = –
-ﺍ
3 )ﺏ( f þﺫﺱ 4 -ﺱ
،
fﺱ
þد )ﺱ( ﺱ = þ ٢د )ﺱ( ﺱ 0
ﺏ
ﺍ
þد )ﺱ( ﺱ =
) (٣ﺴﺘﺨﺪم ا ﺸ
)) (٤ﺍ( ) þﻇﺎ ٢ﺱ ٢ +ﺟﺎ ٢ﺱ ( ﺱ
ا ﺎ
ﺱ ﺱ=٠
)) (٦ﺍ( þﺟﺎ ٥ﺱ ﺟﺘﺎ ﺱ ﺱ ) (١أوﺟﺪ ﻗﻴﻤﺔ
)ﺏ( þﺱ ﺟﺎ ﺱ ﺱ
¤Ù
) ( ١ ، p3أوﺟﺪ ﻣﻌﺎدﻟﺔ ا ﻨﺤ .
p
-
þ pﺫ ﺟﺘﺎ θ θ ﺫ
ا ﻞ
4
0
) (٢إذا ﻧﺖ د داﻟﺔ ﻣﺘﺼﻠﺔ
،ò
1-
4
þد )ﺱ( ﺱ = ، ٢٥٥
1-
4
þد )ﺱ( ﺱ = – ١٥ﻓﺄوﺟﺪ þ :د )ﺱ( ﺱ
ﺫ
· ا ﻈﺮ ﺔ اﻷﺳﺎﺳﻴﺔ ﻠﺘﻔﺎﺿﻞ :
اﻟﻔ ة ] ﺍ ،ﺏ [ ،و ﻧﺖ ت أى
ﰈ
ﻧﻔﺲ اﻟﻔ ة ﻓﺈن :
4
1-
þد )ﺱ( ﺱ =
ﺫ
ﺫ
ا ﻞ 4
þد )ﺱ( ﺱ +ﺫ þد )ﺱ( ﺱ
14
4
ﺇ + (١٥ –) – = ٢٥٥ﺫ þد )ﺱ( ﺱ ﺉ ﺫ þد )ﺱ( ﺱ = ٢٤٠ = ١٥ – ٢٥٥
þد )ﺱ( س = ت )ﺏ( – ت )ﺍ( = ﻋﺪد ﺣﻘﻴ ﻞ ا ﺤﺪد :
ù
ﺫ
اﻟﺔ د
5 1 é
p
4
إذا ﻧﺖ ا اﻟﺔ د ﻣﺘﺼﻠﺔ
1 –( ٤ +ﺫ
ﻥ = ´ 3 úﺫ) (4 +°ﺫ ٦ = ١٢ – ١٨ = ê ë û
)ﺝ( ا ﻘﺪار = é q e ùﺫ ٢ = (١ -) – ١ = pë û
)(٢
ﺱ
1-
)ﺏ( ا ﻘﺪار = ) ٣ þﻥ
ﺣﻴﺚ ﺍ ﺛﺎﺑﺖ ﻓﺈذا ن ا ﻨﺤ ﻳﻤﺮ ﺑﺎ ﻘﻄﺘ ) ، (٥ ، p
ﺍ
ﺍ 1 þﺫ ﺱ ﻷﻧﻬﺎ ﻏ ﻣﺘﺼﻠﺔ ﻋﻨﺪ
ﺫ
) (١٠ﻣﻨﺤ ﻣﻴﻞ ا ﻤﺎس ﻋﻨﺪ أﻳﻦ ﻧﻘﻄﺔ ﻋﻠﻴﻪ ﺴﺎوى ﺍ ﻗﺘﺎ ٢ﺱ
) (٣أوﺟﺪ : )ﺍ(
ﺏ
þد )ﺱ( س = – þد )ﺱ( س ﺍ
-ﺍ
ﻝ
)ﺍ( ا ﻘﺪار = ùûﺱﺫ ١٢ = ( ٣ – ١ ) – ( ٦ + ٤ ) = 1- éë ¤ 3 +
ﺹ = ٥ﻋﻨﺪ ﺱ = ٠
) (١
ﻋﻜﺲ ﺇﺷﺎﺭﺓ ﺍ
٢) þﺱ(٣+ﺱ
)ﺝ(
) (٩إذا ن – ٧ = §Ùﺟﺎ ٢ﺱ أوﺟﺪ ﺹ ﺑﺪﻻﻟﺔ ﺱ إذا ن
ﺏ
٠
ﻣﺜﻞ ﺇﺷﺎﺭﺓ ﺍ
5 3ﻥ ) ﺏ( þ S 0ﻥ 4 +
)ﺏ( þﻇﺘﺎ س ﻗﺘﺎ ٣س س
ﺍ
ﺤﺚ إﺷﺎرة ا اﻟﺔ اﻟ ﻴﻌﻴﺔ :
ﺎ ﻳﺄ :
ﺫ
)ﺍ(
)) (٧ﺍ( / + ١] þﺟﺎ /ﺱ × /ﺟﺘﺎ ﺱ ﺱ
ﺍ
þ
-ﺍ
د )ﺱ( ﺱ
)ﺏ( + ٣ ) þﺟﺎ ﺱ ( ٥ﺟﺘﺎ ﺱ ﺱ
· ﺧﻮاص ا
ﻡ
) (٤ﻻ ﺴﺘﻄﻴﻊ إ ﺎد ﻗﻴﻤﺔ
)ﺏ( þﺱ ٢ﺟﺘﺎ ) ﺱ ( ٥ + ٣ﺱ
)) (٨ﺍ( þﻇﺘﺎ ٣ﺱ ﺱ
-ﺏ
٠
ﻣﺜﻞ ﺇﺷﺎﺭﺓ ﺍ
)) (٥ﺍ( þﺱ ﻗﺎ ) ٢ﺱ ( ٢ + ٢ﺱ
] – ﺍ ،ﺍ [ ﻓﺈن :
ﺍ
ﺍ
)) (٣ﺍ( ) þﺟﺎ ﺱ – ﺟﺘﺎ ﺱ ( ٢ﺱ
ﺏ
-ﺏ
þ
-ﺍ
د )ﺱ( ﺱ
) (٢إذا ﻧﺖ د )ﺱ( زوﺟﻴﺔ وﻣﺘﺼﻠﺔ
)ﺏ( þﻇﺎ ﺱ ﺟﺘﺎ ﺱ ﺱ
ﺸﺘﻘﺔ ﻋﻜﺴﻴﺔ
] – ﺍ ،ﺍ [ ﻓﺈن :
ﺍ
)) (٢ﺍ( + ١ ) þﻇﺎ ٢ﺱ ( ﺟﺘﺎ ٢ﺱ ﺱ
¤g¤i ) ﺏ( þ 1- ¤ i
ﺝ
ﺍ
· ﻼﺣﻈﺎت ﻫﺎﻣﺔ :
أﺟﺐ ﻋﻦ اﻷﺳﺌﻠﺔ اﻵﺗﻴﺔ : ً ﻼت اﻵﺗﻴﺔ : · أوﺟﺪ ﻣﻦ ا )) (١ﺍ( ٣] þﺟﺘﺎ ﺱ ﺱ
ﺏ
)(٣
ﺝ
ﺏ
þد )ﺱ( س = þد )ﺱ( س þ +د )ﺱ( س
ﺍ
) ﺏ(
þد )ﺱ( س = ﺻﻔﺮ
٢٠
4-
-ﺫ
| þﺱ |١+ﺱ | þﺱ | ٤ – ٢ﺱ 3
3-
ا ﻞ
ﻟﻸﺳﺘﺎﺫ ﺍﻟﻘﺪﻳﺮ /ﻋﻠﻰ ﺍﻟﺪﻳﻦ ﻳﺤﻴﻰ ٠١١١٩٦٦٠٦١٦
ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ -ﺍﻟﺼﻒ ﺍﻟﺜﺎﻟﺚ ﺍﻟﺜﺎﻧﻮﻯ - üﺱ 1- > ¤ @ 1 - د )ﺱ( = | ﺱ ý = | ١ + þﺱ 1- £ ¤ @ 1 +
)ﺍ(
ﺫ
ﺇ
4-
þد )ﺱ( ﺱ =
4-
þ
1-
1= ¤ 1 - ùﺫ ¤ 1ù + é¤ -ﺫ é ¤ + û 4- ëﺫ ûﺫ
ﺫ
3
þد )ﺱ( ﺱ =
ﺇ ﺫ
33 ٢
3-
1-
ﺫ
ﰈ د )ﺱ( ﻓﺮدﻳﺔ ﺇ ﺱ ﺫ ﺱ = ﺻﻔﺮ þ -ﺫ +1ﺱ
)(١
p
) (٢ﰈ د )ﺱ( زوﺟﻴﺔ ﺇ ا ﻘﺪار = + ٤ ) þ ٢ﺑﺐ ﺟﺘﺎ ٢ﺱ ( ﺱ ٢
–٢
–
٠
¤ üﺫ - ³ ¤ @ 4 -ﺫ ï د )ﺱ( = ¤ - ýﺫ - @ 4 +ﺫ > > ¤ﺫ ï ¤ þﺫ £ ¤ @ 4 -ﺫ
+ ٠
= e p +¤4ùﺫ ٤ = é ¤ﺑﺐ ٤ +ﺑﺐ = ٨ﺑﺐ ﺫ û 0ë
ﺱ د )ﺱ(
ﺗﻔﻜ ﻧﺎﻗﺪ :
) þﺱ – (٤ﺱ
1 - ù +ﺱé ¤4 + 3 = 1 ùﺱ é ¤4 - 3 ë 3û 3 û 3- ë ﺫ46 7 3 = + 3 + 7 = 3 3 3 -ﺫ
– ) þﺱ ( ٤ + ٢ﺱ +
1 ù +ﺱ é ¤4 - 3 ë 3û
ﺫ -ﺫ
ﺇ
ﺫ
-ﺫ
0
p
) (١ﰈ
) ﺱ ( ٤ – ٢ﺱ +
35
3-
) (٢ﰈ
3 ﺫ
þد )ﺱ( ﺱ =
3
3-
| þﺱ | ٤ – ٢ﺱ :د )– ﺱ( = د )ﺱ( ﺉ ا اﻟﺔ زوﺟﻴﺔ وﻣﺘﺼﻠﺔ 3
ﺫ
ò
ﺫ 3 1 1 ) þﺱ ( ٤ – ٢ﺱ [ = ( éê ¤4 - 3 ¤ ùú + éê 3 ¤ - ¤4ùú ) ٢ ﺫ 3û 0ë 3 û ëﺫ 46 8 8 = = [ ( ٨ – ) – ( ١٢ – ٩ ) + ( – ٨ ) ] ٢ 3 3 3
þد )ﺱ( ﺱ = ٢
40
ﺇ
þد )ﺱ( ﺱ = ١٠
،
ﺫ
4-
þد )ﺱ( ﺱ =
)ﺍ( þﺱ 5 üﺫ ¤ -ﺫ ﺱ 0
ﺫ )ﺏ( þﺱ¤ ü ٣ﺫ 3 +ﺱ -ﺫ
)(٥
ا ﻞ
é3 ù = – × 1ﺫ 5)úﺫ ¤ -ﺫ ( ﺫ ( ٠ – ٢٥ ) – ٠ ] 1 – = ê 5
ﺫ
û 3
0ë
3
0
4-
د )ﺱ( = د )– ﺱ( ﺇ د )ﺱ( ﻓﺮدﻳﺔ وﻣﺘﺼﻠﺔ
) (٥أوﺟﺪ : )ﺍ(
3-
) ﺏ(
þ
p-
3
ﺱ
+1ﺱ ﺫ
4 -6 3 ﺱ ﺱ þ ﺫ
٢ þﺟﺎ ﺑﺐ ﻉ ﻉ
) ﺏ(
4 þﻇﺎ ﻉ ﻗﺎ ٢ﻉ ﻉ
ﺫ¤
1
0
5
| þﺱ –|٣ﺱ
0
p 6 3
p 3þ
¤Ù fﺫ¤
5ﺫ1 = 3
ﺫ
ﺫ
-ﺫ
p
5
þد )ﺱ( ﺱ = ٩ﻣﺎ ﻗﻴﻤﺔ ﺫ
4
،ò
4
þد )ﺱ( ﺱ = ، ٢٤٠
1-
4
1-
þ ، òد )ﺱ( ﺱ = ، ٢٤٠ 4
4
6
þ ، òد )ﺱ( ﺱ = ، ٢٤٠ 1
4
þد )ﺱ( ﺱ = – ١٥ﻓﺄوﺟﺪ þد )ﺱ( ﺱ . 1
ﺧﻂ ﺴﺘﻘﻴﻢ ﻣﻦ ﻧﻘﻄﺔ ﺛﺎﺑﺘﺔ ،و ﻧﺖ
üﺫ 3 ³ k @ 13 + k ﻉ =ý k 3þﺫ 3 < k @ 8 -
5
(٢إذا ﻧﺖ د داﻟﺔ زوﺟﻴﺔ ﻣﺘﺼﻠﺔ
¤Ù 1 ) ﺏ( þ ¤ ﻩ
ﻋﺘﻬﺎ ﺑﺎ ﺴﻢ /ث ﺑﻌﺪ زﻣﻦ ﻗﺪره ﻥ ﺛﺎﻧﻴﺔ ﺗﻌﻄﻰ ﻣﻦ اﻟﻌﻼﻗﺔ :
اﻟﻔ ة ]– ، [ ٥ ، ٣ 3-
0
ﺫ
) (٩ﺗﺘﺤﺮك ﻧﻘﻄﺔ ﻣﺎدﻳﺔ
+ ٤ ) þﺑﺐ ﺟﺘﺎ ٢ﺱ ( ﺱ
0 3
þد )ﺱ( ﺱ = – ١٥ﻓﺄوﺟﺪ þد )ﺱ( ﺱ .
) (٨إذا ﻧﺖ د داﻟﺔ ﻣﺘﺼﻠﺔ
ﺱ
p
٢ ) þﺱ – ٧ﻩﺱ ( ﺱ
ﺫ
1-
þد )ﺱ( ﺱ = ٠
1
Sﺱ 1-
þد )ﺱ( ﺱ = – ١٥ﻓﺄوﺟﺪ þد )ﺱ( ﺱ .
) (٧إذا ﻧﺖ د داﻟﺔ ﻣﺘﺼﻠﺔ ﺫ
) ﺏ(
4
| þﺱ | ٤ – ٢ﺱ
1-
6
(١إذا ﻧﺖ د داﻟﺔ ﻓﺮدﻳﺔ ﻣﺘﺼﻠﺔ
þد )ﺱ( ﺱ ؟
أوﺟﺪ :
) (iإزاﺣﺔ ا ﻘﻄﺔ ا ﺎدﻳﺔ ﺧﻼل ا ﺎﻧﻴﺔ ا ﺎ ﺔ ﻓﻘﻂ .
اﻟﻔ ة ]– ، [ ٤ ، ٤
) (iiإزاﺣﺘﻬﺎ ﺧﻼل ا ﻮا ا ﺎ ﺔ وا ﺮاﺑﻌﺔ وا ﺎ ﺴﺔ . ) (iiiﺑﻌﺪﻫﺎ ﻋﻦ ا ﻘﻄﺔ ا ﺎﺑﺘﺔ ﺑﻌﺪ ٥ﺛﻮان ﻣﻦ ﺑﺪء ا ﺮ ﺔ .
þد )ﺱ( ﺱ = þ ، ٢٠د )ﺱ( ﺱ = ، ٦ﻣﺎ ﻗﻴﻤﺔ
4-
4-
ﺱ 1-ﺱ
) ﺏ(
4
ﺗﻔﻜ ﻧﺎﻗﺪ :
3
0
) (٦إذا ﻧﺖ د داﻟﺔ ﻣﺘﺼﻠﺔ
1 ﺱ ( ٢ﺫ ﺱ
)ﺏ( د )ﺱ( = ﺱ ¤ ü ٣ﺫ ، 3 +د )– س( = – ﺱ ¤ ü ٣ﺫ 3 +ﺉ òﺉ
ﺫ
þد )ﺱ( ﺱ þ +د )ﺱ( ﺱ = ١٦ = ٦ + ١٠
þ
1
)) (٤ﺍ(
3 ﺫ
4-
4-
)) (٣ﺍ(
) (٤أوﺟﺪ :
)ﺍ(
0
þد )ﺱ( ﺱ = ) ٢٠ﻷن ا اﻟﺔ زوﺟﻴﺔ (
4
)) (٢ﺍ(
5
5 þس 5 üﺫ ¤ -ﺫ ﺱ = – ٢ – þ 1ﺱ ) – ٢٥ ﺫ 0 0
3
þد )ﺱ( ﺱ = ﺻﻔﺮ ) ٩ = ٩ +ﻷن ا اﻟﺔ ﻓﺮدﻳﺔ (
)) (١ﺍ(
3
5
þد )ﺱ( ﺱ þ +د )ﺱ( ﺱ
أﺟﺐ ﻋﻦ اﻷﺳﺌﻠﺔ اﻵﺗﻴﺔ :
ﺇ ا ﻘﺪار = | þ ٢ﺱ – | ٤ﺱ = – ٤ ) þ ] ٢ﺱ ( ﺱ + 0
3-
5
٢
0
3
5
ﺣﻞ آﺧﺮ :
٢
ا ﻞ
ﺫ
) þﺱ (١+ﺱ
= ﺫ + 9ﺫ٩ = 9
ً )ﺏ( أوﻻ :ﻧﺒﺤﺚ إﺷﺎرة ا اﻟﺔ د )ﺱ( = ﺱ٤ – ٢ ً ﺛﺎﻧﻴﺎ :ﻧﻌ ﻋﻦ د )ﺱ( ﺑﻔ اﺗﻬﺎ ﻵ : +
þ
4-
) – ﺱ –(١ﺱ + 1- ë
-ﺫ
ﺫ
þد )ﺱ( ﺱ ؟
0
٢١
ﻟﻸﺳﺘﺎﺫ ﺍﻟﻘﺪﻳﺮ /ﻋﻠﻰ ﺍﻟﺪﻳﻦ ﻳﺤﻴﻰ ٠١١١٩٦٦٠٦١٦
ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ -ﺍﻟﺼﻒ ﺍﻟﺜﺎﻟﺚ ﺍﻟﺜﺎﻧﻮﻯ
4 ¤4 4 ﻡ = þد )ﺱ( ﺱ = ) þ ﺫ 0 0 1+ ¤
ً أوﻻ :ﺴﺎﺣﺔ ﻣﻨﻄﻘﺔ ) ﻡ ( ﺪدة ﺑﻤﻨﺤ ا اﻟﺔ د و ﻮر ا ﺴ ﻨﺎت ﺱ=ﺍ ،ﺱ=ﺏ )
وا ﺴﺘﻘﻴﻤ
ﻡ=
ﺍ
þ
ﺏ
4
) (٣أوﺟﺪ ﺴﺎﺣﺔ ا ﻨﻄﻘﺔ ا ﺴﺘﻮ ﺔ ا ﺤﺪدة ﺑﺎ ﻨﺤ ﺹ = ٢ + ٣ﺱ – ﺱ ٢و ﻮر ا ﺴ ﻨﺎت .
.
ﺗﻔﻜ ﻧﺎﻗﺪ :أوﺟﺪ ﺴﺎﺣﺔ ا ﻨﻄﻘﺔ ا ﺴﺘﻮ ﺔ ا ﺤﺪدة ﺑﺎ ﻨﺤ
ﺧﻄﻮات إ ﺎد ا ﺴﺎﺣﺔ :
) (١إ ﺎد أﺻﻔﺎر ا اﻟﺔ ) إن وﺟﺪت ( اﻟ
ﺹ = ٢ + ٣ﺱ – ﺱ٢وا ﺴﺘﻘﻴﻤﺎت ﺱ = – ، ١ﺱ = ، ٤ﺹ = ٠
ﺰئ ﺎل ا اﻟﺔ إ
ﻓ ات ﺟﺰﺋﻴﺔ ] ﺍ ،ﺝ [ ] ،ﺝ ،ﺏ [ .
) (٢دراﺳﺔ إﺷﺎرة ا اﻟﺔ
= ùû ٢ئﻩ ¤ﺫ ٢ = 0éë 1 +ﻮ ٢ – ١٧ﻮ ٢ = ١ﻮ ١٧وﺣﺪة ﺮ ﻌﺔ ﻩ ﻩ ﻩ
اﻟﻔ ة ] ﺍ ،ﺏ [ ( :
د )ﺱ( ﺱ
ﺹ = ٠ﻋﻨﺪﻣﺎ ٢ + ٣ﺱ – ﺱ٠ = ٢
اﻟﻔ ات ا ﺰﺋﻴﺔ ) إن وﺟﺪت ( : ﺝ
3
ﺏ
1-
ﻡ=
ﺍ
þ
ﺏ
د )ﺱ( ﺱ –
ﺍ
ﺭ )ﺱ( ﺱ
ﺗﻔﻜ ﻧﺎﻗﺪ :ﻡ = ٢ + ٣ ) þﺱ – ﺱ ( ﺱ ٢ + ٣ ) þ |+ﺱ – ﺱ ( ﺱ |
ﻡ= ﻼﺣﻈﺎت ﻫﺎﻣﺔ :
) (١ﺣﺪود ا
ﺍ
þ
ﻞ
1-
.
) (٤إذا ﻧﺖ ﺗ ﻠﻔﺔ ﺗﻐﻄﻴﺔ ا
] د )ﺱ( – ﺭ )ﺱ( [ ﺱ
4
3
ا ﺮ ﻊ ا ﻮاﺣﺪ ﻣﻦ أرﺿﻴﺔ ﺮات
اﻟﻔﻨﺪق ﺑﺎ ﺮاﻧﻴﺖ ٤٠٠ﺟﻨﻴﻪ وﺗﻢ ﺗﻐﻄﻴﺔ ٥ﺮات ﻣﺘﻄﺎﺑﻘﺔ
.
ﺑﺎ ﺮاﻧﻴﺖ ﺴﺎﺣﺔ وا ﺴﺘﻘﻴﻤ
ﺣﺎﻟﺔ إ ﺎد ﺴﺎﺣﺔ ﻣﻨﻄﻘﺔ ﺴﺘﻮ ﺔ ﺑ
ﺗﻘﺎﻃﻊ ﻣﻨﺤﻨﻴﺎت
٢
٢
4 ﺫ3 ﺫ3 – ( 64 – ١٦ + ١٢ ) | + ¤ + ¤ 3 ù | +ﺫ = | é 3¤ 1 - = 3 3 3 û 3 3ë ﺫ3 ﺫ3 ١٣ = 7وﺣﺪة ﺮ ﻌﺔ + =| 7 –|+ )=|(٩–٩+٩ 3 3 3 3
أو ﺏ
1-
3
ﺱ = ﺍ ،ﺱ = ﺏ ﺣﻴﺚ د )ﺱ( ﲨﺲ ﺭ )ﺱ( : þ
3
3 ﺫ3 وﺣﺪة ﺮ ﻌﺔ ¤ + ¤ 3 ùﺫ = ( 1 + ١ + ٣ –) – ( ٩ – ٩ + ٩ ) = é 3 ¤ 1 - 3 3 3 û 1- ë
د )ﺱ( ،ﺭ )ﺱ(
ﺏ
١٣ --- +++ --٠ ٠
ﺇ ﻡ = þد )ﺱ( ﺱ = ٢ + ٣ ) þﺱ – ﺱ ( ٢ﺱ =
)ﺏ( د )ﺱ( > ٠ﺉ ﻡ þ | = ٢د )ﺱ( ﺱ | ﺝ
ً ﺛﺎﻧﻴﺎ :ﺴﺎﺣﺔ ا ﻨﻄﻘﺔ ا ﺤﺪدة ﺑﺎ ﻨﺤﻨ
ا ﻞ
أى ﺱ ٢ – ٢ﺱ – ٠ = ٣ﺉ ﺱ = ٣أ ،ﺱ = – ١
)ﺍ( د )ﺱ( < ٠ﺉ ﻡ þ = ١د )ﺱ( ﺱ ﺍ
وا ﺴﺘﻘﻴﻤ
(ﺱ
ﻣﻨﻬﺎ ﺪودة ﺑﻤﻨﺤ ا اﻟﺔ د ،
ﺱ = ، ٠ﺹ = ٠ﺣﻴﺚ د )ﺱ( = 1 – ١٢ﺱ. ٢ 3
أوﺟﺪ ﺗ ﻠﻔﺔ ﺗﻐﻄﻴﺔ ا ﻤﺮات ا ﻤﺴﺔ .
اﻻﺣﺪاﺛﻴﺎت ا ﺴ ﻨﻴﺔ ﻘﻂ ا ﻘﺎﻃﻊ .
ا ﻞ
) (٢ﻻ ﺗﻮﺟﺪ ﺴﺎﺣﺔ ﺳﺎ ﺔ و ﺴﺘﺨﺪم ا ﻘﻴﺎس ﻠﻘﻴﻤﺔ ا ﺴﺎ ﺔ .
ﺹ = ٠ﻋﻨﺪﻣﺎ 1 – ١٢ﺱ ٠ = ٢ﺉ – ٣٦ﺱ٠ = ٢ 3
ﺇ ﺱ = ٦أ ،ﺱ = – ) ٦ﺮﻓﻮض (
6 6 ﺴﺎﺣﺔ ا ﻤﺮ = 1 – ١٢ ) þﺱ ( ٢ﺱ = ùﺫ ٤٨ = é 3¤ 1 - ¤1ﻣ
) (١أوﺟﺪ ﺴﺎﺣﺔ ا ﻨﻄﻘﺔ ا ﺤﺪدة ﺑﻤﻨﺤ ا اﻟﺔ
د )ﺱ( = ٣ﺱ ١ + ٢و ﻮر ا ﺴ ﻨﺎت وا ﺴﺘﻘﻴﻤ
0
د ﻣﺘﺼﻠﺔ
ﺫ
اﻟﻔ ة ]– ، [ ٢ ، ١د )ﺱ( < ٠ ﺫ
ﺱ = –، ١
ﺉ ﺗ ﻠﻔﺔ ا ﻤﺮات ا ﻤﺴﺔ = ٩٦٠٠٠ = ٥ × ١٩٢٠٠ﺟﻨﻴﻪ
) (٥أوﺟﺪ ﺴﺎﺣﺔ ا ﻨﻄﻘﺔ ا ﺤﺪدة ﺑﻤﻨﺤ ا ا
1-
ﺫ é ¤ + 3 ¤ ùû 1- ë
٢
ﺹ =١ﺱ – ، ٢ﺹ )– ٣=٢ﺱ ، (١+ﺹ =١ﺹ ٢ﺉ
ﺱ – ٣ = ٢ – ٢ﺱ ٢ – ٢ﺱ – ١ﺉ ٢ﺱ ٢ + ٢ﺱ – ٠ = ٤ﺉ ﺱ + ٢ﺱ – ٠ = ٢
) (٢أوﺟﺪ ﺴﺎﺣﺔ ا ﻨﻄﻘﺔ ا ﺤﺪدة ﺑﻤﻨﺤ ا اﻟﺔ
ﺇ ) ﺱ ) ( ٢ +ﺱ – ٠ = ( ١وﻣﻨﻬﺎ ﺱ = – ٢أ ،ﺱ = ١
وا ﺴﺘﻘﻴﻢ ﺱ = ٤وﺗﻘﻊ ﻓﻮق ﻮر
ﻡ=
ا ﺴ ﻨﺎت .
د )ﺱ( = ﺻﻔﺮ ﻋﻨﺪﻣﺎ ﺱ = ٠
ا ﻞ
٢
= ) ٨ = ٢ + ٦ = ( ١ – ١ –) – ( ٢ – ٨وﺣﺪات ﺮ ﻌﺔ
¤4 د )ﺱ( = ¤ﺫ 1+
:
د )ﺱ( = ﺱ ، ٢ – ٢ﺭ )ﺱ( = ) – ٣ﺱ ٢( ١ +
ﺱ ﻱ ]– [ ٢ ، ١
ﺇ ﻡ = þد )ﺱ( ﺱ = ٣ ) þﺱ ( ١ + ٢ﺱ = 1-
3
ﺇ ﺗ ﻠﻔﺔ ا ﻤﺮ ا ﻮاﺣﺪ = ١٩٢٠٠ = ٤٨ × ٤٠٠ﺟﻨﻴﻪ
ﺱ=٢ ا ﻞ
û
9
0ë
ﺮﻊ
=
ا ﻞ
1
] þﺭ )ﺱ( – د )ﺱ( [ ﺱ ﺣﻴﺚ ﺭ )ﺱ( ﲨﺲ د )ﺱ(
-ﺫ
1
٢
٢
)– ٣ ] þﺱ ) – ( ١ +ﺱ – [ ( ٢ﺱ
-ﺫ
1 = ٢ – ) þﺱ ٢ – ٢ﺱ ( ٤ +ﺱ = - ùﺫ ¤ - 3 ¤ﺫ é ¤4 + ë 3 û ﺫﺫ16 ﺫ = )– – ٩ = ( ٨ – ٤ – ) – ( ٤ + ١وﺣﺪات ﺮ ﻌﺔ 3 3 1
٠ ﺱ - - - + + +إﺷﺎرة د ٠
٢٢
ﻟﻸﺳﺘﺎﺫ ﺍﻟﻘﺪﻳﺮ /ﻋﻠﻰ ﺍﻟﺪﻳﻦ ﻳﺤﻴﻰ ٠١١١٩٦٦٠٦١٦
ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ -ﺍﻟﺼﻒ ﺍﻟﺜﺎﻟﺚ ﺍﻟﺜﺎﻧﻮﻯ ) (٦ﺗﻘﻮم
ا ﻠﺼﻖ
ﺔ إﻋﻼﻧﺎت ﺑﺎﻧﺘﺎج ﻠﺼﻖ ﻟ ﺴﻮ ﻖ ﺳﻠﻌﺔ ﻣﺎ ﻓﺈذا ن ﺷ
د ،ﺭ ﺣﻴﺚ
ﻣﻨﻄﻘﺔ ﺪدة ﺑﻤﻨﺤ ا ا
د )ﺱ( = ٢ﺱ ، ٢ﺭ )ﺱ( = ﺱ ٢ – ٤ﺱ ، ٢ﺱ ﻣﻘﺪرة
ﺑﺎ ﺴﻴﻤ .أﺣﺴﺐ ا ﺴﺎﺣﺔ ا ﻼزﻣﺔ ﻣﻦ ا ﻮرق ا ﻼﺻﻖ ﻹﻧﺘﺎج ١٠٠٠ﻠﺼﻖ ﺬه ا ﺴﻠﻌﺔ .
