A-LevelMathematicsFunctions9709/12 · May/June 2022 · Q1013 Marks
Functions f and g are defined as follows: f(x) = (2x + 1)/(2x - 1) for x ≠ 1/2, g(x) = x² + 4 for x ∈ ℝ. [Figure 10.1]
- (a)State the domain of f⁻¹.[1]
- (b)Find an expression for f⁻¹(x).[3]
- (c)Find gf⁻¹(3).[2]
- (d)Explain why g⁻¹(x) cannot be found.[1]
- (e)Show that 1 + 2/(2x-1) can be expressed as (2x+1)/(2x-1). Hence find the area of the triangle enclosed by the tangent to the curve y = f(x) at the point where x = 1 and the x- and y-axes.[6]
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Mark scheme
Marking points from the official mark scheme (May/June 2022). The mark or mark type is shown in brackets.
- [B1] x ≠ 1 or x < 1, x > 1 or (-∞,1), (1,∞)
- [M1] y = (2x+1)/(2x-1) leading to (2x-1)y = 2x+1 leading to 2xy - y = 2x + 1
- [DM1] 2xy - 2x = y + 1 leading to 2x(y-1) = y + 1 leading to x = (y+1)/(2(y-1))
- [A1] [f⁻¹(x)] = (x+1)/(2(x-1)) or x/(2(x-1)) + 1/(2(x-1)) or 1/2 + 1/(2(x-1))
- [M1] ('their f⁻¹(3)') leading to ('their f⁻¹(3)')² + 4
- [A1] 5
- [B1] Sight of 'not one to one' or 'many to one' or 'one to many'
- [B1] f(x)=1+ (2x-1+2)/(2x-1)= (2x-1)/(2x-1)+ 2/(2x-1)= (2x+1)/(2x-1) [AG]
- [M1] f'(x) = -4(2x-1)⁻²
- [A1] Gradient m = -4
- [DM1] Equation of tangent is y-3=-4(x-1) [=> y = -4x+7]
- [A1 FT] Crosses axes at (7/4, 0) and (0, 7)
- [A1] [Area =] 49/8
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