A quarter ellipse becomes a self-inverse circle

Sketch the curve y2+ax2=5 y^2+ax^2=5 for x≥0 x\ge0, y≥0 y\ge0, where aa is positive. mark

f:x↦5−ax2 f:x\mapsto\sqrt{5-ax^2}, x∈Rx\in\mathbb R, 0≤x≤5/a0\le x\le\sqrt{5/a}; g:x↦1+e−x g:x\mapsto1+e^{-x}, x∈Rx\in\mathbb R, x≥0 x\ge0.

Show that f−1 f^{-1} exists and define f−1 f^{-1} in a similar form. marks

Given f2(x)=xf^2(x)=x for all x∈Rx\in\mathbb R, 0≤x≤5/a0\le x\le\sqrt{5/a},

(i) show a=1 a=1 without evaluating f2(x) f^2(x), marks

(ii) using (i), show fgfg exists and find its corresponding range. marks

[2011/ACJC/Prelim/I/9]

Total marks: 10

Solution

If this part uses an earlier result, revisit Sketch Sketching conics or open all working together.

Inverse

Inverse functions

f(x)=5−ax2f(x)=\sqrt{5-ax^2} for 0≤x≤5/a0\le x\le\sqrt{5/a}. Show that f−1f^{-1} exists and define it in a similar form.

Hint

  1. Show ff is one-one: any horizontal line meets the graph at most once.

  2. Rearrange y=5−ax2y=\sqrt{5-ax^2} for xx.

  3. Df−1=RfD_{f^{-1}}=R_f.

Answer · Inverse

f−1(x)=(5−x2)/af^{-1}(x)=\sqrt{(5-x^2)/a}, 0≤x≤50\le x\le\sqrt5.

Strictly decreasing original f is one-one.