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Question: Answered & Verified by Expert
If x2+mx+1x2+x+1<3 for all real x, then
MathematicsFunctionsJEE Main
Options:
  • A m<-1
  • B -1<m<6
  • C -1<m<5
  • D m>6
Solution:
1928 Upvotes Verified Answer
The correct answer is: -1<m<5
We have x2+x+1=x+122+34>0
So, -3<x2+mx+1x2+x+1<3
-3x2+x+1<x2+mx+1<3x2+x+1
4x2+m+3x+4>0 and 2x2+3-mx+2>0 for real x
m+32-4×4×4<0 and 3-m2-4×2×2<0
m+3+8m+3-8<0 and m-3+4m-3-4<0
m+11m-5<0 and m+1m-7<0
-11<m<5 and -1<m<7
Hence, -1<m<5

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