Post on 21-Jul-2015
Uji Kompetensi II | 1
1. Diket : π₯2 β 3π₯2 + 5π₯ β 9 βΆ (π₯ β 2)
Ditanya : Sisa =....
Jawab :
Pembuat nol
π₯ β 2 = 0
π₯ = 2
π₯2 β 3π₯2 + 5π₯ β 9
Jadi sisanya adalah 3 (B)
2. Diket : 4π₯4 β 12π₯3 + ππ₯2 + 2 βΆ (2π₯ β 1)
Sisanya nol
Ditanya : P=.....
Jawab :
Pembuat nol
2π₯ β 1 = 0
2π₯ = 1
π₯ =1
2
1
2 4 β12 π 0 2
2 β5 1
2 π β
5
2
1
4 π β
5
4
4 β10 π β 5 1
2 π β
5
2 0
2 + (1
4 π β
5
4) = 0
β2 +5
4=
1
4 π
β3
4=
1
4π
β3 = p (B)
Jadi nilai dari p=-3 (B)
3. Diket : π₯3 β 4π₯2 + 5π₯ + π βΆ (π₯ + 1) = π + π
π₯2 + 3π₯ β 2 βΆ (π₯ + 1) = π + π Sisa kedua operasi perhitungan adalah sama.
Ditanya : p=....
Jawab :
Pembuat nol
(π₯ + 1) = 0
π₯ = 1
β1 1 3 β2
β1 β2
1 2 β4
π ππ πππ¦π β 4, π = β4
ππππ π₯3 β 4π₯2 + 5π₯ + π βΆ (π₯ β 1)
π ππ πππ¦π ππ’ππ βπππ’π β 4
β1 1 β4 5 π
β1 5 β10
1 β5 10 π
π β 10 = π
π β 10 = β4
π = 10 β 4
2 1 β3 5 β9
2 β2 6
1 β1 3 3
Uji Kompetensi II | 2
Maka nilai π = π (D)
4. Diket : Limπ₯βπ
4
πΆππ 2π₯
πΆππ π₯βπππ π₯= β―
Jawab :
Limπ₯βπ
4
πΆππ 2π₯
πΆππ π₯ β πππ π₯
= Limπ₯βπ
4
(πΆππ 2 π₯ + πππ2 π₯)
πΆππ π₯ β πππ π₯
= Limπ₯βπ
4
(πΆππ π₯ + πππ π₯)(πΆππ π₯ β πππ π₯)
(πΆππ π₯ β πππ π₯)
= Limπ₯βπ
4
πΆππ π₯ + πππ π₯
= πΆππ π
4+ πππ
π
4
=1
2β2 +
1
2β2
= β2
Jadi nilai dari ππ’π¦πβπ
π
πͺππ ππ
πͺππ πβπΊππ π= βπ (D)
5. Diket : limπ₯β2
π₯3β2π₯2+4π₯β8
2π₯2β4π₯= β―
Jawab :
limπ₯β2
π₯3 β 2π₯2 + 4π₯ β 8
2π₯2 β 4π₯
= limπ₯β2
(π₯2 + 4)(π₯ β 2)
2π₯(π₯ β 2)
= limπ₯β2
(π₯2 + 4)
2π₯
=22 + 4
2.2
=4 + 4
4
=8
4
= 2
Jadi nilai dari π₯π’π¦πβπ
ππβπππ+ππβπ
πππβππ= π
6. Diket : lim
π₯ β β(β(π₯ + 2)(2π₯ β 1) β (π₯β2 + 1)
Ditanya : Nilai dari limit diatas ?