ا ﻞ
ﻡ= =
4
-ﺫ
= é 3¤ 4 + 5¤ 1 - ù 3 5û
=
8ﺫ 1د ﺴﻢ٢
1 س 2
-1
1
-2
-1
8ﺫ5600 = ١٠٠٠ × 1ﺫ د ﺴﻢ٢
-2
3
) (٧إذا ﻧﺖ ﺍ ) ، ( ٣ ، ٠ﺏ ) ، ( ٤ ، ١ﺝ ) ( ٠ ، ٣رؤوس ﻣﻢ ﺍﺏ ﺝ
ً أوﻻ :ﺣﺠﻢ ﺟﺴﻢ ﻧﺎﺷﺊ ﻣﻦ دوران ﻣﻨﻄﻘﺔ ﺴﺘﻮ ﺔ ﺣﻮل ﻮر :
· إذا ن دوران ا ﻨﻄﻘﺔ ا ﺴﺘﻮ ﺔ ﺣﻮل ﻮر ا ﺴ ﻨﺎت و ﺪودة ﺑﺎ ﻨﺤ
ﻞ ﺴﺎﺣﺔ ﺳﻄﺢ ا ﺜﻠﺚ ﺍﺏ ﺝ .
= òﺑﺐ ﺑﺎ ﻨﺤ
ص
،ﻣﻌﺎدﻟﺔ ﺏ ﺝ ﰐ : - ¤ﺫ -1ﺫ
،ﻣﻌﺎدﻟﺔ ﺍﺝ ﰐ :
ً
2
1
س 4
ﺹ 3 - 3 -0 3 - = = ﺫ ﺱ 0-ﺫ0-
2
1
ﺇ ﺹ = – 3ﺱ ٣+ ﺫ
-1
-1
) ﺱ ، ١ﺹ ) ، ( ١ﺱ ، ٢ﺹ( ٢
0
= ò
ﺫ
ﺫ 1 = ¤ 5 ùﺫ ¤ 5 - ù + éﺫ 5 = 5 + 5 = é ¤5 +وﺣﺪة ﺮ ﻌﺔ 4 4 1ëﺫ 4 û 0ë 4 û
ﺱ1
þ
ﺱﺫ
= ò
ﺹ1
þ
ﺹﺫ
أﺟﺐ ﻋﻦ اﻷﺳﺌﻠﺔ اﻵﺗﻴﺔ : · أوﺟﺪ ) (١ا ﻨﺤ
ﺎ ﻳﺄ
) ﺹ – ٢١ﺹ ( ٢٢ﺱ
.
:
) ﺱ – ٢١ﺱ ( ٢٢ﺹ
.
) (١أوﺟﺪ ﺣﺠﻢ ا ﺴﻢ ا ﺎﺷﺊ ﻣﻦ دوران ا ﻨﻄﻘﺔ ا ﺤﺪدة ﺑﻤﻨﺤ
ﺹ = – ٥ﺱ ٢و ﻮر ا ﺴ ﻨﺎت وا ﺴﺘﻘﻴﻤ :
ا اﻟﺔ د و ﻮر ا ﺴ ﻨﺎت وا ﺴﺘﻘﻴﻤ
ﺹ = 4ﺫ وا ﺴﺘﻘﻴﻤﺎت :
دورة ﻠﺔ ﺣﻮل ﻮر ا ﺴ ﻨﺎت ﻋﻠﻤﺎ ﺑﺄن د )ﺱ( = ﺱ . ً ﻣﺎ إﺳﻢ ا ﺠﺴﻢ ا ﺎﺷﺊ ؟ ﻘﻖ ﻫﻨﺪﺳﻴﺎ ﻣﻦ ﺻﺤﺔ إﺟﺎﺑﺘﻚ
ﺱ
ﺱ = ، ١ﺱ= ، ٤ﺹ =٠ ) (٣ا ﻨﺤ
ﻞ ﻣﻌﺎد ﻴﻬﻤﺎ ﺟ ﺎ و ﻜﻮﻧﺎ :
ﺱ = – ، ٢ﺱ =١ ) (٢ا ﻨﺤ
ً
و ذا ن ا وران ﺣﻮل ﻮر ا ﺼﺎدات :
ﺴﺎﺣﺔ ا ﻨﻄﻘﺔ ا ﺴﺘﻮ ﺔ ا ﺤﺼﻮرة ﺑ
:
) (٢إذا ن ا وران ﺣﻮل ﻮر ا ﺴ ﻨﺎت :
= 5 þﺱﺱ5– ) þ +ﺱ (٥+ﺱ 1
ﺝ
.
ﺛﺎﻧﻴﺎ :ﺣﺠﻢ ﺟﺴﻢ ﻧﺎﺷﺊ ﻣﻦ دوران ﻣﻨﻄﻘﺔ ﺪودة ﺑﻤﻨﺤﻨ
ﺇ ﺴﺎﺣﺔ ﺳﻄﺢ ﻣﻢ ﺍﺏ ﺝ = ﺴﺎﺣﺔ ﻣﻢ ﺍﺏ + ﺴﺎﺣﺔ ﺏ ﺝ [(٣ﺱ
.
:ﺹ = ﺝ ،ﺹ = ﻓﺈن :
Ù þﺱ . ٢ﺹ
) (١ﻧﻮﺟﺪ ﻧﻘﻄ ﺗﻘﺎﻃﻊ ا ﻨﺤﻨ
ﺫ 1 = ) ] þﺱ –) – ( ٣ +ﺫ 3ﺱ [ ٣ +ﺱ ٤ –) ] þ +ﺱ –) – ( ٨ +ﺫ 3ﺱ + 1 0 ﺫ
= òﺑﺐ
3
3
ﺹ . ٢ﺱ
ﺹ = د )ﺱ( و ﺎ ﺴﺘﻘﻴﻤ
4
ﺹ ٤ – = 0-4 = 0-ﺇ ﺹ = – ٤ﺱ ٨ +
1
ﺍ
þ
ﺏ
:ﺱ = ﺍ ،ﺱ = ﺏ ﻓﺈن :
· إذا ن دوران ا ﻨﻄﻘﺔ ا ﺴﺘﻮ ﺔ ﺣﻮل ﻮر ا ﺼﺎدات و ﺪودة
ﺹ ﺱ ١ = 30 --14 = 30 --ﺇ ﺹ – = ٣ﺱ ﺇ ﺹ = ﺱ ٣ +
ﺫ
ﺹ = د )ﺱ( و ﺎ ﺴﺘﻘﻴﻤ
ا ﻞ
ﻣﻌﺎدﻟﺔ ﺍﺏ ﰐ :
ا ﺮ ﻊ ا ﻮاﺣﺪ ﻣﻨﻪ ١٥٠٠
ﻢ ﺗ ﻮن ﺗ ﻠﻔﺔ ا ﺰﺟﺎج ؟
2
15
ﻓﺄوﺟﺪ ﺑﺎﺳﺘﺨﺪام ا
ﺷ
3
ﺇ ا ﺴﺎﺣﺔ ا ﻼزﻣﺔ ﻣﻦ ا ﻮرق ا ﻼﺻﻖ ﻹﻧﺘﺎج ١٠٠٠ﻠﺼﻖ 15
ﺟﻨﻴﻪ .
5
ﺫ
ﻗﻮس ﻣﻌﺎد ﻪ
ُ ﻏﻄﻰ ﻫﺬا ا ﺪﺧﻞ ﺑﺰﺟﺎج ﺗ ﻠﻔﺔ ا
6
– ) þﺱ ٤ + ٤ﺱ ( ٢ﺱ
0
ﺹ = – ) 1ﺱ – ) ( ١ﺱ – ( ٧ﺣﻴﺚ ﺱ ﺑﺎﻷﻣﺘﺎر ﻓﺈذا ﺫ
7
-ﺫ
- ëﺫ
ﺫ
) (٦ﺑﺎﺳﺘﺨﺪام ا ﺴﺎﺣﺔ ﺖ ا ﻨﺤ ﻣﺎ ﻗﻴﻤﺔ : –: ٤ ] þﺱ :٢ :ﺱ
8
ﺫ
ﺫ
ﺱ= ، ٠ﺱ=٣
9
٢ ) þﺱ – ٢ﺱ ٢ + ٤ﺱ ( ٢ﺱ
=
) (٥ا ﻨﺤﻨ
– ٩ :ﺱ ، ٢ﺹ = ﺱ ، ١ + ٢وا ﺴﺘﻘﻴﻤ :
) (٧ﺻﻤﻢ ﻣﻬﻨﺪس ﻣﺪﺧﻞ ﻓﻨﺪق
ص
ﺹ = ١ﺹ ٢ﺉ ﺱ = ٠أ ٢ – ،أ٢ ،
) (٤ا ﻨﺤﻨ
:ﺹ +ﺱ ، ٦ = ٢ﺹ ٢ +ﺱ – ٠ = ٣
ﺹ = ]ﺱ / ٤ /+/وا ﺴﺘﻘﻴﻤﺎت :
ﺱ= ، ٠ﺱ= ، ٥ﺹ=٠
ً
ﺱ = ، ٠ﺱ =٣
ا ﻞ
3 = òﺑﺐ þﺱ ٢ﺱ = ٩ = é 3¤ 1 ùوﺣﺪة ﻜﻌﺒﺔ 3û 0 0ë
3
ا ﺴﻢ ا ﺎﺷﺊ ﻣﻦ ا وران ﺴ
٢٣
ا ﺨﺮوط ا اﺋﺮى اﻟﻘﺎﺋﻢ
ﻟﻸﺳﺘﺎﺫ ﺍﻟﻘﺪﻳﺮ /ﻋﻠﻰ ﺍﻟﺪﻳﻦ ﻳﺤﻴﻰ ٠١١١٩٦٦٠٦١٦
ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ -ﺍﻟﺼﻒ ﺍﻟﺜﺎﻟﺚ ﺍﻟﺜﺎﻧﻮﻯ
ا ﺴ ﻨﺎت .
،ﺣﺠﻤﻪ = 1ﺑﺐ ﻗﻖ ٢ﻉ
3 1 = ﺑﺐ × ) ٣ﻇﺎ ٣ × ( ٤٥ 3 ٢
ﻗﻖ
= ٩وﺣﺪة ﻜﻌﺒﺔ
) (٢ﺑﺎﺳﺘﺨﺪام ا
ا ﻞ
ﺑﻔﺮض ﺹ] = ١ﺱ ، /ﺹ = ٢ﺱ ٢ﺑﻮﺿﻊ ﺹ = ١ﺹ ٢ﺇ ﺱ = ]ﺱ
ﻉ ٤٥ ٣
ﺇ ﺱ – ٤ﺱ = ٠ﺇ ﺱ ) ﺱ ٠ = ( ١ – ٣ﺉ ﺱ = ٠أ ،ﺱ = ، ١ﺹ ١ﲨﺲ ﺹ ) وذ ﻚ ﺑﺎ ﻌﻮ ﺾ ﻋﻦ ﺱ = 1 ﺫ
ﻞ أﺛﺒﺖ أن :
ﺇ = òﺑﺐ ) þﺹ – ٢١ﺹ ( ٢٢ﺱ = ﺑﺐ ] ) ] þﺱ – ٢( /ﺱ [ ٤ﺱ
ﻗ ) ٣ﻗﻖ ﻃﻮل ﻧﺼﻒ ﻗﻄﺮ ا ﻜﺮة ( )ﺍ( ﺣﺠﻢ ا ﻜﺮة = 4ﺑﺐ ﻖ
0
ﻗ ٢ﻉ )ﺏ( ﺣﺠﻢ اﻷﺳﻄﻮاﻧﺔ ا اﺋﺮ ﺔ اﻟﻘﺎﺋﻤﺔ = ﺑﺐ ﻖ
0
ﻮر ا ﺴ ﻨﺎت ﻓﺘﻜﻮن ﻣﻌﺎد ﻪ :
ﻗﻄﺮﻫﺎ ﻗﻖ ﻳﻨﻄﺒﻖ
– :ﺱ. :٢ : ﺱ + ٢ﺹ = ٢ﻗﻖ ٢أى ﺹ = ] ﻗﻖ: : ٢
ا ﺴ ﻨﺎت دورة ﺇ = òﺑﺐ
®-
þﺹ ٢ﺱ = ﺑﺐ
= ﺑﺐ ®ùﺫ é 3¤ 1 - ¤ û 3
®
®- ë
) (٦أوﺟﺪ ﺣﺠﻢ ا ﺴﻢ ا ﺎﺷﺊ ﻣﻦ دوران ا ﻨﻄﻘﺔ ا ﺤﺪدة
ﻗ ﻖ
0
ﻗ ٣وﺣﺪة ﻜﻌﺒﺔ = 43ﺑﺐ ﻖ
ﻉ
ﺱ
ﻉ
4S3 3 = 5
ﺹ
3 Sﺫ
ﻜﻌﺒﺔ .
أﺟﺐ ﻋﻦ اﻷﺳﺌﻠﺔ اﻵﺗﻴﺔ :
) (١أوﺟﺪ ﺣﺠﻢ ا ﺴﻢ ا ﺎﺷﺊ ﻣﻦ دوران ا ﻨﻄﻘﺔ ا ﺤﺪدة ﺑﺎ ﻨﺤ
ﻠﺔ ﺣﻮل ﻮر
ا ﺴ ﻨﺎت .
د )ﺱ( = ﺱ ١ + ٣وا ﺴﺘﻘﻴﻤﺎت :ﺹ = ، ٠ﺱ = ، ٠ﺱ = ١
دورة
ا ﻞ
ﺱ ٢ +ﺹ = ٠وا ﺴﺘﻘﻴﻤﺎت :ﺱ = ، ٠ﺹ = ، ٠ﺹ = ٣
ﺫ
ﺇ = òﺑﺐ ٢ ) þﺱ – ﺱ ٢( ٢ﺱ = ﺑﺐ ٤ ) þﺱ ٤ – ٢ﺱ + ٣ﺱ ( ٤ﺱ 0
0
دورة
ﺫ 16ﺑﺐ وﺣﺪة ﻜﻌﺒﺔ = ﺑﺐ = é 5¤ 1 + 4¤ - 3 ¤ 4 ù 15 0ë 5 3û
ﺹ = ﺱ ، ٢وا ﺴﺘﻘﻴﻢ ﺹ = ٢ﺱ دورة
ﺹ= ، ٠ﺹ=٦
ا ﺴ ﻨﺎت .
ﻧﺼﻒ دورة ﺣﻮل ﻮر ا ﺼﺎدات .
ﺹ = – ٤ﺱ ، ٢وا ﺴﺘﻘﻴﻢ ٢ﺱ +ﺹ = ٤دورة
ا وران ﺣﻮل ﻮر ا ﺼﺎدات ﺉ ﺱ = ٢ﺹ ﺇ = òﺑﺐ þﺱ ٢ﺹ 6
0
ﻮر ا ﺼﺎدات .
6
ûﺫ
ﻠﺔ ﺣﻮل ﻮر
) (٤أوﺟﺪ ﺣﺠﻢ ا ﺴﻢ ا ﺎﺷﺊ ﻣﻦ دوران ا ﻨﻄﻘﺔ ا ﺤﺪدة ﺑﺎ ﻨﺤ
ا ﻞ
6 = ﺑﺐ þﺹ ﺹ = ﺑﺐ § 1ùﺫ ١٨ = éﺑﺐ وﺣﺪة ﻜﻌﺒﺔ
ﻠﺔ ﺣﻮل ﻮر ا ﺼﺎدات .
) (٣أوﺟﺪ ﺣﺠﻢ ا ﺴﻢ ا ﺎﺷﺊ ﻣﻦ دوران ا ﻨﻄﻘﺔ ا ﺤﺪدة ﺑﺎ ﻨﺤ
) (٤أوﺟﺪ ﺣﺠﻢ ا ﺴﻢ ا ﺎﺷﺊ ﻣﻦ دوران ا ﻨﻄﻘﺔ ا ﺤﺪدة ﺑﺎ ﻨﺤ ﺹ = ﺱ ٢و ﻮر ا ﺼﺎدات وا ﺴﺘﻘﻴﻤ
ﻠﺔ ﺣﻮل ﻮر ا ﺴ ﻨﺎت.
) (٢أوﺟﺪ ﺣﺠﻢ ا ﺴﻢ ا ﺎﺷﺊ ﻣﻦ دوران ا ﻨﻄﻘﺔ ا ﺤﺪدة ﺑﺎ ﻨﺤ
ﺑﻮﺿﻊ ﺹ = ٠ﺉ ﺱ ) – ٢ﺱ ( = ٠ﺇ ﺱ = ٠أ٢ ، ﺫ
) ٢ﺹ – ﺹ ( ٤ﺹ
= ﺑﺐ ) S3 1 – 4S3ﺫ = ( 3ﺑﺐ × – ١ ) 4S3ﺫ ( 5 5
0
ﺑﺐ وﺣﺪة
3 Sﺫ
ﻣﻨﻬﻤﺎ (
ﻗ ﻉ وﺣﺪة ﻜﻌﺒﺔ ﺑﺐ ﻖ
ﺹ = ٢ﺱ – ﺱ ٢و ﻮر ا ﺴ ﻨﺎت ،دورة
0ë
:ﺹ = ]ﺱ ،ﺹ = ﺱ ٢دورة
ﻠﺔ ﺣﻮل
) (٥إذا ن ﺣﺠﻢ ا ﺴﻢ ا ورا ا ﺎﺷﺊ ﻋﻦ دوران ا ﻨﻄﻘﺔ
) (٥أوﺟﺪ ﺣﺠﻢ ا ﺴﻢ ا ﺎﺷﺊ ﻣﻦ دوران ا ﻨﻄﻘﺔ ا ﺤﺪدة ﺑﺎ ﻨﺤﻨ
0
0
٢
٤
) (٣أوﺟﺪ ﺣﺠﻢ ا ﺴﻢ ا ﺎﺷﺊ ﻣﻦ دوران ا ﻨﻄﻘﺔ ا ﺤﺪدة ﺑﺎ ﻨﺤ
0
3 Sﺫ
) ﺱ – ٢١ﺱ ( ٢٢ﺹ = ﺑﺐ þ
= ﺑﺐ ùﺹ ﺫ é 5§ 1 - ë 5 û
ﺇ = òﺑﺐ þﺹ ٢ﺱ = ﺑﺐ þﻗﻖ ٢ﺱ = ﺑﺐ
ﺹ = S3ﺫ ،ﺱ ١ﲨﺲ ﺱ ) ٢وذ ﻚ ﺑﺎ ﻌﻮ ﺾ ﻋﻦ ﺱ = ١
) þﻗﻖ – ٢ﺱ ( ٢ﺱ
ﻉ
ﻠﺔ ﺣﻮل ﻮر
ﺱ = ١ﺱ ٢ﺉ ٢ﺹ – ﺹ ٠ = ٤ﺇ ﺹ ) – ٢ﺹ ٠ = ( ٣ﺉ ﺹ = ٠أ،
= òﺑﺐ þ
ﻗ ﻖ
٢ :ﺹ = ﺱ ، ٢ﺹ = ]ﺱ دورة
ا وران ﺣﻮل ﻮر ا ﺼﺎدات ﺇ ﺱ ٢ = ٢١ﺹ ،ﺱ = ٢٢ﺹ
®
ﺗ ﺘﺞ اﻷﺳﻄﻮاﻧﺔ ﻣﻦ دوران ا ﺴﺘﻄﻴﻞ ﺣﻮل ﻮر ا ﺴ ﻨﺎت 0 ﻉ ®ûùﺫ ﺱ = 0 ëé
ﺑﺎ ﻨﺤﻨ
ﺹ
)ﺏ( ﺑﻔﺮض ﺴﺘﻄﻴﻞ ﺑﻌﺪاه :ﻉ ،ﻗﻖ
5
0ë
ا ﻞ
ﻠﺔ ﺣﻮل ﻮر ا ﺴ ﻨﺎت ®-
ûﺫ
ا ﺼﺎدات .
ﺗ ﺘﺞ ا ﻜﺮة ﻣﻦ دوران ا ﻨﻄﻘﺔ ا ﺴﺘﻮ ﺔ ا ﺤﺼﻮرة ﺑ ﻧﺼﻒ ا اﺋﺮة و ﻮر ®
0
1
) ﻗﻖ ﻃﻮل ﻧﺼﻒ ﻗﻄﺮ ﻗﺎﻋﺪة اﻷﺳﻄﻮاﻧﺔ ،ﻉ ارﺗﻔﺎﻋﻬﺎ (
ﺱ
1
1 = ﺑﺐ ) þﺱ – ﺱ ( ٤ﺱ = ﺑﺐ ¤ 1 ùﺫ ٠٣ = é 5¤ 1 -وﺣﺪة ﻜﻌﺒﺔ
3
)ﺍ( ﺑﻔﺮض ﻧﺼﻒ داﺋﺮة ﺮ ﺰﻫﺎ ﻧﻘﻄﺔ اﻷﺻﻞ ،وﻃﻮل ﻧﺼﻒ
٢
ﻣﻨﻬﻤﺎ (
1
ا ﻞ
٢
ا ﺤﺪدة ﺑﺎ ﻨﺤ
ﻠﺔ ﺣﻮل ﻮر
دورة
٢٤
ﺹ = ﺱ ، ٣وا ﺴﺘﻘﻴﻤ
ﺱ = ، ٠ﺹ =١
ﻠﺔ ﺣﻮل ﻮر ا ﺴ ﻨﺎت ﻳﻌﺎدل ﺣﺠﻢ ﺳﻠﻚ اﺳﻄﻮا
ﻟﻸﺳﺘﺎﺫ ﺍﻟﻘﺪﻳﺮ /ﻋﻠﻰ ﺍﻟﺪﻳﻦ ﻳﺤﻴﻰ ٠١١١٩٦٦٠٦١٦
ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ -ﺍﻟﺼﻒ ﺍﻟﺜﺎﻟﺚ ﺍﻟﺜﺎﻧﻮﻯ اﺸ
– ٤أوﺟﺪ ﺣﺠﻢ ا ﺴﻢ ا ﺎﺷﺊ ﻣﻦ دوران ا ﻨﻄﻘﺔ ا ﺤﺪدة ﺑﺎ ﻨﺤ
ﻃﻮ ٤٢وﺣﺪة ﻓﻤﺎ ﻃﻮل ﻧﺼﻒ ﻗﻄﺮ ا ﺴﻠﻚ ؟
ﺹ = ﺱ ٢وا ﺴﺘﻘﻴﻢ ﺹ = ٢ﺱ دورة
) (٦أوﺟﺪ ﺣﺠﻢ ا ﺴﻢ ا ﺎﺷﺊ ﻣﻦ دوران ا ﻨﻄﻘﺔ ا ﺤﺪدة ﺑﺎ ﻨﺤ ﺫ ﺫ ¤ﺫ +ﺹﺫ = ١و ﻮر ا ﺴ ﻨﺎت ،ﺣﻴﺚ ﺍ ،ﺏ ﺛﺎﺑﺘﺎن ،
ﺏ
ﺍ
دورة
– ٥إذا ن د )ﺱ( = ) ﺟﺘﺎ ﺱ ( ﺟﺘﺎ ﺱ ﻓﺈن د ) /ﺻﻔﺮ ( = .......... ﺍ~ – ٣
ﻠﺔ ﺣﻮل ﻮر ا ﺴ ﻨﺎت .
· ﺗﻌﻠﻴﻤﺎت ﻫﺎﻣﺔ :
) (١أﻗﺮأ ا ﺴﺆال ﺑﻌﻨﺎﻳﺔ ،وﻓﻜﺮ ﻓﻴﻪ ﺟﻴﺪا ﻗﺒﻞ ا ﺪء
إﺟﺎﺑﺘﻪ .
) ﺑﻤﻌ أﻧﻪ ﻻ ﺗﻮﺟﺪ أﺳﺌﻠﺔ اﺧﺘﻴﺎر ﺔ (
ا ﻄﻠﻘﺔ
Øأﺳﺌﻠﺔ اﻻﺧﺘﻴﺎر ﻣﻦ ﻣﺘﻌﺪد :
اﻻﺟﺎﺑﺔ ا ﺼﺤﻴﺤﺔ ﺗﻈﻠﻴﻼ
ﺳﺆال وﻻ ﺗﻈﻠﻞ أ
Øاﻷﺳﺌﻠﺔ ا ﻘﺎ ﺔ : أ ﺘﺐ اﺟﺎﺑﺘﻚ
ﻼ
ﻣﻦ داﺋﺮة واﺣﺪة ﺣ ﻻﺗﻔﻘﺪ درﺟﺔ ا ﺴﺆال
ا ن ا ﺨﺼﺺ ً
ﻣﻌﺪل ﺗﻐ
ﺳﺆال .
) (٨ا رﺟﺔ ا ﻴﺔ ﻼﺧﺘﺒﺎر ) ( ٣٠درﺟﺔ .
ﺍ~
9
10¤ -
–٢اﺸ
ﺏ~
ﺍ~ ﻩ - ١٤
9
10¤
~
ﺍ~ – ٤
10
9¤
ﺱ
ﺹ=ﻩ
þ
1-
þ - ١٥
ﺝ~ ١
4
١ - ~
ﺝ~ ﻩ٢
) (
٢ ~ﻩ٢
ﺫ
ﺍ~ ﺻﻔﺮ þ - ١٦ 0
ﺱ
ﻙ
ﻮ ﻩ ) ﺱ ٢ – ٣ﺱ ( ١ +
¤ f +4¤
ﺱ = . ............. ﺝ~ ١
-4 üﺱ ﺫ ﺱ = . ............. ﺏ~ ٢
ﺝ~ ﺑﺐ
٤ ~ ~
p ﺫ
| ﺟﺎ ﺱ | ﺱ = . ............ ﺏ~ ١٠ﺑﺐ
ﺝ~ ٢٠
٢٠ ~ﺑﺐ
) þ – ١٧ﻮ ﻩ ﺱ ( ﺱ = . .............. 1
٢ ~
ﺝ~ 1 ﻩ
~ﺫ ﻩ
ﻩ
ﺍ~ 1 ﻩ
ﻓﺈن د) /ﺱ( = ..........
ﺝ~ ﺻﻔﺮ
ﺏ~ ﻩ٢ ﺱ3
p 10
ﺍ~ ١٠
و ١-
ﺱ 3+
ﺏ~ ﺻﻔﺮ
0
ﻠﺔ ﺣﻮل ﻮر
ﺏ~ – ٢
1
ﺍ~ – ١
ﺹ
ﺴﺎوى ) ﻩ – ١٠ﻩ– ( ٢وﺣﺪة ﻜﻌﺒﺔ .أوﺟﺪ ﻗﻴﻤﺔ ﻙ . – ٣إذا ن د )ﺱ( = ﻩ
ﺏ~ ٢ﻩ
ﺱ¬ ¦
= . ..............
إذا ن ﺣﺠﻢ ا ﺴﻢ ا ﺎﺷﺊ ﻣﻦ دوران ا ﺴ ﻨﺎت وا ﺴﺘﻘﻴﻢ ﺱ = – ، ١ﺱ = ﻙ
ﻞ أوﺟﺪ þﻮ ﻩﺱ ﺱ
) l – ١٣ﺱ ( 5 +ﺱ = . ...........
ا ﺠﺎور :
ا ﻨﻄﻘﺔ ا ﻈﻠﻠﺔ دورة
0
ﺩ) - ( ¤ﺩ p ......... = 4 p – ١٢إذا ن د )ﺱ( = ﻩﻇﺎ ﺱ ﻓﺈن l p ﺱ- ﺱ¬ 4 4
10
ﺏ~ ٢ -
ﺍ~ ﻩ
) (٩ﺴﻤﺢ ﺑﺎﺳﺘﺨﺪام اﻵﻟﺔ ا ﺎﺳﺒﺔ .
