Question

Least value of $\frac{x^{2}y^{2}-2x^{2}y+2x^{2}+2xy-2x+1}{x^{2}y+x}$ is $\lambda$, then {where x,y $\in$ R$^{+}$, x$^{2}$y + x $\neq$ 0}

• A. $\lambda \in (0,1)$
• B. $\lambda \in [1,3)$
• C. $\lambda \in [3,4]$
• D. $\lambda \in (4,7)$

Answer 1

Answer: (A)

(A)

Answer 2

The given expression is $E = \frac{x^{2}y^{2}-2x^{2}y+2x^{2}+2xy-2x+1}{x^{2}y+x}$.

The numerator can be factored as: $x^{2}y^{2}-2x^{2}y+2x^{2}+2xy-2x+1 = (x^2y^2+2xy+1) - 2x^2y - 2x + 2x^2 = (xy+1)^2 - 2x(xy+1) + 2x^2$.

The denominator is $x^2y+x = x(xy+1)$.

So, the expression becomes: $E = \frac{(xy+1)^2 - 2x(xy+1) + 2x^2}{x(xy+1)} = \frac{(xy+1)^2}{x(xy+1)} - \frac{2x(xy+1)}{x(xy+1)} + \frac{2x^2}{x(xy+1)} = \frac{xy+1}{x} - 2 + \frac{2x}{xy+1}$

Let $A = \frac{xy+1}{x}$. Since $x, y \in \mathbb{R}^{+}$, $x > 0$ and $y > 0$. Therefore, $xy > 0$, so $xy+1 > 1$. Since $x > 0$, $A = \frac{xy+1}{x} > 0$.

The expression $E$ can be rewritten as: $E = A - 2 + \frac{2}{A}$

To find the least value of $E$, we apply the AM-GM inequality to the positive terms $A$ and $\frac{2}{A}$: $A + \frac{2}{A} \ge 2\sqrt{A \cdot \frac{2}{A}} = 2\sqrt{2}$

Substituting this back into the expression for $E$: $E \ge 2\sqrt{2} - 2$.

The equality holds when $A = \frac{2}{A}$, which means $A^2 = 2$. Since $A > 0$, $A = \sqrt{2}$.

This value is achievable. For example, if $x=1$, then $y+\frac{1}{1} = \sqrt{2}$, so $y = \sqrt{2}-1$. Since $\sqrt{2} \approx 1.414$, $y \approx 0.414 > 0$. Thus, $x=1, y=\sqrt{2}-1$ are valid positive real numbers.

The least value of the expression is $\lambda = 2\sqrt{2} - 2$. To determine the interval for $\lambda$: We know $\sqrt{2} \approx 1.414$. So, $\lambda \approx 2 \times 1.414 - 2 = 2.828 - 2 = 0.828$. This value is greater than 0 and less than 1. Therefore, $\lambda \in (0,1)$.