Problems

Solution:

1. \(\,\) \begin{align*} && f(x) &= (1+x^2)e^x \\ && f'(x) &= 2xe^x + (1+x^2)e^x \\ &&&= (1+x)^2e^x \geq 0 \end{align*}
   

   + - ↺
   As \(x \to -\infty\), \(f(x) \to 0\) and as \(x \to \infty\), \(f(x) \to +\infty\) and since \(f\) is strictly increasing we have exactly one solution to \(f(x) = k\) on \((0,\infty)\). Since \(f(x) > 0\) there are no solutions if \(k \leq 0\).
2. Considering the function \(g(x) = (e^x-1)-k\tan^{-1} x \) then \(g'(x) = e^x - \frac{k}{1+x^2}\) therefore \(g'(x)\) has exactly one turning point when \(k > 0\) and \(0\) otherwise at the root of \(f(x) = k\) Notice also that \(g(0) = 0\) so we already have one solution, and \(g'(0) = 1 - k > 0\). Notice from our sketch that if \(0 < k < 1\) the root for \(f(x) = k\) has \(x \leq 0\), so our turning point is to the left of the origin and we are interested in the behaviour of \(g(x)\) as \(x \to -\infty\). (ie do we cross the axis again). \(\lim_{x \to -\infty} \left [ e^x - 1 - k \tan^{-1} x \right] = 0 - 1 +k \frac{\pi}{2} = k \frac{\pi}{2} - 1\). if this is positive, ie if \(k > \frac{2}{\pi}\) there are two solutions, otherwise there is only one real root.

Solution: \begin{align*} && y &= \ln x - x \ln a \\ \Rightarrow && y' &= \frac1x - \ln a \\ && y'' &= -\frac{1}{x^2} \end{align*} Therefore the maximum is when \(x = \frac{1}{\ln a}\) and \(y_{max} = -\ln \ln a - 1\). If \(y_{max} < 0\) then \(y \neq 0\). But that's equivalent to \(a > e^{1/e}\). \begin{align*} && 0 &> -\ln \ln a - 1 \\ \Leftrightarrow && 1 &> - \ln \ln a \\ \Leftrightarrow && \ln \ln a &>-1 \\ \Leftrightarrow && \ln a &> e^{-1} \\ \Leftrightarrow && a & > e^{1/e} \end{align*} If \(0 < a < 1\) then, when \(x\) is small, \(\ln x - x \ln a\) is large and negative. When \(x\) is large and positive \(\ln x\) is positive and \(-x \ln a\) is positive. We also notice there is no turning point. Hence exactly one solution

Solution:

1. \(\,\) \begin{align*} && c &= (x-1)^4+(x+1)^4 \\ &&&= 2x^4+12x^2+2 \\ \Rightarrow && 0 &= (x^2+6)^2-\frac{c}{2} - 35 \\ \Rightarrow && \underbrace{x^2+6}_{\geq 6} &= \pm \sqrt{35 + \frac{c}{2}}\\ \end{align*} Therefore there are two solutions if \(c > 2\), one solution if \(c = 2\) and no solutions otherwise.
2. \(\,\) This equation is the same equation if \(y = x-2\), ie there are two solutions if \(c > 2\), one solution if \(c = 2\) and no solutions otherwise.
3. Rewriting as \(|x-1|+|x+1| = c\) we have For \(x < -1\): \(1-x-1-x = -2x\) For \(-1 \leq x \leq 1\): \(1-x+x+1 = 2\) For \(x > 1\): \(x-1+x+1 = 2x\) Therefore there are infinitely many solutions if \(c = 2\) (the interval \([-3,-1]\)), two solutions if \(c > 2\) and none otherwise.
4. Rewriting as \((x-1)^3+(x+1)^3\) we have \(x^3+6x = c\). Notice that \(3x^2+6 > 0\) so the function is increasing, ie there is one solution for all \(c\)