By sketching a suitable pair of graphs, show that the equation ln x = 3x − x² has one real root.
- (a)By sketching a suitable pair of graphs, show that the equation ln x = 3x − x² has one real root.[2]
- (b)Verify by calculation that the root lies between 2 and 2.8.[2]
- (c)Use the iterative formula Xn+1 = √3x - Inx, to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.[3]
Mark scheme
Marking points from the official mark scheme (May/June 2022). The mark or mark type is shown in brackets.
- B1 Sketch a relevant graph, e.g. y = ln x ln(x) : sketch should imply y-axis is an asymptote. Through (1, 0) if marked. Correct shape.
- B1 Sketch a second relevant graph, e.g. y=3x-x², and justify the given statement by marking the root on the sketch or by use of a suitable comment 3x - x²: Symmetrical. Through (0, 0) and (3, 0) if marked. If ln(x) correct accept parabola for +ve y only. If ln(x) incorrect then need parabola in 3 quadrants.
- M1 Calculate the values of a relevant expression or pair of expressions at x = 2 and x = 2.8 Allow for a smaller interval. At least one value correct if comparing with 0. If using pairs then the pairing must be clear.
- A1 Complete the argument correctly with correct calculated values e.g. 0.693 < 2 and 1.03 > 0.56 or 1.307 > 0, -0.47 < 0 using √3x-ln x 0.304>0,-0.085 < 0. Need to have calculated values to at least 2 sf.
- M1 Use the iterative process correctly at least once
- A1 Obtain final answer 2.63
- A1 Show sufficient iterations to at least 4 dp to justify 2.63 to 2 dp or show there is a sign change in the interval (2.625, 2.635) SC Allow M1 A1 A0 to a candidate who starts at a point in the interval and reaches a premature conclusion
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