Arcosine function

Given a function f:[-1, 1] \rightarrow [0, \pi] such that

(1)   \begin{equation*} \cos f(x) = x \quad \text{for all} \;\; x \in [-1 , 1] \end{equation*}

  1. Evaluate f(0).
  2. Prove that f is one to one.
  3. Prove that f \left ( \cos x \right ) = x for all x \in [0, \pi].
  4. Find the range of f.
  5. Sketch the graph of f.

Solution

  1. Setting x=0 at (1) we have that

        \[\cos f(0) =0 \Leftrightarrow x = \frac{\pi}{2} + \kappa \pi \; , \; \kappa \in \mathbb{Z}\]

    However,

        \begin{align*} 0 \leq f(0) \leq \pi &\Rightarrow 0 \leq \kappa \pi + \frac{\pi}{2} \leq \pi \\ &\Rightarrow - \frac{\pi}{2} \leq \kappa \pi \leq \frac{\pi}{2} \\ &\Rightarrow -\frac{1}{2} \leq \kappa \leq \frac{1}{2} \\ &\!\!\!\!\!\!\overset{\kappa \in \mathbb{Z}}{=\! =\! =\!\Rightarrow } \kappa =0 \end{align*}

    Thus f(0)=\frac{\pi}{2}.

  2. Let x_1, x_2 \in [-1, 1] such that f(x_1) = f(x_2). Thus,

        \begin{align*} f(x_1) = f(x_2) &\Leftrightarrow \cos f(x_1) = \cos f(x_2) \\ &\Leftrightarrow x_1 = x_2 \end{align*}

    Hence f is 1-1.

  3. Setting x = \cos x we have

        \begin{align*} \cos f(x) =x &\Leftrightarrow \cos f \left ( \cos x \right ) = \cos x \\ &\Leftrightarrow f \left ( \cos x \right ) = x \end{align*}

    since \cos is strictly decreasing in [0, \pi].

  4. f \left( [-1, 1] \right) = [ 0, \pi] because f^{-1}(x) = \cos x \; , \; x \in [0, \pi].
  5. The graph is seen at the following figure:

    Rendered by QuickLaTeX.com

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Trigonometric inequality

Let 0 \leq \alpha < \frac{\pi}{2}. Prove that

    \[2 \sin \alpha + \tan \alpha \geq 3\alpha\]

Hyperbolic Triangle

The vertices of a triangle lie on the hyperbola y=\frac{1}{x}. Prove that its orthocentre also lies on the hyperbola.

Solution

We are working on the following figure

Rendered by QuickLaTeX.com

Let \mathrm{A} \left( p, \frac{1}{p} \right) , \mathrm{B} \left( q, \frac{1}{q} \right) and \Gamma \left( r , \frac{1}{r} \right). Let us denote as \mathrm{H} its orthocentre. We have that:

    \[\lambda_{\mathrm{B}\Gamma} = \frac{\frac{1}{r} - \frac{1}{q}}{r-q} = - \frac{1}{qr}\]

Hence, the slope of the altitude \mathrm{AZ} is qr. Similarly,

    \[\lambda_{\mathrm{AB}} = \frac{\frac{1}{q} - \frac{1}{p}}{q-p} = - \frac{1}{pq}\]

Hence, the slope of the altitude \Gamma \mathrm{E} is pq. Hence,

    \[\left ( \varepsilon \right )_{\mathrm{B} \Gamma} : y - \frac{1}{p} = qr \left ( x - p \right )\]

and

    \[\left ( \varepsilon \right )_{\Gamma \mathrm{E}} : y - \frac{1}{r} = pq \left ( x - r \right )\]

Solving this linear system we have

    \begin{align*} \left ( \varepsilon \right )_{\mathrm{B} \Gamma} = \left ( \varepsilon \right )_{\Gamma \mathrm{E}} &\Leftrightarrow qr \left ( x- p \right ) + \frac{1}{p} = \frac{1}{r} + pq \left ( x-r \right ) \\ &\Leftrightarrow x = - \frac{1}{pqr} \end{align*}

and finally y=-pqr. So, \mathrm{H} \left( - \frac{1}{pqr} , -pqr \right). This proves the claim.

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Limit of radius of inscribed circle

Consider the points A(0, 0) , B(1, 0) , C(x, 1)  with x>0. Let r(x) be the radius of the inscribed circle of the triangle ABC.

Prove that

    \[\lim_{x \rightarrow +\infty} \rho(x)=0\]

Solution

Since \tan A = \frac{1}{x} we deduce that \hat{A} \rightarrow 0 as x \rightarrow +\infty. Since the incenter I lies on the bisector of A , it follows that if D is the projection of I on the x'x axis

\rho = AD \tan \frac {A}{2}\leq 1\tan \frac {A}{2} \rightarrow 0

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Limit

Evaluate the limit

    \[\ell =\lim_{x\rightarrow a} \frac{x^a-a^x}{x^x-a^a}\]

Solution

Rewrite the limit as

    \[\lim_{x\rightarrow a} \frac{x^a-a^x}{x^x-a^a} = \lim_{x\rightarrow a} \frac{\frac{x^a-a^a}{x-a} - \frac{a^x-a^a}{x-a}}{\frac{x^x-a^a}{x-a}}\]

Using the definition of the derivative we get that limit equals to

    \[\ell = \frac{1- \ln a}{1+\ln a}\]

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