OCR Mechanics 1 2018 March — Question 5

Exam BoardOCR
ModuleMechanics 1 (Mechanics 1)
Year2018
SessionMarch
TopicDifferentiation Applications
TypeFind stationary points

5 In this question you must show detailed reasoning.
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The function f is defined for the domain \(x \geqslant 0\) by $$\mathrm { f } ( x ) = \left( 2 x ^ { 2 } - 3 x \right) \mathrm { e } ^ { - x }$$ The diagram shows the curve \(y = \mathrm { f } ( x )\).
  1. Find the range of f.
  2. The function g is defined for the domain \(x \geqslant k\) by $$\mathrm { g } ( x ) = \left( 2 x ^ { 2 } - 3 x \right) \mathrm { e } ^ { - x } .$$ Given that g is a one-one function, state the least possible value of \(k\).
  3. Find the exact area of the shaded region enclosed by the curve and the \(x\)-axis.
  4. Determine the values of \(p\) and \(q\) for which $$x ^ { 2 } - 6 x + 10 \equiv ( x - p ) ^ { 2 } + q .$$
  5. Use the substitution \(x - p = \tan u\), where \(p\) takes the value found in part (i), to evaluate $$\int _ { 3 } ^ { 4 } \frac { 1 } { x ^ { 2 } - 6 x + 10 } \mathrm {~d} x .$$
  6. Determine the value of $$\int _ { 3 } ^ { 4 } \frac { x } { x ^ { 2 } - 6 x + 10 } \mathrm {~d} x$$ giving your answer in the form \(a + \ln b\), where \(a\) and \(b\) are constants to be determined.