Let f(x)=a2kx2k+a2k−1x2k−1+....+a1x+a0b2kx2k+b2k−1x2k−1+....+b1x+b0 where k is a positive interger, ai,bi∈R and a2k≠0,b2k≠0 such that b2kx2k+b2k−1x2k−1+.....+b1x+b0=0 has no real roots, then
A
f(x) must be one to one
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B
a2kx2k+a2k−1x2k−1+.....+a1x+a0=0 must have real roots
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C
f(x) must be many to one
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D
nothing can be said about the above options
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Solution
The correct option is Cf(x) must be many to one f(x) is continuous ∀xϵR. limx→−∞f(x)=limx→∞f(x)=a2kb2k ⇒f(x) is many to one. Hence, (c) is the correct answer.