Uji kompetensi dan pembahasan matematika semester 2 kelas XI

Uji Kompetensi II | 1

1. Diket : 𝑥2 − 3𝑥2 + 5𝑥 − 9 ∶ (𝑥 − 2)

Ditanya : Sisa =....

Jawab :

Pembuat nol

𝑥 − 2 = 0

𝑥 = 2

𝑥2 − 3𝑥2 + 5𝑥 − 9

Jadi sisanya adalah 3 (B)

2. Diket : 4𝑥4 − 12𝑥3 + 𝑝𝑥2 + 2 ∶ (2𝑥 − 1)

Sisanya nol

Ditanya : P=.....

Jawab :

Pembuat nol

2𝑥 − 1 = 0

2𝑥 = 1

𝑥 =1

2

1

2 4 −12 𝑝 0 2

2 −5 1

2 𝑝 −

5

2

1

4 𝑝 −

5

4

4 −10 𝑝 − 5 1

2 𝑝 −

5

2 0

2 + (1

4 𝑝 −

5

4) = 0

−2 +5

4=

1

4 𝑝

−3

4=

1

4𝑝

−3 = p (B)

Jadi nilai dari p=-3 (B)

3. Diket : 𝑥3 − 4𝑥2 + 5𝑥 + 𝑝 ∶ (𝑥 + 1) = 𝑝 + 𝑎

𝑥2 + 3𝑥 − 2 ∶ (𝑥 + 1) = 𝑞 + 𝑎 Sisa kedua operasi perhitungan adalah sama.

Ditanya : p=....

Jawab :

Pembuat nol

(𝑥 + 1) = 0

𝑥 = 1

−1 1 3 −2

−1 −2

1 2 −4

𝑠𝑖𝑠𝑎𝑛𝑦𝑎 − 4, 𝑎 = −4

𝑚𝑎𝑘𝑎 𝑥3 − 4𝑥2 + 5𝑥 + 𝑝 ∶ (𝑥 − 1)

𝑠𝑖𝑠𝑎𝑛𝑦𝑎 𝑗𝑢𝑔𝑎 ℎ𝑎𝑟𝑢𝑠 − 4

−1 1 −4 5 𝑝

−1 5 −10

1 −5 10 𝑎

𝑝 − 10 = 𝑎

𝑝 − 10 = −4

𝑝 = 10 − 4

2 1 −3 5 −9

2 −2 6

1 −1 3 3


Maka nilai 𝒑 = 𝟔 (D)

4. Diket : Lim𝑥→𝜋

4

𝐶𝑜𝑠 2𝑥

𝐶𝑜𝑠 𝑥−𝑆𝑖𝑛 𝑥= ⋯

Jawab :

Lim𝑥→𝜋

4

𝐶𝑜𝑠 2𝑥

𝐶𝑜𝑠 𝑥 − 𝑆𝑖𝑛 𝑥

= Lim𝑥→𝜋

4

(𝐶𝑜𝑠2 𝑥 + 𝑆𝑖𝑛2 𝑥)

𝐶𝑜𝑠 𝑥 − 𝑆𝑖𝑛 𝑥

= Lim𝑥→𝜋

4

(𝐶𝑜𝑠 𝑥 + 𝑆𝑖𝑛 𝑥)(𝐶𝑜𝑠 𝑥 − 𝑆𝑖𝑛 𝑥)

(𝐶𝑜𝑠 𝑥 − 𝑆𝑖𝑛 𝑥)

= Lim𝑥→𝜋

4

𝐶𝑜𝑠 𝑥 + 𝑆𝑖𝑛 𝑥

= 𝐶𝑜𝑠 𝜋

4+ 𝑆𝑖𝑛

𝜋

4

=1

2√2 +

1

2√2

= √2

Jadi nilai dari 𝐋𝐢𝐦𝒙→𝝅

𝟒

𝑪𝒐𝒔 𝟐𝒙

𝑪𝒐𝒔 𝒙−𝑺𝒊𝒏 𝒙= √𝟐 (D)

5. Diket : lim𝑥→2

𝑥3−2𝑥2+4𝑥−8

2𝑥2−4𝑥= ⋯

Jawab :

lim𝑥→2

𝑥3 − 2𝑥2 + 4𝑥 − 8

2𝑥2 − 4𝑥

= lim𝑥→2

(𝑥2 + 4)(𝑥 − 2)

2𝑥(𝑥 − 2)

= lim𝑥→2

(𝑥2 + 4)

2𝑥

=22 + 4

2.2

=4 + 4

4

=8

4

= 2

Jadi nilai dari 𝐥𝐢𝐦𝒙→𝟐

𝒙𝟑−𝟐𝒙𝟐+𝟒𝒙−𝟖

𝟐𝒙𝟐−𝟒𝒙= 𝟐

6. Diket : lim

𝑥 → ∞(√(𝑥 + 2)(2𝑥 − 1) − (𝑥√2 + 1)

Ditanya : Nilai dari limit diatas ?

