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Q.

Let f:RR be a function defined by f(x)=2x321x2+78x+24. Number of integers in the solution set of x satisfying the inequality fff(x)2x3ff2x3f(x) is 

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answer is 5.

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Detailed Solution

f(x)=2x321x2+78x+24f(x)=6x27x+13f(x)>0xRf(x) is increasing function

Now, fff(x)2x3ff2x3f(x)
ff(x)2x3f2x3f(x)f(x)2x32x3f(x)f(x)2x37x226x80x27,4

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