The curve C has equation x+y=(x−y)2x + y = (x - y)^2. It is given that C has only one turning point.

  1. (a)

    Show that 1−dydx=21+2x−2y1 - \frac{dy}{dx} = \frac{2}{1 + 2x - 2y}.

  2. (b)

    Hence, or otherwise, show that d2ydx2=(1−dydx)3\displaystyle \frac{d^2 y}{dx^2} = \left(1 - \frac{dy}{dx}\right)^3.

  3. (c)

    Hence state, with a reason, whether the turning point is a minimum or a maximum.

  4. (7)
    1. (b)

      The function g−1\text{g}^{-1} exists if the domain of g\text{g} is restricted to x≤kx \le k. State the greatest possible value of kk.

    2. (c)

      In the rest of the question, the domain of g\text{g} is x≤kx \le k, where kk takes the value determined in part (b).

      Find g−1(x)\text{g}^{-1}(x) and state the domain of g−1\text{g}^{-1}.

    3. (d)

      Sketch, on the axes given below, the graphs of y=g(x)y = \text{g}(x) and y=g−1(x)y = \text{g}^{-1}(x). Label the two graphs clearly. Write down the equation of the line in which the graph of y=g(x)y = \text{g}(x) must be reflected in order to obtain the graph of y=g−1(x)y = \text{g}^{-1}(x).

Implicit parametricNature stationary pointsDomain restriction for inverseInverse and compositeGraph of inverseGraphical interpretation derivatives中文
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Worked solution

8(a)

1. Differentiate both sides of the given equation implicitly with respect to $x$.

ddx(x+y)=ddx((x−y)2)\frac{d}{dx}(x + y) = \frac{d}{dx}((x - y)^2)
1+dydx=2(x−y)(1−dydx)1 + \frac{dy}{dx} = 2(x - y)\left(1 - \frac{dy}{dx}\right)

2. Expand and rearrange the equation to make $\frac{dy}{dx}$ the subject.

1+dydx=2x−2y−(2x−2y)dydx1 + \frac{dy}{dx} = 2x - 2y - (2x - 2y)\frac{dy}{dx}
dydx+(2x−2y)dydx=2x−2y−1\frac{dy}{dx} + (2x - 2y)\frac{dy}{dx} = 2x - 2y - 1
dydx(1+2x−2y)=2x−2y−1\frac{dy}{dx}(1 + 2x - 2y) = 2x - 2y - 1
dydx=2x−2y−11+2x−2y\frac{dy}{dx} = \frac{2x - 2y - 1}{1 + 2x - 2y}

3. Substitute the expression for $\frac{dy}{dx}$ into $1 - \frac{dy}{dx}$ and simplify to obtain the required result.

1−dydx=1−2x−2y−11+2x−2y1 - \frac{dy}{dx} = 1 - \frac{2x - 2y - 1}{1 + 2x - 2y}
1−dydx=(1+2x−2y)−(2x−2y−1)1+2x−2y1 - \frac{dy}{dx} = \frac{(1 + 2x - 2y) - (2x - 2y - 1)}{1 + 2x - 2y}
1−dydx=1+2x−2y−2x+2y+11+2x−2y1 - \frac{dy}{dx} = \frac{1 + 2x - 2y - 2x + 2y + 1}{1 + 2x - 2y}
1−dydx=21+2x−2y1 - \frac{dy}{dx} = \frac{2}{1 + 2x - 2y}

Answer

Shown.\text{Shown.}

8(b)

1. Differentiate the result from part (a) implicitly with respect to $x$.

From part (a), we are given:
1−dydx=21+2x−2y=2(1+2x−2y)−11 - \frac{dy}{dx} = \frac{2}{1 + 2x - 2y} = 2(1 + 2x - 2y)^{-1}

Differentiating both sides with respect to xx using the chain rule:
−d2ydx2=−2(1+2x−2y)−2⋅ddx(1+2x−2y)-\frac{d^2y}{dx^2} = -2(1 + 2x - 2y)^{-2} \cdot \frac{d}{dx}(1 + 2x - 2y)
−d2ydx2=−2(1+2x−2y)−2(2−2dydx)-\frac{d^2y}{dx^2} = -2(1 + 2x - 2y)^{-2} \left(2 - 2\frac{dy}{dx}\right)

2. Simplify the differentiated equation.

Factor out 22 from the bracket on the right hand side:
−d2ydx2=−4(1+2x−2y)−2(1−dydx)-\frac{d^2y}{dx^2} = -4(1 + 2x - 2y)^{-2} \left(1 - \frac{dy}{dx}\right)

Divide by −1-1 to isolate d2ydx2\frac{d^2y}{dx^2}:
d2ydx2=4(1+2x−2y)2(1−dydx)\frac{d^2y}{dx^2} = \frac{4}{(1 + 2x - 2y)^2} \left(1 - \frac{dy}{dx}\right)

3. Substitute the expression from part (a) to prove the final result.

Notice that 4(1+2x−2y)2\frac{4}{(1 + 2x - 2y)^2} is the square of 21+2x−2y\frac{2}{1 + 2x - 2y}.

From part (a), we know that 21+2x−2y=1−dydx\frac{2}{1 + 2x - 2y} = 1 - \frac{dy}{dx}. Squaring both sides gives:
4(1+2x−2y)2=(1−dydx)2\frac{4}{(1 + 2x - 2y)^2} = \left(1 - \frac{dy}{dx}\right)^2

Substitute this identity into the second derivative expression:
d2ydx2=(1−dydx)2(1−dydx)\frac{d^2y}{dx^2} = \left(1 - \frac{dy}{dx}\right)^2 \left(1 - \frac{dy}{dx}\right)
d2ydx2=(1−dydx)3\frac{d^2y}{dx^2} = \left(1 - \frac{dy}{dx}\right)^3

Answer

\text{Shown}

8(c)

1. Substitute the condition for a turning point into the expression for the second derivative.

At the turning point, dydx=0\frac{dy}{dx} = 0.
Substitute this into the expression for d2ydx2\frac{d^2 y}{dx^2} given in part (b):
d2ydx2=(1−0)3=1\frac{d^2 y}{dx^2} = \left(1 - 0\right)^3 = 1

2. Use the second derivative test to conclude the nature of the turning point.

Since d2ydx2=1>0\frac{d^2 y}{dx^2} = 1 > 0 at the turning point, the point is a minimum.

Answer

The turning point is a minimum because d2ydx2=1>0\frac{d^2 y}{dx^2} = 1 > 0.

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