Compose through a restricted range, then use symmetry for area

(a) Functions f:x↦(x−2)(x−4) f:x\mapsto(x-2)(x-4) for x∈Rx\in\mathbb R, x<4 x<4; g:x↦x/(x+3) g:x\mapsto x/(x+3) for x∈Rx\in\mathbb R, x≠−3 x\ne-3.

Determine if ff is one-one, justifying your answer. mark

Only one of fgfg and gfgf exists. Give a definition (including domain) of the composite that exists, and explain why the other does not exist. Find the range of the composite that exists. marks

(b) h:x↦a(1+e−x) h:x\mapsto a(1+e^{-x}), x∈R0+ x\in\mathbb R_0^+, where aa is positive. Find h−1(x) h^{-1}(x) and its domain. marks

Given that hh and h−1 h^{-1} intersect where x=bx=b, and ∫0bh(x) dx=I\int_0^b h(x)\, dx=I , find the area bounded by h,h−1 h, h^{-1} and the axes in terms of I,bI, b. marks

[2010/TJC/Prelim/II/3]

Total marks: 12

Solution

If this part uses an earlier result, revisit (b) inverse Inverse functions or open all working together.

(b) area

Area using symmetry

hh and h−1h^{-1} intersect where x=bx=b, and ∫0bh(x) dx=I\int_0^bh(x)\,dx=I. Find the area bounded by hh, h−1h^{-1} and the axes, in terms of II and bb.

Explore this idea

Hint

  1. The two curves are mirror images in y=xy=x, meeting at (b,b)(b,b).

  2. The area is twice the area between hh and y=xy=x on [0,b][0,b].

  3. That area is II minus the triangle under y=xy=x.

Answer · (b) area

2I−b22I-b^2.

Explore the idea

Open interactive exploration

Two mirror-image regions meeting at (b, b)

Each half is the area between a curve and y = x. Doubling gives the whole enclosed area.

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

  • H
  • H⁻¹
  • Y = x
Ideas to try
Which region is I? Which part of it must be removed before doubling?