Problems

Filters
Clear Filters

3 problems found

2017 Paper 1 Q6
D: 1516.0 B: 1484.0
integration proof by contradiction continuous functions intermediate value theorem integration by parts definite integrals polynomial functions inequality

In this question, you may assume that, if a continuous function takes both positive and negative values in an interval, then it takes the value \(0\) at some point in that interval.

  1. The function \(\f\) is continuous and \(\f(x)\) is non-zero for some value of \(x\) in the interval \(0\le x \le 1\). Prove by contradiction, or otherwise, that if \[ \int_0^1 \f(x) \d x = 0\,, \] then \(\f(x)\) takes both positive and negative values in the interval \(0\le x\le 1\).
  2. The function \(\g\) is continuous and \[ \int_0^1 \g(x) \, \d x = 1\,, \quad \int_0^1 x\g(x) \, \d x = \alpha\, , \quad \int_0^1 x^2\g(x) \, \d x = \alpha^2\,. \tag{\(*\)} \] Show, by considering \[ \int_0^1 (x - \alpha)^2 \g(x) \, \d x \,, \] that \(\g(x)=0\) for some value of \(x\) in the interval \(0\le x\le 1\). Find a function of the form \(\g(x) = a+bx\) that satisfies the conditions \((*)\) and verify that \(\g(x)=0\) for some value of \(x\) in the interval \(0\le x \le 1\).
  3. The function \(\h\) has a continuous derivative \(\h'\) and \[ \h(0) = 0\,, \quad \h(1) = 1\,, \quad \int_0^1 \h(x) \, \d x = \beta\,, \quad \int_0^1 x \h(x) \, \d x = \tfrac{1}{2}\beta (2 - \beta) \,. \] Use the result in part (ii) to show that \(\h^\prime(x)=0\) for some value of \(x\) in the interval \(0\le x\le 1\).

Solution:

  1. Claim: If \(f(x)\) non-zero for some \(x \in [0,1]\) and \(\int_0^1 f(x) \d x =0\) then \(f\) takes both positive and negative values in the interval \([0,1]\). Proof: Suppose not, then WLOG suppose \(f(x) > 0\) for some \(x \in [0,1]\). Then notice (since \(f\) is continuous) that there is some interval where \(f(x) > 0\) around the \(x\) we have already shown exists. But then \(\int_{\text{interval}} f(x) \d x > 0\) and since \(f(x) \geq 0\) everywhere \(\int_0^1 f(x) \d x > 0\), which is a contradiction.
  2. \(\,\) \begin{align*} && \int_0^1 (x - \alpha)^2 g(x) \d x &= \int_0^1 x^2g(x) \d x - 2 \alpha \int_0^1 x g(x) \d x + \alpha^2 \int_0^1 g(x) \d x \\ &&&= \alpha^2 - 2\alpha \cdot \alpha + \alpha^2 \cdot 1 \\ &&&= 0 \end{align*} Therefore \(g(x)(x-\alpha)^2\) is a continuous function which is either exactly \(0\) (in which case we've already found our \(0\)) or it is a continuous function which is both positive somewhere and has \(0\) integral, and therefore by part (i) must take both positive and negative values (and therefore takes \(0\) in between those points by continuity). \begin{align*} &&1 &= \int_0^1 a+bx \d x \\ &&&= a + \frac12 b \\ && \alpha &= \int_0^1 ax + bx^2 \d x \\ &&&= \frac12 a + \frac13 b \\ && \alpha^2 &= \int_0^1 ax^2 + bx^3 \d x\\ &&&= \frac13 a + \frac14 b \\ \Rightarrow && \frac1{36}(3a+2b)^2 &= \frac1{12}(4a+3b) \\ \Rightarrow && \frac1{36}(3a+2(2-2a))^2 &= \frac1{12}(4a+3(2-2a)) \\ \Rightarrow && (4-a)^2 &= 3(6-2a) \\ \Rightarrow && 16-8a+a^2 &= 18-6a \\ \Rightarrow && a^2-2a-2 &= 0 \\ \Rightarrow && (a-1)^2 - 3 &= 0 \\ \Rightarrow && a &= \pm \sqrt{3}+1 \\ && b &= \mp 2\sqrt{3} \end{align*} So say \(a = \sqrt{3}+1, b = -2\sqrt{3}\) This has a root at \(-\frac{a}{b} = \frac{1+\sqrt{3}}{2\sqrt{3}} = \frac{\sqrt{3}+3}{6} < 1\) so we have met our condition.
  3. Consider \(h'\), we must have \begin{align*} && \int_0^1 h'(x)\d x &= h(1)-h(0) =1\\ && \int_0^1 xh'(x) \d x &= \left [x h(x) \right]_0^1 - \int_0^1 h(x) \d x \\ &&&= 1 - \beta \\ && \int_0^1 x^2 h'(x) \d x &= \left [ x^2h(x) \right]_0^1 - \int_0^1 2xh(x) \d x \\ &&&= 1 - 2\tfrac12 \beta(2-\beta) \\ &&&= (1-\beta)^2 \end{align*} Therefore \(h'\) satisfies the conditions with \(\alpha = 1-\beta\), so \(h'\) must have a root in our interval.

