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2018 Paper 2 Q2
D: 1600.0 B: 1516.0
concavity Jensen's inequality trigonometric inequality optimisation triangle angles logarithm curve sketching second derivative test sin

A function \(\f(x)\) is said to be concave for \(a< x < b\) if \[ \ t\,\f(x_1) +(1-t)\,\f(x_2) \le \f\big(tx_1+ (1-t)x_2\big) \, ,\] for \(a< x_1 < b\,\), \(a< x_2< b\) and \(0\le t \le 1\,\). Illustrate this definition by means of a sketch, showing the chord joining the points \(\big(x_1, \f(x_1)\big) \) and \(\big(x_2, \f(x_2)\big) \), in the case \(x_1 < x_2\) and \(\f(x_1)< \f(x_2)\,\). Explain why a function \(\f(x)\) satisfying \(\f''(x)<0\) for \(a< x < b\) is concave for \(a< x < b\,\).

  1. By choosing \(t\), \(x_1\) and \(x_2\) suitably, show that, if \(\f(x)\) is concave for \(a< x < b\,\), then \[ \f\Big(\frac{u+ v+w}3\Big) \ge \frac{ \f(u) +\f(v) +\f(w)}3 \, ,\] for \(a< u < b\,\), \(a< v < b\,\) and \(a< w < b\,\).
  2. Show that, if \(A\), \(B\) and \(C\) are the angles of a triangle, then \[ \sin A +\sin B + \sin C \le \frac{3\sqrt3}2 \,. \]
  3. By considering \(\ln (\sin x)\), show that, if \(A\), \(B\) and \(C\) are the angles of a triangle, then \[ \sin A \times \sin B \times \sin C \le \frac {3 \sqrt 3} 8 \,. \]

Solution:

TikZ diagram
Consider the function \(g(t) = f(tx_1 + (1-t)x_2) - tf(x_1) - (1-t)f(x_2)\), notice that \(g(0) = g(1) = 0\). Since \(g''(x) < 0\) over the whole interval, we must have two things: 1. \(g'(x)\) is increasing. 2. It \(g'(x) = 0\) can have at most one solution. Therefore \(g'(x)\) is initially \(0\), we have exactly one turning point. Therefore the function is initially decreasing and then increasing, therefore it is always negative and our inequality holds.
  1. \(\,\) \begin{align*} && f \left ( \frac{u+v+w}{3} \right) &= f \left ( \frac{2}{3}\cdot \frac{u+v}2+\frac{1}{3}w \right) \\ &&&\geq \frac23 f \left ( \frac{u+v}{2} \right) + \frac13 f(w) \\ &&&\geq \frac23 \left (\frac12 f(u) + \frac12 f(v) \right) + \frac13 f(w) \\ &&&= \frac{f(u)+f(v)+f(w)}{3} \end{align*}
  2. Notice that if \(A, B, C\) are angles in a triangle then they add to \(\pi\) \(0 < A,B,C < \pi\). We also have \(f(x) = \sin x \Rightarrow f''(x) = - \sin x < 0\) on this interval. Therefore \(\sin A + \sin B + \sin C \leq 3 \sin \frac{A+B+C}{3} = 3 \sin \frac{\pi}{3} = \frac{3\sqrt{3}}2\)
  3. Also notice that \begin{align*} && f(x) &= \ln ( \sin x) \\ \Rightarrow && f'(x) &= \frac{\cos x}{ \sin x} \\ && f''(x) &= -\textrm{cosec}^2 x < 0 \\ \\ \Rightarrow && \ln( \sin A) + \ln (\sin B) + \ln (\sin C) &\leq 3 \ln \left (\sin \left ( \frac{A + B+ C}{3} \right) \right) \\ &&&= 3 \ln \left ( \frac{\sqrt{3}}{2} \right) = \ln \frac{3\sqrt{3}}{8} \\ \Rightarrow && \sin A \sin B \sin C &\leq \frac{3\sqrt{3}}8 \end{align*}

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2013 Paper 2 Q1
D: 1600.0 B: 1484.0
differentiation tangent logarithmic function curve sketching inequality concavity optimisation geometric proof

  1. Find the value of \(m\) for which the line \(y = mx\) touches the curve \(y = \ln x\,\). If instead the line intersects the curve when \(x = a\) and \(x = b\), where \(a < b\), show that \(a^b = b^a\). Show by means of a sketch that \(a < \e < b\).
  2. The line \(y=mx+c\), where \(c>0\), intersects the curve \(y=\ln x\) when \(x=p\) and \(x=q\), where \(p < q\). Show by means of a sketch, or otherwise, that \(p^q > q^p\).
  3. Show by means of a sketch that the straight line through the points \((p, \ln p)\) and \((q, \ln q)\), where \(\e\le p < q\,\), intersects the \(y\)-axis at a positive value of \(y\). Which is greater, \(\pi^\e\) or \(\e^\pi\)?
  4. Show, using a sketch or otherwise, that if \(0 < p < q\) and \(\dfrac{\ln q - \ln p}{q-p} = \e^{-1}\), then \(q^p > p^q\).

