OSH/3/Solution

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Solution key

  1. Solution
    Let $(a, f(a)) = (a, a/(a + 1))$ be a point on the curve $y = x/(x + 1)$. We have \begin{align*} f^{\prime}(x) = \dfrac{(x + 1) - x}{(x + 1)^2} = \dfrac{1}{(x + 1)^2}. \end{align*} Hence the slope of the tangent line at $x = a$ is \begin{align*} f^{\prime}(a) = \dfrac{1}{(a + 1)^2}. \; \; \text{[Step 1]} \end{align*} Thus the equation of the tangent line at $x = a$ to the curve $y = x/(x + 1)$ is given by \begin{align*} \tag{1} y = f^{\prime}(a)(x - a) + f(a) = \dfrac{1}{(a + 1)^2}(x - a) + \dfrac{a}{a + 1}. \text{[Step 2]} \end{align*} (Note that students might write the tangent line of the form \begin{align*} \dfrac{y - a/(a + 1)}{x - a} = f^{\prime}(a) = 1/(a + 1)^2. \end{align*} The students can get the same marks if they write the equation of the tangent line as above.)
    Assume that the tangent line given by (1) intersects the point $(1, 2)$. This implies that \begin{align*} 2 = \dfrac{1}{(a + 1)^2}(1 - a) + \dfrac{a}{a + 1}, \; \; \text{[Step 3]} \end{align*} and thus \begin{align*} 2 = \dfrac{(1 - a) + a(a + 1)}{(a + 1)^2}. \end{align*} Multiplying both sides by $(a + 1)^2$ and simplifying the right-hand side, we get \begin{align*} 2(a + 1)^2 = a^2 + 1. \end{align*} Therefore we deduce that \begin{align*} a^2 + 4a + 1 = 0. \; \; \text{[Step 4]} \end{align*} Solving the above equation, we get \begin{align*} a = -2 \pm \sqrt{3}. \; \; \text{[Step 5]} \end{align*} Plugging $a = -2 + \sqrt{3}$ into (1), we get the equation of the tangent line at $x = -2 + \sqrt{3}$ given by \begin{align*} y = \dfrac{1}{(-1 + \sqrt{3})^2}(x - (-2 + \sqrt{3})) + \dfrac{-2 + \sqrt{3}}{-1 + \sqrt{3}}. \; \; \text{[Step 6]} \end{align*} Plugging $a = -2 - \sqrt{3}$ into (1), we get the equation of the tangent line at $x = -2 - \sqrt{3}$ given by \begin{align*} y = \dfrac{1}{(-1 - \sqrt{3})^2}(x - (-2 - \sqrt{3})) + \dfrac{-2 - \sqrt{3}}{-1 - \sqrt{3}}. \; \; \text{[Step 7]} \end{align*}
    Students should:
    • Compute correctly the slope of the tangent line at $x = a$ (see [Step 1]) (1 pt).
    • Write correctly the equation of the tangent line at $x = a$ (see [Step 2] and the comments following Step 2) (2 pt).
    • Plug in the point $(1, 2)$ into the equation of the tangent line at $x = a$ (see [Step 3]) (1 pt).
    • Simplify and get the correct quadratic equation of $a$ (see [Step 4]) (2 pt).
    • Get the correct two values of $a$ (see [Step 5]) (2 pt) (1pt for each value of $a$).
    • Get the equation of the 1st tangent line (see [Step 6]) (1 pt).
    • Get the equation of the 2nd tangent line (see [Step 7]) (1 pt).

    Total 10 pts.
  2. Solution
    We have \begin{align*} f^{\prime}(x) = \dfrac{3Ax^2(a^3 + x^3) - Ax^3(3x^2)}{(a^3 + x^3)^2} = \dfrac{3a^3Ax^2}{(a^3 + x^3)^2} \; \; \text{[Step 1]} \end{align*} Hence \begin{align*} f^{\prime \prime}(x) &= \dfrac{6a^3Ax(a^3 + x^3)^2 - 2(a^3 + x^3)3x^2(3a^3Ax^2)}{(a^3 + x^3)^4} \; \; \text{[Step 2]}\\ &= \dfrac{(a^3 + x^3)(6a^3Ax(a^3 + x^3) - 18a^3Ax^4)}{(a^3 + x^3)^4} \\ &= \dfrac{(a^3 + x^3)(6a^6Ax - 12a^3Ax^4)}{(a^3 + x^3)^4} \\ &= \dfrac{6a^3Ax(a^3 + x^3)(a^3 - 2x^3)}{(a^3 + x^3)^4}. \; \; \text{[Step 3]} \end{align*} (In Step 2 above, I assume that students know chain rule to compute the derivative of $(a^3 + x^3)^2$. Another possibility is that students might write $(a^3 + x^3)^2 = (a^3 + x^3)(a^3 + x^3)$, and hence use the product rule to get the derivative of $(a^3 + x^3)^2$ equal to $2(a^3 + x^3)3x^2$ after simplification)
    Consider the equation $f^{\prime \prime}(x) = 0$. We have \begin{align*} f^{\prime \prime}(x) = \dfrac{6a^3Ax(a^3 + x^3)(a^3 - 2x^3)}{(a^3 + x^3)^4} = 0. \; \; \text{[Step 4]} \end{align*} Hence \begin{align*} x = 0, \; a^3 + x^3 = 0, \; \text{or }$a^3 - 2x^3 = 0. \; \; \text{[Step 5]} \end{align*} Thus \begin{align*} x = 0, \; x = - a, \text{or } x = \dfrac{a}{\sqrt[3]{2}}. \; \; \text{[Step 6]} \end{align*} Since the leading coefficient of $6a^3Ax(a^3 + x^3)(a^3 - 2x^3)$ is $-12a^3A < 0$ and $a > 0$, we have the following table for the signs of $f^{\prime \prime}(x)$.
    x (-∞, -a) (-a, 0) (0, a/∛2) (a/∛2,∞)
    f"(x) + - + -
    concavity of f(x) concave up concave down concave up concave down

    [Step 7]


    The reason that we need to check whether f"(x) changes sign is to ensure that the points x=0, -a, a/∛2 are inflection points. Note further that the function f(x) is not defined at x=-a. From the above table, we conclude that x = 0 and x=a/∛2 are inflection points. [Step 8]


    Students should:

    • Get the first derivative (see [Step 1]) (1 pt).
    • Apply correctly the quotient rule to get the second derivative (see [Step 2]) (1 pt).
    • Simplify to get the second derivative of the form in Step 3 (see [Step 3]) (2 pt).
    • Set up the equation of the second derivative equal to zero (see [Step 4]) (1 pt).
    • Solve the equation and deduce that each factor is equal to zero (see [Step 5]) (1 pt).
    • Get all three solutions (see [Step 6]) (1 pt).
    • Get the signs of the second derivative (see [Step 7]) (2 pt).
    • Conclude correctly what the inflection points are (see [Step 8]) (1 pt).
    Total: 10pts.