A logarithmic involution

(a) The curve y=f(x) y=f(x) has horizontal asymptote y=ky=k and cuts the axes at (a,0)(a, 0) and (0,b)(0, b), where a,b,ka, b, k are non-zero constants. Given that f−1 f^{-1} exists, state if possible the coordinates where the following curves cut the axes and their asymptotes: (i) y=f(2x−3) y=f(2x-3), (ii) y=f−1(x) y=f^{-1}(x). marks

(b) g:x↦ln⁡(ex+5ex−1) g:x\mapsto\ln\left(\frac{e^x+5}{e^x-1}\right) for x∈Rx\in\mathbb R, x>0 x>0.

(i) Find g−1(x) g^{-1}(x) and state its domain. marks

The function h:x↦1+9−(x−2)2 h:x\mapsto1+\sqrt{9-(x-2)^2} for x∈Rx\in\mathbb R, −1≤x≤5-1\le x\le5.

(ii) Find the exact solutions of g2021h(x)=ln⁡2 g^{2021}h(x)=\ln2 in simplest form. marks

[2021/NYJC/Prelim/I/5]

Total marks: 8

Solution

(a)i

Transformations of intercepts

y=f(x)y=f(x) has horizontal asymptote y=ky=k and cuts the axes at (a,0)(a,0) and (0,b)(0,b). State, if possible, where y=f(2x−3)y=f(2x-3) cuts the axes and its asymptote.

Hint

  1. f(2x−3)f(2x-3) is f(x)f(x) shifted right 3 units, then scaled by 12\frac12 parallel to the xx-axis.

  2. Follow (a,0)(a,0): it moves to x=a+32x=\frac{a+3}2.

Answer · (a)i

Known x-intercept ((a+3)/2,0)((a+3)/2,0); asymptote y=ky=k.

A y-intercept (0,f(−3)) may exist but cannot be numerically determined.