Solution: \begin{questionparts} \item The tangent to \(y = \ln x\) is \begin{align*} && \frac{y - \ln x_1}{x - x_1} &= \frac{1}{x_1} \\ \Rightarrow && \frac{x_1y -x_1 \ln x_1}{ x- x_1} &= 1 \\ \Rightarrow && x_1 y - x_1 \ln x_1 &= x - x_1 \end{align*} So to run through the origin, we need \(\ln x_1 = 1 \Rightarrow x_1 = e\) so the line will be \(y = \frac1{e} x\) If \(ma = \ln a \Rightarrow m = \frac{\ln a}{a} = \frac{\ln b}{b} \Rightarrow b \ln a = a \ln b \Rightarrow a^b = b^a\). \item

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2007 Paper 2 Q7
D: 1600.0 B: 1516.0
exponentials and logarithms Jensen's inequality concavity AM-GM inequality inequality optimisation logarithms trigonometric triangle angles

A function \(\f(x)\) is said to be concave on some interval if \(\f''(x)<0\) in that interval. Show that \(\sin x\) is concave for \(0< x < \pi\) and that \(\ln x\) is concave for \(x > 0\). Let \(\f(x)\) be concave on a given interval and let \(x_1\), \(x_2\), \(\ldots\), \(x_n\) lie in the interval. Jensen's inequality states that \[ \frac1 n \sum_{k=1}^n\f(x_k) \le \f \bigg (\frac1 n \sum_{k=1}^n x_k\bigg) \] and that equality holds if and only if \(x_1=x_2= \cdots =x_n\). You may use this result without proving it.

  1. Given that \(A\), \(B\) and \(C\) are angles of a triangle, show that \[ \sin A + \sin B + \sin C \le \frac{3\sqrt3}2 \,. \]
  2. By choosing a suitable function \(\f\), prove that \[ \sqrt[n]{t_1t_2\cdots t_n}\; \le \; \frac{t_1+t_2+\cdots+t_n}n \] for any positive integer \(n\) and for any positive numbers \(t_1\), \(t_2\), \(\ldots\), \(t_n\). Hence:
    1. show that \(x^4+y^4+z^4 +16 \ge 8xyz\), where \(x\), \(y\) and \(z\) are any positive numbers;
    2. find the minimum value of \(x^5+y^5+z^5 -5xyz\), where \(x\), \(y\) and \(z\) are any positive numbers.

Solution: \begin{align*} && f(x) &= \sin x \\ \Rightarrow && f''(x) &= -\sin x \end{align*} which is clearly negative on \((0,\pi)\) since \(\sin\) is positive on this interval. \begin{align*} && f(x) &= \ln x \\ \Rightarrow && f''(x) &= -1/x^2 \end{align*} which is clearly negative for \(x > 0\)

  1. Since \(A,B,C\) are angles in a triangle, we must have \(0 < A,B,C< \pi\) and so we can apply Jensen with \(f = \sin\) to obtain: \begin{align*} &&\frac13( \sin A + \sin B + \sin C) &\leq \sin \left ( \frac{A+B+C}{3}\right) \\ &&&= \sin \frac{\pi}{3} = \frac{\sqrt{3}}2 \\ \Rightarrow && \sin A + \sin B + \sin C &\leq\frac{3\sqrt{3}}2 \end{align*}
  2. Suppose \(f(x) = \ln x\), then applying Jensen on the positive numbers \(t_1, \ldots, t_n\) we obtain \begin{align*} && \frac1n \left ( \sum_{i=1}^n \ln t_n \right) &\leq \ln \left ( \frac1n\sum_{i=1}^n t_n \right) \\ \Rightarrow && \frac1n \ln\left (\prod_{i=1} t_n\right)&\leq \ln \left ( \frac1n\sum_{i=1}^n t_n \right) \\ \Rightarrow && \ln\left (\left (\prod_{i=1} t_n\right)^{1/n}\right)&\leq \ln \left ( \frac1n\sum_{i=1}^n t_n \right) \\ \Rightarrow && \left (\prod_{i=1} t_n\right)^{1/n}&\leq\frac1n\sum_{i=1}^n t_n \\ \Rightarrow && \sqrt[n]{t_1t_2 \cdots t_n}&\leq\frac1n(t_1 + t_2 + \cdots + t_n) \tag{AM-GM}\\ \end{align*}
    1. Applying AM-GM with \(t_1 = x^4, t_2 = y^4, t_3 = z^4, t_4 = 2^4\) we have \begin{align*} && \frac{x^4+y^4+z^4+16}{4} & \geq \sqrt[4]{x^4y^4z^42^4} \\ \Rightarrow && x^4+y^4+z^4+16 &\geq 8xyz \end{align*}
    2. Applying AM-GM with \(t_1 = x^5, t_2 = y^5, t_3 = z^5, t_4 = 1^5, t_5 = 1^5\) we have \begin{align*} && \frac{x^5+y^5+z^5+1+1}{5} & \geq \sqrt[5]{x^5y^5z^5} \\ \Rightarrow && x^5+y^5+z^5+2 &\geq 5xyz \\ \Rightarrow && x^5+y^5+z^5 - 5xyz &\geq -2 \end{align*} Therefore the minimum is \(-2\)