View
2016 Paper 2 Q3
D: 1600.0 B: 1517.4
differentiation polynomials exponential approximation roots of polynomials proof by contradiction Rolle's theorem intermediate value theorem curve sketching

For each non-negative integer \(n\), the polynomial \(\f_n\) is defined by \[ \f_n(x) = 1 + x + \frac{x^2}{2!} + \frac {x^3}{3!} + \cdots + \frac{x^n}{n!} \]

  1. Show that \(\f'_{n}(x) = \f_{n-1}(x)\,\) (for \(n\ge1\)).
  2. Show that, if \(a\) is a real root of the equation \[\f_n(x)=0\,,\tag{\(*\)}\] then \(a<0\).
  3. Let \(a\) and \(b\) be distinct real roots of \((*)\), for \(n\ge2\). Show that \(\f_n'(a)\, \f_n'(b)>0\,\) and use a sketch to deduce that \(\f_n(c)=0\) for some number \(c\) between \(a\) and \(b\). Deduce that \((*)\) has at most one real root. How many real roots does \((*)\) have if \(n\) is odd? How many real roots does \((*)\) have if \(n\) is even?

Solution:

  1. \(\,\) \begin{align*} && f'_n(x) &= 0 + 1 + \frac{2x}{2!} + \frac{3x^2}{3!} + \cdots + \frac{nx^{n-1}}{n!} \\ &&&= 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^{n-1}}{(n-1)!} \\ &&&= f_{n-1}(x) \end{align*}
  2. Claim: \(f_n(x) > 0\) for all \(x > 0\) Proof: (By induction) Base case: (\(n = 1\)) \(f_1(x) = 1 + x > 1\) therefore \(f_1(x) > 0\) Suppose it's true for \(n = k\), then consider \(f_{k+1}\), if we differentiate it, we find it is increasing on \((0, \infty)\) by our inductive hypothesis. But then \(f_{k+1}(0) = 1 > 0\). Therefore \(f_{k+1}(x) > 0\) as well. Therefore by the principle of mathematical induction we are done. Since \(f_n(x) > 0\) for non-negative \(x\), if \(a\) is a root it must be negative.
  3. Suppose \(f_n(a) = f_n(b) = 0\) then \(f'_n(a) = -\frac{a^n}{n!}\) and \(f'_n(b) = -\frac{b^n}{n!}\), but then \(f_n'(a) f_n'(b) = \frac{(-a)^n(-b)^n}{(n!)^2} > 0\) since \(a < 0, b < 0\). \(_n'(a) f_n'(b)\) is positive, the two gradients must have the same sign (and not be zero). Therefore if they are both increasing, at some point the curve must cross the axis in between. Therefore there is some root \(c\) between \(a\) and \(b\). But then there is also a root between \(c\) and \(a\) and \(c\) and \(b\), and very quickly we find more than \(n\) roots which is not possivel. Therefore there must be at most \(1\) root. If \(n\) is odd there must be exactly one root, since \(f_n\) changes sign as \(x \to -\infty\) vs \(x = 0\). If \(n\) is even then there can't be any roots, since if it crossed the \(x\)-axis there would be two roots (not possible) and it cannot touch the axis, since \(f'_n(a) \neq 0\) unless \(a = 0\), and we know \(a < 0\)