Jawab :

=lim

𝑥 → ∞( √(𝑥 + 2)(2𝑥 − 1) − 𝑥√2 + 1) ×

(√(𝑥 + 2)(2𝑥 − 1) + 𝑥√2 + 1)

(√(𝑥 + 2)(2𝑥 − 1) + 𝑥√2 + 1)

= lim

𝑥 → ∞

(𝑥+2)(2𝑥−1)−(𝑥√2+1)(𝑥√2+1)

(√(𝑥+2)(2𝑥−1)+𝑥√2+1)

=lim

𝑥 → ∞2𝑥2+3𝑥−2−(2𝑥2+2𝑥√2+1)

(√(𝑥+2)(2𝑥−1)+𝑥√2+1)

= lim

𝑥 → ∞2𝑥2+3𝑥−2−2𝑥2−2𝑥√2−1

(√(𝑥+2)(2𝑥−1)+𝑥√2+1)

=lim

𝑥 → ∞3𝑥−3−2𝑥√2

(√(2𝑥2+3𝑥−2)+𝑥√2+1)

=lim

𝑥 → ∞3𝑥−3−2𝑥√2

(√2𝑥2+3𝑥−2)+𝑥√2+1)


=lim

𝑥 → ∞3𝑥−3−2𝑥√2

(√𝑥2(2+3

𝑥−

2

𝑥2)+𝑥√2+1)

=lim

𝑥 → ∞3𝑥−3−2𝑥√2

(𝑥√2+3

𝑥−

2

𝑥2 +𝑥√2+1)

=lim

𝑥 → ∞3𝑥−3−2𝑥√2

(𝑥√2+3

𝑥−

2

𝑥2 +𝑥√2+1)

=lim

𝑥 → ∞

3𝑥

𝑥−

3

𝑥−

2𝑥

𝑥√2

(𝑥

𝑥√2+

3

𝑥−

2

𝑥2 +𝑥

𝑥√2+

1

𝑥)

=lim

𝑥 → ∞

3− 3

𝑥 −2√2

(√2+3

𝑥−

2

𝑥2 +√2+1

𝑥)

=lim

𝑥 → ∞

3− 3

∞ −2√2

(√2+3

∞−

2

∞2 +√2+1

∞)

=lim

𝑥 → ∞3− 0 −2√2

(√2+0−0 +√2+0)

=lim

𝑥 → ∞3 −2√2

(√2 +√2)

=lim

𝑥 → ∞3 −2√2

2√2

=lim

𝑥 → ∞3 −2√2

2√2 x

2√2

2√2

=lim

𝑥 → ∞6√2 −4.2

4.2

=lim

𝑥 → ∞6√2 −8

8

=lim

𝑥 → ∞3√2 −4

4

=3

4√2 - 1

Jadi nilai dari 𝐥𝐢𝐦

𝒙 → ∞(√(𝒙 + 𝟐)(𝟐𝒙 − 𝟏) − (𝒙√𝟐 + 𝟏) =

𝟑

𝟒√𝟐 – 1 (E)

7. Diket : lim

𝑥 → 0cos4𝑥−1

cos5𝑥−𝑐𝑜𝑠3𝑥


Jawab :

lim𝑥 → 0

cos4𝑥 − 1

cos5𝑥 − 𝑐𝑜𝑠3𝑥

=lim

𝑥 → 0−2 𝑠𝑖𝑛2 2𝑥

1− 2 𝑠𝑖𝑛2 5

2𝑥−(1− 2 𝑠𝑖𝑛2

3

2𝑥)

=lim

𝑥 → 0−2 𝑠𝑖𝑛2 2𝑥

1− 2 𝑠𝑖𝑛2 5

2𝑥−1+ 2 𝑠𝑖𝑛2

3

2𝑥)

=lim

𝑥 → 0−2 𝑠𝑖𝑛2 2𝑥

− 2 𝑠𝑖𝑛2 5

2𝑥+ 2 𝑠𝑖𝑛2

3

2𝑥

=lim

𝑥 → 0−2 𝑠𝑖𝑛2 2𝑥

− 2 𝑠𝑖𝑛2 5

2𝑥+ 2 𝑠𝑖𝑛2

3

2𝑥

=lim

𝑥 → 0−2 𝑠𝑖𝑛2 (2𝑥)

− 2 𝑠𝑖𝑛2 (5

2𝑥−

3

2𝑥)

=lim

𝑥 → 0 2𝑥

5

2𝑥−

3

2𝑥

=lim

𝑥 → 0 2.0

5

2.0−

3

2.0

=0

0

=0


Jadi nilai untuk 𝐥𝐢𝐦

𝒙 → 𝟎𝐜𝐨𝐬 𝟒𝒙−𝟏

𝐜𝐨𝐬 𝟓𝒙−𝒄𝒐𝒔𝟑𝒙= 0 (E)

8. Diket lim𝑥→2

(6−𝑥

𝑥2−4−

1

𝑥−2)


Jawab :

lim𝑥→2

(6 − 𝑥

𝑥2 − 4−

1

𝑥 − 2)

=lim𝑥→2

(6−𝑥

𝑥2−4−

1

𝑥−2)

=lim𝑥→2

((6−𝑥)−(𝑥+2)

𝑥2−4)

=lim𝑥→2

(6−𝑥−𝑥−2

𝑥2−4)

=lim𝑥→2

(−2𝑥+4

𝑥2−4)

=lim𝑥→2

−2(𝑥−2)

(𝑥−2)(𝑥+2)

=lim𝑥→2

−2

(𝑥+2)

=−2

(2+2)

