Question

Let y = f(x) such that xy = x + y + 1, x ∈ R – {1} and g(x) = xf(x)

The minimum value of g(x) is:

• A. 3-√2
• B. 3+√2
• C. 3-2√2
• D. 3+2√2

Answer 1

Answer: (D)

3+2√2

Answer 2

The given relation is $xy = x + y + 1$ for $x \in R - \{1\}$. We need to find an explicit expression for $y = f(x)$.

Rearranging the terms to isolate $y$: $xy - y = x + 1$ $y(x - 1) = x + 1$

Since $x \neq 1$, we can divide by $(x - 1)$: $y = \frac{x + 1}{x - 1}$ So, $f(x) = \frac{x + 1}{x - 1}$.

The function $g(x)$ is defined as $g(x) = xf(x)$. Substituting the expression for $f(x)$: $g(x) = x \left( \frac{x + 1}{x - 1} \right) = \frac{x(x + 1)}{x - 1} = \frac{x^2 + x}{x - 1}$.

To find the minimum value of $g(x)$, we can use calculus. We find the derivative of $g(x)$ with respect to $x$. Using the quotient rule, $\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$, where $u = x^2 + x$ and $v = x - 1$. $u' = \frac{d}{dx}(x^2 + x) = 2x + 1$ $v' = \frac{d}{dx}(x - 1) = 1$ $g'(x) = \frac{(2x + 1)(x - 1) - (x^2 + x)(1)}{(x - 1)^2}$ $g'(x) = \frac{(2x^2 - 2x + x - 1) - (x^2 + x)}{(x - 1)^2}$ $g'(x) = \frac{2x^2 - x - 1 - x^2 - x}{(x - 1)^2}$ $g'(x) = \frac{x^2 - 2x - 1}{(x - 1)^2}$.

To find the critical points, we set $g'(x) = 0$. $\frac{x^2 - 2x - 1}{(x - 1)^2} = 0$ This implies $x^2 - 2x - 1 = 0$. Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ for $ax^2 + bx + c = 0$: $x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2(1)}$ $x = \frac{2 \pm \sqrt{4 + 4}}{2}$ $x = \frac{2 \pm \sqrt{8}}{2}$ $x = \frac{2 \pm 2\sqrt{2}}{2}$ $x = 1 \pm \sqrt{2}$.

These are the critical points. We use the first derivative test to determine whether they correspond to a local minimum or maximum. The denominator $(x - 1)^2$ is positive for $x \neq 1$. The sign of $g'(x)$ is determined by the sign of the numerator $x^2 - 2x - 1$. The roots of $x^2 - 2x - 1 = 0$ are $1 - \sqrt{2}$ and $1 + \sqrt{2}$. The quadratic $x^2 - 2x - 1$ is an upward-opening parabola, so it is positive for $x < 1 - \sqrt{2}$ or $x > 1 + \sqrt{2}$, and negative for $1 - \sqrt{2} < x < 1 + \sqrt{2}$.

The domain of $g(x)$ is $R - \{1\}$. The critical points are $1 - \sqrt{2} \approx 1 - 1.414 = -0.414$ and $1 + \sqrt{2} \approx 1 + 1.414 = 2.414$. Neither of these is equal to 1.

Sign analysis of $g'(x)$:

• For $x < 1 - \sqrt{2}$, $g'(x) > 0$, so $g(x)$ is increasing.
• For $1 - \sqrt{2} < x < 1$, $g'(x) < 0$, so $g(x)$ is decreasing.
• For $1 < x < 1 + \sqrt{2}$, $g'(x) < 0$, so $g(x)$ is decreasing.
• For $x > 1 + \sqrt{2}$, $g'(x) > 0$, so $g(x)$ is increasing.

The function $g(x)$ increases up to $x = 1 - \sqrt{2}$ and then decreases. So $x = 1 - \sqrt{2}$ is a local maximum. The function $g(x)$ decreases up to $x = 1 + \sqrt{2}$ and then increases. So $x = 1 + \sqrt{2}$ is a local minimum.