View
2004 Paper 2 Q7
D: 1600.0 B: 1516.0
solving equations numerically bisection method trigonometric intermediate value theorem root finding interval halving graphical reasoning

The function f is defined by $$\f(x) = 2\sin x - x\,.$$ Show graphically that the equation \(\f(x)=0\) has exactly one root in the interval \([\frac12\pi ,\,{\pi}]\,\). This interval is denoted \(I_0\). In order to determine the root, a sequence of intervals \(I_1\), \(I_2, \,\ldots\) is generated in the following way. If the interval \(I_n=[a_n,b_n]\,\), and \(c_n=(a_n+b_n)/2\,\), then \begin{equation*} I_{n+1}= \begin{cases} [a_n,c_n] & \text{if \(\; \f(a_n)\f(c_n)<0 \,\)}; \\[5pt] [c_n,b_n] & \text{if \(\; \f(c_n)\f(b_n)<0 \,\)}. \end{cases} \end{equation*} By using the approximations \(\displaystyle \frac 1{\sqrt{2}} \approx 0.7\) and \({\pi} \approx \sqrt{10} \,\), show that \(I_2=[\frac12{\pi},\,\frac58{\pi}]\) and find \(I_3\,\).

Solution: \begin{array}{c|c|c|c|c|c} n & a_n & b_n & c_n & f(a_n) & f(c_n) & f(b_n) \\ \hline 0 & \tfrac12 \pi & \pi & \tfrac34\pi & 2\sin(\tfrac12\pi)-\tfrac12\pi = 2-\tfrac12\pi & 2\sin(\tfrac34\pi)-\tfrac34\pi = \frac{2}{\sqrt{2}}-\tfrac34\pi & 2\sin(\pi)-\pi =-\pi \\ 0 & \tfrac12 \pi & \pi & \tfrac34\pi & >0 & 2-\frac{9}{16}10 < 0& <0 \\ \hline 1 & \frac12 \pi & \frac34\pi & \frac58\pi & >0 &2\sin \tfrac58\pi - \tfrac58\pi & < 0\\ 1 & \frac12 \pi & \frac34\pi & \frac58\pi & >0 & \approx 1.4 \cdot \sqrt{1.7} -\frac58\sqrt{10} < 0 & <0 \\ \hline 2 & \frac12 \pi & \frac58\pi & \frac9{16}\pi & >0 & 2\sin \frac{9}{16}\pi-\frac{9}{16}\pi & <0 \\ 2 & \frac12 \pi & \frac58\pi & \frac9{16}\pi & >0 & > 0 & <0 \\ \end{array} Threfore \(I_3 = [\frac9{16}\pi,\frac58\pi]\) \(\sin \frac{5\pi}{8} = \cos \frac{\pi}{8} = \sqrt{\frac12(\cos \frac{\pi}{4}+1)} = \frac{1}{\sqrt{2}}\sqrt{1 + \frac{1}{\sqrt{2}}} \approx 0.7 \cdot \sqrt{1.7}\) \(\sin \frac{9\pi}{16} = \cos \frac{\pi}{16} = \sqrt{\frac12\left ( \cos \frac{\pi}{8}+1 \right)} \) So we are comparing \(2\cos \frac{\pi}{16}\) with \(\frac{9}{16}\pi\) or \(4 \cos^2 \frac{\pi}{16} = 2\cos \frac{\pi}{8}+2\) with \(\frac{90}{16}\)

View