=−2

4

= −1

2

Jadi nilai untuk 𝐥𝐢𝐦𝒙→𝟐

(𝟔−𝒙

𝒙𝟐−𝟒−

𝟏

𝒙−𝟐) = −

𝟏

𝟐 (A)

9. Diket : 𝑓(𝑥) = 3𝑥2 − 2𝑎𝑥 + 7

𝑓1(1) = 0

𝑓1(2) = ⋯

𝐽𝑎𝑤𝑎𝑏

𝑓(𝑥) = 3𝑥2 − 2𝑎𝑥 + 7

𝑓1(𝑥) = 6𝑥 − 2𝑎

𝑓1(1) = 0

𝑓1(1) = 6.1 − 2𝑎 = 0

6 − 2𝑎 = 0

6 = 2𝑎

3 = 𝑎

𝑓1(2) = 6.2 − 2.3

𝑓1(2) = 12 − 6

𝑓1(2) = 6

10. Diket : Jika f(x) = (6x – 3)3 (2x-1)

Ditanya : f 1 (1) ?

Jawab :

𝑓(𝑥) = (6𝑥 − 3)3(2𝑥 − 1)

𝑓(𝑥) = 3(6𝑥 − 3)2. 6(2𝑥 − 1) + (6𝑥 − 3) 3(2)

𝑓(𝑥) = 3(36𝑥2 − 12𝑥 + 9). 6(2𝑥 − 1) + (6𝑥 − 3)3(2)

𝑓(𝑥) = 18(36𝑥2 − 12𝑥 + 9)(2𝑥 − 1) + 2(216𝑥3 − 324𝑥2 + 162𝑥 − 27)

𝑓(𝑥) = 1296𝑥2 − 1944𝑥 + 972𝑥 − 162 + 432𝑥3 − 648𝑥2 + 324𝑥 − 54

𝑓(𝑥) = 1728𝑥3 − 2592𝑥2 + 1296𝑥 + 216

𝑓(1) = 1728(1) 3 − 2592(1) 2 + 1296(1) + 216

𝑓(1) = 1728 − 2592 + 1296 + 216

𝑓(1) = 216

Jadi nilai dari 𝒇(𝟏) = 𝟐𝟏𝟔 (E)


11. Diket : 𝑓(𝑥) =𝑥−1

𝑥+1

𝑔(𝑥) =𝑥+1

2𝑥+1

𝑓1(𝑔(𝑥)) = ⋯

Jawab :

𝑓(𝑥) =𝑥−1

𝑥+1

𝑦 =𝑥−1

𝑥+1

𝑥𝑦 + 𝑦 = 𝑥 − 1

𝑥𝑦 − 𝑥 = −𝑦 − 1

𝑥(𝑦 − 1) = −𝑦 − 1

𝑥 =−𝑦−1

𝑦−1

𝑓1(𝑦) =𝑦+1

1−𝑦

𝑓1(𝑥) =𝑥+1

1−𝑥

𝑓1(𝑔(𝑥)) = 𝑓1 (𝑥+1

2𝑥+1)

𝑓1(𝑔(𝑥)) =( 𝑥+1

2𝑥+1)+1

1−( 𝑥+1

2𝑥+1)

𝑓1(𝑔(𝑥)) =𝑥+1+2𝑥+1

2𝑥+12𝑥+1−𝑥−1

2𝑥+1

𝑓1(𝑔(𝑥)) =3𝑥+2

𝑥

Jadi nilai dari 𝒇𝟏 (𝒈(𝒙)) =𝟑𝒙+𝟐

𝒙 (D)

12. Diket : 𝑓 ∶ 𝑅 → 𝑅

𝑔 ∶ 𝑅 → 𝑅

𝑓(𝑥) = 3𝑥 + 1

𝑔(𝑥) = 2𝑥 + 𝑎

Jawab : Soal tidak/kurang tepat

13. Diket : 𝑓(𝑥) = 𝑥2 − 3𝑥 + 5, (𝑓𝑂𝑔)(𝑥) = 4𝑥2 − 10𝑥 + 9

Ditanya : Nilai dari 𝑔−1(−3) ?

Jawab :