9¤ -
ئﻩ 3
) ﻩ ٢س +ﻩس ( ﺱ
ﺱ = ﻩ ﺴﺎوى . ..............
ﺍ~ ٢
) (٧زﻣﻦ اﻻﺧﺘﺒﺎر ﺳﺎﻋﺘﺎن .
ﺝ~
ﺝ~ ٣
ﻞ أوﺟﺪ þ
٤ ~
– ١١إذا ن د )ﺱ( = ﻮ ﻩﺟﺎ ﺱ – ﻮ ﻩﺟﺘﺎﺱ ﻓﺈن د....... = ( p ) /
) (٦ﺗﺄ ﺪ ﻣﻦ ﺗﺮﻗﻴﻢ اﻷﺳﺌﻠﺔ ،وﻣﻦ ﻋﺪد ا ﺼﻔﺤﺎت ﻗﺒﻞ اﻻﺟﺎﺑﺔ .
– ١إذا ن ﺹ = ﻮ ﻩﺱ ﻓﺈن
ﺏ~ ٢
– ١٠ﺑﺎﺳﺘﺨﺪام أﺣﺪ ﻃﺮق ا
) (٥ﻋﺪد ﺻﻔﺤﺎت ا ﻜﺘﻴﺐ ) ( ١١ﺻﻔﺤﺔ ﻼف اﻟﻐﻼف .
أﺟﺐ ﻋﻦ اﻷﺳﺌﻠﺔ اﻵﺗﻴﺔ :
ﻴﻄﻪ ﻋﻨﺪ ﻫﺬه ا ﻠﺤﻈﺔ ﺴﺎوى .........ﺳﻢ .
– ٩إذا ن د )ﺱ( = ﺱ – ﺱ ﻮ ﻩﺱ ﻓﺈن ﻣﻴﻞ ا ﻤﺎس ﻠﻤﻨﺤ ﻋﻨﺪ
) (٤ﻋﺪد أﺳﺌﻠﺔ ا ﻜﺘﻴﺐ ) ( ٢٠ﺳﺆاﻻ .
§10Ù 10¤Ù
3
– ٨ﺑﺎﺳﺘﺨﺪام أﺣﺪ ﻃﺮق ا ً
وا ﺼﻐﺮى
– ٧ﻣﺜﻠﺚ ﻣ ﺴﺎوى اﻷﺿﻼع ﺿﻠﻌﻪ ﻳ اﻳﺪ ﺑﻤﻌﺪل 1ﺳﻢ /ث ﻓﺈن
ﻫﺬا اﻻﺧﺘﺒﺎر ﻧﻮ ن ﻣﻦ اﻷﺳﺌﻠﺔ :
ﻇﻠﻞ ا اﺋﺮة ا اﻟﺔ
اﻟﺔ .
ﺍ~ ١
ً
ﺝ~ – ١
ا ﺚ ﻓ ات اﻟ اﻳﺪ وا ﻨﺎﻗﺺ ﺛﻢ أوﺟﺪ اﻟﻘﻴﻢ اﻟﻌﻈ
) (٢أﺟﺐ ﻋﻦ ﻴﻊ اﻷﺳﺌﻠﺔ وﻻ ﺗ ك أى ﺳﺆال ﺑﺪون إﺟﺎﺑﺔ . ) (٣ﻳﻮﺟﺪ
ﺏ~ – ٢
~ﺻﻔﺮ
– ٦إذا ن د ، 1 ] :ﻩ [ ﲤﺲ ح ،و ن د )ﺱ( = ﺱ – ﻮ ﻩ ﺱ ﻩ
ً
ﻠﺔ ﺣﻮل ﻮر ا ﺴ ﻨﺎت .
ﺏ~ ﻩ
ﺝ~ ١
– ١٨أوﺟﺪ ﻣﻌﺎدﻟﺔ ا ﻤﺎس واﻟﻌﻤﻮدى ﻠﻤﻨﺤ : + ٢ﻮ ﻩﺹ .ﻮ ﻩﺱ = ﺱ + ٢ﺹ ﻋﻨﺪ ﺱ = . ١
٢٥
١ – ~
ﻟﻸﺳﺘﺎﺫ ﺍﻟﻘﺪﻳﺮ /ﻋﻠﻰ ﺍﻟﺪﻳﻦ ﻳﺤﻴﻰ ٠١١١٩٦٦٠٦١٦
ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ -ﺍﻟﺼﻒ ﺍﻟﺜﺎﻟﺚ ﺍﻟﺜﺎﻧﻮﻯ ا ﺠﺎور :
– ١٩ا ﺸ
ﺹ ٢
أوﺟﺪ ] þد )ﺱ( [ ٢د) /ﺱ( ﺱ 1
0
د )ﺱ (
١
ﺱ
اﺸ
– ٢٠
د )ﺱ( = ﺱ٣
أوﺟﺪ اﻛ
و
ا ﺠﺎور :
ﺍﺏ ﺝ .
٢٠ﺳﻢ ،و ن اﺻﻐﺮﻫﻤﺎ ﻳ اﻳﺪ ﺑﻤﻌﺪل ١ﺳﻢ /دﻗﻴﻘﺔ وأ ﻳ ﻨﺎﻗﺺ ﺑﻤﻌﺪل 1ﺳﻢ /دﻗﻴﻘﺔ ﻓﺄوﺟﺪ ﻣﻌﺪل ا ﻐ ﺫ
ا ﺜﻠﺚ ﺑﻌﺪ ٥دﻗﺎﺋﻖ . ّ – ١١ﻋ ﻓ ات ا ﺤﺪب ﻷ
- ١ﻣﺜﻠﺚ ﻣ ﺴﺎوى اﻷﺿﻼع ﺿﻠﻌﻪ ﻳ اﻳﺪ ﺑﻤﻌﺪل 1ﺳﻢ /ث ﻓﺈن
ﺏ~ ]٣
ﺍ~ ٣
1
þ
1-
ﺍ~ 10 3
٢] ~
ﺝ~ ٣] ٣
45ﺱ = . ............
ﺱ
– ٤إذا ﻧﺖ
ﺏ~ ١
ﺝ~ ﻏ ذ ﻚ
ﺏ~ ٢
ﺝ~ ٣
ﺍ~ ١
~ﺻﻔﺮ
ﺏ~ ٢
٤ ~
ﺝ~ ٣
3
ﺍ~ ٠
١٢ ~
ﺝ~ ٨
ُ – ٦و ء ﻓﺎرغ ﺣﺠﻤﻪ ٧٠ﺳﻢ ٣ﻳﺼﺐ ﻓﻴﻪ ا ﺎء ﺑﻤﻌﺪل ١٠ﺳﻢ / ٣ث ﻓﺈن زﻣﻦ إﻣﺘﻼء ا ﻮ ء ﺴﺎوى .........ﺛﺎﻧﻴﺔ . ﺏ~ ٦
ﺍ~ ٥ p ّ – ٧ﺑ أن + ٣ ) þﺟﺎ ٣ﺱ ( ﺱ ﻳﻘﻊ ﺑ
ﺝ~ ٧
0
ﺍ~ ﺹ
ﺝ~ د) /ﺱ( +ﻣﻢ ﺱ
٨ ~
ﺍ~ ﺟﺘﺎ ﺱ
ﻈﺔ ﻣﺎ ن ﻃﻮﻻ ﺿﻠ
ﺝ~ – ﺟﺘﺎ ٥ﺱ
– ~ﺟﺘﺎ ٥ﺱ +ث
ﺍ~ [ ] ٤ ، ٠
5
ا ﻘﺎﺑﻞ
ص
٣ ٢ 3
~د )ﺱ( ٠ﻣﻢ ﺱ +ﻣﻢ ﺱ /
اﻟﻘﺎﺋﻤﺔ
1
٢ ٣ ٤ ٥
س
5
4
2
3
١
١ -1
1
١٢-1
ﺏ~ [ – ] ٠ ، ٢ﺝ~ [ – ﳘﺲ ، ٢ [ ~ ] ٢ ،ﳘﺲ ]
د )ﺱ( . /
– ١٦ﺴﺎﺣﺔ ا ﻨﻄﻘﺔ ا ﺤﺪدة ﺑﺎ ﻨﺤ
– ~ﺟﺘﺎ ﺱ
ﻣﺜﻠﺚ ﻗﺎﺋﻢ ا ﺰاو ﺔ ، ١٥
ﺹ = ﺱ ٣وا ﺴﺘﻘﻴﻤ
ﺹ = ، ٠ﺱ = ٢ﺴﺎوى ..............وﺣﺪة ﺮ ﻌﺔ ﺏ~ ٤
ﺍ~ ١
– ١٧إذا ﻧﺖ ﺹ = ﻩ ﻙ ﺱ
ﺝ~ ٢
٨ ~
ﻘﻖ ا ﻌﺎدﻟﺔ :
ﺹ ٢ +ﺹ – ٨ﺹ = ٠ﻓﺄوﺟﺪ ﻗﻴﻤﺔ ﻙ . /
//
– ١٨أوﺟﺪ ﺣﺠﻢ ا ﺴﻢ ا ﺎﺗﺞ
ﺱ = ٣ﺹ٢ - ٢
ﻋﻦ دوران ا ﻨﻄﻘﺔ ا ﻈﻠﻠﺔ ا ﺮﻓﻖ دورة
ﻮر ا ﺴ ﻨﺎت .
ﺏ~ ﺹ /ﺱ
ﺝ~ – ﺟﺎ ﺱ
2
– ١٥إذا ﻧﺖ د) /ﺱ( = ﻮ ﻩ ] ] ٦ﺱ ٤ ) /١ /– /ﺱ [ ٣( ٥ +ﻓﺄوﺟﺪ
اﻟﻘﻴﻤﻴ ٢ﺑﺐ ٤ ،ﺑﺐ ا ﺸ
– ٩إذا ن ﺹ = ﺟﺎ ﺱ ﻓﺈن ﺹ). ............ = (٢٩ ﺏ~ ﺟﺎ ﺱ
ﺍ~ – 1ﺟﺘﺎ ٥ﺱ +ث
ﻣﺘﻨﺎﻗﺼﺔ ﻋﻨﺪﻣﺎ ﺱ ﻱ . ............
– ٨ﺗﻔﺎﺿ ا اﻟﺔ ﺹ = د )ﺱ( ﺴﺎوى . .............. /
ﺝ~ ١٧٠
ﺏ~ 1ﺟﺘﺎ ٥ﺱ +ث
– ١٤ﺑﺎﻻﺳﺘﻌﺎﻧﺔ ﺑﺎ ﺸ
= . ...............
٤ ~
| þ1- – ٥ﺱ – | ١ﺱ = . ............ ﺏ~ ٤
ﺏ~ ٢٠
١٣٠ ~
-2
اﻟﺔ د :د )ﺱ( = ﻙ ﺱ ٩ + ٣ﺱ ٢ﻧﻘﻄﺔ اﻧﻘﻼب ﻋﻨﺪ
ﺱ = – ١ﻓﺈن ﻙ = . .............
ﺫ
ا ى ﻳﻤﺜﻞ د) /ﺱ( ﻳ ﻮن د )ﺱ(
Ùﺫ ﺹ – ٣إذا ن ﺱ = ﻗﺎ ﻥ ] ،ﺹ = ﻇﺎ ﻥ ﻓﺈن Ù :ﺱﺫ
ﺍ~ ١
þد )ﺱ( ﺱ = – ٢٠ﻓﺈن þد )ﺱ( ﺱ = . ...............
5
ﺴﺎﺣﺘﻪ ﻋﻨﺪ ا ﻠﺤﻈﺔ اﻟ ﻳ ﻮن ﻓﻴﻬﺎ ﻴﻂ ا ﺜﻠﺚ
= ١٢ﺳﻢ ﺴﺎوى . .............
1-
٥ ) þ – ١٣ﺟﺎ ٢ﺱ ﺟﺘﺎ ٣ﺱ ٥ +ﺟﺎ ٣ﺱ ﺟﺘﺎ ٢ﺱ ( ﺱ = . ......
ﺫ
ﻣﻌﺪل ا ﻐ
4
þد )ﺱ( ﺱ = ، ١٥٠
4
ﺍ~ ١٥٠
أﺟﺐ ﻋﻦ اﻷﺳﺌﻠﺔ اﻵﺗﻴﺔ :
– ١٠
، ò
1-
-٢
اﻟﺔ د ﻋﻨﺪﻣﺎ ﺱ ﻱ ] [ ٦ ، ١
– ١٢إذا ﻧﺖ د داﻟﺔ ﻣﺘﺼﻠﺔ ﺫ
ا ﻄﻠﻘﺔ وا ﺼﻐﺮى
ﺱ
ا ﻄﻠﻘﺔ
ﺴﺎﺣﺔ ﻫﺬا
وا ﺤﺪب ﻷﺳﻔﻞ ﻨﺤ ا اﻟﺔ د
ﺫ ﺣﻴﺚ د )ﺱ( = ﺱ 9 +ﺛﻢ أوﺟﺪ اﻟﻘﻴﻢ اﻟﻌﻈ
ﺴﺎﺣﺔ ﻠﻤﺴﺘﻄﻴﻞ
ﻫﻤﺎ
٢
ﺱ=ﺹ
ﻠﺔ ﺣﻮل
– ١٩أوﺟﺪ ﺴﺎﺣﺔ ا ﻨﻄﻘﺔ ا ﺤﺪودة ﺑﺎ ﻨﺤ
،وا ﺴﺘﻘﻴﻤ
ﻕ )ﺱ( = ﺱ ٤ – ٣ﺱ
:ﺱ = – ، ١ﺱ = ، ٣و ﻮر ا ﺴ ﻨﺎت .
– ٢٠إذا ن ﻣﻨﺤ ا اﻟﺔ د )ﺱ( = ﺍ ﺱ + ٣ﺏ ﺱ + ٢ﺝ ﻷﺳﻔﻞ
اﻟﻔ ة [ – ﳘﺲ ، ] ١ ،و ﺪب ﻷ
ﺪب
ﻋﻨﺪﻣﺎ ﺱ < ، ١و ﻤﺲ
ا ﺴﺘﻘﻴﻢ ﺹ ٩ +ﺱ = ٢٨ﻋﻨﺪ ا ﻘﻄﺔ ) ( ١ ، ٣ﻓﺄوﺟﺪ د )ﺱ( .
٢٦
ﻟﻸﺳﺘﺎﺫ ﺍﻟﻘﺪﻳﺮ /ﻋﻠﻰ ﺍﻟﺪﻳﻦ ﻳﺤﻴﻰ ٠١١١٩٦٦٠٦١٦
ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ -ﺍﻟﺼﻒ ﺍﻟﺜﺎﻟﺚ ﺍﻟﺜﺎﻧﻮﻯ ﺴﺎوى . ........
ﺍ~ ﺑﺐ
أﺟﺐ ﻋﻦ اﻷﺳﺌﻠﺔ اﻵﺗﻴﺔ :
– ١إذا ن ﺭ )ﺱ( = – ٣د )ﺱ( ،و ن ﺭ ٧ = (٥) /ﻓﺈن ﻣﻴﻞ
اﻟﻌﻤﻮدى
ا ﻤﺎس ا ﺮﺳﻮم ﻋﻨﺪ ﺱ = ٥
ﺍ~ ٧
ﺏ~ – ٧
اﻟﺔ د )ﺱ( = . ..........
ﺝ~ 1 7
– ٢دراﺳﺔ ﺳﻠﻮك د )ﺱ( ُوﺟﺪ أن د) /ﺱ( ﺗ اﻳﺪ
وﺗ ﻨﺎﻗﺺ
1 – ~ 7
]– ﳘﺲ ،ﺍ ] ،
[ ﺍ ،ﺏ ] ﻓﺈﻧﻪ ﻳ ﻮن ﻋﻨﺪ ﺍ ﻮﻗﻊ ﻗﻴﻤﺔ . ............
ﺍ~ ﺻﻐﺮى ﻠﻴﺔ
ﺝ~ ﻋﻈ
p3 ﺏ~ ﺫ
ﻠﻴﺔ
üïﺫ¤ﺫ 0£ ¤ @ 1 + – ٣ﻣﻨﺤ ا اﻟﺔ د )ﺱ( = ý - 3 ïþﺱﺫ @ 0> ¤
ﺍ~ [ – ﳘﺲ ] ٠ ،ﺏ~ [ ، ٠ﳘﺲ ]
– ١٣إذا ن د )ﺱ( ،ﻕ )ﺱ( دا ﺎن ﻣﺘﺼﻠﺘﺎن
]،[٣،١و ن:
3
– ١٤ﻧﻘﻄﺔ ﺗﺘﺤﺮك
ﻋﻨﺪ
ò ~
ﺏ~ ﺟﺘﺎ ٢ﺱ +ث
ﺝ~ – ﺟﺘﺎ ٢ﺱ +ث
ﺍ~ ٢
ﺝ~ ٤
3
٤ þ – ٤ﺟﺎ ﺱ ﺟﺘﺎ ﺱ .ﺱ = . .............. ﺍ~ ٢ﺟﺎ ٢ﺱ +ث
ﺏ~ ٣
٦ ~
،ﻓﺄوﺟﺪ ﻗﻴﻤﺔ ٥ þ1د )ﺱ( ﺱ .
ﺱ ﻱ . ............
ﺝ~ +ò
ﺱ = ، ٢ﺹ = ٠ﺑﺎ ﻮﺣﺪات ا ﺮ ﻌﺔ ﺴﺎوى . ...........
3
ﻣﻄﻠﻘﺔ
ﺪب ﻷ
ا اﻟﺔ :د )ﺱ( = ٣ﺱ – ٢ﺱ٣
٤ ] þ1ﻕ )ﺱ( ٧ +د )ﺱ( [ ﺱ = ٣ þ1 ، ١٩ﻕ )ﺱ( ﺱ = ٩
ﺏ~ ﺻﻐﺮى ﻣﻄﻠﻘﺔ ~ﻋﻈ
ﺝ~ ٢ﺑﺐ
– ١٢ﺴﺎﺣﺔ ا ﻨﻄﻘﺔ ا ﺤﺪدة ﺑﻤﻨﺤ
،وا ﺴﺘﻘﻴﻤ
p5 ~ ﺫ
٢ – ~ﺟﺘﺎ ٢ﺱ +ث
ا ﻘﻄﺔ
ا ﻨﺤ
ﺱ ﺹ = ﺱ +ﺹ – ٥أوﺟﺪ ﻮﻗﻊ
ا ﻠﺤﻈﺔ اﻟ ﻳ ﻮن ﻓﻴﻬﺎ ﻣﻌﺪل ﺗﻐ إﺣﺪاﺛﻴﻬﺎ ا ﺴ
ﺑﺎﻟ ﺴﺒﺔ ﻠﺰﻣﻦ ﺴﺎوى ﻣﻌﺪل ﺗﻐ اﺣﺪاﺛﻴﻬﺎ ا ﺼﺎدى ﺑﺎﻟ ﺴﺒﺔ ﻠﺰﻣﻦ
– ١٥إﻧﺎء ﻠﻮء ﺴﺎﺋﻞ ﻳ
ب ﻣﻦ ﺛﻘﺐ ﺻﻐ ﻓﺈذا ن ﺣﺠﻢ ا ﺴﺎﺋﻞ
ا ﻮ ء ﻳﺘﻐ ﺑﻤﻌﺪل ٠٤ﻥ – ٤٠ﺳﻢ / ٣ث و ن ﺣﺠﻢ ا ﺴﺎﺋﻞ
ﺑﻌﺪ ٣٠ﺛﺎﻧﻴﺔ ﻣﻦ ﺑﺪء اﻟ ب ٩٨٠ﺳﻢ . ٣أوﺟﺪ ﺳﻌﺔ اﻹﻧﺎء ، ً ّ و ﺑﻌﺪ ﻢ ﺛﺎﻧﻴﺔ ﻳﺼﺒﺢ اﻹﻧﺎء ﻓﺎر .
– ٥ﻛﺮة ﻣﻦ ا ﻠﺞ ﻳ ﻨﺎﻗﺺ ﻃﻮل ﻧﺼﻒ ﻗﻄﺮﻫﺎ ﻣﻦ ١٥ﺳﻢ إ ١١ﺳﻢ ّ – ١٦ﻋ ﻓ ات اﻟ اﻳﺪ وا ﻨﺎﻗﺺ ﻠﻤﻨﺤ ﺹ = ﺱ )ﺱ – ٢(٣ﺛﻢ ً ً ﻣﺪة ٤٥دﻗﻴﻘﺔ ﻓﺘﻨﺎﻗﺺ ﺗﺒﻌﺎ ﻚ ﺣﺠﻤﻬﺎ ﺑﻤﻌﺪل 40ﺳﻢ / ٣د 9 ارﺳﻢ ا ﺸ اﻟﻌﺎم ﻠﻤﻨﺤ ﻮﺿﺤﺎ ﻋﻠﻴﻪ ﻮاﻗﻊ اﻟﻘﻴﻢ اﻟﻌﻈ ﻓﺈن ﺴﺎﺣﺔ ﺳﻄﺢ ا ﻜﺮة = . .................. وا ﺼﻐﺮى ا ﺤﻠﻴﺔ وﻧﻘﻂ اﻻﻧﻘﻼب إن وﺟﺪت . ﺏ~ ٥٠
ﺍ~ ٤٠
٤٥ ~
ﺝ~ ٢٥
) þ – ٦ﺟﺘﺎ ﺱ +ﻇﺎ ( ١ + ٤٥ ٢ﺱ = + .............ث ﺍ~ ﺟﺎ ﺱ ٢ +
ﺏ~ – ﺟﺎ ﺱ +ﻗﺎ ٢ﺱ +ﺱ
ﺝ~ ﺟﺎ ﺱ ٢ +ﻗﺎ ﺱ +ﺱ
~ﺟﺎ ﺱ ٢ +ﺱ
– ٧إذا ن :ﺱ ﻱ ، +òﺱ < 1 +ﻙ ﻓﺈن ﻗﻴﻤﺔ ﻙ > .............. ﺏ~ – ٢
ﺍ~ ٢
ﺱ
3
١ – ~
ﺝ~ ١ 3
– ٨إذا ن þد )ﺱ( = ٢ﻓﺈن ٣ ] þد )ﺱ( [ ١ +ﺱ = . ..... ﺍ~ ٣
1
1
ﺏ~ ٥
ﺝ~ ٧
Ùﺹ – ٩إذا ن ﺹ = ﻮ ﻩ ) ﻮ ﻩﺱ ( ﻓﺈن Ùﺱ 1 ¤ 1ﺏ~ × 1ﻮ ﻩ ﺱ ﺝ~ ﺍ~ ئﻩ ¤ ﺱ ¤ئﻩ
٨ ~
= . .............. 1 ~ ﺱ
– ١٧ﺧﺰان ﻓﺎرغ ﺳﻌﺘﻪ ٦ﻣ
ﻣ
ﺍ~ 1د )ﺱ( 7
7
دﻗﻴﻘﺔ ﺣﻴﺚ ﻥ ا ﺰﻣﻦ ا ﻼزم ﻻﻣﺘﻼء ا ﺰان .
– ١٨إذا ﻧﺖ ﻡ
ا ﻨﻄﻘﺔ ا ﺤﺪدة ﺑﺎ ﻨﺤ
ﺱ ﺹ = + ٤ﺱ٢
وا ﺴﺘﻘﻴﻤﺎت ﺱ = ، ١ﺱ = ، ٤ﺹ = ٠ﻓﺄوﺟﺪ ﺴﺎﺣﺔ ا ﻨﻄﻘﺔ ﻡ
ﺑﺎ ﻮﺣﺪات ا ﺮ ﻌﺔ ﻷﻗﺮب وﺣﺪة و ﺬ ﻚ أوﺟﺪ ﺣﺠﻢ ا ﺴﻢ ا ﺎﺷﺊ ﻣﻦ دوران ا ﻨﻄﻘﺔ ﻡ دورة
ﻠﺔ ﺣﻮل ﻮر ا ﺴ ﻨﺎت .
