Problems

Filters
Clear Filters

3 problems found

2018 Paper 1 Q8
D: 1500.0 B: 1543.7
integration differential equation first order inverse function substitution product rule quotient rule functional equations proof constant of motion

The functions \(\s\) and \(\c\) satisfy \(\s(0)= 0\,\), \(\c(0)=1\,\) and \[ \s'(x) = \c(x)^2 ,\] \[ \c'(x)=-\s(x)^2. \] You may assume that \(\s\) and \(\c\) are uniquely defined by these conditions.

  1. Show that \(\s(x)^3+\c(x)^3\) is constant, and deduce that \(\s(x)^3+\c(x)^3=1\,\).
  2. Show that \[ \frac{\d }{\d x} \, \Big( \s(x) \c(x) \Big) = 2\c(x)^3-1 \] and find (and simplify) an expression in terms of \(\c(x)\) for $\dfrac{\d }{\d x} \left( \dfrac{\s(x)}{\c(x)} \right) $.
  3. Find the integrals \[ \int \s(x)^2 \, \d x \ \ \ \ \ \ \text{and} \ \ \ \ \ \ \int \s(x)^5 \, \d x \,. \]
  4. Given that \(\s\) has an inverse function, \(\s^{-1}\), use the substitution \(u = \s(x)\) to show that \[ \int \frac{1}{(1-u^3)^{\frac{2}{3}}} \, \d u = \s^{-1}(u) \, + \text{constant}. \]
  5. Find, in terms of \(u\), the integrals \[ \int \frac{1}{{(1-u^3)}^{\frac{4}{3}}} \, \d u \ \ \ \ \ \ \text{and} \ \ \ \ \ \ \int {(1-u^3)}^{\frac{1}{3}} \, \d u \,. \]

Solution: \begin{questionparts} \item \begin{align*} && \dfrac{\d }{\d x} \left( \s(x)^3 + \c(x)^3 \right) &= 3\s(x)^2\s'(x) + 3\c(x)^2 \c'(x) \\ &&&= 3\s(x)^2\c(x)^2 - 3\c(x)^2\s(x)^2 \\ &&&= 0 \\ \\ \Rightarrow && \s(x)^3 + \c(x)^3 &= \text{constant} \\ &&&= \s(0)^3 + \c(0)^3 \\ &&&= 1 \end{align*} \item \begin{align*} \frac{\d }{\d x} \, \Big( \s(x) \c(x) \Big) &= \s'(x) \c(x) + \s(x)\c'(x) \\ &= \c(x)^3 - \s(x)^3 \\ &= \c(x)^3 - (1-\c(x)^3) \\ &= 2\c(x)^3 - 1 \\ \\ \dfrac{\d }{\d x} \left( \dfrac{\s(x)}{\c(x)} \right) &= \frac{\s'(x)\c(x) - \s(x)\c'(x)}{\c(x)^2} \\ &= \frac{\c(x)^3 + \s(x)^3}{\c(x)^2} \\ &= \frac{1}{\c(x)^2} \\ \end{align*} \item \begin{align*} \int \s(x)^2 \d x &= -\int -\s(x)^2 \d x \\ &= -\int \c'(x) \d x \\ &= - \s(x) +C \\ \\ \int \s(x)^5 \, \d x &= \int \s(x)^2 \s(x)^3 \d x \\ &= \int \s(x)^2 (1 - \c(x)^3) \d x \\ &= -\int \c'(x) (1 - \c(x)^3) \d x \\ &= - c(x) + \frac{\c(x)^4}{4} + C \end{align*} \item If \(u = \s(x), \frac{\d u}{\d x} = \c(x)^2\) \begin{align*} \int \frac{1}{(1-u^3)^{\frac{2}{3}}} \, \d u &= \int \frac{1}{(1-\s(x)^3)^{\frac{2}{3}}} \c(x)^2 \d x \\ &= \int 1 \d x \\ &= x + C \\ &= \s^{-1}(u) + C \\ \\ \int \frac{1}{{(1-u^3)^{\frac{4}{3}}}} \d u &= \int \frac1{(1-\s(x)^3)^{\frac43} }\c(x)^2 \d x \\ &= \int \frac1{(\c(x)^3)^{\frac43}} \c(x)^2 \d x \\ &= \int \frac1{\c(x)^2} \d x \\ &= \frac{\s(x)}{\c(x)} + C \\ &= \frac{u}{(1-u^3)^{\frac13}} + C \\ \end{align*} \begin{align*} && \int {(1-u^3)}^{\frac{1}{3}} \, \d u &= \int (1-s(x)^3)^{\frac13} c(x)^2 \d x \\ &&&= \int \c(x)^3 \d x = I\\ &&&= \int \c(x) s'(x) \d x \\ &&&= \left [\c(x) \s(x) \right] + \int \s(x)^2 s(x) \d x \\ &&&= \c(x) \s(x) + \int (1 - \c(x)^3) \d x + C \\ &&&= \c(x) \s(x) + x - I + C \\ \Rightarrow && I &= \frac{x + \c(x) \s(x)}{2} + k \\ \Rightarrow && &= \frac12 \l \s^{-1}(u) + u \sqrt[3](1-u^3)\r + k \end{align*}

