Porisma

Worked solution

A parameter hyperbola and a logarithmic input

(i) Sketch the curve with equation y=x+kx+1y=\dfrac{x+k}{x+1}, where k>1k>1.

State the equations of any asymptotes and the coordinates of the points where the curve crosses the axes. marks

(ii) Hence solve the inequality ln⁡x+kln⁡x+1≥0\dfrac{\ln x+k}{\ln x+1}\geq0. marks

[2016/VJC/Promo/10(a)(modified)]

CJC · Review Problems · Question 14

Answer

Asymptotes x=−1x=-1, y=1y=1; intercepts (−k,0)(-k,0), (0,k)(0,k). (ii) (0,e−k]∪(e−1,∞)(0,e^{-k}]\cup(e^{-1},\infty).

(i)

Write y=1+k−1x+1y=1+\dfrac{k-1}{x+1}. Since k>1k>1, the vertical asymptote is x=−1x=-1 and the horizontal asymptote is y=1y=1. The intercepts are (−k,0)(-k,0) and (0,k)(0,k). The left branch lies below y=1y=1, and the right branch lies above it; both decrease. Since −k<−1-k<-1, the graph is nonnegative for x≤−kx\leq-k or x>−1x>-1.

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(ii)

Replace the graph input by ln⁡x\ln x, with x>0x>0 and ln⁡x≠−1\ln x\ne-1. Then ln⁡x≤−k\ln x\leq-k or ln⁡x>−1\ln x>-1. Since exponentiation preserves order, the answer is 0<x≤e−k0<x\leq e^{-k} or x>e−1x>e^{-1}.

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Compare your sketch

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For k>1, the hyperbola has asymptotes x=−1 and y=1, intercepts (−k,0) and (0,k), a decreasing left branch below y=1 and a decreasing right branch above y=1. The drawing uses k=2 solely to illustrate this family; all stated results retain general k>1.
For k>1, the hyperbola has asymptotes x=−1 and y=1, intercepts (−k,0) and (0,k), a decreasing left branch below y=1 and a decreasing right branch above y=1. The drawing uses k=2 solely to illustrate this family; all stated results retain general k>1.
For k>1, the hyperbola has asymptotes x=−1 and y=1, intercepts (−k,0) and (0,k), a decreasing left branch below y=1 and a decreasing right branch above y=1. The drawing uses k=2 solely to illustrate this family; all stated results retain general k>1.

Common mistakes

The numerator zero is included; the denominator zero and nonpositive logarithm inputs are excluded.

Solution notes

The diagrams are independently reconstructed mathematical equivalents. The plotted geometry is illustrative; use the labelled exact or approximate values and the stated domain.

The source labels the deduction as (i) again in its solution; it answers printed part (ii). The reconstruction preserves the question labels. The illustrative graph uses k=2 without restricting the general result.

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