– ١٩أوﺟﺪ ﻣﻌﺎدﻟﺔ اﻟﻌﻤﻮدى
ا ﻨﺤ
ﺹ = ٢ﺱ ٢ +ﻋﻨﺪ ﻧﻘﻂ
ﺗﻘﺎﻃﻌﻪ ﻣﻊ ا ﺴﺘﻘﻴﻢ ﺹ = ﺱ – ٢٠أوﺟﺪ ﻣﻌﺎدﻟﺔ ا ﻤﺎس ﻠﻤﻨﺤ
ﺹ = ﻮ ﻩ ) ٢] – ٢ﺟﺘﺎ ﺱ (
ﻋﻨﺪ ا ﻘﻄﺔ اﻟ ﺗﻘﻊ ﻋﻠﻴﻪ و ﺣﺪاﺛﻴﻬﺎ ا ﺴ
– ١٠إذا ن د )ﺱ( = ﻩ– ٧ﺱ ﻓﺈن þد )ﺱ( ﺱ = + ...........ث ﺏ~ 1 -د )ﺱ( ﺝ~ د)ﺱ(
ﻜﻌﺐ
ﻜﻌﺐ ﻳﺼﺐ ا ﺎء ﺑﻤﻌﺪل ) ﻥ ( ٢ +
– ~د )ﺱ(
– ١١ﺣﺠﻢ ا ﺴﻢ ا ﺎﺷﺊ ﻣﻦ دوران ا ﻨﻄﻘﺔ ا ﺤﺪدة ﺑﺎ ﻨﺤ ﺹ = ]ﺱ ، /١ /+وا ﺴﺘﻘﻴﻤﺎت :ﺹ = ، ٠ﺱ = – ، ١ﺱ = ١
٢٧
ﺴﺎوى . p 4
ﻟﻸﺳﺘﺎﺫ ﺍﻟﻘﺪﻳﺮ /ﻋﻠﻰ ﺍﻟﺪﻳﻦ ﻳﺤﻴﻰ ٠١١١٩٦٦٠٦١٦
ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ -ﺍﻟﺼﻒ ﺍﻟﺜﺎﻟﺚ ﺍﻟﺜﺎﻧﻮﻯ
) (٩ﺱ = ﻗﺎ qﺉ ﺱ = /ﻗﺎ qﻇﺎ ] ، qﺹ = ﻇﺎ qﺇ ﺹ = ﻗﺎq ٢
§y
= = ٢ﻗﺎ q ﺇ ﺹ ٢ = /ﻇﺎ qﻗﺎ q ٢ﺇ = §Ù ¤Ù y¤ ﺇ Ùﺫ§ = ٢ﻗﺎ qﻇﺎ ٢ = q Ù × qﻗﺎ qﻇﺎ 1 × q ¤Ùﺫ ¤Ù ﺱy 1 =٢ = ٢ﻗﺎ qﻇﺎ × q q gq i
ﺣﻠﻮل ﺗﻤﺎر ﻦ )(١
)(١
ﺱ + ٢ﺹ = ٢ﺱ – ﺹ (١) ......ﺉ ٢ﺱ ٢ +ﺹ §Ù – ١ = §Ù ¤Ù ¤Ù -1 §Ùﺫ ¤ = ﺇ ٢ﺹ ٢ – ١ = §Ù + §Ùﺱ ﺉ ¤Ù ¤Ù + 1 ¤Ùﺫ§
)) (١٠ﺍ(
٣ = §Ùﻗﺘﺎ ) ٣ﺱ – × ( ١ +ﻗﺘﺎ ) ٣ﺱ ( ١ +ﻇﺘﺎ ) ٣ﺱ ٦ × ( ١ +ﺱ ¤Ù
ﻣﻦ ) : (١ﻋﻨﺪﻣﺎ ﺱ = ١ﺇ ﺹ + ٢ﺹ = ٠ﺉ ﺹ = ٠أ١ – ، و
٢
ﻮن ١ – = §Ùأ١ ، ¤Ù
¤Ù Ùﺫ ٢ §Ù § ( ٢ +ﺹ ﺫ = ٦ – ٢ﺱ ﺑﺎﻟﻘﺴﻤﺔ ﺇ )٢ ¤Ù ¤Ù Ùﺫ § ٣ + ٢( §Ù ) +ﺱ = ١ ﺇ ﺹ ¤Ùﺫ ¤Ù ¤Ù
¤Ù
ﻦ – = §Ùﺹ ﺱ ¤Ù
¤Ù ¤Ù Ù ٣ﺫ § ٢ ﺫ = ٢ﺱ ﺹ = ١٤ = ٧ × ٢ و ﺎ ب×ﺱ ﺉ ﺱ ¤Ù Ùﻉ Ùﻉ ﺫ3 - ¤ = × ٢ ) = ¤Ùﺱ – = 1 × ( ٣ )(٤ §Ù ¤Ù §Ù ﺱ ﺫ 1 +ﺱ ﺫ 1+ ﺫ ﺇ Ùﺫ ﻉ = ﺫ) - (1+ ¤ﺫ ¤Ù × (3 - ¤ ` ) ¤ ﺫ §Ùﺫ §Ù ) ¤ﺫ (1+
ﺑﺎﻻﺷﺘﻘﺎق ﺮة أﺧﺮى :
ﺇ – ٩ﺟﺎ ٣ﺱ ٤ +ﺟﺎ ٢ﺹ ٢ - §Ù × §Ùﺟﺘﺎ ٢ﺹ
Ùﺫ§ ¤Ùﺫ
¤Ù =٠ ﺇ – ٩ﺟﺎ ٢ﺹ ٤ +ﺟﺎ ٢ﺹ ) ٢ – ٢( §Ùﺟﺘﺎ ٢ﺹ ¤Ù Ùﺫ § ﺑﺎﻟﻘﺴﻤﺔ ﺟﺎ ٢ﺹ ﺇ ٢ – ٢( §Ù ) ٤ﻇﺘﺎ ٢ﺹ ﺫ = ٩ ¤Ù ¤Ù ) (٧ﰈ ﺹ = ﻗﺎ ﺱ ﺇ = §Ùﻗﺎ ﺱ ﻇﺎ ﺱ ¤Ù Ùﺫ § = ﻗﺎ ﺱ ﻗﺎ ٢ﺱ +ﻗﺎ ﺱ ﻇﺎ ﺱ ﻇﺎ ﺱ ﺇ ¤Ùﺫ
Ùﺫ§ ¤Ùﺫ
¤Ù
٢
¤Ù
ﺇ -1 = §Ùﺱ ،ﺑﻮﺿﻊ ﺹ = ٠ ¤Ù
§ +ﺫ
¤Ù
ﻣﻌﺎدﻟﺔ ا ﻨﺤ ﻳ ﺘﺞ ﻧﻘﻂ ا ﻘﺎﻃﻊ ﻣﻨﻬﻤﺎ _ 3 ﺫ
ً ٣ = §Ùﺱ ) ١ + ٢ﻮﺟﺐ داﺋﻤﺎ ( ﺇ ا ﻤﺎس ﻳﺼﻨﻊ زاو ﺔ ﺣﺎدة ﻣﻊ ﻮر ¤Ù
ا ﺴ ﻨﺎت ،ﻣﻦ ﻣﻌﺎدﻟﺔ ا ﻨﺤ ﻋﻨﺪ ﺱ = ١ﺗ ﻮن ﺹ = ٤ = §Ù ، ٤ ﺉ ﻣﻌﺎدﻟﺔ ا ﻤﺎس )(٣
¤Ù
٤ :ﺱ – ﺹ٠=٨+
= §Ùﺟﺘﺎ ٢ﺱ – ٢ﺱ ﺟﺎ ٢ﺱ +ﺟﺘﺎ ﺱ ﺇ ا ﻴﻞ = ﺟﺘﺎ + ٠ – ٠ﺟﺘﺎ ٠ ¤Ù
:ﺹ =٢ﺱ
ﺇ ا ﻴﻞ = ٢وﻣﻌﺎدﻟﺔ ا ﻤﺎس
) (٤ﺹ = ﻇﺎ ٢ﺱ ٢ = §Ù ،ﻗﺎ ٢ ٢ﺱ ،ﻋﻨﺪ ﺱ = pﻳ ﻮن ﺹ = ]٣ ¤Ù
6
٢٤ :ﺱ – ٣ﺹ – ٤ط ٠ = ٣] ٣ +
وا ﻴﻞ = ، ٨ﻣﻌﺎدﻟﺔ ا ﻤﺎس
ﺫ§ 3 + ¤ ) ٢ (٥ﺱ ٣ +ﺹ ٣ +ﺱ ٢ + §Ùﺹ ٠ = §Ùﺉ – = §Ù ¤Ù ¤Ù ¤Ù + ¤ 3ﺫ§ ٥
) (٦ﻋﻨﺪ ﺱ = ٠ﺉ ﺹ = ٢ = §Ù ، ١ﺟﺘﺎ ﺱ – ﺟﺎ ﺱ ﺉ ا ﻴﻞ = ٢
=٠
ﺉ ﻣﻌﺎدﻟﺔ ا ﻤﺎس
¤Ù
:ﺹ –٢=١ﺱ
) (٧ﻳﻮﺿﻊ ﺹ = ﺱ ﺉ ا ﻘﻄﺔ
)(٣،٣
٣ ) §Ù ،ﺹ ٢ – ٢ﺹ ( = ٣ﺱ ٦ – ٢ﺉ ا ﻴﻞ = ١ﺉ ¤Ù
ﻣﻴﻞ اﻟﻌﻤﻮدى = – ١ﺉ ﻣﻌﺎدﻟﺔ اﻟﻌﻤﻮدى
= ﻗﺎ ٢ﺱ ) ﻗﺎ ﺱ +ﻇﺎ ﺱ ( +ﻗﺎ ﺱ ﻇﺎ ﺱ = ﻗﺎ ﺱ )ﻗﺎ ﺱ ٢ +ﻇﺎ ﺱ(
٢
٢
= ﻗﺎ ٢ﺱ ] ﻗﺎ ٢ﺱ ) ٢ +ﻗﺎ ٢ﺱ – = [ ( ١ﻗﺎ ٢ﺱ ) ٣ﻗﺎ ٢ﺱ – ( ٢
) (٨ﺱ ٦ = /ﺟﺘﺎ – × q ٢ﺟﺎ ٦ – = qﺟﺎ qﺟﺘﺎ ، q ٢ﺹ ١٥ = /ﺟﺎ q ٢ﺟﺘﺎ q y e15ﺫ q f q ﺫ = 5 -ﻇﺎ ، qﻋﻨﺪ ٤٥ = p = q = § = §Ù
4
§Ùﺫ § - ¤ = ) (٨ﻧﻮﺟﺪ - ¤ ¤Ùﺫ§
ﻓﻨﻀﻊ ا ﻘﺎم = ٠
ﺼﻞ
٥
:ﺱ +ﺹ –٠= ٦
= ﻏ ﻣﻌﺮف ) ﻷن ا ﻤﺎس [ ﻮر ا ﺼﺎدات (
ﺱ = ٢ﺹ ﻧﻌﻮض
ﻣﻌﺎدﻟﺔ ا ﻨﺤ
ﺼﻞ
ﺹ = – ١أ ١ ،ﻧﻌﻮض ﺑ ﻞ ﻣﻨﻬﺎ ﻌﺮﻓﺔ ﻗﻴﻢ ﺱ ا ﻘﺎﺑﻠﺔ ﺎ ﺱ = – ٢أ٢ ، ﺉ ا ﻘﻂ ) ، ( ١ – ، ٢ –) ، ( ١ ، ٢ﻣﻴﻞ اﻟﻌﻤﻮدى ﻋﻨﺪ ﺇ ﻣﻌﺎدﻟﺔ اﻟﻌﻤﻮدى ﻋﻨﺪ ) ( ١ ، ٢
= ﺹ ٣ ) ٢ﺹ( ٢ – ٢ ﺫ ¤Ù q fq e 6y¤ ﺉ ﻣﻴﻞ ا ﻤﺎس = × 5 -ﻇﺎ 5 - = ٥٤٥ ﺫ ﺫ
) (١ﺑﺎﻻﺷﺘﻘﺎق :ﺇ ٢ﺱ ٢ +ﺹ ٠ = ٠ – §Ù ٤ + ٢ – §Ù
،ﻋﻨﺪ ا ﻘﻄﺔ ) ( ١ ، ١ﺉ ا ﻴﻞ = – = ١ﻇﺎ ﻩ ﺉ ﻕ ) ﻩ ؟ ( = ١٣٥
Ùﺫ ﺇ ﺹ § = ٢( §Ù ) +ﻗﺎ ﺱ ﻗﺎ ﺱ ) ﻗﺎ ٢ﺱ +ﻇﺎ ٢ﺱ ( +ﻗﺎ ٢ﺱ ﻇﺎ ٢ﺱ ﺫ
٢
= – ٣ﻗﺎ ) ﻇﺘﺎ ٣ﺱ ( ﻗﺘﺎ ٣ﺱ ﻇﺘﺎ ٣ﺱ
ﺣﻠﻮل ﺗﻤﺎر ﻦ )(٢
)(٢
= ﻗﺎ ٣ﺱ +ﻗﺎ ﺱ ﻇﺎ ٢ﺱ = ﻗﺎ ﺱ ) ﻗﺎ ٢ﺱ +ﻇﺎ ٢ﺱ ( ٢
٢
) ( ٠ ، ٢ – ) ، ( ٠ ، ٤وﻣﻴﻞ ا ﻤﺎس ﻋﻨﺪ
ﻳ ﺘﺞ أن Ùﺫ ﻉ = 3 ﺑﺎ ﻌﻮ ﺾ ﻋﻦ ﺱ = 1 = ¤Ù ، ١ §Ùﺫ 4 §Ùﺱ ﺫ 1 + ) ٢ (٥ﺱ ﺹ ٥ = ٣ +ﺱ ٢ﺑﺎﻻﺷﺘﻘﺎق ﺇ ٢ﺹ ٢ +ﺱ ١٠ = ٠ + §Ùﺱ ¤Ù Ùﺫ § ﺑﺎﻟﻘﺴﻤﺔ ٢واﻻﺷﺘﻘﺎق ﺮة أﺧﺮى ﺇ + §Ù + §Ùﺱ ﺫ = ٥ ¤Ù ¤Ù ¤Ù Ùﺫ § ٥ = ( §Ù ) ٢ + ﺇ ﺱ ¤Ùﺫ ¤Ù ) (٦ﺟﺎ ٣ﺱ – ﺟﺎ ٢ﺹ = ٠ﺑﺎﻻﺷﺘﻘﺎق ﺇ ٣ﺟﺘﺎ ﺱ – ٢ﺟﺘﺎ ٢ﺹ ٠ = §Ù ¤Ù
٢
٢
)ﺏ( = §Ùﻗﺎ ) ٢ﻇﺘﺎ ٣ﺱ ( × )– ﻗﺘﺎ ٣ﺱ ﻇﺘﺎ ٣ﺱ ( × ٣ ٢
Ùﺫ § ٠ = §Ù ٢ +وﻟ ) (٣ﺱ + §Ùﺹ = ٠ﺉ ﺱ ﺫ
٢
٢
= – ١٨ﺱ ﻗﺘﺎ ) ٣ﺱ ( ١ +ﻇﺘﺎ ) ٣ﺱ ( ١ +
ﺇ ٢ﺹ ٢ = §Ùﺱ – ٣ﺱ ٢ﺑﺎﻻﺷﺘﻘﺎق ﺮة أﺧﺮى
¤Ù
٢
٣
) (٢ﺹ = ٢ﺱ – ١ ) ٢ﺱ ( = ﺱ – ٢ﺱ ٣ﻳﺎﻻﺷﺘﻘﺎق
¤Ù
٢
٢
،ﻣﻌﺎدﻟﺔ اﻟﻌﻤﻮدى ﻋﻨﺪ )– ( ١ – ، ٢
ﺹ=١
ﻗﻴﻢ
ا ﻘﻄﺘ = ﺻﻔﺮ
ﺹ = –١
٢ ٢ ٣ ) (٩ﺹ = – ٢ﺱ ٩ +ﺱ – ١٢ﺱ ٥ +ﺇ ٦ – = §Ùﺱ ١٨ +ﺱ – ١٢
ﺑﻮﺿﻊ ٠ = §Ùﺉ ﺱ = ٢أ ١ ،و ﺎ ﻌﻮ ﺾ ¤Ù
ﺹ = ١أ ،ﺻﻔﺮ ﺉ ا ﻘﻂ
٢٨
¤Ù
)(٠،١)،(١،٢
ﻣﻌﺎدﻟﺔ ا ﻨﺤ
ﺉ
ﻟﻸﺳﺘﺎﺫ ﺍﻟﻘﺪﻳﺮ /ﻋﻠﻰ ﺍﻟﺪﻳﻦ ﻳﺤﻴﻰ ٠١١١٩٦٦٠٦١٦
ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ -ﺍﻟﺼﻒ ﺍﻟﺜﺎﻟﺚ ﺍﻟﺜﺎﻧﻮﻯ وﻣﻌﺎدﻟ اﻟﻌﻤﻮدى ﻋﻨﺪ ) (١٠ﺹ = ﺱ 3 + ﺱ
ﻣﻨﻬﻤﺎ ﺇ
ﺹ = – ٢ﺱ ﻠﻬﺎ ﻣﻊ ا ﻌﺎدﻟﺔ اﻷﺻﻠﻴﺔ ﻳ ﺘﺞ ﺱ = ٣أ١ – ،
:ﺱ= ، ٢ﺱ=١
3 – ١ = §Ù ¤Ù
وﺗ ﻮن ا ﻘﻂ
ﺱﺫ
ﻣﻴﻞ ا ﻤﺎس = ﻣﻴﻞ ا ﺴﺘﻘﻴﻢ = – ٢ﻋﻮض ا ﺸﺘﻘﺔ اﻷو ﺱ = _ ١ﺛﻢ ﻧﻌﻮض
ﺼﻞ
٢
ا ﺸﺘﻘﺔ -1 = §Ùﺹ ﺉ ﺱ = ٤ﺹ – ٣ﺛﻢ ﻠﻬﺎ ﻣﻊ ﻣﻌﺎدﻟﺔ ا ﻨﺤ 4
ﻣﻌﺎدﻟ ا ﻤﺎﺳ ﻫﻤﺎ ٨ :ﺹ ٢ +ﺱ – ٤ ، ٠ = ١٥ﺹ +ﺱ – ٠ = ١
) ( §Ù ) (١٢ﻠﻤﻨﺤ اﻷول = ٢ﺱ – ( §Ù ) ، ١ﻠﻤﻨﺤ ا ﺎ = ٢ – ٣ﺱ ¤Ù ¤Ù ﻡ = ١ﻡ ، ١ = ٢ﻣﻌﺎدﻟﺔ ا ﻤﺎس ا ﺸ ك :ﺱ – ﺹ ٠ = ١ +
) ( §Ù ) (١٣ﻠﻤﻨﺤ اﻷول = ٦ﺱ – ( §Ù ) ، ٥ﻠﻤﻨﺤ ا ﺎ = ٢ﺱ – ٣ ¤Ù ¤Ù ﻡ ، ١ = ١ﻡ ١ – = ٢ﺉ ﻡ × ١ﻡ ١ – = ٢وﻫﻮ ا ﻄﻠﻮب ﺍ¤ + ﺍ¤ - ( §Ù ) ،ﻠﻤﻨﺤ ا ﺎ = – ) ( §Ù ) (١٤ﻠﻤﻨﺤ اﻷول = ﺹ ¤Ù ﺹ ¤Ù
ﺴﺐ ﻧﻘﻄﺔ ا ﻤﺎس ﻞ ﻣﻌﺎدﻟ ا ﻨﺤﻨ
٢
:
:ﺑﻄﺮح ا ﻌﺎد
= – ١ﺉ
ﺍﺫ - 8ﺍﺫ
٢
وﻧﻌﻮض اﻟﻌﻼﻗﺔ ﻡ × ١ﻡ
)(١
٢
ﺱ¬ ¦
¤Ù
)(٢
ﺹ ¬0
1
٦
٦
ﺱ ¬0
= l 4ﻩ4ﺱ 1 = ١ × 3 - ١ × 4 = ¤ e l 3 - 1 - 5 5 5 ﺱ 5 5 4ﺱ ﺱ ¬0
)(
3ﺱ -ﺫ¤
)l = (٣
ﺱ
= ٢ lﺱ × ) lﺫ( 3 ﺱ ¬0
= ٤٤ﺑﺐ ﺳﻢ / ٣ث
)l = (٥
ﺱ ¬0
= ) ١١ + ١٢ﻥ ٢ +ﻥ ٣ - ١٢ ) ( ٢ﻥ ( ﺇ ٩٦ + ١٤٤ = òﻥ ٩ -ﻥ ٦ - ٢ﻥ
= ١٢ - = ٧٢ - ٣٦ - ٩٦ﺳﻢ / ٣د = ا ﻌﺪل ا ﺼﺎدى واﺧﺘ
ﻓﻴ ﺘﺞ
ﺱ
1-
= ×١ﻮ = 3ﻮ 3 ﻩ ﺫ ﻩ ﺫ
oﻩ )5 +1ﺱ( 3 3¤ 5
5 + 1 )oﺱ ( ﺱ
=5=١× 5 ﺫ ﺫ
× lﺱ ﺱ ¬ 3 0ﺱ 1 -
= 5 + 1 )o l × ٥ﺱ ( × lﺱ 5ﺱ ﺱ ¬ 3 0ﺱ 1 - ﺱ ¬0
٣
= ١٨ - ٩٦ﻥ ١٨ -ﻥ
ﺱ 3 ùﺱ é1- ê ﺫ úûﺫ ١× ë × lﺱ =l ﺱ ﺱ ¬ g 0ﺱ ﺱ ¬0
ﺱ
ﺱ ¬0
أى ﻈﺔ = + ٤ﻥ ،
٢
ﺱ ¬0
)l × 5 = (٤ ﺫ
ﺇ ٢ + ٣ ) = òﻥ ( ) + ٤ﻥ ( ) ٣ - ١٢ﻥ (
) (٤أﺷﺘﻖ ﺛﻢ ﻋﻮض ﻋﻦ ا ﻌﺪل ا ﺴ
1+ 6 ﺹ ﺫ
ﺱ ¬0
ﺱ ¬0
أى ﻈﺔ = ٢ + ٣ﻥ ،ا ﻌﺪ ا ﺎ
BÙ ،ﻋﻨﺪ ﻥ = ٢ﺉ Ùﻥ
+ 3 - 6ﺫ ﺹ ﺫ
=+١) lﺹ(
ﺫ 1- ﺫ ﺹ
¤ e3 ﻩ4ﺱ 1 - l=l 5ﺱ 5ﺱ
ﺱ ¬0
Ùﻥ
BÙ ﺇ Ùﻥ
=
ﲤ ﺻﻔﺮ ﳘ ﻓﺈن ﺹ ﺲ ﲤ ﺲ ،ﻋﻨﺪﻣﺎ ﺱ ﺲ
ﺣﻠﻮل ﺗﻤﺎر ﻦ )(٣
ا ﻌﺪ ا ﺎﻟﺚ أى ﻈﺔ = ٣ - ١٢ﻥ
ﺱ¬ ¦
ﺹ ¬0
ﺫ
) (٣ا ﻌﺪ اﻷول
=ﻩ
٦ ٣ﺱ ٢+
٦ 1
ﺇ ﺴﺎﺣﺔ ا ﺜﻠﺚ = ١ = ١ × ٢ × 1وﺣﺪة ﺮ ﻌﺔ .
)(٢
(
=
3ﺱ´ 4 ﺫ¤ ﻩ
=+١)] lﺹ( ﺹ [ ×)+١ﺹ( ﺫ = ﻩ × =١ﻩ
،ﻣﻴﻞ اﻟﻌﻤﻮدى = – ١ﺇ ﻣﻌﺎدﻟﺔ اﻟﻌﻤﻮدى ﺹ – = ٤ +ﺱ ١ + أى :ﺱ +ﺹ ٠ = ٣ +و ﻘﻄﻊ ﻮر ا ﺼﺎدات ) ( ٣ – ، ٠ ﻃﻮل اﻟﻘﺎﻋﺪة = | – ٢ = | (٥ –) – ٣وﺣﺪة ﻃﻮل ،اﻻرﺗﻔﺎع = ١وﺣﺪة ﻃﻮل
BÙ = òﺑﺐ ﻗﻖ ٢ﻉ ﺛﻢ أﺷﺘﻖ ﺛﻢ ﻋﻮض ﻳ ﺘﺞ Ùﻥ
ﺱ¬ ¦
٣ﺱ ٢+
ﺹ ¬0
ﺇ ﻣﻌﺎدﻟﺔ ا ﻤﺎس ﺹ = ٤ +ﺱ – ١أى :ﺱ – ﺹ – ٠ = ٥و ﻮﺿﻊ ﺱ = ٠ ﺉ ﺹ = – ٥ﺇ ا ﻤﺎس ﻳﻘﻄﻊ ﻮر ا ﺼﺎدات ) ( ٥ – ، ٠
)(١
)ﺍ( = 4 + ١ ) l ﺫ1+ ¤
ﺇ ا ﻘﺪار = + ١ ) lﺹ (
4-
١٠ = §Ù
´5ﺫ
1- 4 = 4ﺹ ﺇ ٢ﺱ 4 =١+ﺇ ﺱ = ﺹ ،ﻧﻔﺮض أن ﺫ ﺹ ﺫ1+ ¤
) (١٥ﰈ ٤ﺱ + ٢ﺹ ٢٠ = ٢ﺑﺎﻻﺷﺘﻘﺎق ﺇ ٨ﺱ ٢ +ﺹ ٠ = §Ù 4 - §Ùﺱ = ﺇ ﺹ ¤Ù
ﺹ
ﺣﻠﻮل ﺗﻤﺎر ﻦ )(٤
= ١-ﺉ ﺍ = _ ٢
ﺇ ﻣﻴﻞ ا ﻤﺎس = ١ = 1´4 -
ﺝ
ﺫ ٣ 4+ 1 + ¤ﺱ ٢ + = ( 4 +١) l ( )ﺏ( = ) l ﺫ1+ ¤ ﺫ1+ ¤
ﺇ ) ﺱ – ﺍ ( ) – ٢ﺱ +ﺍ ( ٠ = ٢ﺉ ﺱ – ٢ﺍ ﺱ +ﺍ – ﺱ – ٢ﺍ ﺱ – ﺍ = ٠ ﺉ ﺍﺱ = ٠ﺇ ﺍ = ) ٠ﺮﻓﻮض ﻷﻧﻪ ﻧﻔﺲ ا ﻨﺤ (
أو ﺱ = ٠ﺉ ﺹ – ٨ = ٢ﺍ ٢ﺛﻢ ﻧﻮﺟﺪ ﻡ ، ١ﻡ
٢
٢
٢
qÙ وﻣﻦ ) (١ﺇ = / ١ – = (١٠ –) × 1ث kÙ
٢٠٠م ﺱ
qÙ ٢ ) (٦ﺹ = ﻇﺎ qﺇ = §Ù × 1ﻗﺎ q kÙ 5 5 k Ù ﺍ q qÙ §Ù 1 (١) ........ × = ﺇ ٥ﻡ i5 kÙﺫ k Ù q ﺏ ،ﻋﻨﺪ ﺏ ٥ = ﻣﱰ ﺇ ﻗﺎ = qﻗﺎ ٢] = ٤٥
ا ﺴﺘﻘﻴﻢ = – ّ ، 1 ﻋﻮض ﻣﻌﺎدﻟﺔ
ﺹ
ف
ﺛﻢ ﻋﻮض ﻳ ﺘﺞ Ùف = 370 13 Ùﻥ
،ﻣﻌﺎدﻟﺔ ا ﻤﺎس ا ﺎ :ﺹ ٢ +ﺱ ٠ = ٦ +
ﺼﻞ
٢
أﺷﺘﻖ ﺛﻢ اﺣﺴﺐ ﻗﻴﻢ ﺱ ،ﺹ ،ف ﺑﻌﺪ ٢ث
،ﻣﻌﺎدﻟﺔ ا ﻤﺎس اﻷول :ﺹ ٢ +ﺱ – ٠ = ٦
ﺱ 1- ¤Ù 1 3 ا ﻘﻂ ) ( ، ١ ) ، ( ، ٦ﺉ ﺫ ﺫ
٢
) (٥ف = ﺱ + ٢٢٠ ) +ﺹ (
ﻣﻌﺎدﻟﺔ ا ﻨﺤ ﻧﻌﺮف ا ﻘﻂ ) ( ٤ – ، ١ –) ، ( ٤ ، ١
) (١١ﻣﻴﻞ ا ﻤﺎس = ﻣﻴﻞ اﻟﻌﻤﻮدى
( ٣ ، ١– ) ، ( ١– ، ٣ ) :
= ×٥ﻮﻩ× ٥= 1ﻮﻩ ﻮﻩ ٣ oﻩ 3 ﻩ3ﺱ 1- ﻩ 3ﺱ 1- ﺱ =l )l = (٦ × lﺫeﺱ ﺱ ﺱ ¬ 0ﺫ eﺫ ﺱ ﺱ ¬S 0 ﺱ ¬0 ﺫ ü ﺫ ﻩ3ﺱ 1 1 ٢] ٣ = ٢] × ١ × ٣ = 1 × =l ×٣ 3ﺱ Sﺫ ´ 0 ﺱ¬ ﺫ
٢٩
ﻟﻸﺳﺘﺎﺫ ﺍﻟﻘﺪﻳﺮ /ﻋﻠﻰ ﺍﻟﺪﻳﻦ ﻳﺤﻴﻰ ٠١١١٩٦٦٠٦١٦
ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ -ﺍﻟﺼﻒ ﺍﻟﺜﺎﻟﺚ ﺍﻟﺜﺎﻧﻮﻯ
Ù ٢ﺫ § ٤ +ﺱ ٢ + §Ùﺹ + ﺇ ﺱ ﺫ
ﺣﻠﻮل ﺗﻤﺎر ﻦ )(٥
¤Ù
Ù ٢ﺫ ¤ﺫ§ + yﺫ ¤ﺫ§ § ٤ +ﺱ ٢ + §Ùﺹ + ﺱﺫ ) (٢ﻋﻦ ﻗﻴﻤﺔ ﺍﺏ ﺇ ﺱ ¤Ùﺫ ¤Ù Ù ٢ﺫ § ٤ +ﺱ ٢ + §Ùﺹ +ﺱ ٢ + §Ùﺹ = ٠ = ٠ﺇ ﺱ ¤Ùﺫ ¤Ù ¤Ù Ù ٢ﺫ § ٥ +ﺱ ٤ + §Ùﺹ = ٠ ﺇ ﺱ ¤Ùﺫ ¤Ù
٢ )) (١ﺍ( ﺹ ٢ ) = /ﺱ – ( ١ﻩﺱ – ﺱ ٢ +ﻗﺎ ٢ﺱ ﻇﺎ ٢ﺱ
)ﺏ( ﺹ ٣ = /ﻗﺎ ﺱ ﻇﺎ ﺱ ﻩﻗﺎ ﺱ –
1 ﺫ¤S
ﻩ
]ﺱ
٢ )ﺝ( ﺹ ٦ ) = /ﺱ – ٣ ) ٧ × ( ٥ﺱ – ٥ﺱ × ( ٥ +ﻮ ٧ ﻩ
) (ﺹ = ﺱ ٣ +ﻩ + ٢ﻩﺱ ﺟﺎ ﺱ
ﺣﻠﻮل ﺗﻤﺎر ﻦ )(٦
ﺹ + ٠ + ١ = /ﻩﺱ ﺟﺎ ﺱ +ﻩﺱ ﺟﺘﺎ ﺱ = + ١ﻩﺱ )ﺟﺎ ﺱ +ﺟﺘﺎ ﺱ (
)) (٢ﺍ( ﺹ ] – ٠ = /ﻗﺎ ٢ﺱ ﻮ ) ٢ﺱ+ ( ٩ + ٣ ﻩ
ﺫ )ﺏ( ﺹ = /ﺱﺱ ﺫ × ﻮ ﻩ ×
ﻩ
= ﺣﻞ آﺧﺮ :
ﺱﺫ
ﻩ
ﺱﺫ
6ﺱ ﺫ
)(١
ﻇﺎ ﺱ [
ﺫ9 + 3 ¤
ﺫ ﺫ ¤ ´ ¤Ú ¤ﺫ -ﺫ Ú ´ ¤
¤ﺫ
× ﺫ¤ ) ¤Ú ¤ﺫ × (1-ﻮ ﻩ = ﺫ) ¤ﺫ(1- ﻮﻩ ﺱ4 ﺱ
٢ ﺹ = ﻮ ﻩﺱ – ﻮ ﺱ
)(٢
)(٣
ﺫ¤ ´ ¤ Ú /ﺫ¤ ﺇ ﺹ = ﻮﻩ– ﺫ ﻮﻩ =)٢ﺱ -ﺫ ( ﻮﻩ ¤ﺫ ﺫ
=
¤3
Ú
) (ﺑﺄﺧﺬ ﻮ ر ﺘﻢ اﻟﻄﺮﻓ ﺑﺎﻻﺷﺘﻘﺎق ﺑﺎﻟ ﺴﺒﺔ إ
ﺱ
ﺱ ﺇ
ﻸﺳﺎس ﻩ ﺇ ﻮ ﺹ = ﻇﺎ ﺱ ﻮ ﺟﺎ ﺱ ﻩ
¤f ﺹ = yﻗﺎ ٢ﺱ ﻮ ﺟﺎ ﺱ +ﻇﺎ ﺱ × ﺹ ﻩ ¤e
ﻩ
)ﻩ( ﺑﺄﺧﺬ ﻮ ر ﺘﻢ اﻟﻄﺮﻓ ﺑﺎﻻﺷﺘﻘﺎق ﺑﺎﻟ ﺴﺒﺔ إ ﺱ ﺇ
ﻩ
ﻸﺳﺎس ه ﺇ ﻮ ﺹ = ﺱ ﻮ ﻩ = ﺱ
ﺹ = yﻩ ﺱ
ﻩ –١
ﻩ
ﺇ ﺹ = /ﻩ ﺱﻩ ×١ -ﻩ
ﺹ 1 ﻮﻩ = ﻮﺱ +ﻮﻩ )) (٣ﺍ( = §Ùﻮ ﺱ +ﺱ × ﺱ ¤Ù ﺇ Ùﺫ§ = 1 = ٠ + 1 ﺱ ¤Ùﺫ ﺱ 4ﻥ )ﺏ( ﺱ ٣ = /ﻩ ٣ﻥ ،ﺹ ٤ = /ﻩ ٤ﻥ ﺇ = §Ùﺹ = yﻩ ﺱ3 y ¤Ù ﺇ Ùﺫ§ = 4ﻩﻥ × Ùﻥ = 4ﻩﻥ × 4 = 1 9ﻩ ﺫﻥ 3ﻩ 3ﻥ 3 ¤Ù ¤Ùﺫ 3
) (٤ﺑﺎﺷﺘﻘﺎق اﻟﻄﺮﻓ ﺑﺎﻟ ﺴﺒﺔ إ
ﺱ :
) (٥ﺑﺎﺷﺘﻘﺎق اﻟﻄﺮﻓ ﺑﺎﻟ ﺴﺒﺔ إ
ﺱ :
ﺱﻩ
)(٦
ﺇ ﺱ ﺇ ﺱ
٢
= – × 1- þ ٢ﻩ ﺫ ¤S
– ]ﺱ
9
= –٢ﻩ
¤ e¤e ﺱ = –þ þﻇﺎ ﺱ ﺱ = þ ¤f ¤f
– ]ﺱ
+ث
ﺱ
ﺫ3 + ¤ ¤6 + 9 ﺱ = ٢ﻮ |ﺱ ٣+ﺱ|+ث ﺱ =þ٢ þ ﻩ ¤ﺫ¤ 3 + ¤ﺫ¤ 3 + ٢
eﺫ + ¤ﺫ¤ f þ ، eﺫ + ¤ﺫ1 - ¤ e
ﺱ :ﻧﻔﺮض أن د )ﺱ( = ﺟﺎ ٢ﺱ ٢ +ﺟﺎ ﺱ – ١