(𝑓𝑂𝑔)(𝑥) = 4𝑥2 − 10𝑥 + 9

𝑓(𝑔(𝑥)) = 4𝑥2 − 10𝑥 + 9

Misal 𝑔(𝑥) = 𝑎𝑥 + 𝑏

(𝑔(𝑥))2

− 3(𝑔(𝑥)) + 5 = 4𝑥2 − 10𝑥 + 9

(𝑎𝑥 + 𝑏) 2 − 3(𝑎𝑥 + 𝑏) + 5 = 4𝑥2 − 10𝑥 + 9

𝑎2𝑥2 + 2𝑎𝑏𝑥 + 𝑏2 − 3𝑎𝑥 − 3𝑏 + 5 = 4𝑥2 − 10𝑥 + 9

𝑎2𝑥2 + (2𝑎𝑏 − 3𝑎)𝑥 + (𝑏2 − 3𝑏 + 5) = 4𝑥2 − 10𝑥 + 9

𝑎2 = 4

𝑎 = ±2

untuk 𝑎 = 2 maka 2𝑎𝑏 − 3𝑎 = −10

2.2. 𝑏 − 3.2 = −10

4𝑏 − 6 = −10

4𝑏 = −4

𝑏 =−4

4

𝑏 = −1

Maka, ax + b = 2x-1

untuk 𝑎 = −2 maka 2𝑎𝑏 − 3𝑎 = −10

2(−2)𝑏 −

3(−2) = −10

−4𝑏 + 6 = −10

−4𝑏 = −16

𝑏 =−16

−4

𝑏 = 4


Maka, ax + b = -2x+4 𝑔1(𝑥) = 2𝑥 − 1

𝑦 = 2𝑥 − 1

2𝑥 = 𝑦 + 1

𝑥 =𝑦+1

2

𝑔1−1(𝑥) =

𝑥+1

2

𝑔1−1(−3) =

−3+1

2

𝑔1−1(−3) =

−2

2

𝑔1−1(−3) = −1

𝑔2(𝑥) = −2𝑥 + 4

𝑦 = −2𝑥 + 4

2𝑥 = 4 − 𝑦

𝑥 =4−𝑦

2

𝑔2−1(𝑥) =

4−𝑥

2

𝑔2−1(−3) =

4+3

2

𝑔2−1(−3) =

7

2

𝑔2−1(−3) = 3

1

2

Jadi nilai untuk 𝒈−𝟏(−𝟑) = −𝟏 (B) atau 𝒈−𝟏(−𝟑) = 𝟑𝟏

𝟐

14. Diket : 𝑓(𝑥) ∶ (𝑥 − 1) = 𝑠𝑖𝑠𝑎 3, 𝑓(𝑥) ∶ (𝑥 − 2) = 𝑠𝑖𝑠𝑎 4

Ditanya : 𝑓(𝑥) ∶ (𝑥2 − 1), sisanya ?

Jawab :

𝑓(𝑥) ∶ (𝑥 − 1) = 𝑠𝑖𝑠𝑎 3

𝑓(1) = 3

Misal 𝑓(𝑥) = 𝑎𝑥 + 𝑏

𝑓(1) = 𝑎 + 𝑏 = 3

𝑓(𝑥) ∶ (𝑥 − 2) = 𝑠𝑖𝑠𝑎 4

𝑓(2) = 4

Misal 𝑓(𝑥) = 𝑎𝑥 + 𝑏

𝑓(2) = 2𝑎 + 𝑏 = 4

𝑎 + 𝑏 = 3

2𝑎 + 𝑏 = 4 aa

−𝑎 = −1

𝑎 = 1

𝑎 + 𝑏 = 3

1 + 𝑏 = 3

𝑏 = 2

Maka 𝑓(𝑥) ∶ (𝑥2 − 1) = 𝑥 + 2

Jadi nilai sisa dari 𝒇(𝒙) ∶ (𝒙𝟐 − 𝟏) = 𝒙 + 𝟐 (C)

15. Diket : Jika V(x) dibagi (x2 - x) dan (x2 + x) sisanya V(x) dibagi (x2-1)

Ditanya : sisa


16. Diket : (8𝑥3 − 9𝑥 + 7) ∶ (2𝑥 − 1)

Ditanya : Hasil baginya ?

Jawab :

2𝑥 − 1 = 0

𝑥 =1

2

1

2 8 0 −9 7

4 2 −31

2


8 4 −7 31

2

(8𝑥2+4𝑥−7)

2= 4𝑥2 + 2𝑥 − 3

1

2

Jadi hasil bagi dari (𝟖𝒙𝟑 − 𝟗𝒙 + 𝟕) ∶ (𝟐𝒙 − 𝟏) = 𝟒𝒙𝟐 + 𝟐𝒙 − 𝟑𝟏

𝟐 (C)

17. Diket : f(x) = x – 3

g(x) = x2 + 5

(gof)(x) = (gof)(x)

Ditanya : Nilai x ?

Jawab : (𝑔𝑂𝑓)(𝑥) = (𝑓𝑂𝑔)(𝑥)

𝑔(𝑓(𝑥)) = 𝑓(𝑔(𝑥))

(𝑥 − 3)2 + 5 = (𝑥2 + 5) − 3

𝑥2 − 6𝑥 + 9 + 5 = 𝑥2 + 2

𝑥2 − 6𝑥 + 14 = 𝑥2 + 2

−6𝑥 = −12

𝑥 =−12

−6

𝑥 = 2

Jadi nilai dari 𝒙 = 𝟐 (B)

18. Diket : 𝑓(𝑥) =2𝑥+1

𝑥−3, 𝑥 ≠ 3

Ditanya : maka nilai 𝑓−1(𝑥 + 1) ?

Jawab :

𝑓(𝑥) =2𝑥+1

𝑥−3

𝑦 =2𝑥+1

𝑥−3

𝑦(𝑥 − 3) = 2𝑥 + 1

𝑥𝑦 − 3𝑦 = 2𝑥 + 1

𝑥𝑦 − 2𝑥 = 1 + 3𝑦

𝑥(𝑦 − 2) = 1 + 3𝑦

𝑥 =1+3𝑦

𝑦−2

𝑓−1(𝑥) =1+3𝑦

𝑥−2

𝑓−1(𝑥 + 1) =1+3(𝑥+1)

(𝑥+1)−2

𝑓−1(𝑥 + 1) =1+3𝑥+3

𝑥−1

𝑓−1(𝑥 + 1) =3𝑥+4

𝑥−1

Jadi nilai dari 𝒇−𝟏 (𝒙 + 𝟏) =𝟑𝒙+𝟒

𝒙−𝟏 (D)

19. Diket : f(x) = 2x – 1

g(x) = cos x

Ditanya : Nilai dari (fog)(x) ?

