If x and y are real numbers with xy and x y=1 then the minimum possible value of (x2+y2/x-y) is

Question

If x and y are real numbers with x>y and x y=1 then the minimum possible value of (x2+y2/x-y) is (A) 4 √2 (B) 2 √2 (C) 8 √2 (D) √2

Tardigrade Admin · Accepted Answer

Z =( x 2+(1/ x 2)/ x -(1) x )=(( x -(1/ x ))2+2/ x -(1) x )= x -(1/ x )+(2/ x -(1) x )= t +(2/ t )( t >0) =(√t-√(2/t))2+2 √2 ∴ Z min =2 √2 text when t=√2 text i.e. x-(1/x)=√2

Q. If $x$ and $y$ are real numbers with $x > y$ and $x y = 1$ then the minimum possible value of $\frac{x ^{2} + y ^{2}}{x - y}$ is

$42$

$22$

$82$

$2$

$Z = \frac{x ^{2} + \frac{1}{x ^{2}}}{x - \frac{1}{x}} = \frac{( x - \frac{1}{x} ) ^{2} + 2}{x - \frac{1}{x}} = x - \frac{1}{x} + \frac{2}{x - \frac{1}{x}} = t + \frac{2}{t} (t > 0)$
$= (t - \frac{2}{t})^{2} + 22$
$∴ Z_{m i n} = 22 when t = 2 i.e. x - \frac{1}{x} = 2$

Q. If x and y are real numbers with x>y and xy=1 then the minimum possible value of x−yx2+y2​ is

42​

22​

82​

2​

Z=x−x1​x2+x21​​=x−x1​(x−x1​)2+2​=x−x1​+x−x1​2​=t+t2​(t>0) =(t​−t2​​)2+22​ ∴Zmin​=22​ when t=2​ i.e. x−x1​=2​

Q. If $x$ and $y$ are real numbers with $x > y$ and $x y = 1$ then the minimum possible value of $\frac{x ^{2} + y ^{2}}{x - y}$ is

$42$

$22$

$82$

$2$

$Z = \frac{x ^{2} + \frac{1}{x ^{2}}}{x - \frac{1}{x}} = \frac{( x - \frac{1}{x} ) ^{2} + 2}{x - \frac{1}{x}} = x - \frac{1}{x} + \frac{2}{x - \frac{1}{x}} = t + \frac{2}{t} (t > 0)$
$= (t - \frac{2}{t})^{2} + 22$
$∴ Z_{m i n} = 22 when t = 2 i.e. x - \frac{1}{x} = 2$