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Question

f:RR, f(x)=3x2+mx+nx2+1. If the range of f(x) is [4,3], then
  1. m=0
  2. n=4
  3. m=2
  4. n=4


Solution

The correct options are
A m=0
D n=4
y=3x2+mx+nx2+1
x2(y3)mx+yn=0
xRD0
m24(y3)(yn)0
m24(y2ny3y+3n)0
4y24y(n+3)+12nm20   (1)

Given range is [4,3]
So, y2+y120   (2)

Compare (1) and (2), we get
41=4(n+3)1=12nm212
m=0 and n=4
   

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