Jawab :
=lim
π₯ β β( β(π₯ + 2)(2π₯ β 1) β π₯β2 + 1) Γ
(β(π₯ + 2)(2π₯ β 1) + π₯β2 + 1)
(β(π₯ + 2)(2π₯ β 1) + π₯β2 + 1)
= lim
π₯ β β
(π₯+2)(2π₯β1)β(π₯β2+1)(π₯β2+1)
(β(π₯+2)(2π₯β1)+π₯β2+1)
=lim
π₯ β β2π₯2+3π₯β2β(2π₯2+2π₯β2+1)
(β(π₯+2)(2π₯β1)+π₯β2+1)
= lim
π₯ β β2π₯2+3π₯β2β2π₯2β2π₯β2β1
(β(π₯+2)(2π₯β1)+π₯β2+1)
=lim
π₯ β β3π₯β3β2π₯β2
(β(2π₯2+3π₯β2)+π₯β2+1)
=lim
π₯ β β3π₯β3β2π₯β2
(β2π₯2+3π₯β2)+π₯β2+1)
Uji Kompetensi II | 3
=lim
π₯ β β3π₯β3β2π₯β2
(βπ₯2(2+3
π₯β
2
π₯2)+π₯β2+1)
=lim
π₯ β β3π₯β3β2π₯β2
(π₯β2+3
π₯β
2
π₯2 +π₯β2+1)
=lim
π₯ β β3π₯β3β2π₯β2
(π₯β2+3
π₯β
2
π₯2 +π₯β2+1)
=lim
π₯ β β
3π₯
π₯β
3
π₯β
2π₯
π₯β2
(π₯
π₯β2+
3
π₯β
2
π₯2 +π₯
π₯β2+
1
π₯)
=lim
π₯ β β
3β 3
π₯ β2β2
(β2+3
π₯β
2
π₯2 +β2+1
π₯)
=lim
π₯ β β
3β 3
β β2β2
(β2+3
ββ
2
β2 +β2+1
β)
=lim
π₯ β β3β 0 β2β2
(β2+0β0 +β2+0)
=lim
π₯ β β3 β2β2
(β2 +β2)
=lim
π₯ β β3 β2β2
2β2
=lim
π₯ β β3 β2β2
2β2 x
2β2
2β2
=lim
π₯ β β6β2 β4.2
4.2
=lim
π₯ β β6β2 β8
8
=lim
π₯ β β3β2 β4
4
=3
4β2 - 1
Jadi nilai dari π₯π’π¦
π β β(β(π + π)(ππ β π) β (πβπ + π) =
π
πβπ β 1 (E)
7. Diket : lim
π₯ β 0cos4π₯β1
cos5π₯βπππ 3π₯
Ditanya : Nilai dari limit diatas ?
Jawab :
limπ₯ β 0
cos4π₯ β 1
cos5π₯ β πππ 3π₯
=lim
π₯ β 0β2 π ππ2 2π₯
1β 2 π ππ2 5
2π₯β(1β 2 π ππ2
3
2π₯)
=lim
π₯ β 0β2 π ππ2 2π₯
1β 2 π ππ2 5
2π₯β1+ 2 π ππ2
3
2π₯)
=lim
π₯ β 0β2 π ππ2 2π₯
β 2 π ππ2 5
2π₯+ 2 π ππ2
3
2π₯
=lim
π₯ β 0β2 π ππ2 2π₯
β 2 π ππ2 5
2π₯+ 2 π ππ2
3
2π₯
=lim
π₯ β 0β2 π ππ2 (2π₯)
β 2 π ππ2 (5
2π₯β
3
2π₯)
=lim
π₯ β 0 2π₯
5
2π₯β
3
2π₯
=lim
π₯ β 0 2.0
5
2.0β
3
2.0
=0
0
=0
Uji Kompetensi II | 4
Jadi nilai untuk π₯π’π¦
π β πππ¨π¬ ππβπ
ππ¨π¬ ππβπππππ= 0 (E)
8. Diket limπ₯β2
(6βπ₯
π₯2β4β
1
π₯β2)
Ditanya : Nilai dari limit diatas ?
Jawab :
limπ₯β2
(6 β π₯
π₯2 β 4β
1
π₯ β 2)
=limπ₯β2
(6βπ₯
π₯2β4β
1
π₯β2)
=limπ₯β2
((6βπ₯)β(π₯+2)
π₯2β4)
=limπ₯β2
(6βπ₯βπ₯β2
π₯2β4)
=limπ₯β2
(β2π₯+4
π₯2β4)
=limπ₯β2
β2(π₯β2)
(π₯β2)(π₯+2)
=limπ₯β2
β2
(π₯+2)
=β2
(2+2)
=β2
4
= β1
2
Jadi nilai untuk π₯π’π¦πβπ
(πβπ
ππβπβ
π
πβπ) = β
π
π (A)
9. Diket : π(π₯) = 3π₯2 β 2ππ₯ + 7
π1(1) = 0
π1(2) = β―
π½ππ€ππ
π(π₯) = 3π₯2 β 2ππ₯ + 7
π1(π₯) = 6π₯ β 2π
π1(1) = 0
π1(1) = 6.1 β 2π = 0
6 β 2π = 0
6 = 2π
3 = π
π1(2) = 6.2 β 2.3
π1(2) = 12 β 6
π1(2) = 6
10. Diket : Jika f(x) = (6x β 3)3 (2x-1)
Ditanya : f 1 (1) ?
