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 Bm<7124 mx2+3x+4x2+2x+2<5Weknowthatforf(x)=ax2+bx+cifa>0andD=b2−4ac<0Then,f(x)>0x2+2x+2>0D=−4Somx2+3x+4<5(x2+2x+2)(5−m)x2+7x+6>05−m>0AndD<0m<5and49−24(5−m)<0m<7124..