Express f(x + 3) in the form x2 + bx + c where b and c are integers.
3 marks
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x2 + 8x + 15
6
f(x) = 3x + 2 g(x) = x2 − 4
Find f−1(x)
Solve gf(x) = 0
5 marks
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a) f-1(x) = (x - 2)/3b) x = 0 or x = -4/3
7
f(x) = 4x − 1 g(x) = x2 + 2
Find fg(−3)
Find gf(x), giving your answer in the form ax2 + bx + c
Find f−1(x)
5 marks
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a) 43b) 16x2 - 8x + 3c) f-1(x) = (x + 1)/4
8
f(x) = x2g(x) = x + 2
Find fg(x), giving your answer in the form x2 + bx + c
Solve fg(x) = gf(x)
5 marks
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a) x2 + 4x + 4b) x = -1/2
9
f(x) = 2x + 1x − 3, x ≠ 3
Show that f−1(x) = 3x + 1x − 2
g(x) = x + 4
Solve fg(x) = 5
6 marks
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a) Rearranging y = (2x + 1)/(x - 3) gives x = (3y + 1)/(y - 2), so f-1(x) = (3x + 1)/(x - 2)b) x = 4/3
10
f(x) = x + 2 g(x) = x2 − 3x
Find fg(1)
Find f−1(x)
Solve gf(x) = 10
7 marks
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a) 0b) f-1(x) = x - 2c) x = -4 or x = 3
11
f(x) = 1x − 4 and g(x) = √(x + 3)
State the value of x that must be excluded from the domain of f.
State the largest possible domain of g.
State the range of g.
fg(x) is not defined for one value of x in the domain of g. Find this value.
4 marks
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a) x = 4b) x >= -3c) g(x) >= 0d) x = 13
12
f(x) = 3x + 10 and g(x) = sin x°
Work out fg(30).
Work out gf(20), correct to 3 decimal places.
Ahmed solves gf(x) = 0.5 and writes “3x + 10 = 30, so x = 203 is the only solution.” Show that Ahmed is wrong by finding another solution between 40 and 50.
5 marks
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a) 11.5b) 0.940c) x = 140/3 (46.67)
13
f(x) = x2 − 6x + 11 for all real x.
Write f(x) in the form (x − a)2 + b.
Hence state the range of f.
Explain why f does not have an inverse function for all real x.
4 marks
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a) (x - 3)2 + 2b) f(x) >= 2c) f is many-to-one, e.g. f(2) = f(4) = 3
Every question here is original and written by Robinson Tuition. The grade follows the Edexcel grade descriptors and suits AQA and OCR too; this is not an exam-board paper.