Jawab :
π(π₯) = (6π₯ β 3)3(2π₯ β 1)
π(π₯) = 3(6π₯ β 3)2. 6(2π₯ β 1) + (6π₯ β 3) 3(2)
π(π₯) = 3(36π₯2 β 12π₯ + 9). 6(2π₯ β 1) + (6π₯ β 3)3(2)
π(π₯) = 18(36π₯2 β 12π₯ + 9)(2π₯ β 1) + 2(216π₯3 β 324π₯2 + 162π₯ β 27)
π(π₯) = 1296π₯2 β 1944π₯ + 972π₯ β 162 + 432π₯3 β 648π₯2 + 324π₯ β 54
π(π₯) = 1728π₯3 β 2592π₯2 + 1296π₯ + 216
π(1) = 1728(1) 3 β 2592(1) 2 + 1296(1) + 216
π(1) = 1728 β 2592 + 1296 + 216
π(1) = 216
Jadi nilai dari π(π) = πππ (E)
Uji Kompetensi II | 5
11. Diket : π(π₯) =π₯β1
π₯+1
π(π₯) =π₯+1
2π₯+1
π1(π(π₯)) = β―
Jawab :
π(π₯) =π₯β1
π₯+1
π¦ =π₯β1
π₯+1
π₯π¦ + π¦ = π₯ β 1
π₯π¦ β π₯ = βπ¦ β 1
π₯(π¦ β 1) = βπ¦ β 1
π₯ =βπ¦β1
π¦β1
π1(π¦) =π¦+1
1βπ¦
π1(π₯) =π₯+1
1βπ₯
π1(π(π₯)) = π1 (π₯+1
2π₯+1)
π1(π(π₯)) =( π₯+1
2π₯+1)+1
1β( π₯+1
2π₯+1)
π1(π(π₯)) =π₯+1+2π₯+1
2π₯+12π₯+1βπ₯β1
2π₯+1
π1(π(π₯)) =3π₯+2
π₯
Jadi nilai dari ππ (π(π)) =ππ+π
π (D)
12. Diket : π βΆ π β π
π βΆ π β π
π(π₯) = 3π₯ + 1
π(π₯) = 2π₯ + π
Jawab : Soal tidak/kurang tepat
13. Diket : π(π₯) = π₯2 β 3π₯ + 5, (πππ)(π₯) = 4π₯2 β 10π₯ + 9
Ditanya : Nilai dari πβ1(β3) ?
Jawab :
(πππ)(π₯) = 4π₯2 β 10π₯ + 9
π(π(π₯)) = 4π₯2 β 10π₯ + 9
Misal π(π₯) = ππ₯ + π
(π(π₯))2
β 3(π(π₯)) + 5 = 4π₯2 β 10π₯ + 9
(ππ₯ + π) 2 β 3(ππ₯ + π) + 5 = 4π₯2 β 10π₯ + 9
π2π₯2 + 2πππ₯ + π2 β 3ππ₯ β 3π + 5 = 4π₯2 β 10π₯ + 9
π2π₯2 + (2ππ β 3π)π₯ + (π2 β 3π + 5) = 4π₯2 β 10π₯ + 9
π2 = 4
π = Β±2
untuk π = 2 maka 2ππ β 3π = β10
2.2. π β 3.2 = β10
4π β 6 = β10
4π = β4
π =β4
4
π = β1
Maka, ax + b = 2x-1
untuk π = β2 maka 2ππ β 3π = β10
2(β2)π β
3(β2) = β10
β4π + 6 = β10
β4π = β16
π =β16
β4
π = 4
Uji Kompetensi II | 6
Maka, ax + b = -2x+4 π1(π₯) = 2π₯ β 1
π¦ = 2π₯ β 1
2π₯ = π¦ + 1
π₯ =π¦+1
2
π1β1(π₯) =
π₯+1
2
π1β1(β3) =
β3+1
2
π1β1(β3) =
β2
2
π1β1(β3) = β1
π2(π₯) = β2π₯ + 4
π¦ = β2π₯ + 4
2π₯ = 4 β π¦
π₯ =4βπ¦
2
π2β1(π₯) =
4βπ₯
2
π2β1(β3) =
4+3
2
π2β1(β3) =
7
2
π2β1(β3) = 3
1
2
Jadi nilai untuk πβπ(βπ) = βπ (B) atau πβπ(βπ) = ππ
π
14. Diket : π(π₯) βΆ (π₯ β 1) = π ππ π 3, π(π₯) βΆ (π₯ β 2) = π ππ π 4
Ditanya : π(π₯) βΆ (π₯2 β 1), sisanya ?
