You visited us 1 times! Enjoying our articles? Unlock Full Access!

Terms

If x > 0, x...

Question

If $x > 0, x y = 1$ then the minimum value of $x + y$ is?

2

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

\sqrt{2}

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

3

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

\sqrt{3}

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

Open in App

Solution

The correct option is D $2$
$y = \frac{1}{x}$
$z = x + y$
$z = x + \frac{1}{x}$
$\frac{d z}{d x} = 1 - \frac{1}{x^{2}}$
$x^{2} = 1$
$x = \pm 1$ as $x > 0$
$x = 1$ and $y = 1$
$x + y = 2$ minimum value.

Suggest Corrections

Similar questions

2 x^{2} + 3 x y - 2 y^{2} = 0

2 x^{2} + 3 x y - 2 y^{2} + 3 x + y + 1 = 0

A

20 A

{log}_{10} (x^{3} + y^{3}) - {log}_{10} (x^{2} - x y + y^{2}) \leq 2 \forall x > 0, y > 0

x + y

(x^{3} y^{3} + x y) \frac{d y}{d x}, y (1) = 0

y = \frac{1}{2}

x^{2}

If 2x+3y=12; x, y $>$ 0, then the maximum possible value of $x y^{2}$ is ___.

{log}_{10} (x^{3} + y^{3}) - {log}_{10} (x^{2} + y^{2} - x y) \leq 2

x y

x \geq 0, y \geq 0

Join BYJU'S Learning Program