Jawab :

(𝑓𝑂𝑔)(𝑥) = 𝑓(𝑔(𝑥))

(𝑓𝑂𝑔)(𝑥) = 2(𝐶𝑜𝑠 𝑥) 2 − 1

(𝑓𝑂𝑔)(𝑥) = 2 𝐶𝑜𝑠2 𝑥 − 1

Jadi nilai dari (𝒇𝑶𝒈)(𝒙) = 𝟐 𝑪𝒐𝒔𝟐 𝒙 − 𝟏 (A)

20. Diket : f(x) = 1+ 2x, g(x) = cos(x + 𝜋

6) dan (fog)(x) = 0

Ditanya : Nilai x ? (0<x<2π)


Jawab :

(𝑓𝑂𝑔)(𝑥) = 0

𝑓(𝑔(𝑥)) = 0

1 + 2 (𝐶𝑜𝑠 (𝑥 +𝜋

6)) = 0

2 𝐶𝑜𝑠 (𝑥 +𝜋

6) = −1


𝐶𝑜𝑠 (𝑥 +𝜋

6) = −

1

2 (kuadran II)


6) = 𝐶𝑜𝑠

2𝜋

3

𝑥 +𝜋

6=

2𝜋

3

𝑥 =2𝜋

3−

𝜋

6

𝑥 =4𝜋−𝜋

6

𝑥 =3𝜋

6

𝑥 =1𝜋

2


6) = −

1

2 (kuadran III)


6) = 𝐶𝑜𝑠

8𝜋

6

𝑥 +𝜋

6=

8𝜋

6

𝑥 =8𝜋

6−

𝜋

6

𝑥 =7𝜋

6

Jadi nilai dari 𝒙 =𝟏𝝅

𝟐 dan 𝒙 =

𝟕𝝅

𝟔 (A)

21. Diket : lim

𝑥 →1

2

√𝑥2+2𝑥−𝑥

1−𝑥2


Jawab :

lim

𝑥 →1

2

√𝑥2+2𝑥−𝑥

1−𝑥2

=√(1

2)

2+2(1

2)−(1

2)

1−(1

2)

2

=√

1

4+1−(1

2)

1−1

4

=√5

4−(2

4)

3

4

=1

4√5−(2

4)

3

4

=√5−2

3


𝒙 →𝟏

𝟐

√𝒙𝟐+𝟐𝒙−𝒙

𝟏−𝒙𝟐 =

√𝟓−𝟐

𝟑

22. Diket : lim

𝑥 → ∞√2𝑥2+2𝑥−3−√2𝑥2−2𝑥−3

2


Jawab :

lim𝑥 → ∞

√2𝑥2+2𝑥−3−√2𝑥2−2𝑥−3

2

=lim

𝑥 → ∞√2𝑥2+2𝑥−3−√2𝑥2−2𝑥−3

2 x

√2𝑥2+2𝑥−3+√2𝑥2−2𝑥−3

√2𝑥2+2𝑥−3+√2𝑥2−2𝑥−3

=lim

𝑥 → ∞2𝑥2+2𝑥−3−(2𝑥2−2𝑥−3)

2(√2𝑥2+2𝑥−3+√2𝑥2−2𝑥−3)

=lim

𝑥 → ∞2𝑥2+2𝑥−3−2𝑥2+2𝑥+3)

2(√2𝑥2+2𝑥−3+√2𝑥2−2𝑥−3)

=lim

𝑥 → ∞4𝑥

2(√2𝑥2+2𝑥−3+√2𝑥2−2𝑥−3)

=lim

𝑥 → ∞4𝑥

2(√𝑥2(2+2

𝑥−

3

𝑥2)+√𝑥2(2−2

𝑥−

3

𝑥2))

=lim

𝑥 → ∞4𝑥

2𝑥√2+2

𝑥−

3

𝑥2 +2𝑥√2−2

𝑥−

3

𝑥2

=lim

𝑥 → ∞

4𝑥

𝑥

2𝑥

𝑥√2+

2

𝑥−

3

𝑥2 +2𝑥

𝑥√2−

2

𝑥−

3

𝑥2


=lim

𝑥 → ∞4

2√2+2

𝑥−

3

𝑥2 + 2√2−2

𝑥−

3

𝑥2

=2

√2+2

∞−

3

∞2 + √2−2

∞−

3

∞2

=2

√2 + √2

=2

2√2

=1

√2

=1

2 √2


𝒙 → ∞

√𝟐𝒙𝟐+𝟐𝒙−𝟑−√𝟐𝒙𝟐−𝟐𝒙−𝟑

𝟐 =

𝟏

𝟐 √𝟐 (C)

23. Diket : lim

𝑥 → −21−cos (𝑥+2)

𝑥2+4𝑥+4


Jawab :

lim𝑥 → −2

1−cos (𝑥+2)

𝑥2+4𝑥+4

Misal 𝑥 + 2 = 𝑎

𝑥 = 𝑎 − 2

𝑥 → −2 , 𝑎 → 0

lim𝑥 → −2

1−cos (𝑥+2)

𝑥2+4𝑥+4

=lim

𝑥 → −21−cos (𝑥+2)

(𝑥+2)(𝑥+2)

=lim

𝑎 → 01−cos 𝑎

𝑎2

=lim

𝑎 → 0

1−(1−2 𝑆𝑖𝑛21

2𝑎)

𝑎2

=lim

𝑎 → 0

1−1+2 𝑆𝑖𝑛21

2𝑎

𝑎2

=lim

𝑎 → 0

2 𝑆𝑖𝑛21

2𝑎

𝑎2

= 2lim

𝑎 → 0

𝑆𝑖𝑛1

2𝑎

𝑎

𝑆𝑖𝑛1

2𝑎

𝑎

= 2.1

2.