Jawab :
π(π₯) βΆ (π₯ β 1) = π ππ π 3
π(1) = 3
Misal π(π₯) = ππ₯ + π
π(1) = π + π = 3
π(π₯) βΆ (π₯ β 2) = π ππ π 4
π(2) = 4
Misal π(π₯) = ππ₯ + π
π(2) = 2π + π = 4
π + π = 3
2π + π = 4 aa
βπ = β1
π = 1
π + π = 3
1 + π = 3
π = 2
Maka π(π₯) βΆ (π₯2 β 1) = π₯ + 2
Jadi nilai sisa dari π(π) βΆ (ππ β π) = π + π (C)
15. Diket : Jika V(x) dibagi (x2 - x) dan (x2 + x) sisanya V(x) dibagi (x2-1)
Ditanya : sisa
Jawab : Soal tidak/kurang tepat
16. Diket : (8π₯3 β 9π₯ + 7) βΆ (2π₯ β 1)
Ditanya : Hasil baginya ?
Jawab :
2π₯ β 1 = 0
π₯ =1
2
1
2 8 0 β9 7
4 2 β31
2
Uji Kompetensi II | 7
8 4 β7 31
2
(8π₯2+4π₯β7)
2= 4π₯2 + 2π₯ β 3
1
2
Jadi hasil bagi dari (πππ β ππ + π) βΆ (ππ β π) = πππ + ππ β ππ
π (C)
17. Diket : f(x) = x β 3
g(x) = x2 + 5
(gof)(x) = (gof)(x)
Ditanya : Nilai x ?
Jawab : (πππ)(π₯) = (πππ)(π₯)
π(π(π₯)) = π(π(π₯))
(π₯ β 3)2 + 5 = (π₯2 + 5) β 3
π₯2 β 6π₯ + 9 + 5 = π₯2 + 2
π₯2 β 6π₯ + 14 = π₯2 + 2
β6π₯ = β12
π₯ =β12
β6
π₯ = 2
Jadi nilai dari π = π (B)
18. Diket : π(π₯) =2π₯+1
π₯β3, π₯ β 3
Ditanya : maka nilai πβ1(π₯ + 1) ?
Jawab :
π(π₯) =2π₯+1
π₯β3
π¦ =2π₯+1
π₯β3
π¦(π₯ β 3) = 2π₯ + 1
π₯π¦ β 3π¦ = 2π₯ + 1
π₯π¦ β 2π₯ = 1 + 3π¦
π₯(π¦ β 2) = 1 + 3π¦
π₯ =1+3π¦
π¦β2
πβ1(π₯) =1+3π¦
π₯β2
πβ1(π₯ + 1) =1+3(π₯+1)
(π₯+1)β2
πβ1(π₯ + 1) =1+3π₯+3
π₯β1
πβ1(π₯ + 1) =3π₯+4
π₯β1
Jadi nilai dari πβπ (π + π) =ππ+π
πβπ (D)
19. Diket : f(x) = 2x β 1
g(x) = cos x
Ditanya : Nilai dari (fog)(x) ?
