O-LevelAdditional MathematicsFunctions4037/22 · May/June 2017 · Q127 Marks
The function g is defined, for x > -1/2, by g(x) = 3/(2x + 1).
- (i)Show that g'(x) is always negative.[2]
- (ii)Write down the range of g.[1]
- (iii)The function h is defined, for all real x, by h(x) = kx + 3, where k is a constant. Find an expression for hg(x).[1]
- (iv)Given that hg(0) = 5, find the value of k.[2]
- (v)State the domain of hg.[1]
Mark scheme
Marking points from the official mark scheme (May/June 2017). The mark or mark type is shown in brackets.
- [B1] -6(2x+1)⁻² or -6/(2x+1)² oe isw
- [B1] Denominator or (2x+1)² is positive [and numerator negative therefore g'(x) is always negative] oe
- [B1] g > 0
- [B1] 3k/(2x+1) + 3 oe isw
- [B1] 3k/(2(0)+1) + 3 = 5
- [B1] k = 2/3 isw implies the first B1
- [B1] x > -1/2
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