By sketching a suitable pair of graphs, show that the equation |x-2|= 2sinx has only one root in the interval 0 < x < π.

Mark scheme

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

1. [B1] Sketch a relevant graph for 0<x<π For y = |x-2| graph should be symmetrical and have correct intercepts on the axes For y = 2sin(x/2), graph should pass through the origin, have correct curvature and max y = 2 when x = π
2. [B1] Sketch second relevant graph and confirm root. The vertex of the modulus graph in roughly correct position relative to π and/or π/2 If no mark on the graph, check to see if they have written something below the graph. SC A sketch y = x − 2 and y = ±2sin(x/2) (above and below the x-axis) scores B1. A clear indication of the root scores second B1.
3. [M1] Calculate the values of a relevant expression or pair of expressions at x = 1 and x = 1.5 e.g. f(x) = |x-2| - 2sin(x/2) f(1) = 0.0411... > 0 f(1.5) = -0.863... < 0 e.g. f(x)=(x-2)^2-4sin^2(x/2) → f(1)=0.0806..., f (1.5) = -1.60... Or comparing |x - 2| and 2sin(x/2). Using 1: |x − 2| =1, 2sin(x/2)=0.958... ⇒1>0.958... Using 1.5: |x - 2| = 0.5 2sin(x/2)=1.36... ⇒ 0.5 < 1.36... Need all values but condone one error. If the solution involves 4 values, the pairing must be clear. Embedded values are not sufficient, e.g. f1(1) = ... and f2(1) =... etc. M0 if working in degrees (gives 0.98... and 0.47... if using f(x) = 0). Allow if working on a smaller interval.
4. [A1] Complete the argument correctly with correct calculated values Values correct. They must have a conclusion in words or symbols, but they do not need to say that the function is continuous. A correct statement with correct inequalities is sufficient.
5. [M1] Use the iterative process correctly at least once starting at 1.03 (get as far as 1.0281) M0 if working in degrees.
6. [A1] Obtain final answer 1.02 No working seen at all scores 0/3.
7. [A1] Show sufficient iterations to 4 d.p. to justify 1.02 to 2 d.p. or show there is a sign change in the interval (1.015, 1.025) (or a smaller interval containing the root) 1.03, 1.0149, 1.0281, 1.0166, 1.0266, 1.0179, 1.0255, 1.0189, 1.0246 Incorrect starting point is M0. Once the convergence is established ISW.