Jawab :
(πππ)(π₯) = π(π(π₯))
(πππ)(π₯) = 2(πΆππ π₯) 2 β 1
(πππ)(π₯) = 2 πΆππ 2 π₯ β 1
Jadi nilai dari (ππΆπ)(π) = π πͺπππ π β π (A)
20. Diket : f(x) = 1+ 2x, g(x) = cos(x + π
6) dan (fog)(x) = 0
Ditanya : Nilai x ? (0<x<2Ο)
Uji Kompetensi II | 8
Jawab :
(πππ)(π₯) = 0
π(π(π₯)) = 0
1 + 2 (πΆππ (π₯ +π
6)) = 0
2 πΆππ (π₯ +π
6) = β1
Uji Kompetensi II | 9
πΆππ (π₯ +π
6) = β
1
2 (kuadran II)
πΆππ (π₯ +π
6) = πΆππ
2π
3
π₯ +π
6=
2π
3
π₯ =2π
3β
π
6
π₯ =4πβπ
6
π₯ =3π
6
π₯ =1π
2
πΆππ (π₯ +π
6) = β
1
2 (kuadran III)
πΆππ (π₯ +π
6) = πΆππ
8π
6
π₯ +π
6=
8π
6
π₯ =8π
6β
π
6
π₯ =7π
6
Jadi nilai dari π =ππ
π dan π =
ππ
π (A)
21. Diket : lim
π₯ β1
2
βπ₯2+2π₯βπ₯
1βπ₯2
Ditanya : Nilai dari limit diatas ?
Jawab :
lim
π₯ β1
2
βπ₯2+2π₯βπ₯
1βπ₯2
=β(1
2)
2+2(1
2)β(1
2)
1β(1
2)
2
=β
1
4+1β(1
2)
1β1
4
=β5
4β(2
4)
3
4
=1
4β5β(2
4)
3
4
=β5β2
3
Jadi nilai dari π₯π’π¦
π βπ
π
βππ+ππβπ
πβππ =
βπβπ
π
22. Diket : lim
π₯ β ββ2π₯2+2π₯β3ββ2π₯2β2π₯β3
2
Ditanya : Nilai dari limit diatas ?
Jawab :
limπ₯ β β
β2π₯2+2π₯β3ββ2π₯2β2π₯β3
2
=lim
π₯ β ββ2π₯2+2π₯β3ββ2π₯2β2π₯β3
2 x
β2π₯2+2π₯β3+β2π₯2β2π₯β3
β2π₯2+2π₯β3+β2π₯2β2π₯β3
=lim
π₯ β β2π₯2+2π₯β3β(2π₯2β2π₯β3)
2(β2π₯2+2π₯β3+β2π₯2β2π₯β3)
=lim
π₯ β β2π₯2+2π₯β3β2π₯2+2π₯+3)
2(β2π₯2+2π₯β3+β2π₯2β2π₯β3)
=lim
π₯ β β4π₯
2(β2π₯2+2π₯β3+β2π₯2β2π₯β3)
=lim
π₯ β β4π₯
2(βπ₯2(2+2
π₯β
3
π₯2)+βπ₯2(2β2
π₯β
3
π₯2))
=lim
π₯ β β4π₯
2π₯β2+2
π₯β
3
π₯2 +2π₯β2β2
π₯β
3
π₯2
=lim
π₯ β β
4π₯
π₯
2π₯
π₯β2+
2
π₯β
3
π₯2 +2π₯
π₯β2β
2
π₯β
3
π₯2
Uji Kompetensi II | 10
=lim
π₯ β β4
2β2+2
π₯β
3
π₯2 + 2β2β2
π₯β
3
π₯2
=2
β2+2
ββ
3
β2 + β2β2
ββ
3
β2
=2
β2 + β2
=2
2β2
=1
β2
=1
2 β2
Jadi nilai dari π₯π’π¦
π β β
βπππ+ππβπββπππβππβπ
π =
π
π βπ (C)
23. Diket : lim
π₯ β β21βcos (π₯+2)
π₯2+4π₯+4
Ditanya : Nilai dari limit diatas ?
