A-LevelMathematicsFunctions9709/12 · Feb/Mar 2025 · Q1110 Marks

Functions f and g are defined for all real values of x by f(x) = 4x² - c and g(x) = 2x+k, where c and k are positive constants. It is given that g⁻¹(3k+ 1) = c.

(a)Show that gf(x) = 8x² - k - 1.[4]
(b)The curve with equation y = 8x²-k-1 is transformed to the curve with equation y = h(x) by the following sequence of transformations. Translation of (2 3), Stretch in the y-direction by scale factor k, Reflection in the x-axis. Find an expression for h(x) in terms of x and k.[3]
(c)The range of h is given by h(x) ≤ 15. Find the values of c and k.[3]

Mark scheme

Marking points from the official mark scheme (Feb/Mar 2025). The mark or mark type is shown in brackets.

[B1] State or imply g⁻¹(x) = ½(x-k) or equivalent
[B1] Obtain ½(3k+1-k) = c and hence 2k+1=2c OE
[B1] g(c) = 3k+1
[B1] [2c+k=3k+1⇒ ] 2c = 2k+1 OE
[M1] [gf(x) =] 2(4x²-c)+k Allow ± errors only.
[A1] [8x² - 2c+k → 8x² - (2k+1)+k>] gf(x)=8x² -k-1 AG All necessary detail needed.
[B1] 8(x-2)²-k-1+3 OE Translation coming before the stretch.
[B1] k(their (8(x-2)²-k-1+3)) FT OE Stretch.
[B1] [h(x)=]-(k(their (8(x-2)²-k-1+3))) FT OE Reflection.
[B1] k²-2k=15 or k²-2k ≤ 15 OE
[B1] k=5 only
[B1] c = 11/2 only OE

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