View
2016 Paper 3 Q8
D: 1700.0 B: 1484.0
simultaneous equations functional equations substitution involution rational function algebraic manipulation self-inverse function

  1. The function f satisfies, for all \(x\), the equation \[ \f(x) + (1- x)\f(-x) = x^2\, . \] Show that \(\f(-x) + (1 + x)\f(x) = x^2\,\). Hence find \(\f(x)\) in terms of \(x\). You should verify that your function satisfies the original equation.
  2. The function \({\rm K}\) is defined, for \(x\ne 1\), by \[{\rm K}(x) = \dfrac{x+1}{x-1}\,.\] Show that, for \(x\ne1\), \({\rm K(K(}x)) =x\,\). The function g satisfies the equation \[ \g(x)+ x\, \g\Big(\frac{ x+1 }{x-1}\Big) = x \ \ \ \ \ \ \ \ \ \ \ ( x\ne 1) \,. \] Show that, for \(x\ne1\), \(\g(x)= \dfrac{2x}{x^2+1}\,\).
  3. Find \(\h(x)\), for \(x\ne0\), \(x\ne1\), given that \[ \h(x)+ \h\Big(\frac 1 {1-x}\Big)= 1-x -\frac1{1-x} \ \ \ \ \ \ ( x\ne0, \ \ x\ne1 ) \,. \]

Solution:

  1. \(\,\) Let \(P(x)\) mean the proposition that \(f(x) + (1-x)f(-x) = x^2\) so \begin{align*} P(x): && f(x) + (1-x)f(-x) &= x^2 \\ P(-x): && f(-x)+(1+x)f(x) &= (-x)^2 = x^2 \\ \Rightarrow && f(x)+(1-x)\left (x^2-(1+x)f(x) \right) &= x^2 \\ \Rightarrow && f(x) \left (1 -(1-x^2) \right) &= x^2 + (x-1)x^2 \\ \Rightarrow && f(x)x^2 &= x^3 \\ \Rightarrow && f(x) &= x \end{align*} Notice that \(x + (1-x)(-x) = x^2\) so it does satisfy the functional equation.
  2. Let \(K(x) = \frac{x+1}{x-1}\) if \(x \neq 1\) so \begin{align*} && K(K(x)) &= \frac{K(x)+1}{K(x)-1} \\ &&&= \frac{\frac{x+1}{x-1}+1}{\frac{x+1}{x-1}-1} \\ &&&= \frac{\frac{2x}{x-1}}{\frac{2}{x-1}} \\ &&&= x \end{align*} Let \(Q(x)\) denote the proposition that \(g(x) + xg(K(x)) = x\) so \begin{align*} Q(x): && g(x) + xg(K(x)) &= x \\ Q(K(x)): && g(K(x)) + K(x)g(x) &= K(x) \\ \Rightarrow && g(x) +xK(x)[1-g(x)] &= x \\ \Rightarrow && g(x)[1-xK(x)] &= x(1-K(x)) \\ \Rightarrow && g(x) \frac{x-1-x^2-x}{x-1} &= \frac{-2x}{x-1} \\ \Rightarrow && g(x) &= \frac{2x}{x^2+1} \end{align*}. And notice that \(\frac{2x}{x^2+1} + x \frac{2\frac{x+1}{x-1}}{\left( \frac{x+1}{x-1}\right)^2+1} = \frac{2x}{x^2+1} + \frac{2x(x^2-1)}{2x^2+2} = x\)
  3. Consider \(H(x) = \frac{1}{1-x}\) then notice that \(H(H(x)) = \frac{1}{1-\frac{1}{1-x}} = \frac{x-1}{x}\) and \(H^3(x) = \frac{\frac{1}{1-x}-1}{\frac{1}{1-x}} = 1-(1-x) = x\). So So letting \(S(x)\) be the statement that \(h(x) + h(H(x)) = 1 - x - \frac{1}{1-x}\) we have \begin{align*} S(x): && h(x) + h(H(x)) &= 1 - x - H(x) \\ S(H(x)): && h(H(x)) + h(H^2(x)) &= 1 - H(x) - H^2(x) \\ S(H^2(x)): && h(H^2(x)) + h(x) &= 1 - H^2(x) - x \\ S(x) - S(H(x)) + S(H^2(x)): && 2h(x) &= 1 - 2x \\ \Rightarrow && h(x)& = \frac12 - x \end{align*} and notice that \(\frac12 -x +\frac12 - \frac{1}{1-x} = 1 - x - \frac{1}{1-x}\) so it does satisfy the equation.