Jawab :
limπ₯ β β2
1βcos (π₯+2)
π₯2+4π₯+4
Misal π₯ + 2 = π
π₯ = π β 2
π₯ β β2 , π β 0
limπ₯ β β2
1βcos (π₯+2)
π₯2+4π₯+4
=lim
π₯ β β21βcos (π₯+2)
(π₯+2)(π₯+2)
=lim
π β 01βcos π
π2
=lim
π β 0
1β(1β2 πππ21
2π)
π2
=lim
π β 0
1β1+2 πππ21
2π
π2
=lim
π β 0
2 πππ21
2π
π2
= 2lim
π β 0
πππ1
2π
π
πππ1
2π
π
= 2.1
2.
1
2
=1
2
Jadi nilai dari π₯π’π¦
π β βππβππ¨π¬ (π+π)
ππ+ππ+π =
π
π (C)
24. Diket : π‘π2π₯β π ππ2π₯8π₯
π₯2 sin4π₯
Ditanya :
Jawab : Soal tidak/kurang tepat
25. Diket : lim
π₯ β β(3π₯ β 2 β β9π₯2 β 2π₯ + 5)
Ditanya : Nilai dari limit diatas ?
Jawab : lim
π₯ β β(3π₯ β 2 β β9π₯2 β 2π₯ + 5)
=lim
π₯ β β(3π₯ β 2 β β9π₯2 β 2π₯ + 5) x
3π₯β2+β9π₯2β2π₯+5
3π₯β2+β9π₯2β2π₯+5
=lim
π₯ β β9π₯2β12π₯+4β(9π₯2β2π₯+5)
3π₯β2+β9π₯2β2π₯+5
Uji Kompetensi II | 11
=lim
π₯ β β9π₯2β12π₯+4β9π₯2+2π₯β5
3π₯ β2+β9π₯2β2π₯+5
=lim
π₯ β ββ10π₯β1
3π₯β2+β9π₯2β2π₯+5
=lim
π₯ β ββ10π₯β1
3π₯β2+βπ₯2(9β2
π₯+
5
π₯2)
=lim
π₯ β ββ10π₯β1
3π₯β2+π₯β9β2
π₯+
5
π₯2
=lim
π₯ β β
β10π₯
π₯β
1
π₯
3π₯
π₯β
2
π₯+
π₯
π₯β9β
2
π₯+
5
π₯2
=lim
π₯ β β
β10β1
π₯
3β2
π₯+β9β
2
π₯+
5
π₯2
=β10β
1
β
3β2
β+β9β
2
β+
5
β2
=β10
3+β9
=β10
6
=β5
3
= β12
3
Jadi nilai dari π₯π’π¦
π β β(ππ β π β βπππ β ππ + π) = βπ
π
π
26. Diket : Jika nilai stasioner dari f(x) = x3- px2-px -1 adalah x = p
Ditanya : maka nilai dari p ?
Jawab :
π(π₯) = π₯3 β ππ₯2 β ππ₯ β 1
π1(π₯) = 3π₯2 β 2ππ₯ β π
Syarat stasioner π1(π₯) = 0
3π₯2 β 2ππ₯ β π = 0
π₯ = π maka
3π2 β 2π2 β π = 0
π2 β π = 0
π(π β 1) = 0
π = 0 atau π = 1
Jadi niali p adalah 0 atau 1 (A)
27. Diket : grafik turun dari π¦ = π₯4 β 8π₯2 β 9
Ditanya : nilai x
Jawab :
π¦ = π₯4 β 8π₯2 β 9
π¦1 = 4π₯3 β 16π₯
Grafik turun, syaratnya π¦1 < 0
4π₯3 β 16π₯ < 0
4π₯(π₯2 β 4) < 0
Pembuat nol
4π₯ = 0 (π₯2 β 4) = 0
π₯ = 0 (π₯ β 2)(π₯ + 2) = 0
π₯ = β2
- + - +
-2 0 1 2
Uji Kompetensi II | 12
Misal x=1 β 4.13 β 16.1 =
4 β 16 =negatif (-)
Maka nilai x < -2 atau 0 < x < 2 (D)
28. Diket : PGS π¦ = 2π₯ βπ₯ + 2 di (2,8) memotong sumbu x di (a,0) dan di sumbu y pada (0,b)
Ditanya : a+b=....
