Halo semua, kali ini RumusHitung akan memberikan lagi beberapa soal matematika tentang invers fungsi komposisi dan pembahasannya. Seperti biasa, rumushitung hanya memberikan 15 soal saja. Bagi yang belum belajar materi ini, dipelajari dulu yaa. Bisa cari di laman RumusHitung.com atau klik tulisan yang berwarna merah. Langsung saja kita mulai membahas satu per satu soalnya.
1.) Diketahui f(x) = 8x – 3a, dengan a ≠ 0. Jika f-1(5) = 4, maka nilai √a adalah . . .
A. 3
B. -3
C. 4
D. -5
E 5
Pembahasan :
f(x) = 8x – 3a
y = 8x – 3a
8x = y + 3a
x = (y + 3a)/8
f-1(x) = (x + 3a)/8
Maka,
f-1(x) = (x + 3a)/8
f-1(5) = (5 + 3a)/8
4 = (5 + 3a)/8
32 = 5 + 3a
3a = 27
a = 9
Jadi,
= √a
= √9
= 3 (A)
2.) Jika g(x + 5) = (2x – 1)/(x + 3), maka g-1(-2) = . . .
A. 15/4
B. 5
C. 15/2
D. 15
E. 3
Pembahasan :
g(x + 5) = (2x – 1)/(x + 3)
misal : a = x + 5 → x = a – 5
g(a) = (2(a – 5) – 1)/(a – 5 + 3)
g(a) = (2a – 11)/(a – 2)
Maka,
g(x) = (2x – 11)/(x – 2)
Buat invers :
y = (2x – 11)/(x – 2)
y(x – 2) = 2x – 11
xy – 2y = 2x – 11
xy – 2x = 2y – 11
x(y – 2) = 2y – 11
x = (2y – 11)/(y – 2)
g-1(x) = (2x – 11)/(x – 2)
g-1(-2) = (2(-2) – 11)/(-2 – 2)
g-1(-2) = -15/-4
g-1(-2) = 15/4 (A)
3.) Diketahui f(x) = 2x – 7, invers dari f(4x + 3) = . . .
A. 2x – 5
B. 5 – 2x
C. 5x – 2
D. 2 – 5x
E. 2x + 5
Pembahasan :
f(x) = 2x – 7
y = 2x – 7
2x = y + 7
x = (y + 7)/2
f-1(x) = (x + 7)/2
Maka,
f-1(4x + 3) = (4x + 3 + 7)/2
f-1(4x + 3) = (4x + 10)/2
f-1(4x + 3) = 2x + 5 (E)
4.) Diketahui f(x) = (4x – 2)/(x – 1) dengan f-1(x) merupakan invers dari f(x). Jika f(x) = f-1(x), maka persamaannya adalah . . .
A. x2 – 5x – 2 = 0
B. x2 + 5x + 2 = 0
C. x2 – 5x + 2 = 0
D. 2x2 – 5x + 2 = 0
E. 2x2 + 5x – 2 = 0
Pembahasan :
f(x) = (3x – 2)/(x – 1)
y = (3x – 2)/(x – 1)
y(x – 1) = 4x – 2
xy – y = 4x – 2
xy – 4x = y – 2
x(y – 4) = y – 2
x = (y – 2)/(y – 4)
f-1(x) = (x – 2)/(x – 4)
Maka,
f(x) = f-1(x)
(4x – 2)/(x – 1) = (x – 2)/(x – 4)
(4x – 2)(x – 4) = (x – 2)(x – 1)
4x2 – 18x + 8 = x2 – 3x + 2
4x2 – x2 – 18x + 3x + 8 – 2 = 0
3x2 – 15x + 6 = 0
x2 – 5x + 2 = 0 (C)
5.) Diketahui f(x) = 2x – 6 dan g(x) = 5 – x. Invers dari f-1(g-1(-1)) = . . .
A. 6
B. 8
C. 5
D. 2
E. 3
Pembahasan :
f(x) = 2x – 6
y = 2x – 6
2x = y + 6
x = (y + 6)/2
f-1(x) = (x + 6)/2
g(x) = 5 – x
y = 5 – x
x = 5 – y
g-1(x) = 5 – x
Maka,
f-1(g-1(x)) = 2(5 – x) – 6
f-1(g-1(x)) = 10 – 2x – 6
f-1(g-1(x)) = 4 – 2x
f-1(g-1(-1)) = 4 – 2(-1)
f-1(g-1(-1)) = 4 + 2
f-1(g-1(-1)) = 6 (A)
6.) Invers dari f(x) = x2 – 6 adalah . . .
A. x + 6
B. √(x – 6)
C. x – 6
D. √(x + 6)
E. (x + 6)2
Pembahasan :
f(x) = x2 – 6
y = x2 – 6
x2 = y + 6
x = √(y + 6)
f-1(x) = √(x + 6) (D)
7.) Diketahui fungsi p(x) = 9x – 2 dan q(x) = (2x – 1)/x . Invers dari (q o p)-1(1) = . . .
