1. Nilai x yang memenuhi persamaan
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Friday, May 31, 2024
Soal Matematika Kelas XII no 1
Thursday, May 9, 2024
Kisi Kisi PAT Matematika Kelas XI no 150
150. Diketahui fungsi π(π₯) = β2π₯ + 2 , π(π₯) = βπ₯^2 , dan β(π₯) = π₯^2 + 2π₯ + 1 . Tentukan (π β β β π)(β2) !
a. 22
b. 32
c. 18
d. 17
e. 49
Jawaban :
π(π₯) = β2π₯ + 2
π(π₯) = βπ₯^2
β(π₯) = π₯^2 + 2π₯ + 1
(π β β β π)(β2)
= g (h (f (-2 )))
maka perlu dihitung f(-2) terlebih dahulu
π(-2) = β2(-2) + 2
f(-2) = 4 + 2
f (-2) = 6
selanjutnya,
(π β β β π)(β2)
= g (h (f (-2 )))
= g(h (6))
sehingga, perlu dihitung h (6)
β(π₯) = π₯^2 + 2π₯ + 1
h(6) = 6^2 + 2(6) + 1
h(6) = 36 + 12 + 1
h(6) = 49
selanjutnya,
(π β β β π)(β2)
= g (h (f (-2 )))
= g(h (6))
= g(49)
π(π₯) = βπ₯^2
π(49) = β(49)^2
g(49) = 49
Jawaban : E
Kisi Kisi PAT Matematika Kelas XI no 149
149. Diketahui fungsi f(π₯) = (1 β 6π₯) / (1 β 2π₯) , π(π₯) = π₯^2 β 3π₯ + 6 , dan β(π₯) = 2π₯ + 3 . Tentukan (β β π β π)(1)
a. 15
b. 25
c. 35
d. 45
e. 55
Jawaban :
f(π₯) = (1 β 6π₯) / (1 β 2π₯)
π(π₯) = π₯^2 β 3π₯ + 6
β(π₯) = 2π₯ + 3
(β β π β π)(1)
= h (g (f (1)))
sehingga perlu dihitung f(1) terlebih dahulu
f(1) = (1 β 6(1)) / (1 β 2(1))
f(1) = (1 - 6) / (1 - 2)
f(1) = - 5 / - 1
f(1) = 5
(β β π β π)(1)
= h (g (f (1)))
= h (g (5))
sehingga perlu dihitung g(5)
π(5) = (5)^2 β 3(5) + 6
g(5) = 25 - 15 + 6
g(5) = 10 + 6
g(5) = 16
selanjutnya,
(β β π β π)(1)
= h (g (f (1)))
= h (g (5))
= h (16)
β(π₯) = 2π₯ + 3
h(16) = 2(16) + 3
h(16) = 32 + 3
h(16) = 35
Jawaban : C
Kisi Kisi PAT Matematika Kelas XI no 148
148. Tentukan (β β π)(β4) dengan π(π₯) = β(π₯^2 + 9) dan β(π₯) = π₯^2 β 2π₯ + 5 !
a. 15
b. 20
c. 19
d. 12
e. 10
Jawaban :
π(π₯) = β(π₯^2 + 9)
β(π₯) = π₯^2 β 2π₯ + 5
g ( - 4) = β(4^2 + 9)
g (- 4) = β(16 + 9)
g (- 4)= β25
g (- 4) = 5
(β β π)(β4)
= h (g ( - 4)
= h (5)
= (5)^2 - 2 (5) + 5
= 25 - 10 + 5
= 20
(β β π)(β4) = 20
Jawaban : B
Wednesday, May 8, 2024
Kisi Kisi PAT Matematika Kelas XI no 147
147. Diketahui π(π₯) = π₯^2 + 2π₯ β 3 dan g(π₯) = π₯ + 1 , tentukan (π β π)(β4) !
a. -22
b. 3
c. 12
d. 0
e. -13
Jawaban :
π(π₯) = π₯^2 + 2π₯ β 3
g(π₯) = π₯ + 1
g (- 4) = - 4 + 1
g (4) = - 3
(π β π)(β4)
= f (g(-4))
= f (-3)
= (-3)^2 + 2 (-3) - 3
= 9 - 6 - 3
= 9 - 9
= 0
(π β π)(β4) = 0
Jawaban : D
Kisi Kisi PAT Matematika Kelas XI no 146
146. Diketahui π(π₯) = 2π₯ β 5 dan g(π₯) = π₯^2 + 4 , tentukan (π β π)(3) !
a. 10
b. 15
c. 5
d. 1
e. 0
Jawaban :
π(π₯) = 2π₯ β 5
g(π₯) = π₯^2 + 4
f(3) = 2(3) - 5
f(3) = 6 - 5
f(3) = 1
(π β π)(3)
= g(f (3))
= g (1)
= (1)^2 + 4
= 1 + 4
= 5
(π β π)(3) = 5
Jawaban : C
Kisi Kisi PAT Matematika Kelas XI no 145
145. Diketahui π(π₯) = 2π₯ β 5 dan g(π₯) = π₯^2 + 4 , tentukan (π β π)(π₯)!
a. π₯^2 + 3
b. π₯^2 β 3
c. 3π₯^2 + 2
d. 2π₯^2 β 2
e. 2π₯^2 + 3
Jawaban :
π(π₯) = 2π₯ β 5
g(π₯) = π₯^2 + 4
(π β π)(π₯)
= f (g (x))
= f ( π₯^2 + 4 )
= 2 (π₯^2 + 4) - 5
= 2x^2 + 8 - 5
= 2x^2 + 3
(π β π)(π₯) = 2x^2 + 3
Jawaban : E
Tuesday, May 7, 2024
Kisi Kisi PAT Matematika Kelas XI no 144
144. Diketahui β(π₯) = (4π₯ + 2) / (π₯ + 4) , tentukan invers ββ1(π₯) !
a. ββ1(π₯) = (β4π₯ + 2) / (π₯ β 4)
b. ββ1(π₯) = (4π₯ + 2) / (π₯ + 4)
c. ββ1(π₯) = (β4π₯ + 4) / (π₯ β 2)
d.ββ1(π₯) = (β2π₯ + 4) / (π₯ β 4)
e. ββ1(π₯) = (β2π₯ + 2) / (π₯ β 2)
Jawaban :
β(π₯) = (4π₯ + 2) / (π₯ + 4)
h(x) berbentuk : (ax + b) / (cx + d)
untuk mencari fungsi inversnya, terdapat cara cepatnya yaitu :
h β1(x) = (- dx + b) / (cx - a)
dalam soal ini,
a = 4
b = 2
c = 1
d = 4
h β1(x) = (-4x + 2) / (x - 4)
Jawaban : A
Kisi Kisi PAT Matematika Kelas XI no 143
143. Tentukan invers dari π(π₯) = 8π₯ + 4 !
a. πβ1(π₯) = π₯ + 4
b. πβ1(π₯) = (π₯ β 4) / 8
c. πβ1(π₯) = π₯ β 4
d. πβ1(π₯) = (π₯ β 8) / 4
e. πβ1(π₯) = (βπ₯ β 4) / 8
Jawaban :
π(π₯) = 8π₯ + 4
y = 8x + 4
y - 4 = 8x
x = (y - 4) / 8
πβ1(y) = (y - 4) / 8