A) 4
B) 6
C) 2
D) 1
Correct Answer: D
Solution :
[1] We have \[{{(x+1)}^{2}}>5x-1\] \[\Rightarrow {{x}^{2}}-3x+2>0\] \[\Rightarrow \,(x-1)(x-2)>0\] \[\Rightarrow x<1\,or\,x>2\] ...(i) And \[{{(x+1)}^{2}}<7x-3\] \[\Rightarrow \,{{x}^{2}}-5x+4<0\] \[\Rightarrow \,(x-1)(x-4)<0\] \[\Rightarrow \,1<x<4\] ...(ii) From (i) and (ii), common values are, \[x\in (2, 4)\] therefore, there is only one integral value of x which is 3.
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