ﺇ د ) /ﺱ( = ٢ﺟﺎ ﺱ ﺟﺘﺎ ﺱ ٢ +ﺟﺘﺎ ﺱ = ﺟﺎ ٢ﺱ ٢ +ﺟﺘﺎ ﺱ
ﺇ ا ﻘﺪار = ﻮ | ﺟﺎ ٢ﺱ ٢ +ﺟﺎ ﺱ – + | ١ث ﻩ
)(٧
ﺇ ﻩﺱ ﺹ ) ﺹ +ﺱ ٢ = ( §Ùﺱ §Ù + ¤Ù ¤Ù ﺇ ﺹ ﻩﺱ ﺹ +ﺱ §Ùﻩﺱ ﺹ = ٢ﺱ §Ù + ¤Ù ¤Ù ﺱﺹ ﺇ ) §Ùﺱ ﻩﺱ ﺹ – ٢ = ( ١ﺱ – ﺹ ﻩ ¤Ù
٢
ﺱ
) ٩ þﻩ × (١ +ﻩ ﺱ = × ٩
) ﻩ ﺱ (1+
1+8
ﺱ
٩
+ث = ) ﻩ + (١ +ث
ﺫÚ +ث +ﺱ (ﺱ=٢ﻩ ﻮﻩ|ﺱ|+ )) þ (٥ ﺫÚ ﺱ 4ﻩ ﻩﺱ ﻩﺱ +ﺱ ﻮ (٣ﺱ +ﺱ ﻮ (٣ﺱ=) þ )þ ، ﻩ ﻩ ¤ﻩ ¤S 1´ ¤ﻩ ¤ ﺫ = ) þﺫ +ﺱ ﻮ ( ٣ﺱ = ٢ﻮ | ﺱ | 1 +ﺱ ٢ﻮ + ٣ث ﺫ ﺱ ﻩ ﻩ ﻩ
ﻩ
ﻩ
ﺱ
ﺫ1 + Ú
3
ﺱﺫ
= ﻗﺎ ٢ﺱ ﻮ ﺟﺎ ﺱ ١ +ﺇ ﺹ ) = /ﺟﺎ ﺱ (ﻇﺎ ﺱ ) ﻗﺎ ٢ﺱ ﻮ ﺟﺎ ﺱ ( ١ + ﻩ
٨
ﺱ ﺫ1+ Ú
1 +ﻩ ٣ﺱ +ﺑﺐ ٤ﺱ +ث
¤ g+ ¤ i ) þ (٤ﻗﺎ ﺱ ﺱ :ﺑﺎ ب × ¤ g+ ¤ i iﺫ ﺱ ¤ g ¤ i + ﺱ = ﻮ ﻩ | ﻗﺎ ﺱ +ﻇﺎ ﺱ | +ث ﺇ ا ﻘﺪار = þ ¤g+¤i ¤h+¤j þ ،ﻗﺘﺎ ﺱ ﺱ :ﺑﺎ ب × ¤h+¤j jﺫﺱ ¤ h ¤ j - ﺱ = – ﻮ ﻩ | ﻗﺘﺎ ﺱ +ﻇﺘﺎ ﺱ | +ث ﺇ ا ﻘﺪار = – þ ¤h+¤j
ﺫ
ﻩ
3
3
9
= – ﻮ ﻩ | ﺟﺘﺎ ﺱ | +ث = ﻮ ﻩ | ﻗﺎ ﺱ | +ث
)(Ú ¤1 - ¤ 3 ) ( ¤
3
¤Ù þ ، ¤Sﻩ ¤S
٢
¤ Ú 1 ¤ ¤ 3 3 Ú ´ Ú- 3 ´ Ú ´ ¤ ¤ )ﺝ( ﺹ = / ﺫ ) ( ¤
þﺫ ﻩ ٣ﺱ – ٢ﺱ = ﺫ × ٣ þ 1ﻩ ٣ﺱ – ٢ﺱ = ﺫ ﻩ ٣ﺱ – + ٢ث
) þ ،ﺱ ٢ﻩ +ﻩ ٣ﺱ +ﺑﺐ ( ٤ﺱ =
ﺱ4
ﺫ
¤Ù
ﺍB
ﺱﺫ
= (٢) ........... ٠ﺑﺎ ﻌﻮ ﺾ ﻣﻦ )(١
3 þ ﺫ ¤ﻩ ¤
ﺱ = ﺫ1 ) þ 3ﺱ (
٣
1ﺱ = ﻮ ﻩ| ﻮ ﻩﺱ|+ث ﻩﺱ
þ ،ﻇﺘﺎ ﺱ ﺱ = þﻇﺘﺎ ﺱ ) ﻇﺘﺎ ٢ﺱ ( ﺱ = þﻇﺘﺎ ﺱ ) ﻗﺘﺎ ٢ﺱ – ( ١ﺱ = ) þﻇﺘﺎ ﺱ ﻗﺘﺎ ٢ﺱ – ﻇﺘﺎ ﺱ ( ﺱ
¤f = – þﻇﺘﺎ ﺱ )– ﻗﺘﺎ ٢ﺱ ( ﺱ – þ ¤e = – 1ﻇﺘﺎ ٢ﺱ – ﻮ ﻩ | ﺟﺎ ﺱ | +ث ﺫ
ﺍB ٢ + §Ùﺱ ﺹ = ﺱ ¤Ù Ùﺫ ﺍB§Ù §Ù § ¤Ùﺫ ٢ +ﺱ ٢ + ¤Ùﺱ ٢ + ¤Ùﺹ = ﺱ ﺫ
(١) .........ﺑﺎﻻﺷﺘﻘﺎق ﺮة أﺧﺮى :
ﺣﻞ آﺧﺮ ﻵﺧﺮ ﺧﻄﻮﺗ ٢
¤f ﺱ þ – = :ﻗﺘﺎ ﺱ )– ﻗﺘﺎ ﺱ ﻇﺘﺎ ﺱ ( ﺱ – þ ¤e
= – 1ﻗﺘﺎ ﺱ – ﻮ ﻩ | ﺟﺎ ﺱ | +ث ﺫ
٣٠
ﺱ
ﻟﻸﺳﺘﺎﺫ ﺍﻟﻘﺪﻳﺮ /ﻋﻠﻰ ﺍﻟﺪﻳﻦ ﻳﺤﻴﻰ ٠١١١٩٦٦٠٦١٦
ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ -ﺍﻟﺼﻒ ﺍﻟﺜﺎﻟﺚ ﺍﻟﺜﺎﻧﻮﻯ
د ) /ﺱ( ﻏ ﻣﻌﺮﻓﺔ ﻋﻨﺪ ﺱ = – ٢ – ، ٢ﻳﻲ ا ﺠﺎل
ﺣﻠﻮل ﺗﻤﺎر ﻦ )(٧ )(١
ﺇ ﻻ ﺗﻮﺟﺪ ﻧﻘﻂ ﺣﺮﺟﺔ
ا ﺠﺎل = ، òد )ﺱ( = ٩ﺱ – ﺱ ٣ﺉ د ) /ﺱ( = ٣ – ٩ﺱ٢
وﻣﻦ ا ﺮﺳﻢ :
،د ) /ﺱ( = ٠ﻋﻨﺪﻣﺎ ٣ – ٩ﺱ ٠ = ٢ﺉ ﺱ = ] ٣أن ﺱ = – ]٣ و ﺎ ﻌﻮ ﺾ
ﻣﻌﺎدﻟﺔ ا ﻨﺤ
ﺇ ﺹ = ٣] ٦أ ،ﺹ = – ٣] ٦
]٣
ﺱ
ﺇ ا ﻘﻂ ا ﺮﺟﺔ
اﻟﺔ .
٢-
ﳘﺲ
ﳘﺲ
+++
( ٣] ٦ – ، ٣] –) ، ( ٣] ٦ ، ٣] ) :
ا اﻟﺔ ﻣ اﻳﺪة
ﺱ
د ) /ﺱ(
+++
ﺳﻠﻮك د )ﺱ(
ﻣﻦ [ – ﳘﺲ ، ٢ – [ ، ] ٢ – ،ﳘﺲ ]
) (٧د ) /ﺱ( = ﺟﺘﺎ ﺱ – ﺟﺎ ﺱ ،ﺑﻮﺿﻊ د ) /ﺱ( = ٠ﺇ ﺟﺘﺎ ﺱ = ﺟﺎ ﺱ – ]٣
+++
–––
ا اﻟﺔ ﻣﺘﻨﺎﻗﺼﺔ
،ا اﻟﺔ ﻣ اﻳﺪة )(٢
ﺑﺎﻟﻘﺴﻤﺔ وﻣﻦ ا ﺮﺳﻢ :
د ) /ﺱ(
–––
،د )ﺱ( = ٠ﻋﻨﺪﻣﺎ ﺱ = – ، ٤د )– ٠ = (٤ﺇ ا ﻘﻄﺔ ا ﺮﺟﺔ ً ا اﻟﺔ ﻣ اﻳﺪة òﻷن د ) /ﺱ( ﻮﺟﺒﺔ داﺋﻤﺎ
ﺇ ا اﻟﺔ ﻣ اﻳﺪة
[ ، ٢ﳘﺲ ]
[ – ﳘﺲ ] ٢ ،وﻣﺘﻨﺎﻗﺼﺔ
ﺇ ٢ﺏ +ﺝ = – ٤ ، ٣ﺏ +ﺝ = – ٢١
ﺱ
د ) /ﺱ(
و ﻞ ﻫﺎﺗ ا ﻌﺎد
ﺳﻠﻮك د )ﺱ(
ﺑﻮﺿﻊ د ) /ﺱ( = ٠ﺇ ٦ﺱ – ٠ = ٦ﺉ ﺱ = ١ﻱ ﺡ ﺇ ﺗﻮﺟﺪ ﻧﻘﻄﺔ ﺣﺮﺟﺔ ﺇ ﻋﻨﺪ ﺱ = ٠ﻻﺗﻮﺟﺪ ﻧﻘﻄﺔ ﺣﺮﺟﺔ ﻣﻦ ا ﺮﺳﻢ ﺪ أن :ﳘﺲ
٠
+++
ا اﻟﺔ ﻣﺘﻨﺎﻗﺼﺔ
ﻳ +ò ﻦ ٠ﻲ
) ، ( ٦ ، ١د ) /ﺱ( ﻏ ﻣﻌﺮﻓﺔ ﻋﻨﺪ ﺱ = ٠وﻟ ١
[ ، ] ١ ، ٠وﻣ اﻳﺪة
--[ ، ١ﳘﺲ ]
) (١د )ﺱ( = ﺍ ﺱ + ٢ﺏ ﺱ ، ٢ +د ) /ﺱ( = ٢ﺍ ﺱ +ﺏ
د ) ٤ = (١ﺉ ﺍ +ﺏ = ، ٢د ٠ = (١) /ﺉ ٢ﺍ +ﺏ = ٠ﺇ ﺍ = – ، ٢ﺏ = ٤
د) //ﺱ( = ٢ﺍ ﺉ د ٠ > ٤ – = (١) //ﺉ ) ( ٤ ، ١ﻧﻘﻄﺔ ﻗﻴﻤﺔ ﻋﻈ ) (٢د )ﺱ( = ﺱ – ٤ﺱ
ﺱ
د ) /ﺱ( ﺳﻠﻮك د )ﺱ(
ا اﻟﺔ ﻣﺘﻨﺎﻗﺼﺔ
[ – ﳘﺲ ، ] ١ ،وﻣ اﻳﺪة
٢
---
ﻋﻨﺪ ﺱ = ٠ﻧﻘﻄﺔ ﺣﺮﺟﺔ ﻟ ﺴﺖ ﻋﻈ
د ) /ﺱ(
ﺳﻠﻮك د )ﺱ(
) (٣د )ﺱ( = - 3ﺱ ¤ﺫ - ¤ -ﺫ
= ﺫ
---
أو ﺻﻐﺮى ﻠﻴﺔ
د )ﺱ( ﺳﻠﻮك د )ﺱ( /
،ﻋﻨﺪ ﺱ = 3ﺗﻮﺟﺪ ﻧﻘﻄﺔ ﺣﺮﺟﺔ ﺻﻐﺮى ﻠﻴﺔ 4 7 - ٣ 3ﺫ 3 3 ٤ ،اﻟﻘﻴﻤﺔ ا ﺼﻐﺮى ا ﺤﻠﻴﺔ = د ) ( = ) ( – ) ( = 4 4 4 56ﺫ
[ ، ١ﳘﺲ ]
) + ¤ﺫ( - ¤ ) -ﺫ( 4 ) (٦ا ﺠﺎل = ، { ٢ – } – òد ) /ﺱ( = = ﺫ ) + ¤ﺫ( ) + ¤ﺫ(
ﺎ ﺎ = ، òد ) /ﺱ( = ٤ﺱ ٣ – ٣ﺱ
٢
،ﺑﻮﺿﻊ د ) /ﺱ( = ٠ﺉ ﺱ ٤ ) ٢ﺱ – ٠ = ( ٣ﺇ ﺱ = ٠أ ،ﺱ = 3 4 3 وﻣﻦ ا ﺮﺳﻢ : 4 ٠ ﺱ
+++
ﺇ ﺗﻮﺟﺪ ﻧﻘﻄﺔ ﺣﺮﺟﺔ ﻋﻨﺪ ﺱ = ، ٠ﰈ د ) ٢ = (٠ﺇ ) ( ٢ ، ٠ﻧﻘﻄﺔ ﺣﺮﺟﺔ ٠ ﺱ ﳘﺲ ﳘﺲ
)(٤) ، (٣
ﺉ ﺏ = – ، ٩ﺝ = ١٥
٣
ﺉ ٤ﻩ ٢ﺱ – ٠ = ٤ﺇ ﻩ ٢ﺱ = ١ﺉ ﺱ = ٠
+++
ﻘﻖ ﻣﻌﺎد ﺔ
ﺣﻠﻮل ﺗﻤﺎر ﻦ )(٨
) (٥ا ﺠﺎل = ، òد ) /ﺱ( = ٤ﻩ ٢ﺱ ، ٤ -ﺑﻮﺿﻊ د ) /ﺱ( = ٠
---
٩ :ﺱ +ﺹ = ٢٠أى أن :د )٩ – = (٢ /
ﻣﻦ ) (٥) ، (٣ﺑﺎﻟﻄﺮح ﺉ ﺍ = ، ١ﺑﺎ ﻌﻮ ﺾ
6 - ¤6 6 ) (٤ا ﺠﺎل = ، +òد ) /ﺱ( = 1 × ٣ – ٦ﺫ × ٢ﺱ = = – ٦ ﺱ ﺱ ¤
ﻋﻨﺪ ﺱ = – ١
أى أن :د ) ٠ = (١وﻣﻦ ) : (٢ﺇ ٣ﺍ ٢ +ﺏ +ﺝ = (٣) ........... ٠ /
وﻣﻦ ) (١ﺇ ٨ﺍ ٤ +ﺏ ٢ +ﺝ = ٢أى أن ٤ :ﺍ ٢ +ﺏ +ﺝ = ، (٥) ............ ١
،د ) /ﺱ( ﻵ ٠ﻷن – ٢ﻵ ، ٠د )ﺱ( ﻏ ﻣﻌﺮﻓﺔ ﻋﻨﺪﻣﺎ ﺱ – ٠ = ٢
---
[ ] p45 ،
د) ٢ = ٢× ٩ – ٢٠ = (٢ﺇ ﻧﻘﻄﺔ ا ﻤﺎس ) ( ٢ ، ٢ﻱ ا ﻨﺤ ﻓ
/
+++
ﺳﻠﻮك د )ﺱ(
وﻣﻦ ) : (٢ﺇ ١٢ﺍ ٤ +ﺏ +ﺝ = – ، (٤) ........... ٩وﻣﻦ ﻣﻌﺎدﻟﺔ ا ﻤﺎس
ﺫ ﺇ د ) /ﺱ( = – - ¤S3ﺫ
-ﳘﺲ
د ) /ﺱ(
ﻘﻖ ﻣﻌﺎد ﻪ وﻣﻦ ) : (١ﺇ ، ٠ = ﰈ ﻋﻨﺪ ﺱ = ١ﺗﻮﺟﺪ ﻧﻘﻄﺔ ﺣﺮﺟﺔ
،ﰈ ﻣﻌﺎدﻟﺔ ا ﻤﺎس ﻋﻨﺪ ﺱ = ٢
ﺫ 3
)(٠،٢ ٢
ﺇ
)– ( ٠ ، ٤
/
ﺱ
،د ) /ﺱ( = ٣ﺍﺱ ٢ + ٢ﺏ ﺱ +ﺝ ، (٢) ..........ﰈ ا ﻨﺤ ﻳﻤﺮ ﺑﺎ ﻘﻄﺔ ) ( ٠ ، ٠
ا ﺠﺎل = ، òد )ﺱ( = ) ﺱ ٣( ٤ +ﺉ د ) /ﺱ( = ) ٣ﺱ ٢( ٤ +
1 3
---
٠
+++
) (٨د )ﺱ( = ﺍ ﺱ + ٣ﺏ ﺱ + ٢ﺝ ﺱ (١) ........ +
اﻟﻔ ة [ – ]] ٣] ، ٣
) (٣ا ﺠﺎل = ، òد )ﺱ( = – ) ٣ﺱ – ( ٢
p 4
[ ٢ ، p45 [ ، ] p4 ، ٠ﺑﺐ ] ،وﻣﺘﻨﺎﻗﺼﺔ
ا اﻟﺔ ﻣ اﻳﺪة
ﻣﻦ – [ :ﳘﺲ ، ٣] [ ، ] ٣] – ،ﳘﺲ ] .
ﺉ ﻋﻨﺪ ﺱ = ٢ﻧﻘﻄﺔ ﺣﺮﺟﺔ ﳘﺲ
٢ﺑﺐ
ﺳﻠﻮك د )ﺱ(
٣] ٦
3
p5 4
+++
– ٣] ٦
ﺇ د ) /ﺱ( = – × ٣ﺫ ) ﺱ – -( ٢
ﺟﺘﺎ ﺱ ﺉ ﻇﺎ ﺱ = ١ﺇ ﺱ = pأ ،ﺱ = p 5 4 4
¤ﺫ5 + ¤6 -
) ¤ﺫ - ¤ -ﺫ(
ﺫ
=
،د ) /ﺱ( =
) (5 - ¤ )(1- ¤
) ¤ﺫ - ¤ -ﺫ(
ﺫ
¤ ) -ﺫ - ¤ -ﺫ( ) -ﺫ ( ¤ - 3 )(1- ¤
) ¤ﺫ - ¤ -ﺫ(
ﺫ
،ﺑﻮﺿﻊ د )ﺱ( = ٠ﺉ /
) ﺱ – ) ( ١ﺱ – ٠ = ( ٥ﺇ ﺗﻮﺟﺪ ﻧﻘﻂ ﺣﺮﺟﺔ ﻋﻨﺪ ﺱ = ، ١ﺱ = ٥
٣١
ﻟﻸﺳﺘﺎﺫ ﺍﻟﻘﺪﻳﺮ /ﻋﻠﻰ ﺍﻟﺪﻳﻦ ﻳﺤﻴﻰ ٠١١١٩٦٦٠٦١٦
ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ -ﺍﻟﺼﻒ ﺍﻟﺜﺎﻟﺚ ﺍﻟﺜﺎﻧﻮﻯ وﻣﻦ ا ﺮﺳﻢ : +++
٥
---
٢
١ ---
+++
١-
+++
ﻋﻨﺪ ﺱ = ، ٢ﺱ = – ٢ﺗﻮﺟﺪ ﻧﻘﻂ اﻧﻘﻼب
ﺱ
) (٣د )ﺱ( = ﺱ ٧ – ٢ﺇ د ) /ﺱ( = ٢ﺱ ﺇ د) //ﺱ( = ٢ ً ً ﺇ د) //ﺱ( < ٠داﺋﻤﺎ ﺇ ﻣﻨﺤ ا اﻟﺔ ﺪب ﻷﺳﻔﻞ داﺋﻤﺎ
د ) /ﺱ( ﺳﻠﻮك د )ﺱ(
،ﻻ ﺗﻮﺟﺪ ﻧﻘﻂ اﻧﻘﻼب ﻟﻌﺪم ﺗﻐ
üïﺱ ) 4 £ ¤ @ (4 - ¤ üïﺱ ﺫ ¤4 - = ) (٤د )ﺱ( = ý ý -ïþﺱ ﺫ 4 > ¤ @ ¤4 + - ïþﺱ ) 4 > ¤ @ (4 - ¤ ﰈ د ، ٨ - = ٤ × ٤ – ٤ × ٢ = +(٤) /د ٠ = ١٦ + ١٦ - = -(٤) /
ﻋﻨﺪ ﺱ = ١ﺗﻮﺟﺪ ﻗﻴﻤﺔ ﺻﻐﺮى ﻠﻴﺔ = د )١ – = (١ ،ﻋﻨﺪ ﺱ = ٥ﺗﻮﺟﺪ ﻗﻴﻤﺔ ﻋﻈ
ﻠﻴﺔ = د )1 – = (٥ 9
) (٤د )ﺱ( = ﺟﺎ ﺱ +ﺟﺘﺎ ﺱ ﺇ د ) /ﺱ( = ﺟﺘﺎ ﺱ – ﺟﺎ ﺱ ،د ) /ﺱ( = ٠ﻋﻨﺪﻣﺎ ﺟﺘﺎ ﺱ – ﺟﺎ ﺱ = ٠ﺇ ﺟﺘﺎ ﺱ = ﺟﺎ ﺱ ﺇ ﻇﺎ ﺱ = ١
ﺫ4 - ¤
4 ،ﻋﻨﺪ ﺱ = p5ﺗﻮﺟﺪ ﻗﻴﻤﺔ ﺻﻐﺮى ﻣﻄﻠﻘﺔ = – ]٢ 4
ﻩ – ﺱ = ﻩ – ﺱ ) – ١ﺱ ( ،د ) /ﺱ( = ٠ﻋﻨﺪﻣﺎ ﻩ – ﺱ ) – ١ﺱ ( = ٠ﺉ ﺱ = ١ ﻱ ] ، [ ٢ ، ٠د ) ، ٠ = (٠د ) = (١ﻩ ، 1د )= (٢ ﻗﻴﻤﺔ ﻋﻈ
ﺫ
ﺉ ﻋﻨﺪ ﺱ = ١ﺗﻮﺟﺪ
ﻣﻄﻠﻘﺔ = ، 1ﻋﻨﺪ ﺱ = ٠ﺗﻮﺟﺪ ﻗﻴﻤﺔ ﺻﻐﺮى ﻣﻄﻠﻘﺔ = ٠ ﻩ
ü ï ﺇ د) //ﺱ( = ýﻏ ـ ﻮﺟــﻮدة @ 4 = ¤ ï @ 4> ¤ ﺫþ
ﺇ ا ﻨﺤ
) (٥ﺹ = ٠ﺉ ﺱ = ١ ، ٣و ﺒﺤﺚ اﺷﺎرﺗﻬﺎ ﺪ أن : ا اﻟﺔ ﻣ اﻳﺪة
+++
،
ﺱ > ، ٠د ) /ﺱ( = ٢ﺱ – ٢
ﺱ<٠
ﺇ د (٠) /ﻏ ﻣﻌﺮﻓﺔ ﺉ ﻋﻨﺪ ﺱ = ٠ﺗﻮﺟﺪ ﻧﻘﻄﺔ ﺣﺮﺟﺔ @ 0> ¤
¤3 üﺫ ¤6 + ï ﺇ د ) /ﺱ( = ýﻏ ـ ﻮﺟــﻮدة @ ، 0= ¤ﺑﻮﺿﻊ د )ﺱ( = ٠ ï þﺫ - ¤ﺫ @ 0< ¤
وﻣﻴﻞ
،ا ﻨﺤ
،د )– ، ٠ = ٢٧ + ٢٧ - = ( ٣د )٣ = ٦ – ٩ = (٣
[ – ﳘﺲ [ ٢ ،ا ﺤﺪب ﻷ
-2
-4
-6
-8
] ، ٢ﳘﺲ ] ا ﺤﺪب ﻷﺳﻔﻞ ( ٢ ، ٢ ) ،ﻧﻘﻄﺔ اﻧﻘﻼب . ا ﻨﺤ
ﻣﻨﻬﻤﺎ = ١ ﺪب ﻷ
ﻋﻨﺪ ﺱ > ٠
ﺣﻠﻮل ﺗﻤﺎر ﻦ )(١٠ )(١
3 -17ﺱ وﻣﻨﻬﺎ ﺹ = ﺫ
) (٢ﻧﻔﺮض ﺏ ﻩ = ﺱ ،ﺍ ﻥ = ﺹ ﺉ ﻣـ ) ﻣﻢ ﺍ و ﺏ ( = ) 1ﺱ ) ( ٤ +ﺹ ( ٣ + ﺫ ﺱ 3 = و ا ﻬﺎﻳﺔ ﻳ ﺘﺞ وﻣﻦ ﺸﺎﺑﻪ ﻣﻢ ﻣﻢ :ﺝ ﻩ ﺏ ،ﺍ ﻥ ﻩ ﺇ 4 ﺹ
ﺣﻠﻮل ﺗﻤﺎر ﻦ )(٩
ﺱ = ، ٤ﺹ = ٣وا ﺴﺎﺣﺔ ا ﻄﻠﻮ ﺔ = ٢٤وﺣﺪة ﺴﺎﺣﺔ
[ – ﳘﺲ [ ٣ ،
] ، ٣ﳘﺲ ] ﺛﻢ ﺹ ٠ = /ﺉ ﺱ = ٤ ، ٢ﺉ
) – ( ٤٤ – ، ١ﺻﻐﺮى ﻣﻄﻠﻘﺔ ( ١٠ ، ٢ ) ،ﻋﻈ
ﻴﻂ ا ﺴﺘﻄﻴﻞ +ﻴﻂ ا ﺮ ﻊ = ٣٤ﺉ ٢ ) ٢ﺱ +ﺱ ( ٤ +ﺹ = ٣٤
ﻳ ﺘﺞ ﺱ = ٣وﻋﻨﺪﻫﺎ ﺗ ﻮن ﻗﻴﻤﺔ ﺻﻐﺮى
ا ﻄﻠﻘﺔ = ، ٤اﻟﻘﻴﻤﺔ ا ﺼﻐﺮى ا ﻄﻠﻘﺔ = – ١
ﻣﻄﻠﻘﺔ
) (٢ﺹ ٠ = //ﺉ ﺱ = _ – [ ، ٢ﳘﺲ [ ٢ – ،ﺪب ﻷﺳﻔﻞ ] ٢ ، ٢ – [ ،ﺪب ﻷ
س 5
4
3
2
1
ﻤﻮع ا ﺴﺎﺣﺘ = ﻡ = ٢ﺱ + ٢ﺹّ ٢ ﻋﻮض ﺛﻢ أﺷﺘﻖ وﺳﺎو ﻬﺎ ﺑﺎ ﺼﻔﺮ
د ) ، ٠ = (٠د ) ، ١ – = ٢ – ١ = (١د )– ٤ = ١٢ + ٨ – = (٢
،ا ﺤﺪب ﻷﺳﻔﻞ
2
-1
،ﺪب ﻷﺳﻔﻞ ﻋﻨﺪ ﺱ < ٠
ﻋﻨﺪ ﺱ = ٢ – ، ١ ، ٠
) (١ﺹ ٠ = //ﺉ ﺱ = ٣ﺉ ا ﺤﺪب ﻷ
4
) ( ٠ ، ١ ) ، ( ٢ ، ٠ ) ، ( ٤ ، ١ –) (٦ﻧﻘﺎط
،ﻋﻨﺪ ﺱ < ) ٢ : ٠ﺱ – ٠ = ( ١ﺉ ﺱ = ١
ﺇ اﻟﻘﻴﻤﺔ اﻟﻌﻈ
6
ﻠﻴﺔ ( ٠ ، ٣ ) ،ﺻﻐﺮى ﻠﻴﺔ ،
/
ﻋﻨﺪ ﺱ > ٣ : ٠ﺱ ) ﺱ ٠ = ( ٢ +ﺉ ﺱ = ) ٠ﺮﻓﻮض ( أ ،ﺱ = – ٢
ﺇ ا ﻘﻂ ا ﺮﺟﺔ
8
،ا ﻤﺎس ﻋﻨﺪ ﺱ = [ ١ا ﻤﺎس ﻋﻨﺪ ﺱ = – ١
ﺇ د ، ٠ = -(٠) /د ٢ – = ٢ – (٠) ٢ = +(٠) /ﺇ د +(٠) /ﻵ د- (٠) /
اﻟﺔ
ص
][٣،١
//
د )ﺱ( ﺳﻠﻮك د )ﺱ(
[ – ﳘﺲ ، ٣ ] ، [ ١ ،ﳘﺲ ] ،
ﺹ = ٠ﺉ ﺱ = ٢ﺉ
/
ﻋﻨﺪ ﺱ = ١ﺗﻮﺟﺪ ﻗﻴﻤﺔ ﺻﻐﺮى ﻠﻴﺔ = د )١ = (١ ) (٧د ) /ﺱ( = ٣ﺱ ٦ + ٢ﺱ
( ٤ ، ١ ) ،ﻋﻈ
ﺱ
---
[ – ﳘﺲ ، ] ٤ ،و ﺪب ﻷﺳﻔﻞ
/
ا اﻟﺔ ﻣﺘﻨﺎﻗﺼﺔ
،ﺑﻮﺿﻊ د ) /ﺱ( = ٠ﺉ ﺱ – ٠ = ١ﺇ ﺱ = ١
وﻣﻦ ا ﺮﺳﻢ :
ﺪب ﻷ
[ ، ٤ﳘﺲ ]
،وﻻ ﺗﻮﺟﺪ ﻧﻘﻂ إﻧﻘﻼب ﻷن ا اﻟﺔ ﻏ ﻗﺎﺑﻠﺔ ﻼﺷﺘﻘﺎق ﻋﻨﺪ ﺱ = ٤
) (٦د )ﺱ( = ﺱ – ﻮ ﺱ ،ﺱ < ، ٠د ) /ﺱ( = = 1 – ١ﺱ 1 - ﺱ ﺱ ﻩ
١
@ 4< ¤
ﺫ
ﺇ د ) /ﺱ( = ﺱ ﻩ – ﺱ × – + ١ﻩ – ﺱ = – ﺱ ﻩ – ﺱ +
ﻩﺫ
@ 4< ¤
ü ï ﺇ د ) /ﺱ( = ýﻏـــ ﻗﺎﺑﻠـــﺔ ﻼﺷـــﺘﻘﺎق @ 4 = ¤ ï @ 4> ¤ ﺫ4 + ¤þ
، ٢] – = ( p5د ) ٢ﺑﺐ ( = ١ ﺇ د ) ، ١ = (٠د ) ، ٢] = ( 4pد ) 4 ﺇ ﻋﻨﺪ ﺱ = pﺗﻮﺟﺪ ﻗﻴﻤﺔ ﻋﻈ ﻣﻄﻠﻘﺔ = ]٢
) (٥د )ﺱ( = ﺱ ﻩ
@ 4£ ¤
ﺇ د +(٤) /ﻵ د -(٤) /ﺇ ا اﻟﺔ ﻏ ﻗﺎﺑﻠﺔ ﻼﺷﺘﻘﺎق ﻋﻨﺪ ﺱ = ٤
p5ﻱ ] ٢ ، ٠ﺑﺐ [ ﺇ ﺱ = 4p = ٥٤٥ﻱ ] ٢ ، ٠ﺑﺐ [ أ ،ﺱ = = ٥٢٢٥ 4
–ﺱ
ﺪب ا ﻨﺤ ﻋﻨﺪ أى ﻧﻘﻄﺔ
ﻗ ﻉ وﻣﻨﻬﺎ ﻉ = ) (٣ﻡ = ط ﻗﻖ ٢ + ٢ط ﻖ ّ ﻋﻮض
ﻗﺎﻧﻮن ا ﺠﻢ واﻷﺑﻌﺎد
®- 75ﺫ
ﺫ®
٥ :ﻡ ٥،ﻡ
) (٤ف ) = ٢ﺱ – + ٢( ٢ﺹّ ٢ ﻋﻮض ﺛﻢ أﻓﺮض ف = ٢ل وأﺷﺘﻖ ﺉ ﺱ = ، ٠ﺹ = ] ٣ﺇ ا ﻘﻄﺔ
، ٢ ] ،ﳘﺲ ] ﺪب ﻷﺳﻔﻞ ،
٣٢
) ( ٣] ، ٠
ﻟﻸﺳﺘﺎﺫ ﺍﻟﻘﺪﻳﺮ /ﻋﻠﻰ ﺍﻟﺪﻳﻦ ﻳﺤﻴﻰ ٠١١١٩٦٦٠٦١٦
ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ -ﺍﻟﺼﻒ ﺍﻟﺜﺎﻟﺚ ﺍﻟﺜﺎﻧﻮﻯ ا اﺋﺮﺗ = ﺱ
) (٥ﻧﻔﺮض ﻃﻮل ﻧﺼﻒ ﻗﻄﺮ ﻧﺼ
1
ﻓﻴﻜﻮن ﺑﻌﺪى ا ﻠﻌﺐ ﻫﻤﺎ ٢ﺱ ،ﺹ
ﰈ ا ﺤﻴﻂ = ٤٢٠ﺇ ٢ﺹ ٢ +ﺑﺐ ﺱ = ٤٢٠
ﺱ
ﺹ
ﺱ
ﺇ ﺹ = – ٢١٠ﺑﺐ ﺱ
،ﺴﺎﺣﺔ ا ﻠﻌﺐ = ﻡ = ٢ﺱ ﺹ +ﺑﺐ ﺱ ٢ = ٢ﺱ ) – ٢١٠ﺑﺐ ﺱ ( +ﺑﺐ ﺱ
٢
( ﺇ ﺗ ﻮن ا ﺴﺎﺣﺔ أ
أى أن ا ﻌﺪ ﺹ ﻳﺘﻼ
ﻣﺎ ﻳﻤ ﻦ ﻋﻨﺪﻣﺎ
10ﺫ ا ﻠﻌﺐ داﺋﺮة ﻃﻮل ﻧﺼﻒ ﻗﻄﺮﻫﺎ p
) (٦ﻣﻢ ﺍﻡ ﰲ ﻣﻢ ﺏ ﺝ ﻡ إ ﺹ =® ﺉ ﺹ =
®ﺫ
.
