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2022 Paper 2 Q5
D: 1500.0 B: 1500.0
inequality coordinate geometry triangle perpendicular distance optimisation area algebraic identity cauchy-schwarz inequality

  1. Given that \(a > b > c > 0\) are constants, and that \(x\), \(y\), \(z\) are non-negative variables, show that \[ax + by + cz \leqslant a(x + y + z).\]
In the acute-angled triangle \(ABC\), \(a\), \(b\) and \(c\) are the lengths of sides \(BC\), \(CA\) and \(AB\), respectively, with \(a > b > c\). \(P\) is a point inside, or on the sides of, the triangle, and \(x\), \(y\) and \(z\) are the perpendicular distances from \(P\) to \(BC\), \(CA\) and \(AB\), respectively. The area of the triangle is \(\Delta\).
    1. Find \(\Delta\) in terms of \(a\), \(b\), \(c\), \(x\), \(y\) and \(z\).
    2. Find both the minimum value of the sum of the perpendicular distances from \(P\) to the three sides of the triangle and the values of \(x\), \(y\) and \(z\) which give this minimum sum, expressing your answers in terms of some or all of \(a\), \(b\), \(c\) and \(\Delta\).
    1. Show that, for all real \(a\), \(b\), \(c\), \(x\), \(y\) and \(z\), \[(a^2+b^2+c^2)(x^2+y^2+z^2) = (bx-ay)^2 + (cy-bz)^2 + (az-cx)^2 + (ax+by+cz)^2.\]
    2. Find both the minimum value of the sum of the squares of the perpendicular distances from \(P\) to the three sides of the triangle and the values of \(x\), \(y\) and \(z\) which give this minimum sum, expressing your answers in terms of some or all of \(a\), \(b\), \(c\) and \(\Delta\).
  3. Find both the maximum value of the sum of the squares of the perpendicular distances from \(P\) to the three sides of the triangle and the values of \(x\), \(y\) and \(z\) which give this maximum sum, expressing your answers in terms of some or all of \(a\), \(b\), \(c\) and \(\Delta\).

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2017 Paper 2 Q4
D: 1600.0 B: 1500.0
integration Schwarz inequality Cauchy-Schwarz inequality inequality definite integrals proof choosing functions strategically

The Schwarz inequality is \[ \left( \int_a^b \f(x)\, \g(x)\,\d x\right)^{\!\!2} \le \left( \int_a^b \big( \f(x)\big)^2 \d x \right) \left( \int_a^b \big( \g(x)\big)^2 \d x \right) . \tag{\(*\)} \]

  1. By setting \( \f(x)=1\) in \((*)\), and choosing \(\g(x)\), \(a\) and \(b\) suitably, show that for \(t> 0\,\), \[ \frac {\e^t -1}{\e^t+1} \le \frac t 2 \,. \]
  2. By setting \( \f(x)= x\) in \((*)\), and choosing \( \g(x)\) suitably, show that \[ \int_0^1\e^{-\frac12 x^2}\d x \ge 12 \big(1-\e^{-\frac14})^2 \,. \]
  3. Use \((*)\) to show that \[ \frac {64}{25\pi} \le \int_0^{\frac12\pi} \!\! {\textstyle \sqrt{\, \sin x\, } } \, \d x \le \sqrt{\frac \pi 2 } \,. \]

Solution:

