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Question

The number of integral value(s) of y for which (y25y+3)(x2+x+1)2x<0 for all xR is

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Solution

(y25y+3)(x2+x+1)<2x, xR
y25y+3<2xx2+x+1(x2+x+1>0 xR)

L.H.S. must be less than the least value of R.H.S.
Now, finding the least value of R.H.S.

Let 2xx2+x+1=p
px2+(p2)x+p=0
As xR, so
D0(p2)24p203p24p+403p2+4p403p2+6p2p40(3p2)(p+2)0p[2,23]

Therefore, the least value of R.H.S. is 2, so
y25y+3<2y25y+5<0y(552,5+52)

Hence, the possible intergral values are 2,3

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