® ﺱ 1 ،ﺴﺎﺣﺔ ﺷﺒﻪ ا ﻨﺤﺮف = ﻡ = ) ﺱ +ﺹ ( × ٢ﻗﻖ ﺫ ®3 ﻗﺱ+ ﻗﺹ= ﻖ ﻗﺱ +ﻖ = ﻖ ﺱ ® ¤ﺫ 3®- 3 3 =٠ ﻗ ® -ﺫ = ٠ﺇ ﻗ ® -ﺫ ،ﺑﻮﺿﻊ ﻡ ٠ = /ﺉ ﻖ ﺇ ﻡ = /ﻖ ﺱﺫ ﺱ ﺱ ﺫ®3 ﻗ ،ﻡ ) ٠ < 3 = //ﺻﻐﺮى ( ﻗ ) ﺱ – ٢ﻗﻖ ٠ = ( ٢ﺉ ﺱ = ﻖ ﺇ ﻖ ﺱ ﺫ ﻗ = ٢ﻗﻖ ٢وﺣﺪة ﺮ ﻌﺔ . ﺇ أﺻﻐﺮ ﺴﺎﺣﺔ = ﻡ = ) 1ﻗﻖ ٢ × ( ® +ﻖ ® ﺫ ﺱ
ﺱ
ﻗ ﻡ ﻗ ﻖ ﻖ
ﺏ
ﺣﻠﻮل ﺗﻤﺎر ﻦ )(١١
) (١ﺑﻮﺿﻊ ﻉ = Sﺱ 1+ﺇ ﺱ = ﻉ ١ – ٢ﺇ ﺱ = ٢ﻉ .ﻉ
ﺇ ا ﻘﺪار = ) ] þﻉ × [ ١ – ٢( ١ – ٢ﻉ × ٢ﻉ .ﻉ = ٢ ) þﻉ ٤ – ٦ﻉ ( ٤ﻉ = ﺫ 7
)(٢ ﺇ ا
=1 ) þﻉ 8
= ×1 8
5 – ) 4ﺱ (١+ﺫ+ث
1 3 ﺫ 1+ﻉ ﺫ
4
5 ﺫ ﻉ ﺫ ×1+
5
= ٢) 1ﺱ 0ﺫ
4
5 – (١ﺫ
1+ﻉ 8
3 ﺫ
ﺫﻉ ٢×1+
3
8
1 ﻉ ﺫ+ث=
٢) 1+ﺱ –(١ 6
)(٤
3 ﺫ
1 0ﺫ
5 ﻉ ﺫ1 +
6
٢ ) 1+ﺱ 4
ﻉ
1 – (١ﺫ
4
٢ﺱ
ﺇ ﺹ = ٢ﺱ ﻩ ٢ﺱ +ﻩ ٢ﺱ = ﻩ ٢ﺱ ) ٢ﺱ ( ١ +ﺱ ٢ﺱ
٢ﺱ –٢
)٢ﺱ (١+ﺱ
ﺉ ﺹ=ﻩ þﺉ ﻉ = ٢ ) 1ﺱ ١– ( ١ +ﺱ
þﻮ ﺱﺱ: ﻩ
ﻧﻔﺮض ﺹ = ﻮ ﻩ ﺱ
ﺉ ﺹ= 1ﺱ ﺱ þﺉ ﻉ =ﺱ
،ﻉ =١ﺱ
ﺇ ا ﻘﺪار = ﺱ ﻮ ﻩ ﺱ – þﺱ = ﺱ ﻮ ﻩ ﺱ – ﺱ +ث þ ،ﺱ ٣ﻮ ﺱ ﺱ : ﻩ
ﻧﻔﺮض ﺹ = ﻮ ﻩ ﺱ ﺉ ﺹ= 1ﺱ ﺱ þ٤ 1 ،ﻉ = ﺱ ٣ﺱ ﺉ ﻉ = ﺱ 4 ٤ 1 1 ﺇ ا ﻘﺪار = 1ﺱ ٤ﻮ ﻩ ﺱ – þ 1ﺱ ٣ﺱ = 1ﺱ ٤ﻮ ﻩ ﺱ – × ﺱ 4 4 4 4 4 1 ٤ ٤ 1 ﺱ = ﺱ ﻮﻩﺱ – 16 4 ) þﻮ ﺱ ( ٢ﺱ ﻩ
٢
،ﻉ =ﺱ
ﺉ ﺹ= ﺫ ﻮﻩﺱ.ﺱ ﺱ þﺉ ﻉ =ﺱ
ﺇ ا ﻘﺪار = ﺱ ) ﻮ ﻩ ﺱ ( ٢ þ – ٢ﻮ ﻩ ﺱ .ﺱ : ﻧﻔﺮض ﻡ = ﻮ ﻩ ﺱ
(ﻉ 3 ﺫ 1+
+ث
ﺱ:
ﻧﻔﺮض ﺹ = ) ﻮ ﻩ ﺱ (
(
1ﺫ
) ﺫ(1+ ¤
+ ) ü 5 10ﺫ( ¤S - 11
11
ﺫ٢ﺱ ﺇ ا ﻘﺪار = -ﺱ ) ٢ﺱ ١ – ( ١ +ﻩ ٢ﺱ – ) þ ( 1-ﻩ ) ٢ﺱ × ( ١ + ﺫ ﺫ 1 ﺱ ٢ﺱ – ٢ ١ﺱ ) ٢ﺱ ١– ( ١ +ﺱ = ) ٢ﺱ ( ١ +ﻩ þ +ﻩ ﺱ ﺫ ﺫ ﺱ Úﺫ ¤ 1 +ﻩ ٢ﺱ +ث = -ﺱ ) ٢ﺱ ١ – ( ١ +ﻩ ٢ﺱ 1 +ﻩ ٢ﺱ +ث = ﺫ) ﺫ4 (1+ ¤ 4 ﺫ
)(٥
ﻉ 4–٧ﻉ+٥ث= ﺫ )ﺱ (١+ﺫ 5 7 5 1 1 1 ﺑﻮﺿﻊ ﻉ = ٢ﺱ – ١ﺇ ﺱ = ﻉ +ﺇ ﺱ = ﻉ ﺫ ﺫ ﺫ ﺫ 1ﻉ 1+ 1ﻉ ﺫ 1+ ¬ 1 + 1 1 ﻘﺪار = þﺫ ﺫ × ﻉ = 4 þ 1ﺫ × 4ﻉ ﺫ ﺫ Sﻉ ﻉﺫ
)
Ú ¤ﺫ ¤
ﺫ
+ث
،ﻉ =)٢ﺱ (١+
ﺝ
ﻩ
ﺹ ﺍ
7
6
ﺇ ﺹ = ﺱﻩ
= ﺻﻔﺮ
و ﺼﺒﺢ ﺷ
11 ﻉ 5
ﻧﻔﺮض ﺹ = ﺱ ﻩ
10ﺫ ﺇ ﻡ ٢ – ٤٢٠ = /ﺑﺐ ﺱ ،ﺑﻮﺿﻊ ﻡ ٠ = /ﺉ ﺱ = p 10ﺫ 10ﺫ ،ﺹ = – ٢١٠ﺑﺐ × ﺱ= p p
6
= – 10ﻉ 10 + 5 11 3 6 = – 10 ) ü 5 3ﺫ( ¤S -
þ ،
= ٤٢٠ﺱ – ٢ﺑﺐ ﺱ + ٢ﺑﺐ ﺱ ٤٢٠ = ٢ﺱ – ﺑﺐ ﺱ ٢و ﺎﻻﺷﺘﻘﺎق
ﻡ ٢ - = //ﺑﺐ > ) ٠ﻋﻈ
6
= ٤ –) þﻉ ٢ + 5ﻉ . ( 5ﻉ = – 5 × ٤ﻉ 5 × ٢ + 5 11 6
1 ﻉ ﺫ+ث
ﺉ ﻡ= 1ﺱ ﺱ þﺉ ﻥ =٢ﺱ
،ﻥ =٢ﺱ
ﺇ ا ﻘﺪار = ﺱ ) ﻮ ﻩ ﺱ ( ٢ ] – ٢ﺱ ﻮ ﻩ ﺱ – ٢ þﺱ [ = ﺱ ) ﻮ ﻩ ﺱ ( ٢ – ٢ﺱ ﻮ ﻩ ﺱ ٢ +ﺱ +ث
+ث
þ ،ﺱ ) ﺱ – ( ١ﺱ :ﺑﻮﺿﻊ ﻉ = ﺱ – ١ﺇ ﺱ = ﻉ ١ +
þﻗﺘﺎ ) ٢ﺱ ( 3 +ﺱ = – þ ٢ﻇﺘﺎ ) ﺱ + ( 3 +ث
ﺇ ٢ﺱ .ﺱ = ﻉ ﺇ ا ﻘﺪار = þ 1ﺱ ) ٢ﺱ ٢ × ٥( ١ – ٢ﺱ .ﺱ
) ٦ þ (٦ﻗﺎ ٢ﺱ ) ﻇﺎ ٢ﺱ +ﺟﺘﺎ ٢ ٢ﺱ ( ﺱ
٣
٢
٥
٢
٢
ﺫ 1 ٦ 1 =) þ 1ﻉ ×(١+ﻉ.٥ﻉ = ) þﻉ +ﻉ (.ﻉ ﺫ ﺫ ﺫ = 1 ×1ﻉ 1 ×1 + ٧ﻉ + ٦ث = 1ﻉ 1 + ٧ﻉ + ٦ث ﺫ1 14 ﺫ 6 ﺫ 7 1 ٦ ٢ ٧ ٢ = ) 1ﺱ –) + (١ﺱ –+ (١ث ﺫ1 14
ﺫ
1
ﺫ
= ٦ ) þﻗﺎ ٢ﺱ ﻇﺎ ٢ﺱ ٦ +ﺟﺘﺎ ٢ﺱ ( ﺱ
٥
) (٣ﻧﻀﻊ ﻉ = ] – ٢ﺱ ﺇ ﺱ = ) – ٢ﻉ ( ٤ – ٤ = ٢ﻉ +ﻉ
11 ﻉ 5
+ث
= 1 × ٦ﻗﺎ ٢ﺱ 1 × ٦ +ﺟﺎ ٢ﺱ +ث = ٣ﻗﺎ ٢ﺱ ٣ +ﺟﺎ ٢ﺱ +ث ﺫ
þ ،ﺱ ﺟﺎ ﺱ .ﺱ : ﻧﻔﺮض ﺹ = ﺱ
٢
ﺇ ﺱ = )– ٢ + ٤ﻉ ( .ﻉ ﺇ ا ﻘﺪار = þﻉ ٢ + ٤ –) 5ﻉ ( .ﻉ
،ﻉ = ﺟﺎ ﺱ .ﺱ
ﺫ
ﺉ ﺹ =ﺱ þﺉ ﻉ = – ﺟﺘﺎ ﺱ .
ﺇ ا ﻘﺪار = – ﺱ ﺟﺘﺎ ﺱ þ +ﺟﺘﺎ ﺱ .ﺱ = – ﺱ ﺟﺘﺎ ﺱ +ﺟﺎ ﺱ +ث
٣٣
ﻟﻸﺳﺘﺎﺫ ﺍﻟﻘﺪﻳﺮ /ﻋﻠﻰ ﺍﻟﺪﻳﻦ ﻳﺤﻴﻰ ٠١١١٩٦٦٠٦١٦
ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ -ﺍﻟﺼﻒ ﺍﻟﺜﺎﻟﺚ ﺍﻟﺜﺎﻧﻮﻯ f - 1ﺫ) ﺫ (3 - ¤ eﺫ ) ﺫ( 3 - ¤ ﺱ=þ )þ (٧ ) f - 1ﺫ (3 - ¤ ) f -1ﺫ ( 3 - ¤ ) f -1ùûﺫ ) f + 1ùû éë(3 - ¤ﺫé(3 - ¤ =þ ëﺱ ) f - 1ûùﺫ ëé(3 - ¤
)) (٥ﺍ( ا ﻘﺪار = ٢ þ 1ﺱ ﻗﺎ ) ٢ﺱ ( ٢ + ٢ﺱ = 1ﻇﺎ ) ﺱ + ( ٢ + ٢ث
ﺱ
ﺫ
)ﺏ( ا ﻘﺪار = ٣ þ 1ﺱ ٢ﺟﺘﺎ ) ﺱ ( ٥ + ٣ﺱ = 1ﺟﺎ ) ﺱ + ( ٥ + ٣ث 3
= + ١ ] þﺟﺘﺎ ) ٢ﺱ [ ( ٣ -ﺱ = ﺱ 1 +ﺟﺎ ) ٢ﺱ – + ( ٣ث ﺫ
) þ ،ﻇﺎ ٢ﺱ ٢ +ﺟﺎ ٢ﺱ ( ﺱ = ) þﻗﺎ ٢ﺱ – – ١ + ١ﺟﺘﺎ ٢ﺱ ( ﺱ = ) þﻗﺎ ٢ﺱ – ﺟﺘﺎ ٢ﺱ ( ﺱ = ﻇﺎ ﺱ – 1ﺟﺎ ٢ﺱ +ث ﺫ
) þ (٨ﺱ ٢ﺟﺘﺎ ) ﺱ ( ٥ + ٣ﺱ = ٣ þ 1ﺱ ٢ﺟﺘﺎ ) ﺱ ( ٥ +ﺱ 3
= 1ﺟﺎ ) ﺱ + ( ٥ + ٣ث
¤f = ) þﻇﺘﺎ ﺱ ﻗﺘﺎ ٢ﺱ – ﻇﺘﺎ ﺱ ( .ﺱ = ) þﻇﺘﺎ ﺱ ﻗﺘﺎ ٢ﺱ – ¤e
(.ﺱ
= 1 -ﻇﺘﺎ ٢ﺱ – ﻮ | ﺟﺎﺱ | +ث ﻩ
ﺱ ) (٩ﰈ = §Ù § f +1 ¤Ù ﺇ ﺹ +ﺟﺎ ﺹ = 1ﺱ + ٢ث ،ﰈ ا ﻨﺤ ﻳﻤﺮ ﺑﺎ ﻘﻄﺔ ) ( ٠ ، ٠ﺉ ث = ٠ ﺫ ٢ 1 ﺇ ﻣﻌﺎدﻟﺔ ا ﻨﺤ :ﺹ +ﺟﺎ ﺹ = ﺱ ﺫ
ﺇ + ١ ) þﺟﺘﺎ ﺹ ( ﺹ = þﺱ ﺱ
Ùﺫ § = – ٢ ) ٢ﺟﺎ ﺱ ﺟﺘﺎ ﺱ ( = – ٢ﺟﺎ ٢ﺱ ) (١٠ﰈ ﺫ
¤Ù 1 ﺇ þ ٢ – = §Ùﺟﺎ ٢ﺱ .ﺱ = – ( ) × ٢ﺟﺘﺎ ٢ﺱ +ث ﺫ ¤Ù
= ﺟﺘﺎ ٢ﺱ +ث وﻟ ﻦ د ٢ = (٠) /ﺉ ﺟﺘﺎ + ٠ث = ٢ﺉ ث = ١
¤Ù
ﺇ ﺹ = ) þﺟﺘﺎ ٢ﺱ ( ١ +ﺱ
= 1ﺟﺎ ٢ﺱ +ﺱ +ث ، /ﰈ ا ﻘﻄﺔ ) ( ٢ ، ٠ﻱ ا ﻨﺤ ﺫ
ﻓ
ﻘﻖ ﻣﻌﺎد ﻪ
ﺇ = ٢ﺟﺎ ) + ٠ + ( ٠ × ٢ث /ﺉ ث٢ = / :ﺹ = 1ﺟﺎ ٢ﺱ +ﺱ ٢ +
ﺇ ﻣﻌﺎدﻟﺔ ا ﻨﺤ
٦
)ﺏ( ا ﻘﺪار = + ٣ ) 61ﺟﺎ ﺱ ( +ث )) (٧ﺍ( ا ﻘﺪار = þ
1 ) + ١ﺟﺎ ﺱ ( ﺫ
= ﺫ ) + ١ﺟﺎ ﺱ ( 3
) ﺟﺘﺎ ﺱ ( ﺱ 3 ﺫ
+ث
)ﺏ( ا ﻘﺪار = – þﻗﺘﺎ ٢ﺱ )– ﻗﺘﺎ ﺱ ﻇﺘﺎ ﺱ ( ﺱ = – 1ﻗﺘﺎ ٣ﺱ +ث 3
3
ﺇ = §Ùﺟﺘﺎ ٢ﺱ ١ +
3
)) (٦ﺍ( ا ﻘﺪار = 61ﺟﺎ ٦ﺱ +ث
٣
þ ،ﻇﺘﺎ ٣ﺱ .ﺱ = þﻇﺘﺎ ٢ﺱ ﻇﺘﺎ ﺱ .ﺱ = ) þﻗﺘﺎ ٢ﺱ – ( ١ﻇﺘﺎ ﺱ .ﺱ
ﺫ
ﺫ
ﺫ
ﺣﻠﻮل ﺗﻤﺎر ﻦ )(١٢
)) (٨ﺍ( ا ﻘﺪار = þﻇﺘﺎ ٢ﺱ ﻇﺘﺎ ﺱ ﺱ = ) þﻗﺘﺎ ٢ﺱ – ( ١ﻇﺘﺎ ﺱ ﺱ = ) þﻇﺘﺎ ﺱ ﻗﺘﺎ ٢ﺱ – ﻇﺘﺎ ﺱ ( ﺱ = – 1ﻇﺘﺎ ٢ﺱ – ﻮ | ﺟﺎ ﺱ | +ث ﺫ
)ﺏ( ﺑﻔﺮض أن :ﺹ = ﺱ
ﺇ ﺹ=ﺱ
،ﻉ = ﺟﺎ ﺱ ﺱ ﺇ ﻉ = – ﺟﺘﺎ ﺱ
ا ﻘﺪار = – ﺱ ﺟﺘﺎ ﺱ – – þﺟﺘﺎ ﺱ ﺱ = – ﺱ ﺟﺘﺎ ﺱ +ﺟﺎ ﺱ +ث ) (٩ﰈ – ٧ = §Ùﺟﺎ ٢ﺱ ﺇ ﺹ = – ٧ ) þﺟﺎ ٢ﺱ ( ﺱ ¤Ù
ﺇ ﺹ = ٧ﺱ 1 +ﺟﺘﺎ ٢ﺱ +ث ،ﰈ ) ( ٥ ، ٠ﻱ ا ﻨﺤ ﺫ
ﺇ
ﻘﻖ ﻣﻌﺎد ﻪ
9 ﺇ + (٠) ٧ = ٥ 1ﺟﺘﺎ ) + (٠ث ﺇ ث = ﺫ ﺫ 9 1ﺟﺘﺎ ٢ﺱ + :ﺹ =٧ﺱ+ ﺇ ﻣﻌﺎدﻟﺔ ا ﻨﺤ ﺫ ﺫ ) (١٠ﰈ = §Ùﺍ ﻗﺘﺎ ٢ﺱ ﺇ ﺹ = þﺍ ﻗﺘﺎ ٢ﺱ ﺱ ¤Ù
ﺇ ﺹ = – ﺍ ﻇﺘﺎ ﺱ +ث ،ﰈ ا ﻨﺤ ﻳﻤﺮ ﺑﺎ ﻘﻄﺔ ) ( ٥ ، 4p
( ١ ، p3ﻱ ا ﻨﺤ ﻘﻖ ﻣﻌﺎد ﻪ ﺉ – = ٥ﺍ +ث ، (١) ..........ﰈ ) 4
ﺇ
ﻘﻖ ﻣﻌﺎد ﻪ ﺉ = ١ﺍ +ث (٢) ........