  1. Let \(f(x) = 1, g(x) = e^x, a = 0, b = t\), so \begin{align*} && \left ( \int_0^t e^x \d x \right)^2 &\leq \left (\int_0^t 1^2 \d x \right) \cdot \left (\int_0^t (e^x)^2 \d x \right) \\ \Rightarrow && (e^t-1)^2 &\leq t \cdot (\frac12e^{2t} - \frac12) \\ \Rightarrow && \frac{e^t-1}{e^t+1} & \leq \frac{t}{2} \end{align*}
  2. Let \(f(x) = x, g(x) = e^{-\frac14 x^2}, a = 0, b = 1\) \begin{align*} && \left ( \int_0^1 xe^{-\frac14 x^2} \d x \right)^2 &\leq \left (\int_0^1 x^2 \d x \right) \cdot \left (\int_0^1 (e^{-\frac14x^2})^2 \d x \right) \\ \Rightarrow && \left ( \left [-2e^{-\frac14x^2} \right]_0^1 \right)^2 & \leq \frac{1}{3} \int_0^1 e^{-\frac12 x^2} \d x \\ \Rightarrow && \int_0^1 e^{-\frac12 x^2} \d x & \geq 12(1-e^{-\frac14})^2 \end{align*}
  3. Let \(f(x) = 1, g(x) = \sqrt{\sin x}, a = 0, b = \tfrac12 \pi\), then \begin{align*} && \left ( \int_0^{\frac12 \pi} \sqrt{\sin x} \d x \right)^2 &\leq \left (\int_0^{\frac12 \pi} 1^2 \d x \right) \cdot \left (\int_0^{\frac12 \pi}|\sin x| \d x \right) \\ &&&= \frac{\pi}{2} \cdot 1 \\ \Rightarrow && \int_0^{\frac12 \pi} \sqrt{\sin x} \d x & \leq \sqrt{\frac{\pi}{2}} \end{align*} Let \(f(x) =(\sin x)^{\frac14}, g(x) = \cos x, a = 0, b = \tfrac12 \pi\), so \begin{align*} && \left ( \int_0^{\frac12 \pi} (\sin x)^{\frac14} \cos x \d x \right)^2 &\leq \left (\int_0^{\frac12 \pi} \cos^2 x \d x \right) \cdot \left (\int_0^{\frac12 \pi}\sqrt{\sin x} \d x \right) \\ \Rightarrow &&\left ( \left [\frac45 (\sin x)^{\frac54} \right]_0^{\frac12 \pi} \right)^2 & \leq \frac{\pi}{4} \int_0^{\frac12 \pi}\sqrt{\sin x} \d x \\ \Rightarrow && \frac{64}{25\pi} &\leq \int_0^{\frac12 \pi}\sqrt{\sin x} \d x \end{align*}

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2006 Paper 2 Q6
D: 1600.0 B: 1516.0
vectors scalar product Cauchy-Schwarz inequality inequality equality condition proof

By considering a suitable scalar product, prove that \[ (ax+by+cz)^2 \le (a^2+b^2+c^2)(x^2+y^2+z^2) \] for any real numbers \(a\), \(b\), \(c\), \(x\), \(y\) and \(z\). Deduce a necessary and sufficient condition on \(a\), \(b\), \(c\), \(x\), \(y\) and \(z\) for the following equation to hold: \[ (ax+by+cz)^2 = (a^2+b^2+c^2)(x^2+y^2+z^2) \,. \]

  1. Show that \((x+2y+2z)^2 \le 9(x^2+y^2+z^2)\) for all real numbers \(x\), \(y\) and \(z\).
  2. Find real numbers \(p\), \(q\) and \(r\) that satisfy both \[ p^2+4q^2+9r^2 = 729 \text{ and } 8p+8q+3r = 243\,. \]

Solution: Consider \(\begin{pmatrix} a \\ b \\ c \end{pmatrix}\), \(\begin{pmatrix} x \\ y \\ z \end{pmatrix}\), then we know that \begin{align*} && \begin{pmatrix} a \\ b \\ c \end{pmatrix} \cdot \begin{pmatrix} x \\ y \\ z \end{pmatrix} &= \sqrt{a^2+b^2+c^2} \sqrt{x^2+y^2+z^2} \cos \theta \\ \Rightarrow && (ax+by+cz)^2 &= (a^2+b^2+c^2)(x^2+y^2+z^2) \cos^2 \theta \\ &&&\leq (a^2+b^2+c^2)(x^2+y^2+z^2) \end{align*} For equality to hold, we must have that the vectors are parallel, ie \(\begin{pmatrix} a \\ b \\ c \end{pmatrix} = \lambda \begin{pmatrix} x \\ y \\ z \end{pmatrix}\)

  1. By applying our inequality from the first part with \(a=1, b = 2, c=2\) we have \((x+2y+2z)^2 \leq (1+2^2+2^2)(x^2+y^2+z^2) = 9(x^2+y^2+z^2)\)
  2. Since \begin{align*} && (p^2+(2q)^2+(3r)^2)\left (8^2 +4^2+1^2 \right) &\geq (8p+8q+3r)^2 \\ \Leftrightarrow && 729 \cdot 81 &\geq 243^2 \\ &&3^6 \cdot 3^4 &\geq 3^{10} \end{align*} Therefore we must be in the equality case, ie \(p = 8\lambda, 2q = 4\lambda, 3r = \lambda\) as well as \(64\lambda + 16\lambda +\lambda = 243 \Rightarrow 81\lambda = 243 \Rightarrow \lambda = 3\) so we have \[ (p,q,r) = \left (24, 6, 1 \right) \]

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