Jawab :
π¦ = 2π₯ βπ₯ + 2
π¦1 = 2(βπ₯ + 2) + 2π₯ (1
2(π₯ + 2)β
1
2)
π¦1 = 2βπ₯ + 2 +π₯
βπ₯+2
π¦1 =2(π₯+2)
βπ₯+2+
π₯
βπ₯+2
π¦1 =3π₯ +4
βπ₯+2
π¦1 = π
π =3π₯+4
βπ₯+2
Di titik (2,8), x = 2
π1(2) = π
π =3.2+4
β2+2
π =10
2
π = 5
π¦ β 8 = 5(π₯ β 2)
π¦ β 8 = 5π₯ β 10
π¦ = 5π₯ β 2
Memotong sumbu x di (a,0)
π¦ = 5π₯ β 2
0 = 5π β 2
2 = 5π 2
5= π
Memotong sumbu y di (0,b)
π¦ = 5π₯ β 2
π = 5. π β 2
π = β2
π + π =2
5β 2
π + π = β8
5
π + π = β13
5
Jadi nilai dari π + π = βππ
π (B)
29. Diket : π(π₯) = πΆππ 3 (5 β 4π₯)
Ditanya : π1(π₯)
Jawab :
π(π₯) = πΆππ 3 (5 β 4π₯)
π1(π₯) = 3πΆππ 2 (5 β 4π₯) (β4)(βπππ(5 β 4π₯))
π1(π₯) = 12πΆππ 2 (5 β 4π₯) πππ(5 β 4π₯)
Jadi nilai ππ(π) = πππͺπππ (π β ππ) πΊππ(π β ππ) (B)
Uji Kompetensi II | 13
31. Diket : (π₯4 β 7π₯3 + ππ₯2 + ππ₯ + 7): (π₯2 β 5π₯ + 6) = π»ππ ππ + (π₯ + 1)
Ditanya : a dan p =...
Jawab :
Pembaginya (π₯2 β 5π₯ + 6)
Pembuat nol
(π₯2 β 5π₯ + 6) = 0 (π₯ + 1)(π₯ β 6) = 0
Maka x1 = 1 dan x2=-6
β1 1 β7 π π 7
β1 8 βπ β 8 βπ + π + 8
6 1 β8 π + 8 π β π β 8 βπ + π + 15
6 β12 6π β 24
1 β2 π β 4 π + 5π β 32
Sisa = (π + 5π β 32)(π₯ + 1) + (βπ + π + 15) (π₯ + 1) = (ππ₯ + π + 5ππ₯ + 5π β 32π₯ β 32) + (βπ + π + 15)
(π₯ + 1) = (ππ₯ + π + 5ππ₯ + 5π β 32π₯ β 32 β π + π + 15)
(π₯ + 1) = (π + 5π β 32)π₯ + (6π β 17)
1 = π + 5π β 32 6π β 17 = 1
33 = π + 5π 6π = 18
33 = π + 5.3 π = 3
18 = π
Jadi nilai π = π dan nilai ππ = π
32. Diket : Limπ₯β0βπ₯2+2π₯+3β βπ₯2β2π₯+3
βπ₯+3ββ3βπ₯
Jawab :
Limπ₯β0βπ₯2+2π₯+3β βπ₯2β2π₯+3
βπ₯+3ββ3βπ₯
= Limπ₯β0βπ₯2+2π₯+3β βπ₯2β2π₯+3
βπ₯+3ββ3βπ₯ x
βπ₯+3+β3βπ₯
βπ₯+3+β3βπ₯
= limπ₯β0
(βπ₯2+2π₯+3β βπ₯2β2π₯+3)(βπ₯+3+β3βπ₯)
π₯+3β3+π₯
= limπ₯β0
(βπ₯2+2π₯+3β βπ₯2β2π₯+3)(βπ₯+3+β3βπ₯)
2π₯
= limπ₯β0
β(π₯2+2π₯+3)(π₯+3)+ β(π₯2+2π₯+3)(3βπ₯)β β(π₯2β2π₯+3)(π₯+3)β β(π₯2β2π₯+3)(3βπ₯)
2π₯
= limπ₯β0