ﺇ ،
ﻤﻊ ) (٢) + (١ﺉ ٢ث = ٦ﺇ ث = ، ٣وﻣﻦ ) (٢ﺇ ﺍ = – ٢ :ﺹ = ٢ﻇﺘﺎ ﺱ ٣ +
ﺇ ﻣﻌﺎدﻟﺔ ا ﻨﺤ
ﺣﻠﻮل ﺗﻤﺎر ﻦ )(١٣
)) (١ﺍ( ا ﻘﺪار = ] ٣ﺟﺎ ﺱ +ث )ﺏ( ا ﻘﺪار = – 1ﺟﺘﺎ ) ٣ﺱ – + ( ٥ث 3
٢
ﻩ
٢
)) (٢ﺍ( ا ﻘﺪار = ) þﻗﺎ ﺱ × ﺟﺘﺎ ﺱ ( ﺱ = þﺱ = ﺱ +ث )ﺏ( ا ﻘﺪار = þﺟﺎ ﺱ ﺱ = – ﺟﺘﺎ ﺱ +ث )) (٣ﺍ( ا ﻘﺪار = ) þﺟﺎ ٢ﺱ +ﺟﺘﺎ ٢ﺱ – ٢ﺟﺎ ﺱ ﺟﺘﺎ ﺱ ( ﺱ = – ١ ) þﺟﺎ ٢ﺱ ( ﺱ = ﺱ 1 +ﺟﺘﺎ ٢ﺱ +ث ﺫ
)ﺏ( ا ﻘﺪار = ) þﺟﺘﺎ ﺱ – ٤ﻗﺎ ٢ﺱ ( ﺱ = ﺟﺎ ﺱ – ٤ﻇﺎ ﺱ +ث )) (٤ﺍ( ا ﻘﺪار = ) þﻗﺎ ٢ﺱ – – ١ + ١ﺟﺘﺎ ٢ﺱ ( ﺱ = ) þﻗﺎ ٢ﺱ – ﺟﺘﺎ ٢ﺱ ( ﺱ = ﻇﺎ ﺱ – 1ﺟﺎ ٢ﺱ +ث
ﺫ iﺱ i ¤ gﺱ i ¤ gﺱ 1+ iﺱ 1+ )ﺏ( ﺑﺎ ب × × = ﺇ iﺱ i 1 -ﺱ i 1 -ﺱ 1+ iﺱ 1+ ﺫ ﺫ iﺱ iﺫﺱ i ¤ g ¤ i + ¤ gﺱ ¤ g ¤ i + ¤ g = iﺱ + = = gﺱ gﺱ gﺫﺱ iﺫﺱ 1- ﺫ iﺱ 1 = ﻻﺣﻆ أن : ﺇ ا ﻘﺪار = i ) þﺱ +ﻗﺘﺎ ﺱ ( ﺱ gﺱ fﺱ gﺱ fﺱ 1 × = ﻮ | ﻇﺎ ﺱ | – ﻮ | ﻗﺘﺎ ﺱ +ﻇﺘﺎ ﺱ | +ث = = ﻗﺘﺎ ﺱ ﻩ ﻩ eﺱ eﺱ
)) (١ﺍ( ا ﻘﺪار = ٣ ) 3 þﺱ – – ٢ﺱ ( ٢ﺱ = é 3¤ 1 -1- ¤ 3 - ù 6 û 1 1ë 7 – = ( 19 –) – 11 =– 3 6 ﺫ
3
)ﺏ( ا ﻘﺪار = þ 1
4
) (1+ ¤S )(1- ¤S ) (1- ¤S
4
ﺱ = ] ) þﺱ ( ١ +ﺱ 1
3 8 éﺫ 3 5ﺫ ù = úﺫ ¤ﺫ = – = ê¤ + 3 3 3 3û 1ë 4
1 4 )ﺍ( ا ﻘﺪار = - ùﺫ – = é ¬ p fﺫ )– = ( ١ – ١ p p p û 0ë
)(٢
)ﺏ( ا ﻘﺪار =
p g 1ùﺫ ﻉ = 4 é ûﺫ 0ë
1 =٠– ١ ×1 ﺫ ﺫ
-üﺱ 3 > ¤ @ 3 + )) (٣ﺍ( د )ﺱ( = | ﺱ – ý = | ٣ þﺱ 3£¤ @ 3 - 5
5
ﺇ þد )ﺱ( ﺱ = þد )ﺱ( ﺱ þ +د )ﺱ( ﺱ 0
3
3
0
5 3 13 = ¤ 1 - ùﺫ 3 +ﺱ ¤ 1ù + éﺫ 3 -ﺱ = ٢ + 9 = é ﺫ 3ëﺫ û 0ëﺫ ûﺫ
)ﺏ( ا ﻘﺪار = ùûﺱ ﺫ ٧ – ٩ ) = 0éë ¤Ú 7 -ﻩ ٧ – ١٦ = ( ٧ –) – ( ٢ﻩ 3
٣٤
٢
ﻟﻸﺳﺘﺎﺫ ﺍﻟﻘﺪﻳﺮ /ﻋﻠﻰ ﺍﻟﺪﻳﻦ ﻳﺤﻴﻰ ٠١١١٩٦٦٠٦١٦
ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ -ﺍﻟﺼﻒ ﺍﻟﺜﺎﻟﺚ ﺍﻟﺜﺎﻧﻮﻯ ¤Ù 3 )) (٤ﺍ( þ p fﺫ¤ 6 6 3Sﺫ 3S = = ]– ٣ 3 3 ¤Ù 1 1 )ﺏ( þ = ûùئﻩ ëé ¤ﻩ = ﻮ ﻩ – ١ﻮ ﻩﻩ = ١ – = ١ – ٠ ﻩ ¤ p
= þp
p p ٢ 3 3 ﻗﺎ ﺱ ﺱ = p ëé¤ gûù 6
٠
3
)(٥
| þﺱ | ٤ – ٢ﺱ :
4-
üﺱ ﺫ - ³ ¤ @ 4-ﺫ ï ﺇ | ﺱ ¤ - 4 ý = | ٤ – ٢ﺫ @ -ﺫ > ³ ¤ﺫ ﺉ ï ¤ þﺫ < ¤ @ 4 -ﺫ
=
4-
-ﺫ
þ
ﺫ= é 3 ¤ 1 - ¤4ù + é ¤4 - 3 ¤ 1 ù ëê 3 ûú ëê 3 ûú
٠
3
4-
| þﺱ | ٤ – ٢ﺱ 3
4-
-ﺫ
ﺫ
é ¤4 - 3 ¤ 1 ù + ëê 3 ûú
3 ﺫ
8 8 648+ [ ( + ٨ –) – ( – ٨ ) ] + [ ( ١٦ + )–(٨+ =]) 3 3 3 3 ﺫ 3ﺫ71 7 3 8 ] ) = + + = [ ( ٨ – ) – ( ١٢ – ٩ 3 3 3 3 3 1-
4
)(٦
4
þد )ﺱ( ﺱ = þد )ﺱ( ﺱ þ +د )ﺱ( ﺱ
ﺫ
= – ٢٢٥ = ٢٤٠ + ١٥ 1-
4
)(٧
ﺫ
٢
٢
3
1-
1-
4
4
þد )ﺱ( ﺱ = þد )ﺱ( ﺱ þ +د )ﺱ( ﺱ
1
4
4
1
4
6
4
4
þ = ٢٤٠د )ﺱ( ﺱ – )– ( ١٥ﺇ þد )ﺱ( ﺱ = ٢٢٥ = ١٥ – ٢٤٠ 1
1
ﺫ
= ٢ – ) þﺱ ( ٨ + ٢ﺱ ٢ ) þ +ﺱ ( ٨ – ٢ﺱ 3
0
ﺫ
46وﺣﺪة ﺮ ﻌﺔ = - ùﺫ ù + é ¤ 8 + 3 ¤ﺫ = é ¤ 8 - 3¤ ëﺫ 3 3 û 0ë 3 û ﺫ
3
) (٦ﺑﻔﺮض ﺹ = ]: – ٤ﺱ :٢ :ﺇ ﺹ – ٤ = ٢ﺱ ٢ﺇ ﺱ + ٢ﺹ٤ = ٢ ﻣﻌﺎدﻟﺔ داﺋﺮة ﺮ ﺰﻫﺎ ﻧﻘﻄﺔ اﻷﺻﻞ وﻃﻮل ﻧﺼﻒ ﻗﻄﺮﻫﺎ ٢
و
ﺇ ﻡ = ﺴﺎﺣﺔ ر ﻊ ا اﺋﺮة = 41ﺑﺐ ) = ٢(٢ﺑﺐ وﺣﺪة ﺮ ﻌﺔ . ) (٧ﺑﻮﺿﻊ ﺹ = ٠ﺉ ﺱ = ١أ٧ ، 7 ) 1ﺱ –)(١ﺱ –(٧ﺱ ﺴﺎﺣﺔ ا ﻤﺮ ا ﻮاﺣﺪ = – þ1 ﺫ
= – ) 7 þ 1ﺱ ٨ – ٢ﺱ ( ٧ +ﺱ = – ¤4 - 3 ¤ 1 ù 1ﺫ é ¤ 7 + ﺫ 1 ﺫ 3û
ﺮ ﻊ ﺇ ﺗ ﻠﻔﺔ ا ﺪﺧﻞ = ٢٧٠٠٠ = ١٥٠٠ × ١٨ﺟﻨﻴﻪ .
= ٢ ) þﻥ . ( ١٣ +ﻥ = ùûﻥ ﺫ 13 +ﻥ éëﺫ = ) ١٨ = (٣٠) – (٤٨ﻣ 3
ﺫ
) (iiاﻹزاﺣﺔ ﺧﻼل ا ﻮا = ٥ ، ٤ ، ٣ف + ٣ف + ٤ف٥ 3
0
١٤٨ = ١٠٠ + ٤٨ = ١٠٠ﻣ
ﺣﻠﻮل ﺗﻤﺎر ﻦ )(١٤ ﻡ = – ٥ ) 1þﺱ ( ٢ﺱ = é 3¤ 1 - ¤ 5ù 3 û -ﺫ
1
- ëﺫ
ﻡ= þ
4
) (٣ﻡ = þ 0
= ١٢وﺣﺪة ﺮ ﻌﺔ
4 4 4ﺫ ﺱ = ٤ þ1ﺱ ٢ -ﺱ = 4 - ùûﺱ ٣ = 1éë 1-وﺣﺪة ﺮ ﻌﺔ
ﺱ
1
5
1
0
0
ﺫ 5 ]ﺱ /٤ /+/ﺱ = ùúﺫ ) ﺱ 38 = éê 3 (4 +وﺣﺪة ﺮ ﻌﺔ 3 0ë 3û
٢
ﺹ – = ١ﺱ ، ٦ +ﺹ ٢ – = ٢ﺱ ، ٣ +ﺑﻮﺿﻊ ﺹ = ١ﺹ
٢
٢ ﺇ ﺱ – ٢ﺱ – ٠ = ٣ﺉ ﺱ = ٣أ ١ – ،ﺇ ﻡ = ) þﺹ – ١ﺹ ( ٢ﺱ 3
1-
1
) (٢ﺱ = – ٢ﺹ ﺇ = òﺑﺐ þﺱ ٢ﺹ = ﺑﺐ ٤ þﺹ ٢ﺹ 3
3
0
0
= ﺑﺐ ٣٦ = é 3§ 4 ùﺑﺐ وﺣﺪة ﻜﻌﺒﺔ 3û 0ë
) (٣ﺹ = ١ﺱ ، ٢ﺹ ٢ = ٢ﺱ ،ﺑﻮﺿﻊ ﺹ = ١ﺹ ٢ﺇ ﺱ ) ﺱ – ٠ = ( ٢ ﺇ ﺱ = ٠أ ، ٢ ،و ﻫﺬه اﻟﻔ ة ﺪ أن ﺹ ٢ﲨﺲ ﺹ ١ ﺫ
0
) (iiiا ﻌﺪ ﺑﻌﺪ ٥ﺛﻮان ﻣﻦ ﺑﺪء ا ﺮ ﺔ = ٢ ) þﻥ . ( ١٣ +ﻥ + 5
1
ﺫ
5
٣ ) þﻥ . ( ٨ – ٢ﻥ =
1ë
ﺇ = òﺑﺐ ) þﺹ – ٢٢ﺹ ( ٢١ﺱ = ﺑﺐ ٤ ) þﺱ – ٢ﺱ ( ٤ﺱ
= ùû + ١٨ﻥ ﺫ 8 -ﻥ ١٠٠ = ( ٣ – ٨٥ ) + ١٨ = 3 éëﻣ
3 ûùﻥ ﺫ 13 +ﻥ+ 0 ëé
7
) = ò (١ﺑﺐ þﺹ ٢ﺱ = ﺑﺐ ) þﺱ ٢( ١ + ٣ﺱ
3
) (i) (٩اﻹزاﺣﺔ ﺧﻼل ا ﺎﻧﻴﺔ ا ﺎ ﺔ ﻓﻘﻂ = ف٣ 3
٢
= ﺑﺐ ) 1þﺱ ٢ + ٦ﺱ ( ١ + ٣ﺱ = ﺑﺐ é ¤ + 4¤ 1 + 7 ¤ 1 ù ﺫ 7û 0 0ë 3ﺫ = 14ﺑﺐ وﺣﺪة ﻜﻌﺒﺔ
6
1
0
)ﺱ [(١+ﺱ )] þ +ﺱ – ٩)–(١+ﺱ ([ﺱ
1-
= þد )ﺱ( ﺱ – þد )ﺱ( ﺱ ﺉ
)(٤
ﺫ
4
ﺫ
6
)(٨
)(٢
٢ ﺇ ﺱ – ٠ = ٤ﺉ ﺱ = ٢أ ) ٢ – ،ﺮﻓﻮض ( ﺇ ﻡ = – ٩ ) ] þﺱ– ( ٢
þد )ﺱ( ﺱ = þد )ﺱ( ﺱ þ +د )ﺱ( ﺱ ﺫ
)(١
ﺫ3
ﺣﻠﻮل ﺗﻤﺎر ﻦ )(١٥
= þد )ﺱ( ﺱ – þد )ﺱ( ﺱ = – ٢٥٥ – = ٢٤٠ – ١٥
3
= + 3¤ 1 - ùﺱ ﺫ 3 = é ¤ 3 +وﺣﺪة ﺮ ﻌﺔ 3 û 1- ë
= ١٨ﻣ
1-
ﺫ
1-
) (٥ﺹ – ٩ = ١ﺱ ، ٢ﺹ = ٢ﺱ ، ١ + ٢ﺑﻮﺿﻊ ﺹ = ١ﺹ ٢ﺇ ٢ﺱ٠ = ٨ – ٢
– ٤ ) þﺱ ( ٢ﺱ ) þ +ﺱ ( ٤ – ٢ﺱ ﺫ
3
1-
ﺫ
ﺫ
-ﺫ
3
3
+++ --- +++ ٢٢
ﺑﻮﺿﻊ ﺱ ٠ = ٤ – ٢ﺉ ﺱ = – ٢أ ،ﺱ = ٢
) ﺱ ( ٤ – ٢ﺱ +
= ﻇﺎ – ٦٠ﻇﺎ ٣٠
= – ) þﺱ ٢ + ٦ + ٢ﺱ – ( ٣ﺱ = – ) þﺱ ٢ + ٢ﺱ ( ٣ +ﺱ
0
64 é 5 1 3 4 ù 15ﺑﺐ وﺣﺪة ﻜﻌﺒﺔ . = ﺑﺐ = ¤ - ¤ 5 3û 0ë
)(٤
ﺫ
ﺹ =–٤ﺱ
١
٢
ﺇ ﺱ – ٤ = ٢١ﺹ ٢ ،ﺱ + ٢ﺹ = ٤
٢
ﺇ ﺱ 1 – ٢ = ٢ﺹ ،ﺑﻮﺿﻊ ﺱ = ٢١ﺱ ٢٢ﺇ – ٤ﺹ = ) 1 – ٢ﺹ ( ﺫ ﺫ 1 1 ﺇ – ٤ﺹ = ٢ – ٤ﺹ +ﺹ ٢ﺇ ﺹ ) – ١ﺹ ( = ٠ﺉ ﺹ = ٠أ٤ ، 4 4
،و ﻫﺬه اﻟﻔ ة ﺪ أن ﺱ ١ﲨﺲ ﺱ ٢ﺇ = òﺑﺐ ) þﺱ – ٢١ﺱ ( ٢٢ﺹ 4
0
1ﺹ( [ ٢ﺹ = ﺑﺐ – ٤ ) ] þﺹ ( – ) – ٢ ﺫ 4
0
٢
4 4 = ﺑﺐ – ٤ ) þﺹ – ٢ + ٤ﺹ – 41ﺹ ( ٢ﺹ = ﺑﺐ ) þﺹ – 41ﺹ ( ﺹ 0 0
8ﺑﺐ وﺣﺪة ﻜﻌﺒﺔ . = ﺑﺐ § 1ùúﺫ = é 3 § 1 - ﺫ3 0êë 1 ûﺫ 4
٣
) (٥ﺹ = ١ﺱ ،ﺹ ، ١ = ٢ﺹ ٢ﲨﺲ ﺹ ١ ٢
٢
اﻟﻔ ة ] [ ١ ، ٠ ٦
ﺇ = òﺑﺐ ) þﺹ – ٢ﺹ ( ١ﺱ = ﺑﺐ – ١ ) þﺱ ( ﺱ 1
0
٣٥
1
0
ﻟﻸﺳﺘﺎﺫ ﺍﻟﻘﺪﻳﺮ /ﻋﻠﻰ ﺍﻟﺪﻳﻦ ﻳﺤﻴﻰ ٠١١١٩٦٦٠٦١٦
ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ -ﺍﻟﺼﻒ ﺍﻟﺜﺎﻟﺚ ﺍﻟﺜﺎﻧﻮﻯ 1 6ﺑﺐ ﻗ= ٤٢ × ٢ 6ﺑﺐ وﺣﺪة ﻜﻌﺒﺔ ﺉ ﺑﺐ ﻖ = ﺑﺐ ùﺱ = é 7 ¤ 1 - 7 7 0ë 7 û
1وﺣﺪة ﻃﻮل ﻗ= ﺉ ﻖ 7
ﺫ ) (٦ا وران ﺣﻮل ﻮر ا ﺴ ﻨﺎت ﺉ ﺹﺫ = – ١
ﺏ
ﺫ ﺇ ﺹ = ٢ﺏ¤ – ١ ) ٢ﺫ ( و ﻌﺮﻓﺔ ﺣﺪود ا
)) (٧ﺍ( ﺑﻔﺮض ﻃﻮل ﺿﻠﻊ ا ﺜﻠﺚ = ﺱ ،ﻃﻮل ﻴﻂ ا ﺜﻠﺚ = ﻡ ﺇ ﻡ = ٣ﺱ
ﻞ ﻧﻮﺟﺪ اﻻﺣﺪاﺛﻴﺎت ا ﺴ ﻨﻴﺔ ﻘﻂ
ﺍ
ﺗﻘﺎﻃﻊ ا ﻨﺤ ﻣﻊ ﻮر ا ﺴ ﻨﺎت ﺑﻮﺿﻊ ﺹ = ٠ﺇ – ١ ﺍ
٢
= ٠ﺇ ﺍ –ﺱ
٢
-ﺍ
ﺍ ﺍ ﺫ = ﺑﺐ þﺏ¤ – ١ )٢ﺫ ( ﺱ = ٢ﺑﺐ ﺏ– ١ ) þ ٢
ﺍ
-ﺍ
زوﺟﻴﺔ ( ﺇ ò
3ﺍ é ¤ ù٢ = ٢ﺑﺐ ﺏ 3 - ¤ úﺍﺫ ê û 0ë
0
( ﺱ ) ﻷن ا اﻟﺔ
= ٢ﺑﺐ ﺏ ) ٢ﺍ – 1ﺍ ( 3
0
ﻉ = 1ﺱ ﺱ
ﺉ
ص=س
ﺉ
ﻉ =ﺱ
)) (١١ﺍ( د )ﺱ( = ﻮ eﺱ = ﻮ ﻇﺎ ﺱ ﺇ د ) /ﺱ( = iﺫ ﺱ ﻩ gﺱ ﻩ fﺱ
ﺇ د
0 1 / )) (١ﺍ( ﺹ = = ﺱ ﺱ ﺱ 9 3 )(١٠ ...... ، 4ﺉ ﺹ ﺹ )= 6 - = (٤ = -ﺱ10 ﺱ - 4ﺱ 1
،ﺹ - = //ﺫ= 1
ﻙ
ﺱﺫ
،ﺹ) = (٣ﺫ= 3 ﺱ
ﺫ
ﺱ3
،
)) (١٥ﺝ( ﺑﻮﺿﻊ ﺹ = -4S
و
) (٤ﺹ = ١ﺹ ٢ﺇ ﺱ ٢ – ٢ﺱ = ٠ﺉ ﺱ = ٠أ ،ﺱ = ٢ ﺫ
ﺫ ﺫ 3ﺫ3 ( = 64ﺑﺐ وﺣﺪة ﻜﻌﺒﺔ = ﺑﺐ = é 5¤ 1 - 3¤ 4 ùﺑﺐ ) – 15 5 3 5 3û 0ë
× ﺟﺘﺎ ﺱ
)) (١٦ﺝ( ا ﻘﺪار = þ ١٠ﺟﺎ ﺱ ﺱ = ٢٠ = ( ١ + ١ ) ١٠ = 0éë ¤ f - ùû ١٠ p
p
0
ﻉ = 1ﺱ ﺱ
ﺉ
ص=س
ﺉ
ﻉ =ﺱ
ﺇ ا ﻘﺪار = ] ¤ئﻩ ) = Ú1[ ¤ - ¤ﻩ ﻮ ﻩﻩ – ﻩ ( – ) × ١ﻮ ﻩ ( ١ – ١ = ) ﻩ – ﻩ ( – )– ١ = ( ١
ﻣﻌﺎدﻟﺔ ا ﻨﺤ ﻋﻦ ﺱ = ١ﺇ + ١ = ٠ + ١ﺹ
ﺇ ﺹ = ١ﺇ ﻧﻘﻄﺔ ا ﻤﺎس
) ، ( ١ ، ١ﺑﺎﻻﺷﺘﻘﺎق ﺑﺎﻟ ﺴﺒﺔ إ ﺱ ﺉ
ﺹ . yﻮ ﻩﺱ . 1 +ﻮ ﻩﺹ = ٢ﺱ +ﺹ ، /ﻋﻨﺪ )( ١ ، ١
= ) -ﺟﺘﺎ ﺱ (ﺟﺘﺎ ﺱ ﺟﺎ ﺱ ) ﻮ ﻩ ﺟﺘﺎ ﺱ ( ١ +
ﺹ
ﺉ د × ١ – = (٠) /ﺻﻔﺮ ) ﻮ ﻩ = ( ١ + ١ﺻﻔﺮ
ﺱ
ﺇ ﻣﻴﻞ ا ﻤﺎس = – ٢ﺇ ﻣﻴﻞ اﻟﻌﻤﻮدى = 1ﺉ
) (٦د )ﺱ( = ﺱ – ﻮ ﻩﺱ ﺇ د ) /ﺱ( = = 1 – ١ﺻﻔﺮ ﻋﻨﺪﻣﺎ ﺱ = ١ ﺱ
ﳘﺲ ++++
١
ﻣﻄﻠﻘﺔ (
،د ) – ١ = (١ﻮ ﻩ ) ١ = ٠ – ١ = ١ﻗﻴﻤﺔ ﺻﻐﺮى ﻣﻄﻠﻘﺔ (
ﻣﻌﺎدﻟﺔ ا ﻤﺎس
ﳘﺲ ﺱ - - - -إﺷﺎرة ﺹ
ﺗﺰاﻳﺪ
د ) - 1 = ( 1ﻮ ﻩ + 1 = 1ﻮ ﻩ ﻩ = ١٤ = ١ + 1 ﻩ ﻩ ﻩ ﻩ ﻩ
،د )ﻩ( = ﻩ – ﻮ ﻩ ﻩ = ﻩ – ) ١٧ = ١ﻗﻴﻤﺔ ﻋﻈ
ﺱﺫ
ﺇ ا ﻘﺪار ا ﻄﻠﻮب = ﺴﺎﺣﺔ ر ﻊ داﺋﺮة = 1ﻗﻖ ﺑﺐ = ﺑﺐ 4
) (١٨ﺑﺎ ﻌﻮ ﺾ
ﺇ ﺹ – = /ﺹ ﺟﺎ ﺱ ) ﻮ ﻩ ﺟﺘﺎ ﺱ ( ١ +
د )ﺱ( ﻣ اﻳﺪة اﻟﻔ ة [ ، ١ﳘﺲ ]
ﺇ ﺹ =–٤ﺱ ﺇ ﺱ +ﺹ =٤
ﺇ þﺹ ﻉ = ﺹ ﻉ – þﻉ ﺹ = ﺱ ﻮ ﻩ ﺱ – þﺱ = ﺱ ﻮ ﻩ ﺱ – ﺱ +ث
ﻸﺳﺎس ﻩ ﺇ ﻮ ﻩ ﺹ = ﺟﺘﺎ ﺱ ﻮ ﻩ ﺟﺘﺎ ﺱ
اﻟﻔ ة [ – ﳘﺲ ] ١ ،
٢
٢
/
ﺗﻨﺎﻗﺺ
٢
ﻣﻌﺎدﻟﺔ داﺋﺮة ﺮ ﺰﻫﺎ ﻧﻘﻄﺔ اﻷﺻﻞ وﻃﻮل ﻧﺼﻒ ﻗﻄﺮﻫﺎ = ٢
،
0
eﺱ ﺇ ﺹ – = yﺟﺎ ﺱ × ﻮ ﻩ ﺟﺘﺎ ﺱ – ﺹ fﺱ
٢
)) (١٧ﺝ( ﺑﻔﺮض ﺹ = ﻮ ﻩﺱ
ﺇ = òﺑﺐ ) ] þﺹ ) – ٢( ٢ﺹ [ ٢( ١ﺱ = ﺑﺐ ٤ ) þﺱ – ٢ﺱ ( ٤ﺱ
) () (٥ﺑﺄﺧﺬ ﻮ ر ﺘﻢ اﻟﻄﺮﻓ
(ﺱ =
ﺱ´ ﺫ ﺱ ﻩ
=ﻩ
٢
٢
)) (٣ﺏ( د )ﺱ( = ﺱ ٢ – ٣ﺱ ١ +ﺇ د ) /ﺱ( = ٣ﺱ ٢ – ٢ﺇ د ٢ – = (٠) /
0
4
pg 4
)) (١٤ﺏ( ﰈ ا اﻟﺔ ﻓﺮدﻳﺔ ﺇ ا ﻘﺪار = ﺻﻔﺮ
ﺉ ٢ﻙ = ١٠ﺇ ﻙ = ٥
ﺫ
4
)) (١٣ﺏ( ا ﻘﺪار = ) ) ﺱ + ( 3 +ﺫ (ﺱ = ) + ١ﺫ ) ﺱ (3 + ﺱ3+
1-
ﻙ = ﺑﺐ × Ú ùû 1ﺫ ) p = 1- éë ¤ﻩ – ﻩ ( ﺫ ﺫ p –٢ ٢ك –٢ ١٠ ﺇ ) pﻩ –ﻩ (= )ﻩ –ﻩ ( ﺫ ﺫ
4
= ) ] × ( ٢ﻩ = ٢ﻩ
ﻙ
٢ك
1
٢
) (٢ﰈ = òﺑﺐ ] ) þﻩﺱ [ ٢ﺱ = ﺑﺐ þﻩ ٢ﺱ ﺱ –٢
/
) S ) = ( pﺫ (
ﺫ
=٢
)) (١٢ﺏ( ا ﻘﺪار = د = ( p ) /ﻗﺎ ٢ﺱ × ﻩ ﻇﺎ ﺱ = ﻗﺎ × ( p ) ٢ﻩ
ﺣﻠﻮل اﻻﺧﺘﺒﺎر اﻷول
د )ﺱ( ﻣﺘﻨﺎﻗﺼﺔ
ﺫ
ﺫ
ﺇ þﺹ ﻉ = ﺹ ﻉ – þﻉ ﺹ = ﺱ ﻮ ﻩ ﺱ – þﺱ = ﺱ ﻮ ﻩ ﺱ – ﺱ +ث
ﻣﻦ ا ﺮﺳﻢ ا ﺠﺎور :
ئﻩ 3
= ) 1ﻩ ﺫئﻩ + 3
ﻩ ئﻩ 3
( – ) 1ﻩ + ٠ﻩ( ٠
)) (٩ﺍ( د ) /ﺱ( = × ١ ) – ١ﻮ ﻩﺱ × 1 +ﺱ ( = – ١ﻮ ﻩﺱ – – = ١ﻮ ﻩﺱ ﺱ
،
3
ﺑﺎﺷﺘﻘﺎق اﻟﻄﺮﻓ
) (٨ا ﻘﺪار = Ú 1ùﺫé ¤ Ú + ¤ ë ûﺫ =) ٦=(١+ 1 )– (٣+ 9 ﺫ ﺫ
) (١٠ﺑﻔﺮض ﺹ = ﻮ ﻩﺱ
= 4ﺑﺐ ﺏ ٢ﺍ وﺣﺪة ﻜﻌﺒﺔ
1-
3
،ﻋﻨﺪ ﺱ = ﻩ ﺉ ﻣﻴﻞ ا ﻤﺎس = – ﻮ ﻩﻩ = – ١
= ٠ﺉ ﺱ = – ﺍ أ ،ﺱ = ﺍ ﺇ = òﺑﺐ þﺹ ٢ﺱ ¤ﺫ ﺍﺫ
Ùﻥ
Ùﻥ
ﺇ
¤ﺫ ﺍﺫ ¤ﺫ ﺍﺫ
Ùﻡ = ١ = 1 × ٣ = ¤Ù × ٣ﺳﻢ /ث
ﺳﻠﻮك ﺹ
،ﻣﻌﺎدﻟﺔ اﻟﻌﻤﻮدى
ﺫ
:ﺹ – ) ٢ – = ١ﺱ – ( ١أى ٢ :ﺱ +ﺹ = ٣ :ﺹ – ) 1 = ١ﺱ – ( ١أى :ﺱ – ٢ﺹ = – ١ ﺫ