A. 1/3
B. 1/4
C. 1/5
D. 1/6
E. 1/7
Pembahasan :
p(x) = 9x – 2
y = 9x – 2
9x = y + 2
x = (y + 2)/9
p-1(x) = (x + 2)/9
q(x) = (2x – 1)/x
y = (2x – 1)/x
xy = 2x – 1
xy – 2x = -1
x(y – 2) = -1
x = -1/(y – 2)
q-1(x) = -1/(x – 2)
Maka,
(q o p)-1(x) = p-1(q-1(x))
= [(-1/(x – 2) + 2)]/9
p-1(q-1(1)) = [(-1/(1 – 2) + 2)]/9
= (1 + 2)/9
= 3/9
= 1/3 (A)
8.) Diketahui fungsi f(x) = x2 + 6x + 9. Invers dari f(9) = . . .
A. 4
B. 3
C. 2
D. 1
E. 0
Pembahasan :
f(x) = x2 + 6x + 9
y = x2 + 6x + 9
y = (x + 3)2
√y = x + 3
x = √y – 3
f-1(x) = √x – 3
Maka,
f-1(9) = √9 – 3
f-1(9) = 3 – 3
f-1(9) = 0 (E)
9.) Jika fungsi f(x) = √3x – 6. Maka invers dari f-1(3) = . . .
A. 25
B. 26
C. 27
D. 28
E. 29
Pembahasan :
f(x) = √3x – 6
y = √3x – 6
√3x = y + 6
3x = (y + 6)2
3x = y2 + 12y + 36
x = (y2 + 12y + 36)/3
f-1(x) = (x2 + 12x + 36)/3
Maka,
f-1(3) = ((3)2 + 12(3) + 36)/3
= (9 + 36 + 36)/3
= 3 + 12 + 12
= 27 (C)
10.) Diketahui f(2x – 1) = 1 – 2x dan g(x +1) = x – 4. Invers (f o g)-1(x) = . . .
A. -x + 5
B. x – 5
C. -x – 5
D. x + 5
E. 2x – 5
Pembahasan :
f(2x – 1) = 1 – 2x
misal,
a = 2x – 1
x = (a + 1)/2
f(a) = 1 – 2((a + 1)/2)
f(a) = 1 – a + 1
f(a) = -a
f(x) = -x
f(x) = -x
f-1(x) = -x
g(x + 1) = x – 4
misal,
a = x + 1
x = a – 1
g(a) = a – 1 – 4
g(a) = a – 5
g(x) = x – 5
g(x) = x – 5
y = x – 5
x = y + 5
g-1(x) = x + 5
Maka,
(f o g)-1(x) = (g-1 o f-1)(x)
(f o g)-1(x) = (-x) + 5
(f o g)-1(x) = -x + 5 (A)
11.) Diketahui fungsi f(x) = √(2x – 3). Invers f(-3) = . . .
A. 3
B. 4
C. 5
D. 6
E. 7
Pembahasan :
f(x) = √(2x – 3)
y = √(2x – 3)
y2 = 2x – 3
2x = y2 + 3
x = (y2 + 3)/2
Maka,
f-1(x) = (x2 + 3)/2
f-1(-3) = ((-3)2 + 3)/2
f-1(-3) = (9 + 3)/2
f-1(-3) = 12/2
f-1(-3) = 6 (D)
12.) Diketahui f(x) = -2x + 3. Invers f(x) adalah . . .
A. (3 + x)/2
B. (x – 3)/2
C. (3 – x)/3
D. (3 + x)/3
E. (3 – x)/2
Pembahasan :
f(x) = -2x + 3
y = -2x + 3
2x = 3 – y
x = (3 – y)/2
Jadi,
f-1(x) = (3 – x)/2 (E)
13.) Diketahui g(x + 1) = (x – 1)/(x + 1). Invers g(0) = . . .
A. 4
B. -4
C. -2
D. 1
E. 2
Pembahasan :
g(x + 1) = (x – 1)/(x + 1)
misal,
a = x + 1
x = a – 1
g(a) = (a – 2)/a
g(x) = (x – 2)/x
Maka,
g(x) = (x – 2)/x
y = (x – 2)/x
xy = x – 2
x – xy = 2
x(1 – y) = 2
x = 2/(1 – y)
g-1(x) = 2/(1 – x)
g-1(0) = 2/(1 – 0)
g-1(0) = 2 (E)
14.) Diberikan f(x) = x + a. Jika f(x) = f-1(x), maka nilai a adalah . . .
A. 2
B. 1
C. 0
D. -1
E. -2
Pembahasan :
f(x) = x + a
y = x + a
x = y – a
f-1(x) = x – a
Maka,
f(x) = f-1(x)
x + a = x – a
a + a = x – x
2a = 0
a = 0 (C)
15.) Diberikan fungsi g(x) = √2x + √6. Jika g-1(x) merupakan invers dari g(x), maka g-1(2√6) = . . .
A. 2
B. 3
C. 5
D. 6
E. 8
Pembahasan :
g(x) = √2x + √6
y = √2x + √6
√2x = y – √6
2x = (y – √6)2
x = (y – √6)2/2
Maka,
g-1(x) = (x – √6)2/2
g-1(2√6) = (2√6 – √6)2/2
g-1(2√6) = (√6)2/2
g-1(2√6) = 6/2
g-1(2√6) = 3 (B)
Itulah 15 soal + pembahasan matematika tentang invers fungsi komposisi. Semoga bermanfaat dan sekian terima kasih.