If x is real, then the value of the expression x2+14x+9x2+2x+3 lies between
A
4 and 5
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B
−4 and 5
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C
−5 and 4
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D
None of these
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Solution
The correct option is C−5 and 4 Let y=x2+14x+9x2+2x+3 ⇒x2+14x+9=x2y+2xy+3y ⇒x2(y−1)+2x(y−7)+(3y−9)=0 Since, x is real, we have 4(y−7)2−4(3y−9)(y−1)>0 ⇒(y2+49−14y)−(3y2+9−12y)>0 ⇒(y+5)(y−4)<0 Therefore, y lies between −5 and 4.