Moderate -0.8 This is a straightforward application of the sign change method requiring only two function evaluations at x=0 and x=1, with no algebraic manipulation or problem-solving insight needed. The calculations are simple (e^0=1, e^1≈2.7) and the conclusion is immediate, making this easier than a typical A-level question.
Attempting to evaluate the function at both values
When \(x=1\): \(e^1 - 5\times1^3 = e-5 < 0\)
So [as the function is continuous and there is a change of sign] there is a root between 0 and 1
E1 [2]
Conclusion from correct values
## Question 2:
| Answer | Marks | Guidance |
|--------|-------|----------|
| When $x=0$: $e^0 - 5\times0^3 = 1 > 0$ | M1 | Attempting to evaluate the function at both values |
| When $x=1$: $e^1 - 5\times1^3 = e-5 < 0$ | | |
| So [as the function is continuous and there is a change of sign] there is a root between 0 and 1 | E1 [2] | Conclusion from correct values |
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2 By considering a change of sign, show that the equation $\mathrm { e } ^ { x } - 5 x ^ { 3 } = 0$ has a root between 0 and 1 .
\hfill \mbox{\textit{OCR MEI Paper 1 2018 Q2 [2]}}