We need to find the minimum value of $g(x)$, which corresponds to the local minimum at $x = 1 + \sqrt{2}$. Substitute $x = 1 + \sqrt{2}$ into the expression for $g(x) = \frac{x^2 + x}{x - 1}$: $g(1 + \sqrt{2}) = \frac{(1 + \sqrt{2})^2 + (1 + \sqrt{2})}{(1 + \sqrt{2}) - 1}$ $g(1 + \sqrt{2}) = \frac{(1 + 2\sqrt{2} + 2) + (1 + \sqrt{2})}{\sqrt{2}}$ $g(1 + \sqrt{2}) = \frac{3 + 2\sqrt{2} + 1 + \sqrt{2}}{\sqrt{2}}$ $g(1 + \sqrt{2}) = \frac{4 + 3\sqrt{2}}{\sqrt{2}}$ Rationalize the denominator by multiplying the numerator and denominator by $\sqrt{2}$: $g(1 + \sqrt{2}) = \frac{(4 + 3\sqrt{2})\sqrt{2}}{\sqrt{2} \cdot \sqrt{2}}$ $g(1 + \sqrt{2}) = \frac{4\sqrt{2} + 3(\sqrt{2})^2}{2}$ $g(1 + \sqrt{2}) = \frac{4\sqrt{2} + 6}{2}$ $g(1 + \sqrt{2}) = 2\sqrt{2} + 3 = 3 + 2\sqrt{2}$.

Alternatively, we can rewrite $g(x)$ using algebraic manipulation: $g(x) = \frac{x^2 + x}{x - 1} = \frac{x^2 - x + 2x - 2 + 2}{x - 1} = \frac{x(x - 1) + 2(x - 1) + 2}{x - 1} = x + 2 + \frac{2}{x - 1}$. Let $u = x - 1$. Then $x = u + 1$. As $x \in R - \{1\}$, $u \in R - \{0\}$. $g(x) = (u + 1) + 2 + \frac{2}{u} = u + 3 + \frac{2}{u}$. So, $g(x) - 3 = u + \frac{2}{u}$. We need to find the minimum value of $g(x)$. This is equivalent to finding the minimum value of $u + \frac{2}{u} + 3$. Consider the expression $u + \frac{2}{u}$. If $u > 0$, by AM-GM inequality, $\frac{u + \frac{2}{u}}{2} \ge \sqrt{u \cdot \frac{2}{u}} = \sqrt{2}$. So, $u + \frac{2}{u} \ge 2\sqrt{2}$ for $u > 0$. Equality holds when $u = \frac{2}{u}$, i.e., $u^2 = 2$, so $u = \sqrt{2}$ (since $u > 0$). If $u = \sqrt{2}$, then $x - 1 = \sqrt{2}$, so $x = 1 + \sqrt{2}$. This is the critical point where the local minimum occurs. The minimum value of $u + \frac{2}{u}$ for $u > 0$ is $2\sqrt{2}$. So, the minimum value of $g(x)$ for $x - 1 > 0$ (i.e., $x > 1$) is $3 + 2\sqrt{2}$.

If $u < 0$, let $u = -v$ where $v > 0$. $u + \frac{2}{u} = -v + \frac{2}{-v} = -(v + \frac{2}{v})$. By AM-GM inequality for $v > 0$, $v + \frac{2}{v} \ge 2\sqrt{2}$. So, $-(v + \frac{2}{v}) \le -2\sqrt{2}$. The maximum value of $u + \frac{2}{u}$ for $u < 0$ is $-2\sqrt{2}$. This occurs when $v = \sqrt{2}$, so $u = -\sqrt{2}$. If $u = -\sqrt{2}$, then $x - 1 = -\sqrt{2}$, so $x = 1 - \sqrt{2}$. This is the critical point where the local maximum occurs. The maximum value of $g(x)$ for $x - 1 < 0$ (i.e., $x < 1$) is $3 - 2\sqrt{2}$.

The function $g(x)$ approaches $+\infty$ as $x \to 1^+$ and $x \to +\infty$. The function $g(x)$ approaches $-\infty$ as $x \to 1^-$ and $x \to -\infty$. In the interval $(1, \infty)$, the minimum value is $3 + 2\sqrt{2}$ at $x = 1 + \sqrt{2}$. In the interval $(-\infty, 1)$, the maximum value is $3 - 2\sqrt{2}$ at $x = 1 - \sqrt{2}$. The function goes to $-\infty$ as $x$ approaches $1^-$ and $-\infty$.

Since the function goes to $-\infty$, there is no global minimum over the entire domain $R - \{1\}$. However, the options provided are specific values, and one of them matches the local minimum value we found. In such contexts, the question is likely asking for the local minimum value.

The local minimum value is $3 + 2\sqrt{2}$.