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17 Sep 2025, 03:19
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B
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Difficulty:
65%
(hard)
Question Stats:
57% (01:44) correct
43%
(01:52)
wrong
based on 30
sessions
History
Date
Time
Result
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If f(x) = ax + b, where a and b are constants and f(f(f(f(x)))) = x + 8, Find the value of a + b.
A. 1
B. 3
C. 5
D. 4
E. 2
A. 1
B. 3
C. 5
D. 4
E. 2
Attachment:
Screenshot 2025-09-17 12.17.24 PM.png
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18 Sep 2025, 22:40
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If f(x) = ax + b, where a and b are constants and f(f(f(f(x)))) = x + 8, Find the value of a + b.
f(x) = ax+b
f(f(x)) = a(ax+b) + b = a^2x + ab + b = a^2x + b(1+a)
f(f(f(x))) = a(a(ax+b) + b) = a^3x + b(1+a+a^2)
f(f(f(f(x)))) = a^4x + b(1+a+a^2+a^3) = x + 8
a^4 = 1; a = 1 or - 1
Case 1: a = 1
b(1+a+a^2+a^3) = 8; 4b = 8; b = 2
a+b = 1+2 = 3
Case 2: a=-1
a^2b(a+1) + ab(a+1) + b(a+1) = 0; Not feasible
a+b = 3
To check:
f(x) = x+2
f(f(x)) = x+4
f(f(f(x))) = x + 6
f(f(f(f(x)))) = x + 8
IMO B
f(x) = ax+b
f(f(x)) = a(ax+b) + b = a^2x + ab + b = a^2x + b(1+a)
f(f(f(x))) = a(a(ax+b) + b) = a^3x + b(1+a+a^2)
f(f(f(f(x)))) = a^4x + b(1+a+a^2+a^3) = x + 8
a^4 = 1; a = 1 or - 1
Case 1: a = 1
b(1+a+a^2+a^3) = 8; 4b = 8; b = 2
a+b = 1+2 = 3
Case 2: a=-1
a^2b(a+1) + ab(a+1) + b(a+1) = 0; Not feasible
a+b = 3
To check:
f(x) = x+2
f(f(x)) = x+4
f(f(f(x))) = x + 6
f(f(f(f(x)))) = x + 8
IMO B
30 Sep 2025, 07:28
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A simpler way to solve:
f(x)=ax+b.
f^4(x)=a^4 x + b(1+a+a^2+a^3)
= x + 8.
Compare coefficients:
a^4 = 1.
Real solutions: a = 1 or a = −1.
If a = 1 → sum 1+a+a^2+a^3 = 4, so 4b = 8 → b = 2 → a + b = 3.
If a = −1 → 1+a+a^2+a^3 = 0, so constant term = 0 ≠ 8, impossible.
Answer: 3
f(x)=ax+b.
f^4(x)=a^4 x + b(1+a+a^2+a^3)
= x + 8.
Compare coefficients:
a^4 = 1.
Real solutions: a = 1 or a = −1.
If a = 1 → sum 1+a+a^2+a^3 = 4, so 4b = 8 → b = 2 → a + b = 3.
If a = −1 → 1+a+a^2+a^3 = 0, so constant term = 0 ≠ 8, impossible.
Answer: 3
Kinshook
