A squared inequality is a region between two branches

The curve C1C_1 has equation y24−x2=1\frac{y^2}{4}-x^2=1 and the curve C2C_2 has equation y=k−1xy=k-\frac1x, where kk is a positive constant. The horizontal asymptote of C2C_2 is a tangent to C1C_1.

(i) Sketch C1C_1, giving the equations of any asymptotes and the coordinates of any turning points. marks

(ii) Explain why k=2k=2. mark

(iii) On the same diagram as C1C_1, sketch C2C_2 showing the coordinates of the points of intersection of C1,C2C_1,C_2. marks

(iv) It is given that if [f(x)]2≤g(x)[f(x)]^2\le g(x), then −g(x)≤f(x)≤g(x)-\sqrt{g(x)}\le f(x)\le\sqrt{g(x)}. Using this result and the graphs in part (iii), solve the inequality (2−1x)2≤4(x2+1)(2-\frac1x)^2\le4(x^2+1). marks

[2016/DHS/Promo/10]

Total marks: 9

Solution

(i)

Hyperbolas

The curve C1C_1 has equation y24−x2=1\dfrac{y^2}{4}-x^2=1. Sketch C1C_1, giving the equations of any asymptotes and the coordinates of any turning points.

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Hint

  1. Which variable has the plus sign? That tells you the branches open up and down.

  2. Asymptotes: replace 1 by 0.

  3. Vertices: put x=0x=0.

Q17 vertical hyperbola
Q17 vertical hyperbolaScroll across to see the full diagram.

Answer · (i)

Vertices (0,±2)(0,\pm2); asymptotes y=±2xy=\pm2x.

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A hyperbola that opens up and down

Switch on the asymptotes and vertices.

Switch on a feature to add it to the sketch and see how it is found. A full-mark sketch shows all of them.

  • Two branches of C₁
  • Asymptotes y = ±2x
Ideas to try
Which branch is above the other? Where do they come closest?
(ii)

Tangency

The curve C2C_2 has equation y=k−1xy=k-\dfrac1x, where kk is a positive constant. The horizontal asymptote of C2C_2 is a tangent to C1C_1. Explain why k=2k=2.

Hint

  1. What is the horizontal asymptote of C2C_2?

  2. Which horizontal lines are tangent to C1C_1?

Answer · (ii)

k=2k=2.

(iii)

Intersections

On the same diagram as C1C_1, sketch C2C_2, showing the coordinates of the points of intersection of C1C_1 and C2C_2.

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Hint

  1. C2C_2 is y=2−1xy=2-\frac1x: asymptotes x=0x=0 and y=2y=2.

  2. Find where it meets the hyperbola with a graphing calculator.

Q17 hyperbola and reciprocal
Q17 hyperbola and reciprocalScroll across to see the full diagram.

Answer · (iii)

Intersections (−1.07235…,2.9325…)(-1.07235\ldots,2.9325\ldots) and (0.24632…,−2.0598…)(0.24632\ldots,-2.0598\ldots).

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Add the reciprocal curve

C₂ has asymptotes x = 0 and y = 2.

Switch on a feature to add it to the sketch and see how it is found. A full-mark sketch shows all of them.

  • Hyperbola C₁
  • C₂
  • Two intersections
Ideas to try
Which parts of C2C_2 lie between the two branches of C1C_1?
(iv)

Inequalities from graphs

Given that if [f(x)]2≤g(x)[f(x)]^2\le g(x) then −g(x)≤f(x)≤g(x)-\sqrt{g(x)}\le f(x)\le\sqrt{g(x)}, use this result and the graphs in part (iii) to solve the inequality (2−1x)2≤4(x2+1)\left(2-\dfrac1x\right)^2\le4(x^2+1).

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Hint

  1. Take f(x)=2−1xf(x)=2-\frac1x and g(x)=4(x2+1)g(x)=4(x^2+1).

  2. Then g=2x2+1\sqrt g=2\sqrt{x^2+1}, which is the upper branch of C1C_1.

  3. Read the inequality as: C2C_2 lies between the two branches of C1C_1.

Answer · (iv)

x≤αx\le\alpha or x≥βx\ge\beta, where α≈−1.07235\alpha\approx-1.07235, β≈0.24632\beta\approx0.24632.

α,β are the real roots of 4x⁴+4x−1=0; x=0 is outside these intervals.

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Where does C₂ lie between the two branches?

The vertical guides mark the two roots α ≈ −1.072 and β ≈ 0.246.

Switch on a feature to add it to the sketch and see how it is found. A full-mark sketch shows all of them.

  • Branches y = ±2√(x²+1)
  • C₂
Ideas to try
Test a point in the gap, such as x=−0.5x=-0.5: is C2C_2 between the branches?