View
2013 Paper 2 Q5
D: 1600.0 B: 1484.0
differentiation chain rule functional equations symmetry curve sketching rational function stationary points inequality

  1. A function \(\f(x)\) satisfies \(\f(x) = \f(1-x)\) for all \(x\). Show, by differentiating with respect to \(x\), that \(\f'(\frac12) =0\,\). If, in addition, \(\f(x) = \f(\frac1x)\) for all (non-zero) \(x\), show that \(\f'(-1)=0\) and that \(\f'(2)=0\).
  2. The function \(\f\) is defined, for \(x\ne0\) and \(x\ne1\), by \[ \f(x) = \frac {(x^2-x+1)^3}{(x^2-x)^2} \,. \] Show that \(\f(x)= \f(\frac 1 x)\) and \(\f(x) = \f(1-x)\). Given that it has exactly three stationary points, sketch the curve \(y=\f(x)\).
  3. Hence, or otherwise, find all the roots of the equation \(\f(x) = \dfrac {27} 4\,\) and state the ranges of values of \(x\) for which \(\f(x) > \dfrac{27} 4\,\). Find also all the roots of the equation \(\f(x) = \dfrac{343}{36}\,\) and state the ranges of values of \(x\) for which \(\f(x) > \dfrac{343}{36}\).

Solution:

  1. \(\,\) \begin{align*} && f(x) &= f(1-x) \\ \Rightarrow && f'(x) &= -f'(1-x) \\ \Rightarrow && f'(\tfrac12) &= -f'(\tfrac12) \\ \Rightarrow && f'(\tfrac12) &= 0 \\ \\ && f(x) &= f(\tfrac1x) \\ \Rightarrow && f'(x) &= f'(\tfrac1x) \cdot \frac{-1}{x^2} \\ \Rightarrow && f'(-1) &= -f'(-1) \\ \Rightarrow && f'(-1) &= 0 \\ \\ && f'(2) &= -\frac{1}{4}f'(\tfrac12) \\ &&&= 0 \end{align*}
  2. Suppose \begin{align*} && f(x) &= \frac{(x^2-x+1)^3}{(x^2-x)^2} \\ && f(1/x) &= \frac{(x^{-2}-x^{-1}+1)^3}{(x^{-2}-x^{-1})^2} \\ &&&= \frac{(1-x+x^2)^3/x^6}{((x-x^2)^2/x^6} \\ &&&= f(x) \\ \\ && f(1-x) &= \frac{((1-x)^2-(1-x)+1)^3}{((1-x)^2-(1-x))^2} \\ &&&= \frac{(1-x+x^2)^3}{(x^2-x)^2} = f(x) \end{align*}
    TikZ diagram
  3. Clearly \(x = -1\) is a root of \(f(x) = \frac{27}{4}\), so we must also have \(x=2\) and \(x = \frac12\), therefore \(f(x) > \frac{27}{4}\) if \(x \in \mathbb{R} \setminus \{-1, 2, \tfrac12, 0, 1 \}\). Clearly \(x = 3\) and \(x = -2\) are solutions so we also have: \(\frac13, -\frac12, \frac32, \frac23\) and these must be all solutions so we must have: \(f(x) > \frac{343}{36} \Leftrightarrow x \in (-\infty, -2) \cup (-\frac12, 0) \cup (0, \frac13) \cup (\frac23, 1) \cup (1, \frac32) \cup (3, \infty)\)

View