1

2

=1

2


𝒙 → −𝟐𝟏−𝐜𝐨𝐬 (𝒙+𝟐)

𝒙𝟐+𝟒𝒙+𝟒 =

𝟏

𝟐 (C)

24. Diket : 𝑡𝑔2𝑥− 𝑠𝑖𝑛2𝑥8𝑥

𝑥2 sin4𝑥

Ditanya :


25. Diket : lim

𝑥 → ∞(3𝑥 − 2 − √9𝑥2 − 2𝑥 + 5)


Jawab : lim

𝑥 → ∞(3𝑥 − 2 − √9𝑥2 − 2𝑥 + 5)

=lim

𝑥 → ∞(3𝑥 − 2 − √9𝑥2 − 2𝑥 + 5) x

3𝑥−2+√9𝑥2−2𝑥+5

3𝑥−2+√9𝑥2−2𝑥+5

=lim

𝑥 → ∞9𝑥2−12𝑥+4−(9𝑥2−2𝑥+5)

3𝑥−2+√9𝑥2−2𝑥+5


=lim

𝑥 → ∞9𝑥2−12𝑥+4−9𝑥2+2𝑥−5

3𝑥 −2+√9𝑥2−2𝑥+5

=lim

𝑥 → ∞−10𝑥−1

3𝑥−2+√9𝑥2−2𝑥+5

=lim

𝑥 → ∞−10𝑥−1

3𝑥−2+√𝑥2(9−2

𝑥+

5

𝑥2)

=lim

𝑥 → ∞−10𝑥−1

3𝑥−2+𝑥√9−2

𝑥+

5

𝑥2

=lim

𝑥 → ∞

−10𝑥

𝑥−

1

𝑥

3𝑥

𝑥−

2

𝑥+

𝑥

𝑥√9−

2

𝑥+

5

𝑥2

=lim

𝑥 → ∞

−10−1

𝑥

3−2

𝑥+√9−

2

𝑥+

5

𝑥2

=−10−

1

∞

3−2

∞+√9−

2

∞+

5

∞2

=−10

3+√9

=−10

6

=−5

3

= −12

3


𝒙 → ∞(𝟑𝒙 − 𝟐 − √𝟗𝒙𝟐 − 𝟐𝒙 + 𝟓) = −𝟏

𝟐

𝟑

26. Diket : Jika nilai stasioner dari f(x) = x3- px2-px -1 adalah x = p

Ditanya : maka nilai dari p ?

Jawab :

𝑓(𝑥) = 𝑥3 − 𝑝𝑥2 − 𝑝𝑥 − 1

𝑓1(𝑥) = 3𝑥2 − 2𝑝𝑥 − 𝑝

Syarat stasioner 𝑓1(𝑥) = 0

3𝑥2 − 2𝑝𝑥 − 𝑝 = 0

𝑥 = 𝑝 maka

3𝑝2 − 2𝑝2 − 𝑝 = 0

𝑝2 − 𝑝 = 0

𝑝(𝑝 − 1) = 0

𝑝 = 0 atau 𝑝 = 1

Jadi niali p adalah 0 atau 1 (A)

27. Diket : grafik turun dari 𝑦 = 𝑥4 − 8𝑥2 − 9

Ditanya : nilai x

Jawab :

𝑦 = 𝑥4 − 8𝑥2 − 9

𝑦1 = 4𝑥3 − 16𝑥

Grafik turun, syaratnya 𝑦1 < 0

4𝑥3 − 16𝑥 < 0

4𝑥(𝑥2 − 4) < 0

Pembuat nol

4𝑥 = 0 (𝑥2 − 4) = 0

𝑥 = 0 (𝑥 − 2)(𝑥 + 2) = 0

𝑥 = −2

- + - +

-2 0 1 2


Misal x=1 → 4.13 − 16.1 =

4 − 16 =negatif (-)

Maka nilai x < -2 atau 0 < x < 2 (D)

28. Diket : PGS 𝑦 = 2𝑥 √𝑥 + 2 di (2,8) memotong sumbu x di (a,0) dan di sumbu y pada (0,b)

Ditanya : a+b=....

