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2002 Paper 2 Q4
D: 1600.0
B: 1484.0
integration
inequality
quadratic
definite integral
positivity
trigonometric
logarithm
substitution
discriminant
Give a sketch to show that, if \(\f(x) > 0\) for \(p < x < q\,\), then \(\displaystyle \int_p^{q} \f(x) \d x > 0\,\).
- By considering \(\f(x) = ax^2-bx+c\,\) show that, if \(a > 0\) and \(b^2 < 4ac\), then \(3b < 2a+6c\,\).
- By considering \(\f(x)= a\sin^2x - b\sin x + c\,\) show that, if \(a > 0\) and \(b^2< 4ac\), then \(4b < (a+2c)\pi\)
- Show that, if \(a > 0\), \(b^2 < 4ac\) and \(q > p > 0\,\), then $$ b\ln(q/p) < a\left(\frac1p -\frac1q\right) +c(q-p)\;. $$
Solution:
- If \(a > 0\) and \(b^2 < 4ac \Rightarrow \Delta < 0\) then \(f(x) = ax^2-bx+c > 0\) for all \(x\). Therefore \begin{align*} && 0 & < \int_0^1 (ax^2-bx+c) \d x\\ &&&= \frac13 a-\frac12b+c \\ \Rightarrow && 3b &< 2a+6c \end{align*}
- Similar logic tells us this must also be always positive, therefore \begin{align*} && 0 &< \int_0^{\pi} (a \sin^2 x - b \sin x +c ) \d x\\ &&&= \frac{\pi}{2}a - 2b+\pi c \\ \Rightarrow && 4b &< (a+2c)\pi \end{align*}
- Similar logic shows that \(f(x) = \frac{a}{x^2}-\frac{b}{x} +c>0\), therefore \begin{align*} && 0 &< \int_p^q \left (\frac{a}{x^2} - \frac{b}{x} + c\right) \d x \\ &&&=a\left (\frac{1}{p} - \frac{1}{q} \right) - b(\ln q - \ln p)+c(q-p) \\ \Rightarrow && b \ln (q/p) &,< a\left (\frac{1}{p} - \frac{1}{q} \right) +c(q-p) \end{align*}