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Question

Find for what real values of x the expressions
(a) x22x3
(b) 2x2+5x3 are positive or negative.
(c) If x2+2ax+103a>0, for all xϵR, then
(a) a<5
(b) 5<a<2
(c) a>5
(d) 2<a<5

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Solution

(a) x22x3=(x3)(x+1)=[x(1)](x3)
a=1 and b=3,a<b.
By §5 the above expression is +ive if x does not lie between a and b i.e. if x does not lie between 1 and 3. Again it will be ive if x lies between a and b i.e. x lies between 1 and 3.
(b) +ive if x does not lie between 3 and 12.
(c) Ans.(b). x2+2ax+103a=+ive real values of x if disc 4a24(103a)=ive and sign of the first term i.e., 1 is +ive.
a2+3a10<0
or (a+5)(a2)<0 5<a<2

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