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Question

If the inequality mx2+3x+4x2+2x+2<5 is satisfied ∀x∈R, then

A
m<12
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B
m<913
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C
m<7124
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D
m>913
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Solution

The correct option is B m<7124
mx2+3x+4x2+2x+2<5Weknowthatforf(x)=ax2+bx+cifa>0andD=b24ac<0Then,f(x)>0x2+2x+2>0D=4Somx2+3x+4<5(x2+2x+2)(5m)x2+7x+6>05m>0AndD<0m<5and4924(5m)<0m<7124..

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