1 ) (١٩ا ﻘﺪار = ûù 1ﺩ ) ûù ù 1 = éë 3 ( ¤ﺩ ) ûù - ëé 3 (1ﺩ )é ëé 3 (0 ë û 3 0 3 = 8 – =(٨–٠) 1 3 3
) (٢٠ﺑﻔﺮض ﺴﺎﺣﺔ ا ﺴﺘﻄﻴﻞ = ﻡ
ﺇ ﻡ = ﺱ ) – ٣٢ﺱ ٣٢ = ( ٣ﺱ – ﺱ
ﺇ ﻡ ٤ – ٣٢ = /ﺱ ، ٣ﻡ ١٢ – = //ﺱ
٣٦
٤
٢
ﻟﻸﺳﺘﺎﺫ ﺍﻟﻘﺪﻳﺮ /ﻋﻠﻰ ﺍﻟﺪﻳﻦ ﻳﺤﻴﻰ ٠١١١٩٦٦٠٦١٦
ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ -ﺍﻟﺼﻒ ﺍﻟﺜﺎﻟﺚ ﺍﻟﺜﺎﻧﻮﻯ
– 18 ٣ ﺇ د) //ﺱ( = ١٨ﺱ = ﺱ3
،ﻡ ٠ = /ﻋﻨﺪﻣﺎ ٣٢ﺱ – ٤ﺱ٠ = ٣ ﺇ ﺱ ٨ = ٣ﺉ ﺱ = ٢
اﻟﻔ ة [ – ﳘﺲ : ] ٠ ،د) //ﺱ( > ٠ﺉ ا ﻨﺤ
،ﻡ٠ > ٤٨ – = ٤ × ١٢ – = (٢) // ﺇ ا ﺴﺎﺣﺔ ﺗ ﻮن أ
اﻟﻔ ة [ ، ٠ﳘﺲ ] :د) //ﺱ( < ٠ﺉ ا ﻨﺤ
ﻣﺎﻳﻤ ﻦ ﻋﻨﺪﻣﺎ ﺱ = ٢
9
)) (١ﺏ( ﺑﻔﺮض ﻃﻮل ﺿﻠﻊ ا ﺜﻠﺚ = ﺱ ﺳﻢ ،ﺴﺎﺣﺘﻪ =
ﻡ ﺳﻢ٢
ﺇ ٣ﺱ = ١٢ﺇ ﺱ = ٤ﺉ
ﺫ
٣
) (١٥د )ﺱ( = ﻮ ﻩ ] ٦ﺱ + / ١/ /–/ /ﻮ ﻩ ) ٤ﺱ ( ٥ +
ﺹ = yﻗﺎ ٢ﻥ
= 1ﻮﻩ)٦ﺱ – ٣+(١ﻮﻩ)٤ﺱ (٥+
ﺫ §S
Ùﺫﺹ 1 Ùﻥ ﺱ = ٢ﻗﺎ ﻥ ﻇﺎ ﻥ × × ﻥ ﻇﺎ ﻥ ﻗﺎ ٢ = ﺫ Ù iﻥ gﻥ Ùﺱ
ﺫ
1 ﺇ د ) /ﺱ( = × ٣ + ٦ × 1 × 1 4ﺱ 5 + ﺫ 6ﺱ 1- 3 + ¤ 84 ﺫ1 = = + 3 6ﺱ 4 1-ﺱ 6) 5 +ﺱ (5 + ¤4)(1 -
=٢
ﺫ
)(٥
3
) (١٧ﺹ = ﻩ
1
= ¤ 1 - ¤ ùﺫ ¤ 1 ù + éﺫ= é ¤ -
ﺫ û 1- ëﺫ û 1 3 3 1 1 9 1 1 ] ) ٤ = + + + = [ ( ١ – ) – ( ٣ – ) ] + [ ( – ١ –) – ( – ١ ﺫ ﺫ ﺫ ﺫ ﺫ ﺫ ﺫ ﺫ 1ë
)) (٦ﺝ( ﰈ = òﻥ × ا ﻌﺪل ا ﺰﻣ )(٧
3 3 +¤ﺫ ﺑﻮﺿﻊ ﺹ = ١ﺹ ٢ﺇ = ﺱ ﺇ ﺱ ٣=٢+ﺱ ﺉ ﺱ =١ 3 3
ﺇ ٣ﲪﺲ + ٣ﺟﺎ ٣ﺱ ﲪﺲ ٤
= òﺑﺐ
ﲪ ٤ þﺱ ﲪ + ٣ ) þﺟﺎ ٣ﺱ ( ﺱ ﺲ ﺇ ٣ þﺱ ﺲ
= ﺑﺐ
p
0
p
ﲪ ٤ﺑﺐ ﺉ ٣ﺑﺐ ﲪﺲ + ٣ ) þﺟﺎ ٣ﺱ ( ﺱ ﺲ
p
0
)) (٨ﺏ( ﺹ = د ) /ﺱ( ﺱ = ﺹ /ﺱ )) (٩ﺍ( ﺹ = /ﺟﺘﺎ ﺱ ،ﺹ – = //ﺟﺎ ﺱ ،ﺹ – = ///ﺟﺘﺎ ﺱ
1 = 3
ﻥط ،ﺹ) = (٤ﺟﺎ ﺱ ﺉ ﺹ)ﻥ( = ﺟﺎ ) ﺱ + ﺫ 9ﺫط ( = ﺟﺎ ) ﺱ +ط ( = ﺟﺘﺎ ﺱ ﺉ ﺹ) = (٢٩ﺟﺎ ) ﺱ + ﺫ ﺫ
(
ﺫ 5ﺫ ﺇ ﺴﺎﺣﺔ ﺳﻄﺢ ا ﺜﻠﺚ :ﻡ = 1ﻥ ( = 4 + ١٥٠ﻥ – + ١٥ ) 1ﻥ ( ) – ٢٠ ﺫ ﺫ 5ﺫ 1ﻥ، 41ﻥ ٢ﺉ ﻣﻌﺪل ﺗﻐ ا ﺴﺎﺣﺔ = م – = / 4ﺫ 5ﺫ ﻋﻨﺪﻣﺎ ﻥ = ٥دﻗﺎﺋﻖ ﺇ ﻣﻌﺪل ا ﻐ ا ﺴﺎﺣﺔ = – 4ﺫ ٣٧٥ = 5ﺳﻢ / ٢د
) (١١ا اﻟﺔ ﻏ ﻣﺘﺼﻠﺔ ﻋﻨﺪ ﺱ = ٠ د )ﺱ( = ﺱ ٩ +ﺱ
ص
–١
ﺇ د ) /ﺱ( = ٩ – ١ﺱ
–٢
س
þﺹ ٢١ﺱ +ﺑﺐ ) þﺹ – ٢١ﺹ ( ٢٢ﺱ 0
1
ﺫ0
0
) þﺱ +ﺫ ( ﺱ +ﺑﺐ ) þﺱ +ﺫ – ﺱ ( ﺱ 1
3 3 3 ﺫ0 ﺫ - 1ﺫ 0ﺱ ﺫ = ﺑﺐ ( 3 + 3 ) þﺱ +ﺑﺐ 3 ) þﺱ ( 3 +ﺱ ﺫ0 0 1 ﺫ = 31ﺑﺐ ) þﺱ ( ٢ +ﺱ – 3ﺑﺐ ) þﺱ – ( ١ﺱ ﺫ0
0
) (١٠ﻃﻮﻻ ﺿﻠ اﻟﻘﺎﺋﻤﺔ ﺑﻌﺪ ﻥ دﻗﻴﻘﺔ ﻫﻤﺎ ) + ١٥ﻥ ( 1 – ٢٠ ) ،ﻥ (
،ﺹ = //ﻙ ﻩ
+¤ ٢ﺫ = ﺱ +ﺫ ،ﺹ = ٢٢ﺱ ) (١٨ﺹ= ١
ﰈ ﺻﻔﺮ ﲪﺲ ﺟﺎ ﺱ ﲪﺲ ١ﺇ ﺻﻔﺮ ﲪﺲ ﺟﺎ ٣ﺱ ﲪﺲ ١
0
٢ﻙﺱ
ﺉ
ﰈ ﻩﻙ ﺱ ﻵ ٠ﺇ ) ﻙ ) ( ٤ +ﻙ – ٠ = ( ٢ﺉ ﻙ = – ٤أ ،ﻙ = ٢
ﺇ ١٠ = ٧٠ﻥ ﺉ ﻥ = ٧
p
ﻙﺱ
،ﺹ = /ﻙ ﻩ
ﻙﺱ
0ë
ﻙ ٢ﻩﻙ ﺱ ٢ +ﻙ ﻩﻙ ﺱ – ٨ﻩﻙ ﺱ = ٠ﺇ ﻩ ﻙ ﺱ ) ﻙ ٢ + ٢ﻙ – ٠ = ( ٨
3
1
4û
0
)ﺏ( ا ﻘﺪار = – ١ ) þﺱ ( ﺱ ) þ +ﺱ – ( ١ﺱ 1-
×٤
ﺫ )) (١٦ﺏ( ﻡ = þﺱ . ٣ﺱ = ٤ = ٠ – ١٦ × 41 = é4¤ 1ùوﺣﺪة ﺮ ﻌﺔ
)) (٤ﺝ( ﺹ ٦ = //ﻙ ﺱ ٠ = ١٨ +ﻋﻨﺪ ﺱ = – ١ﺉ ﻙ = ٣ 1
ﺫ
)) (١٤ﺍ( ﺗ ﻮن ا اﻟﺔ ﻣﺘﻨﺎﻗﺼﺔ إذا ن ﻣﻴﻞ ا ﻤﺎس= د ) /ﺱ( = ﺳﺎﻟﺐ
Ù ﺇ ﺹ ] ٢ = /ﺹ /ﻗﺎ ٢ﻥ = ٢ﻇﺎ ﻥ ﻗﺎ ٢ﻥ ﺇ ﺹ = ﺹ ٢ = yﻗﺎ ﻥ Ùﺱ ﺱy
ﺉ
4
þد )ﺱ( ﺱ = ١٣٠
= – ﺟﺘﺎ ٥ﺱ +ث
)) (٢ﺝ( ﻷن ا اﻟﺔ ﻏ ﻣﺘﺼﻠﺔ ﻋﻨﺪ ﺱ = ﺻﻔﺮ )) (٣ﺏ( ﺱ = /ﻗﺎ ﻥ ﻇﺎ ﻥ ،
1-
ﺫ
3S Ùﻡ × ٣] = 1 × ٤ﺳﻢ / ٢ث =
Ùﻥ
14
ﺫ
) () (١٣ا ﻘﺪار = ٥ þﺟﺎ ٥ﺱ ﺱ = – 1 × ٥ﺟﺘﺎ ٥ﺱ +ث 5
Ùﻥ
ﺫ
4
þد )ﺱ( ﺱ = þد )ﺱ( ﺱ þ +د )ﺱ( ﺱ
ﺇ þ + ٢٠ = ١٥٠د )ﺱ( ﺱ ﺇ
3 3 Ùﻡ = ٢ × Sﺱ × ، ¤Ùﻋﻨﺪﻣﺎ ا ﺤﻴﻂ = ١٢ ﺇ ﻡ = Sﺱ ٢ﺇ 4
ﺫ
4
) () (١٢ﰈ
ﺣﻠﻮل اﻻﺧﺘﺒﺎر ا ﺎ
4
9
،د ) ، ١٠ = ٩ + ١ = (١د ) ، ٦ = 3 + ٣ = (٣د )٧٥ = 6 + ٦ = (٦ ﺇ اﻟﻘﻴﻤﺔ اﻟﻌﻈ ا ﻄﻠﻘﺔ = د ) ، ١٠ = (١اﻟﻘﻴﻤﺔ ا ﺼﻐﺮى ا ﻄﻠﻘﺔ = د )٦ = (٣
٤٨ = ٤٢ – ٢ × ٣٢وﺣﺪة ﺮ ﻌﺔ
Ùﻥ
ﺪب ﻷﺳﻔﻞ .
اﻟﻔ ة ] : [ ٦ ، ١د ) /ﺱ( = ٠ﻋﻨﺪ ﺱ = ٣
ﺉ ﺴﺎﺣﺔ ا ﺴﺘﻄﻴﻞ ﻋﻨﺪﺋﺬ =
ﺪب ﻷ
1 = 3 ﺫ = 3
3
0 ﺑﺐ ¤ 1 ùﺫ +ﺫ – é ¤ﺫ ﺑﺐ ¤ 1 ùﺫ +ﺫé ¤ 3 ë ﺫ û ûﺫ 0 -ëﺫ [٠– (١–1 ﺑﺐ ] – [ ( ٤ – ٢ ) – ٠ﺫ 3ﺑﺐ ] ) ﺫ ﺑﺐ 1 +ﺑﺐ = ﺑﺐ وﺣﺪة ﻜﻌﺒﺔ . 3
1
) (١٩ﺑﻮﺿﻊ ﺹ = ٠ﻹ ﺎد ﻧﻘﻂ ﺗﻘﺎﻃﻊ ا اﻟﺔ ﻣﻊ ﻮر ا ﺴ ﻨﺎت
ص
17
ﺇ ﺱ ٤ – ٣ﺱ = ٠ﺇ ﺱ ) ﺱ٠ = ( ٤ – ٢
16 15 14 13 12 11
ﺇ ﺱ = ٠أ ،ﺱ = – ٢أ ،ﺱ = ٢
ﻡ = ﻡ + ١ﻡ + ٢ﻡ =
10 9 8 7 6 5
٣
س
ﺫ
0
1-
4
3
3
þﺹﺱ þ |+ﺹﺱ| þ +ﺹﺱ
= - 4¤ 1ùﺫ ¤ﺫ é 4û
ﺫ
0
0
1- ë
+
٢
٣
٠
ﻡ٣
4 3 2 1 2
ﻡ٢ 1
١-
-1 -2
ﻡ١
-1
-2
-3
-4
-3
٢-
ﺱ + + + + + + - - - - - + + + + + +إﺷﺎرة د )ﺱ(
ﺫ | -4¤ 1ùﺫ ¤ﺫ -4¤ 1ù + | éﺫ ¤ﺫ é ë 4 û 4û ﺫ 0ë
3
٣٧
18
ﻟﻸﺳﺘﺎﺫ ﺍﻟﻘﺪﻳﺮ /ﻋﻠﻰ ﺍﻟﺪﻳﻦ ﻳﺤﻴﻰ ٠١١١٩٦٦٠٦١٦
ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ -ﺍﻟﺼﻒ ﺍﻟﺜﺎﻟﺚ ﺍﻟﺜﺎﻧﻮﻯ = ] [ ( ٨ – ٤ ) – ( ١٨ - 81 ) ] + | ٠ – ( ٨ – ٤ ) | + [ ( ٢ – 1 ) – ٠
)(١٤
) (٢٠د)ﺱ( = ﺍ ﺱ + ٣ﺏ ﺱ + ٢ﺝ ،د ) /ﺱ( = ٣ﺍ ﺱ ٢ + ٢ﺏ ﺱ
) ﺑﺎﻟﻘﺴﻤﺔ
4 15 9 = = ( ٤ – 4 ) – ٤ + 47ﺫ وﺣﺪة ﺮ ﻌﺔ .
،د) //ﺱ( = ٦ﺍ ﺱ ٢ +ﺏ
ﰈ ا اﻟﺔ ﻣﺘﺼﻠﺔ ﻷﻧﻬﺎ ﻛﺜ ة ﺣﺪود ،
§Ù ¤Ù §Ù + = ﺑﺎﻻﺷﺘﻘﺎق ﺑﺎﻟ ﺴﺒﺔ ﻟـ ﻥ :ﺹ + ¤Ùﺱ Ùﻥ Ùﻥ Ùﻥ Ùﻥ
4
( §Ù = ¤Ùﺇ ﺹ +ﺱ = ٢ Ùﻥ
ﺑﺎ ﻌﻮ ﺾ
ﺪﺑﺔ ﻷﺳﻔﻞ ﻋﻨﺪ ﺱ > ، ١و ﺪﺑﺔ ﻷ
Ùﻥ
ﻋﻨﺪ ﺱ < ١ﺉ ﻋﻨﺪ ﺱ = ١ﺗﻮﺟﺪ ﻧﻘﻄﺔ اﻧﻘﻼب ﺇ د )٠ = (١
ﺑﺎ ﻌﻮ ﺾ
ﺉ ٦ﺍ ٢ +ﺏ = ٠ﺉ ٣ﺍ +ﺏ = ٠ﺇ ﺏ = – ٣ﺍ (١) ..........
٢
وﻣﻨﻬﺎ ﺱ = ٣أ ،ﺱ = – ١
) : (٢ﺇ ﺹ = – ١أ ،ﺹ = ٣
ﺇ ا ﻮاﻗﻊ
،ﻣﻴﻞ ا ﻤﺎس ﺹ ٩ +ﺱ = ٢٨ﻫﻮ – ٩ﻋﻨﺪ ﺱ = ٣ﺉ د ٩ – = (٣) /
) ( ٣ ، ١ –) ، ( ١ – ، ٣
) ٠٤ = BÙ (١٥ﻥ – ٤٠ Ùﻥ
ﺇ ٢٧ﺍ ٦ +ﺏ = – ٩ﺉ ٩ﺍ ٢ +ﺏ = – (٢) .......... ٣
وﻣﻨﻬﺎ ﺹ = – ٢ﺱ (٢) ........
) : (١ﺱ ) – ٢ﺱ ( = ﺱ – ٢ +ﺱ – ٥ﺇ ٢ﺱ – ﺱ = – ٣
ﺇ ﺱ ٢ – ٢ﺱ – ٠ = ٣
//
،ﰈ ) ( ١ ، ٣ﻱ ﻣﻨﺤ ا اﻟﺔ ﻓ
ﺱ ﺹ = ﺱ +ﺹ – (١) ........................... ٥
٢
٠٢ = òﻥ – ٤٠ﻥ +ث
ﺉ
ﺇ + ( ٣٠ ) ٤٠ – ٢( ٣٠ ) ٠٢ = ٩٨٠ث وﻣﻨﻬﺎ ث = ٢٠٠٠
ﻘﻖ ﻣﻌﺎد ﻪ ﺇ ٢٧ﺍ ٩ +ﺏ +ﺝ = (٣) .. ١
ﺑﺎ ﻌﻮ ﺾ ﻣﻦ ) (٢) (١ﺉ ٩ﺍ ٣ –) ٢ +ﺍ( = – ٣ﺉ ﺍ = – ١ﺉ ﺏ = ٣
ﺇ ٠٢ = òﻥ ٤٠ – ٢ﻥ ٢٠٠٠ +
ﺇ د )ﺱ( = – ﺱ ٣ + ٣ﺱ١ + ٢
ﻹ ﺎد ا ﺰﻣﻦ ا ى ﻳﻔﺮغ ﻓﻴﻪ اﻹﻧﺎء ﻧﻀﻊ ٠ = òﺇ ٠٢ﻥ ٤٠ – ٢ﻥ ٠ = ٢٠٠٠ +
) (٣ﺉ – + ٢٧ + ٢٧ﺝ = ١ﺉ ﺝ = ١
و ﺎ ﻌﻮ ﺾ
ﻹ ﺎد ﺳﻌﺔ اﻹﻧﺎء ﻧﻀﻊ ﻥ = ٠ﺇ ﺑﺎ
ﺣﻠﻮل اﻻﺧﺘﺒﺎر ا ﺎﻟﺚ )) (٢ﺝ( ﻋﻈ
٢
ﺹ = ﺱ – ٦ﺱ ٩ +ﺱ (١) .........................
ﻠﻴﺔ
٢
)(٣ )(٤
)ﺝ( ا ﻘﺪار = ٢ þﺟﺎ ٢ﺱ .ﺱ = -ﺟﺘﺎ ٢ﺱ +ث
ﺇ
٣
- 11 ®Ù = ﻗ ٣ﺉ 43 = ò ، 445- = 1545ﺑﺐ ﻖ Ùﻥ
= 409-ﺴﺎﺣﺔ ﺳﻄﺢ ا ﻜﺮة × 445-ﺇ
ﺹ ٦ = //ﺱ – (٣) ..................................... ١٢ ٢
،ﺹ ٠ = /ﻋﻨﺪﻣﺎ ﺱ – ٤ﺱ ٠ = ٣ +ﺉ ﺱ = ٣أ ،ﺱ = ١
®Ù ٢ pÙ ﻗ × = ٤ﺑﺐ ﻖ Ùﻥ Ùﻥ
،ﺹ ٠ = //ﻋﻨﺪﻣﺎ ﺱ = ٢ ٣
ﺴﺎﺣﺔ ﺳﻄﺢ ا ﻜﺮة = ٥٠ﺳﻢ٢
+++++
) () (٦ا ﻘﺪار = ) þﺟﺘﺎ ﺱ ( ٢ +ﺱ = ﺟﺎ ﺱ ٢ +ﺱ +ث
1
ﺪب ﻷﺳﻔﻞ
)) (١٠ﺏ( = – ٧ – þ 71ﻩ – ٧ﺱ ﺱ = – 71ﻩ – ٧ﺱ +ث = – 71د )ﺱ( +ث )) (١١ﺝ( = òﺑﺐ =ﺑﺐ ¤ 1 ùﺫ +ﺱ é ûﺫ
1
1- ë
þﺹ ٢ﺱ = ﺑﺐ 1
1-
1
1-
ﺫ
û
0
4
0ë
ﺪب ﻷ
٠
٢
ﺩ )ﺱ(
٤
ا اﻟﺔ ﻣ اﻳﺪة
اﻟﻔ ﺗ [ – ﳘﺲ ، ٣ [ ، ] ١ ،ﳘﺲ ]
7
6 5
اﻟﻔ ة [ ] ٣ ، ١
ﻣﻨﺤ ا اﻟﺔ ﺪب ﻷ
ﺫ )) (١٢ﺝ( ﻡ = ٣ ) þﺱ – ٢ﺱ ( ٣ﺱ = ùﺱ ٤ = ١٦ × 41 – ٨ = é4¤ 1 - 3
ا ﺤﺪب
ص
ا اﻟﺔ ﻣﺘﻨﺎﻗﺼﺔ
٢ = [ ( ١ – 1ﺑﺐ )– (١+1 = ﺑﺐ ] ) ﺫ ﺫ
ﻣ اﻳﺪة
ﻠﻴﺔ
)) (٩ﺍ( ﺹ 1 = 1 × 1 = / ئﻩ ﺱ ﺱ ¤ئﻩ ﺱ
) þﺱ (١+ﺱ
+++++
ﻗﻴﻤﺔ ﺻﻐﺮى ﻠﻴﺔ
1
وﺻﻒ ا اﻟﺔ
ﻗﻴﻤﺔ ﻋﻈ
) () (٨ا ﻘﺪار = þ × ٣د )ﺱ( ﺱ ١ þ +ﺱ
إﺷﺎرة د )ﺱ( /
- - - - - - - - - - - -إﺷﺎرة د) //ﺱ(
إﻧﻘﻼب
3
------ ------
+++++
= ] + ٢ × ٣ﺱ[3 1
ﺱ +++++
ﻣﺘﻨﺎﻗﺼﺔ
ﺱ
3
١
٢
ﻣ اﻳﺪة
1 ﺱ ،ﰈ د )ﺱ( < ﻙ ﺇ ﻙ ﺻﻐﺮى ﻠﻴﺔ )) (٧ﺍ( ﺑﻔﺮض د )ﺱ( = ﺱ + ﺉ د ) /ﺱ( = ٠ﺇ 1 – ١ﺫ = ٠ﺉ ﺱ = ١ﺇ ﻙ > ١ + ١أى :ﻙ > ٢
=٨=(١– ٣)+٦
٢
ﺹ ٣ = /ﺱ – ١٢ﺱ (٢) ........................ ٩ +
)ﺍ( د) //ﺱ( = – ٠ > ٢ﻋﻨﺪﻣﺎ ﺱ ﻱ [ – ﳘﺲ ] ٠ ،
)(٥
ب × ٥ﺇ ﻥ ٢٠٠ – ٢ﻥ ٠ = ١٠٠٠٠ +ﺇ ) ﻥ – ٠ = ٢( ١٠٠ﺉ ﻥ = ١٠٠ث
) (١٦ﺹ = ﺱ ) ﺱ – ( ٣
)) (١ﺏ( ﺭ ) /ﺱ( = – د ) /ﺱ( ﺇ د – = (٥) /ﺭ ٧ – = (٥) /
ﺳﻌﺔ اﻹﻧﺎء = ٢٠٠٠ﺳﻢ٣
ﻣﻨﺤ ا اﻟﺔ ﺪب ﻷﺳﻔﻞ
) ( ٤ ، ١ﻧﻘﻄﺔ ﻗﻴﻤﺔ ﻋﻈ
4 3 2
اﻟﻔ ة [ – ﳘﺲ ] ٢ ،
1 س 6
5
4
3
2
1
اﻟﻔ ة [ ، ٢ﳘﺲ ]
-1
-2
-3
-1 -2 -3
ﻠﻴﺔ ( ٠ ، ٣ ) ،ﻧﻘﻄﺔ ﻗﻴﻤﺔ ﺻﻐﺮى ﻠﻴﺔ ،
) ( ٢ ، ٢ﻧﻘﻄﺔ اﻧﻘﻼب .
) (١٣ﰈ ٣ þ1ﻕ )ﺱ( ﺱ = ٩ﺇ þ1 ٣ﻕ )ﺱ( = ٩ﺉ þ1ﻕ )ﺱ( = ٣
) = BÙ (١٧ﻥ ٢ +ﺉ = òﻥ ٢ + ٢ﻥ +ث ،ﻋﻨﺪ ﻥ = ٠ﻳ ﻮن ٠ = ò
ﺇ þ1 ٤ﻕ )ﺱ( ﺱ þ1 ٧ +د )ﺱ( ﺱ = ١٩
ﺉ ث = ٠ﺇ = òﻥ ٢ + ٢ﻥ ،و
3
3
3
،ﰈ ٤ ] þ1ﻕ )ﺱ( ٧ +د )ﺱ( [ ﺱ = ١٩ 3
3
3
ﺉ þ1 ٧ + ٣ × ٤د )ﺱ( ﺱ = ١٩ﺉ þ1 ٧د )ﺱ( ﺱ = ٧ 3
3 7 ﺉ þ1د )ﺱ( ﺱ = 7
3
Ùﻥ
٢
ﺇ =٦ﻥ ٢+ﻥ
= ١ﺇ þ1 ٥د )ﺱ( ﺱ = ٥ = ١ × ٥ 3
٣٨
٢
ﻳﻤﺘﻠﺊ ا ﺰان ﺐ أن ﻳ ﻮن ٦ = ò
ﺉ ﻥ ٤ +ﻥ – ٠ = ١٢وﻣﻨﻬﺎ ﻥ = ٢دﻗﻴﻘﺔ
ﻟﻸﺳﺘﺎﺫ ﺍﻟﻘﺪﻳﺮ /ﻋﻠﻰ ﺍﻟﺪﻳﻦ ﻳﺤﻴﻰ ٠١١١٩٦٦٠٦١٦
ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ -ﺍﻟﺼﻒ ﺍﻟﺜﺎﻟﺚ ﺍﻟﺜﺎﻧﻮﻯ 4 4 + 4ﺱﺫ 4 ﺱ +ﺱ(ﺱ ﺱ+ﺱ ﺇ ﻡ= ) þ ) (١٨ﺹ = ﺱ = 1
= 4ùئﻩ ¤ 1 + ¤ﺫ ٤ = éﻮ ﻩ ٤ – ٨ + ٤ﻮ ﻩ ٤ = 1 - ١ﻮ ﻩ ١٣ = 15 + ٤وﺣﺪة ﺫ ﺫ 4
ﺫ û 0ë 16 4 4 4 ٢ ﺮ ﻌﺔ = ò ،ﺑﺐ ) þﺱ +ﺱ ( ﺱ = ﺑﺐ ) þﺫ + ٨ +ﺱ ( ﺱ 1 1 ﺱ ٢
4 3 é ¤
ﺱ -ﺫ1+ ù = ﺑﺐ êë 3 + ¤ 8 + 1 - ´16úû 1 = ﺑﺐ ] )– ٥٧ = [ ( 1 + ٨ + ١٦ –) – ( 64 + ٣٢ + ٤ﺑﺐ وﺣﺪة ﻜﻌﺒﺔ 3 3
) (١٩ﺹ = ٢ﺱ ، ٢ +ﺹ = ﺱ ) ﺑﺎ ﻌﻮ ﺾ ﻣﻦ )( (١) (٢ ﺇ ﺱ – ٢ﺱ ٠ = ٢ +وﻣﻨﻬﺎ ﺱ = ٢أ١ – ،
ﺇ ﻧﻘﻂ ا ﻘﺎﻃﻊ
( ١ – ، ١ –) ، ( ٢ ، ٢ ) :
ﺑﺎﺷﺘﻘﺎق اﻟﻄﺮﻓ ﺑﺎﻟ ﺴﺒﺔ إ ﺱ ٢ :ﺹ ١ = §Ù ¤Ù 1 §Ù = ﻋﻨﺪ ) : ( ٢ ، ٢ 4 ¤Ù
ﻣﻌﺎدﻟﺔ اﻟﻌﻤﻮدى
:ﺹ – ) ٤ – = ٢ﺱ – ( ٢أى ٤ :ﺱ +ﺹ – ٠ = ١٠
ﻋﻨﺪ )– – = : ( ١ – ، ١ ﻣﻌﺎدﻟﺔ اﻟﻌﻤﻮدى
ﺉ ﻣﻴﻞ اﻟﻌﻤﻮدى = – ٤
! ؛٢
ﺉ ﻣﻴﻞ اﻟﻌﻤﻮدى = ٢
:ﺹ ) ٢ = ١ +ﺱ ( ١ +أى ٢ :ﺱ – ﺹ ٠ = ١ +
) (٢٠ﺹ = ﻮ ﻩ) ٢] - ٢ﺟﺘﺎ ﺱ ( ﺇ
Sﺫ¤ e = §Ù ¤Ùﺫ S -ﺫ¤ f
،ﻋﻨﺪﻣﺎ ﺱ = p4ﺉ ﺹ = ﻮ ﻩ) ٢] – ٢ﺟﺘﺎ = ( ٤٥ﻮ ﻩ = ١ﺻﻔﺮ
Sﺫ45 e = ١= 1 ،ﻣﻴﻞ ا ﻤﺎس = ﺫ 1-
ﺫ S -ﺫ45 f p p ﺇ ﻣﻌﺎدﻟﺔ ا ﻤﺎس :ﺹ – ) × ١ = ٠ﺱ – ( 4أى :ﺱ – ﺹ – = 4ﺻﻔﺮ
٣٩