1px2+qx+r=p(x-A)(x-B)
lim x→n |p||(x-A)(x-B)|/p(x-A)(x-B) =1
|p|/p x(-1)=1 , if n lies between A and B --------------------1
|p|=-p
p is negative, p<0-----------------2
so from 1 and 2 option a
if A and B are the roots of px2+qx+r=0
where 1<A<B then lim x→n |px2+qx+r|/px2+qx+r =1
when
a) p<0 and A<n<B
b)p>0 and n>1
c)p<0 and n>1
d)|p|/p=1 and n>A
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2 Answers
106as the roots of the eqn are positive.. hence product = r/p >0
==> r and p are of same sign
and -q/p >0
==> q and p are of opposite sign
CASE I p>0
... now... p>0 ==> r>0 and q<0 .. so when x lies between the roots i.e. when A<x<B then f(x) = px2 + qx + r is <0
So. the limit when A<n<B = -1
and when x is outside the roots then f(x) >0 so the limit is 1
Now.. option (b) is not clear whether n lies between or outside the roots so discarded
Option (d) .. here p=1 i.e. >0 and n>A but it doesnt say whether n>B or n<B so discarded again
CASE II p<0
then here .. q>0 and r<0 ..
when x lies between the roots .. the f(x) > 0
So. option (a) is correct..
Again option (c) is not clear whether n lies between or outside the roots so discarded
So answer is (a)
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