examday

Community mocks

Sed (CoderLovesEggs) Paper 2 with Solutions · Q10

Paper 2Logic & proofFunctions & graphs

Consider the following statement:

A function f(x)f(x) is uniformly continuous if and only if for all ϵ>0\epsilon>0, there exists δ>0\delta>0 such that for all uu and vv, if ∣u−v∣<δ|u-v|<\delta, then ∣f(u)−f(v)∣<ϵ|f(u)-f(v)|<\epsilon.

It follows that f(x)f(x) is not uniformly continuous if and only if:

AFor all ϵ>0\epsilon>0, there exist δ>0\delta>0 and u,vu,v such that ∣u−v∣≥δ|u-v|\ge\delta and ∣f(u)−f(v)∣≥ϵ|f(u)-f(v)|\ge\epsilon.
BFor all ϵ>0\epsilon>0, there exist δ>0\delta>0 and u,vu,v such that ∣u−v∣<δ|u-v|<\delta and ∣f(u)−f(v)∣≥ϵ|f(u)-f(v)|\ge\epsilon.
CFor all ϵ,δ>0\epsilon,\delta>0, there exist u,vu,v such that ∣u−v∣<δ|u-v|<\delta and ∣f(u)−f(v)∣<ϵ|f(u)-f(v)|<\epsilon.
DFor all ϵ,δ>0\epsilon,\delta>0 and u,vu,v, ∣u−v∣≥δ|u-v|\ge\delta and ∣f(u)−f(v)∣≥ϵ|f(u)-f(v)|\ge\epsilon.
EThere exists ϵ>0\epsilon>0 where for all δ>0\delta>0, there exist u,vu,v such that ∣u−v∣≥δ|u-v|\ge\delta and ∣f(u)−f(v)∣≥ϵ|f(u)-f(v)|\ge\epsilon.
FThere exists ϵ>0\epsilon>0 where for all δ>0\delta>0, there exist u,vu,v such that ∣u−v∣<δ|u-v|<\delta and ∣f(u)−f(v)∣≥ϵ|f(u)-f(v)|\ge\epsilon.
GThere exist ϵ,δ>0\epsilon,\delta>0 and u,vu,v such that ∣u−v∣≥δ|u-v|\ge\delta and ∣f(u)−f(v)∣<ϵ|f(u)-f(v)|<\epsilon.
HThere exist ϵ,δ>0\epsilon,\delta>0 such that for all uu and vv, ∣u−v∣≥δ|u-v|\ge\delta and ∣f(u)−f(v)∣≥ϵ|f(u)-f(v)|\ge\epsilon.

Read the questions freely. Sign in to check your answer, see the worked solution and save your progress.

Sign in to check your answerPractise the whole paper →