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Question

f:RR,f(x)=3x2+mx+nx2+1. If the range of this functionis [4,3), then find the value of |m+n|.

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Solution

Given, y=3x2+mx+nx2+1

x2(y3)mx+yn=0

xR

D0

m24(y3)(yn)0

m24(y2ny3y+3n)0


4y24y(n+3)+12nm20

Also given (y+4)(y3)0

y2+y120

Compare (1) and (2), we get 41=4(n+3)1=12nm212

m=0 and n=4

|m+n|=|4+0|=4

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