Jawab :

𝑦 = 2𝑥 √𝑥 + 2

𝑦1 = 2(√𝑥 + 2) + 2𝑥 (1

2(𝑥 + 2)−

1

2)

𝑦1 = 2√𝑥 + 2 +𝑥

√𝑥+2

𝑦1 =2(𝑥+2)

√𝑥+2+

𝑥

√𝑥+2

𝑦1 =3𝑥 +4

√𝑥+2

𝑦1 = 𝑚

𝑚 =3𝑥+4

√𝑥+2

Di titik (2,8), x = 2

𝑓1(2) = 𝑚

𝑚 =3.2+4

√2+2

𝑚 =10

2

𝑚 = 5

𝑦 − 8 = 5(𝑥 − 2)

𝑦 − 8 = 5𝑥 − 10

𝑦 = 5𝑥 − 2

Memotong sumbu x di (a,0)

𝑦 = 5𝑥 − 2

0 = 5𝑎 − 2

2 = 5𝑎 2

5= 𝑎

Memotong sumbu y di (0,b)

𝑦 = 5𝑥 − 2

𝑏 = 5. 𝑜 − 2

𝑏 = −2

𝑎 + 𝑏 =2

5− 2

𝑎 + 𝑏 = −8

5

𝑎 + 𝑏 = −13

5

Jadi nilai dari 𝑎 + 𝒃 = −𝟏𝟑

𝟓 (B)

29. Diket : 𝑓(𝑥) = 𝐶𝑜𝑠3 (5 − 4𝑥)

Ditanya : 𝑓1(𝑥)

Jawab :

𝑓(𝑥) = 𝐶𝑜𝑠3 (5 − 4𝑥)

𝑓1(𝑥) = 3𝐶𝑜𝑠2 (5 − 4𝑥) (−4)(−𝑆𝑖𝑛(5 − 4𝑥))

𝑓1(𝑥) = 12𝐶𝑜𝑠2 (5 − 4𝑥) 𝑆𝑖𝑛(5 − 4𝑥)

Jadi nilai 𝒇𝟏(𝒙) = 𝟏𝟐𝑪𝒐𝒔𝟐 (𝟓 − 𝟒𝒙) 𝑺𝒊𝒏(𝟓 − 𝟒𝒙) (B)


31. Diket : (𝑥4 − 7𝑥3 + 𝑎𝑥2 + 𝑝𝑥 + 7): (𝑥2 − 5𝑥 + 6) = 𝐻𝑎𝑠𝑖𝑙 + (𝑥 + 1)

Ditanya : a dan p =...

Jawab :

Pembaginya (𝑥2 − 5𝑥 + 6)

Pembuat nol

(𝑥2 − 5𝑥 + 6) = 0 (𝑥 + 1)(𝑥 − 6) = 0

Maka x1 = 1 dan x2=-6

−1 1 −7 𝑎 𝑝 7

−1 8 −𝑎 − 8 −𝑝 + 𝑎 + 8

6 1 −8 𝑎 + 8 𝑝 − 𝑎 − 8 −𝑝 + 𝑎 + 15

6 −12 6𝑎 − 24

1 −2 𝑎 − 4 𝑝 + 5𝑎 − 32

Sisa = (𝑝 + 5𝑎 − 32)(𝑥 + 1) + (−𝑝 + 𝑎 + 15) (𝑥 + 1) = (𝑝𝑥 + 𝑝 + 5𝑎𝑥 + 5𝑎 − 32𝑥 − 32) + (−𝑝 + 𝑎 + 15)

(𝑥 + 1) = (𝑝𝑥 + 𝑝 + 5𝑎𝑥 + 5𝑎 − 32𝑥 − 32 − 𝑝 + 𝑎 + 15)

(𝑥 + 1) = (𝑝 + 5𝑎 − 32)𝑥 + (6𝑎 − 17)

1 = 𝑝 + 5𝑎 − 32 6𝑎 − 17 = 1

33 = 𝑝 + 5𝑎 6𝑎 = 18

33 = 𝑝 + 5.3 𝑎 = 3

18 = 𝑝

Jadi nilai 𝒂 = 𝟑 dan nilai 𝟏𝟖 = 𝒑

32. Diket : Lim𝑥→0√𝑥2+2𝑥+3− √𝑥2−2𝑥+3

√𝑥+3−√3−𝑥

Jawab :

Lim𝑥→0√𝑥2+2𝑥+3− √𝑥2−2𝑥+3

√𝑥+3−√3−𝑥

= Lim𝑥→0√𝑥2+2𝑥+3− √𝑥2−2𝑥+3

√𝑥+3−√3−𝑥 x

√𝑥+3+√3−𝑥

√𝑥+3+√3−𝑥

= lim𝑥→0

(√𝑥2+2𝑥+3− √𝑥2−2𝑥+3)(√𝑥+3+√3−𝑥)

𝑥+3−3+𝑥

= lim𝑥→0

(√𝑥2+2𝑥+3− √𝑥2−2𝑥+3)(√𝑥+3+√3−𝑥)

2𝑥

= lim𝑥→0

√(𝑥2+2𝑥+3)(𝑥+3)+ √(𝑥2+2𝑥+3)(3−𝑥)− √(𝑥2−2𝑥+3)(𝑥+3)− √(𝑥2−2𝑥+3)(3−𝑥)

2𝑥

= lim𝑥→0

(𝑥3+3𝑥2+2𝑥2+6𝑥+3𝑥+9) +( 3𝑥2−𝑥3+6𝑥−2𝑥2+9−3𝑥)−(𝑥3+3𝑥2−2𝑥2−6𝑥+3𝑥+9)− ( 3𝑥2−𝑥3−6𝑥+2𝑥2+9−3𝑥)

