JEE Advanced 2016JEE Adv 2016 Paper 1, Paper 1MathematicsHardUnclassified (maths)The least value of α∈R\alpha \in \mathbb{R}α∈R for which 4αx2+1x≥14 \alpha x^{2}+\frac{1}{x} \geq 14αx2+x1≥1, for all x>0x>0x>0, isAOption A. 164\frac{1}{64}641BOption B. 132\frac{1}{32}321COption C. 127\frac{1}{27}271DOption D. 125\frac{1}{25}251CheckStep-by-step hintSkip · show solution
