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1)

The  least value of  αϵR for which  4αx2+1x1, , for all x >0, is 


A) 164

B) 132

C) 127

D) 125

Answer:

Option C

Explanation:

 Here, to find the least value of   αϵR . for which  

                           4αx2+1x1, , for all x >0

 i.e. to find the minimum value of α when

                  y=4αx2+1x;x>0  attains minimum value of α

                            \frac{\text{d}y}{\text{d}x}=8\alpha x-\frac{1}{x^{2}}    .....(i)

           Now,      \frac{d^{2}y}{dx^{2}}=8\alpha +\frac{2}{x^{3}}       ........(ii)

                        when       \frac{\text{d}y}{\text{d}x}=0,

         then                8x^{3}\alpha =1

           \frac{d^{2}y}{dx^{2}}=8\alpha+16\alpha=24\alpha , Thus, y attains minimum when

                            x=(\frac{1}{8\alpha})^{\frac{1}{3}}: \alpha >0.

     \therefore y attains minimum when   x=(\frac{1}{8\alpha})^{\frac{1}{3}}

 i.e,   4\alpha(\frac{1}{8\alpha})^{\frac{2}{3}}+(8\alpha)^{\frac{1}{3}}\geq1

    \Rightarrow \alpha^{\frac{1}{3}}+2\alpha^{\frac{1}{3}}\geq1\Rightarrow 3\alpha^{\frac{1}{3}}\geq1

\Rightarrow                      \alpha \geq\frac{1}{27}

     Hence, the least value of \alpha   is  \frac{1}{27}