2𝑥

= lim𝑥→0

𝑥3+3𝑥2+2𝑥2+6𝑥+3𝑥+9 +3𝑥2−𝑥3+6𝑥−2𝑥2+9−3𝑥−𝑥3−3𝑥2+2𝑥2+6𝑥−3𝑥−9− 3𝑥2+𝑥3+6𝑥−2𝑥2−9+3𝑥

2𝑥

= lim𝑥→0

8𝑥

2𝑥

=4

Jadi nilai dari 𝐋𝐢𝐦𝒙→𝟎√𝒙𝟐+𝟐𝒙+𝟑− √𝒙𝟐−𝟐𝒙+𝟑

√𝒙+𝟑−√𝟑−𝒙= 𝟒

33. Diket : 𝑓(𝑥) =𝑥2+2𝑥−3

𝑥2−2𝑥+3

Ditanya : Titik stasioner dan jenisnya

Jawab :

𝑓(𝑥) =𝑥2+2𝑥−3

𝑥2−2𝑥+3

𝑓1(𝑥) =(2𝑥+2)(𝑥2−2𝑥+3)−(𝑥2+2𝑥−3)(2𝑥−2)

(𝑥2−2𝑥+3)2

𝑓1(𝑥) =(2𝑥+2)(𝑥2−2𝑥+3)−(𝑥2+2𝑥−3)(2𝑥−2)

(𝑥2−2𝑥+3)2


𝑓1(𝑥) =2𝑥3−4𝑥2+6𝑥+2𝑥2−4𝑥+6−(2𝑥3−2𝑥2+4𝑥2−4𝑥−6𝑥+6)

(𝑥2−2𝑥+3)2

𝑓1(𝑥) =2𝑥3−2𝑥2+2𝑥+6−(2𝑥3+2𝑥2−10𝑥+6)

(𝑥2−2𝑥+3)2

𝑓1(𝑥) =2𝑥3−2𝑥2+2𝑥+6−2𝑥3−2𝑥2+10𝑥−6

(𝑥2−2𝑥+3)2

𝑓1(𝑥) =−4𝑥2+12𝑥

(𝑥2−2𝑥+3)2

Syarat stasioner𝑓1(𝑥) = 0

𝑓1(𝑥) =−4𝑥2+12𝑥

(𝑥2−2𝑥+3)2

−4𝑥2+12𝑥

(𝑥2−2𝑥+3)2= 0

−4𝑥2 + 12𝑥 = 0 4𝑥(−𝑥 + 3) = 0

4𝑥 = 0 −𝑥 + 3 = 0

𝑥 = 0 𝑥 = 3

𝑥 𝑥 = 0 𝑥 = 3

0− 0 0+ 3− 3 3+

𝑓(𝑥) =𝑥2+2𝑥−3

𝑥2−2𝑥+3 − 0 + + 0 +

grafik

Titik balik minimum (0,-1) Titik belok(3,2)

34. Diket : 𝑓(𝑥) = 2𝑥 + 4

𝑔(𝑥) = 2𝑥+5

𝑥−4

ℎ(𝑥) = (𝑔𝑜𝑓−1)(𝑥)

Ditanya : ℎ−1(𝑥)

Jawab :

𝑓(𝑥) = 2𝑥 + 4

𝑦 = 2𝑥 + 4

2𝑥 = 𝑦 − 4

𝑥 =𝑦−4

2

𝑓1(𝑥) =𝑥−4

2

(𝑔𝑜𝑓−1)(𝑥) =2(𝑥−4

2)+5

𝑥−4

2−4

(𝑔𝑜𝑓−1)(𝑥) =2𝑥−8+10

2𝑥−4−8

2

(𝑔𝑜𝑓−1)(𝑥) =2𝑥+2

2𝑥−12

2

(𝑔𝑜𝑓−1)(𝑥) =2𝑥+2

𝑥−12

ℎ(𝑥) =2𝑥+2

𝑥−12

𝑦 =2𝑥+2

𝑥−12

𝑥𝑦 − 12𝑦 = 2𝑥 + 2

𝑥𝑦 − 2𝑥 = 12𝑦 + 2

𝑥(𝑦 − 2) = 12𝑦 + 2

𝑥 =12𝑦+2

𝑦−2

ℎ−1(𝑥) =12𝑥+2

𝑥−2

Jadi nilai dari 𝒉−𝟏(𝒙) =𝟏𝟐𝒙+𝟐

𝒙−𝟐


35. Diket : 𝑟 = 𝑥2 + 𝑦2 − 8𝑥 − 4𝑦 + 11 salah satu akar 𝑥3 − 2𝑥2 − 𝑎𝑥 + 6 = 0

Ditanya : a. P dan r

b. nilai a

Jawab :

a. Pusatnya pada (4,2)

𝑟 = √42 + 22 − 11

𝑟 = √20 − 11

𝑟 = √9

𝒓 = 𝟑

b. 3 adalah akar 𝑥3 − 2𝑥2 − 𝑎𝑥 + 6 = 0

𝑥 = 3 → 33 − 2.32 − 𝑎3 + 6 = 0

27 − 18 − 𝑎3 + 6 = 0

5 = 3𝑎

𝟓 = 𝒂

Education

Uji kompetensi dan pembahasan matematika semester 2 kelas XI