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2001 Paper 1 Q7
D: 1500.0 B: 1516.0
differential equation first order cosmological model curve sketching concavity tangent inequality separable ODE integration

In a cosmological model, the radius \(\rm R\) of the universe is a function of the age \(t\) of the universe. The function \(\rm R\) satisfies the three conditions: $$ \mbox{\({\rm R}(0)=0\)}, \ \ \ \ \ \ \ \ \ \mbox{\({\rm R'}(t)>0\) for \(t>0\)}, \ \ \ \ \ \ \ \ \ \ \mbox{\({\rm R''}(t)<0\) for \(t>0\)}, \tag{*} $$ where \({\rm R''}\) denotes the second derivative of \(\rm R\). The function \({\rm H}\) is defined by \[ {\rm H} (t)= \frac{{\rm R}'(t)}{{\rm R}( t)}\;. \]

  1. Sketch a graph of \({\rm R} (t)\). By considering a tangent to the graph, show that \(t<1/{\rm H}(t)\).
  2. Observations reveal that \({\rm H}(t) = a/t\), where \(a\) is constant. Derive an expression for \({\rm R}(t)\). What range of values of \(a\) is consistent with the three conditions \((*)\)?
  3. Suppose, instead, that observations reveal that \({\rm H}(t)= b t^{-2}\), where \(b\) is constant. Show that this is not consistent with conditions \((*)\) for any value of \(b\).

Solution:

  1. \(\,\)
    TikZ diagram
    Notice the tangent must hit the \(y\)-axis above the origin, ie \begin{align*} && 0 &< R'(t)(0-t) + R(t) \\ \Rightarrow && R'(t) t &< R(t) \\ \Rightarrow && t &< \frac{R(t)}{R'(t)} = \frac{1}{H(t)} \end{align*}
  2. Suppose \(H(t) = a/t\) then \begin{align*} && \frac{R'}{R} &= \frac{a}{t} \\ \Rightarrow && \int \frac{1}{R} \d R &= \int \frac{a}{t} \d t \\ \Rightarrow && \ln R &= a \ln t + C \tag{t, R > 0} \\ \Rightarrow && R &= Kt^a \end{align*} Since we need \(R(t) > 0\), \(K > 0\), since \(R'(t) > 0\) we need \(a > 0\), since \(R''(t) < 0\) we need \(a(a-1) < 0\) ie \(0 < a < 1\)
  3. Suppose instead \(H(t) = bt^{-2}\) then \begin{align*} && \frac{R'}{R} &= \frac{b}{t^2} \\ \Rightarrow && \int \frac{1}{R} \d R &= \int \frac{b}{t^2} \d t \\ \Rightarrow && \ln R &= -bt^{-1} + C \tag{R > 0} \\ \Rightarrow && R &= Ke^{-b/t} \end{align*} Since \(R > 0\) we must have \(K > 0\). \begin{align*} R' > 0: && R' &= K(b/t^2)e^{-b/t} > 0 \\ \Rightarrow && b &> 0 \\ R'' < 0: && R'' &= K(b^2/t^4)e^{-b/t} -K2b/t^3 e^{-b/t} \\ &&&= Kb/t^4 (b-2t)e^{-b/t} < 0 \\ \Rightarrow && b &< 2t\\ \Rightarrow && b &< 2t \end{align*} which cannot be true for all \(t\), ie there is no \(b\) which satisfies this.

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