Lf A-x2-1-x2-B-x-1-x then minimum value of A-B equals

The correct option is D 2√(2) A=(x 1x)^2+2 B=(x 1x) AB=(x 1x)+2x 1x Use AM amp; GM (x 1x)+2x 1x2≥√(2) (x 1x)+2x 1x≥ 2√(2) ...

You visited us 46 times! Enjoying our articles? Unlock Full Access!

Local Maxima

lf A=x2+1/x...

Question

lf $A = x^{2} + \frac{1}{x^{2}}, B = x - \frac{1}{x}$ then minimum value of $\frac{A}{B}$ equals

\sqrt{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 \sqrt{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{1}{\sqrt{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

Similar questions

lim x \to \infty (\frac{x^{2} - 1}{x + 1} - a x - b) = 2

x^{2} + \frac{1}{x^{2}} = A

x - \frac{1}{x} = B

\frac{A}{B}

\int \frac{2 x^{2} + 3}{(x^{2} - 1) (x^{2} + 4)} d x = a log (\frac{x - 1}{x + 1}) + b {tan}^{- 1} (\frac{x}{2}) + c

b

x^{2} + x + 1 = 0

(x + \frac{1}{x}) + (x^{2} + \frac{1}{x^{2}}) + \dots + (x^{52} + \frac{1}{x^{52}})

sin [{t a n}^{- 1} \frac{1 - x^{2}}{2 x} + {cos}^{- 1} \frac{1 - x^{2}}{1 + x^{2}}] = 1

{sin}^{- 1} (x - \frac{x^{2}}{2} + \frac{x^{3}}{4} - . . . . .) + {cos}^{- 1} (x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} - . . . . . . .) = \frac{π}{2}

0 < | x | < \sqrt{2}

Join BYJU'S Learning Program