(π₯3+3π₯2+2π₯2+6π₯+3π₯+9) +( 3π₯2βπ₯3+6π₯β2π₯2+9β3π₯)β(π₯3+3π₯2β2π₯2β6π₯+3π₯+9)β ( 3π₯2βπ₯3β6π₯+2π₯2+9β3π₯)
2π₯
= limπ₯β0
π₯3+3π₯2+2π₯2+6π₯+3π₯+9 +3π₯2βπ₯3+6π₯β2π₯2+9β3π₯βπ₯3β3π₯2+2π₯2+6π₯β3π₯β9β 3π₯2+π₯3+6π₯β2π₯2β9+3π₯
2π₯
= limπ₯β0
8π₯
2π₯
=4
Jadi nilai dari ππ’π¦πβπβππ+ππ+πβ βππβππ+π
βπ+πββπβπ= π
33. Diket : π(π₯) =π₯2+2π₯β3
π₯2β2π₯+3
Ditanya : Titik stasioner dan jenisnya
Jawab :
π(π₯) =π₯2+2π₯β3
π₯2β2π₯+3
π1(π₯) =(2π₯+2)(π₯2β2π₯+3)β(π₯2+2π₯β3)(2π₯β2)
(π₯2β2π₯+3)2
π1(π₯) =(2π₯+2)(π₯2β2π₯+3)β(π₯2+2π₯β3)(2π₯β2)
(π₯2β2π₯+3)2
Uji Kompetensi II | 14
π1(π₯) =2π₯3β4π₯2+6π₯+2π₯2β4π₯+6β(2π₯3β2π₯2+4π₯2β4π₯β6π₯+6)
(π₯2β2π₯+3)2
π1(π₯) =2π₯3β2π₯2+2π₯+6β(2π₯3+2π₯2β10π₯+6)
(π₯2β2π₯+3)2
π1(π₯) =2π₯3β2π₯2+2π₯+6β2π₯3β2π₯2+10π₯β6
(π₯2β2π₯+3)2
π1(π₯) =β4π₯2+12π₯
(π₯2β2π₯+3)2
Syarat stasionerπ1(π₯) = 0
π1(π₯) =β4π₯2+12π₯
(π₯2β2π₯+3)2
β4π₯2+12π₯
(π₯2β2π₯+3)2= 0
β4π₯2 + 12π₯ = 0 4π₯(βπ₯ + 3) = 0
4π₯ = 0 βπ₯ + 3 = 0
π₯ = 0 π₯ = 3
π₯ π₯ = 0 π₯ = 3
0β 0 0+ 3β 3 3+
π(π₯) =π₯2+2π₯β3
π₯2β2π₯+3 β 0 + + 0 +
grafik
Titik balik minimum (0,-1) Titik belok(3,2)
34. Diket : π(π₯) = 2π₯ + 4
π(π₯) = 2π₯+5
π₯β4
β(π₯) = (πππβ1)(π₯)
Ditanya : ββ1(π₯)
Jawab :
π(π₯) = 2π₯ + 4
π¦ = 2π₯ + 4
2π₯ = π¦ β 4
π₯ =π¦β4
2
π1(π₯) =π₯β4
2
(πππβ1)(π₯) =2(π₯β4
2)+5
π₯β4
2β4
(πππβ1)(π₯) =2π₯β8+10
2π₯β4β8
2
(πππβ1)(π₯) =2π₯+2
2π₯β12
2
(πππβ1)(π₯) =2π₯+2
π₯β12
β(π₯) =2π₯+2
π₯β12
π¦ =2π₯+2
π₯β12
π₯π¦ β 12π¦ = 2π₯ + 2
π₯π¦ β 2π₯ = 12π¦ + 2
π₯(π¦ β 2) = 12π¦ + 2
π₯ =12π¦+2
π¦β2
ββ1(π₯) =12π₯+2
π₯β2
Jadi nilai dari πβπ(π) =πππ+π
πβπ
Uji Kompetensi II | 15
35. Diket : π = π₯2 + π¦2 β 8π₯ β 4π¦ + 11 salah satu akar π₯3 β 2π₯2 β ππ₯ + 6 = 0
Ditanya : a. P dan r
b. nilai a
Jawab :
a. Pusatnya pada (4,2)
π = β42 + 22 β 11
π = β20 β 11
π = β9
π = π
b. 3 adalah akar π₯3 β 2π₯2 β ππ₯ + 6 = 0
π₯ = 3 β 33 β 2.32 β π3 + 6 = 0
27 β 18 β π3 + 6 = 0
5 = 3π
π = π