It is given that the equation e^(2x) = 5 + cos 3x has only one root. Show by calculation that this root lies in the interval 0.7 < x < 0.8 .

Mark scheme

Marking points from the official mark scheme (May/June 2024). The mark or mark type is shown in brackets.

1. [M1] Calculate the value of a relevant expression or values of a pair of expressions at x=0.7 and x=0.8. Allow if working with a smaller interval, e.g. (0.72, 0.78). Need all relevant values but condone one error. Pairings must be clear for solutions involving four values (do not accept embedded values). M0 if working in degrees e.g -1.94..., 1.04...
2. [A1] Complete the argument correctly with correct calculated values. (can be using the equation in the rubric or the equation in (b) or equivalent) E.g. 4.95...>4.26... and 4.06... <4.49... -0.439...< 0, 0.690...> 0. 1.4 <1.5029..., 1.6 > 1.449... 1.1 > 1, 0.86 <1. 0.0515 > 0, -0.075 < 0. Allow values rounded or truncated to 2sf.
3. [B1] State 2x = ln (5+ cos3x) and take exponential of both sides to obtain e^(2x) = 5 + cos3x. Given answer requires fully correct working or work vice versa. If working in reverse, must get to the iterative formula, including subscripts.
4. [M1] Use the iterative process correctly at least once. M0 if working in degrees (e.g. values heading for 0.89....).
5. [A1] Obtain final answer 0.740.
6. [A1] Show sufficient iterations to at least 5dp to justify 0.740 to 3dp, or show that there is a sign change in the interval (0.7395, 0.7405). E.g. 0.7,0.75150,0.73719,0.74105,0.74000,0.74028. 0.75,0.73759,0.74094, 0.74003,0.74028. 0.8,0.72494, 0.74443,0.73909,0.74053,0.74014,0.74025. Allow recovery. Allow truncation or rounding and condone small differences in the final decimal place.