Q.The least value of the function $f(x) = ax + \dfrac{b}{x}$ ($a > 0$, $b > 0$, $x > 0$) is ______.

Write $f(x)=\dfrac{(x+1)^{2}+1}{x+1}=(x+1)+\dfrac{1}{x+1}$. Then

• At $x=-2$ (where $x+1<0$): local maximum, $l=f(-2)=-2$, so $\alpha=-2$. …

We want to maximize $x^3 y^5$ given $x+y=24$ and $x,y>0$ (positive integers). The exponents 3 and 5 suggest a weighted approach: the product is maximized when the numbers are split in proportion to the exponents. This is a classic application of the weighted AM–GM inequality, which tells us that for fixed sum, a product of powers is maximized when the variables are in the ratio of their exponents.

Let’s work through it carefully.

1. Set up the weighted AM–GM For positive $x, y$, the weighted arithmetic mean–geometric mean inequality says:

But more directly, we can write:

The left side simplifies to $\frac{x+y}{8} = \frac{24}{8} = 3$.

2. Equality condition Equality in weighted AM–GM occurs when all terms are equal:

So $5x = 3y$.

3. Solve for $x$ and $y$ Using $x+y=24$ and $5x=3y$, substitute $y = \frac{5}{3}x$: $$x + \frac{5}{3}x = 24 \quad\Rightarrow\quad \frac{8}{3}x = 24 \quad\Rightarrow\quad x = 9 …$$

We are asked for the minimum value of

over all real $x$. The denominator $1 + x + x^2$ is always positive (its discriminant $1 - 4 = -3 < 0$), so the expression is defined for every real $x$. The problem is to find the smallest possible $y$ as $x$ varies over $\mathbb{R}$.

The direct approach — differentiating and setting the derivative to zero — works, but there is a cleaner algebraic method that avoids calculus and gives the entire range at once. That method is the discriminant technique for rational functions.

1. Set up the equation. Let

Since the denominator is never zero, we can cross-multiply:

2. Rearrange into a quadratic in $x$. Expand and bring all terms to one side:

This is a quadratic equation in $x$ (unless $y = 1$, which we will treat separately).

3. Apply the condition that $x$ is real. For a given $y$, if there exists a real $x$ satisfying the equation, then the discriminant of this quadratic must be non-negative. The discriminant $\Delta$ is:

Simplify:

4. Simplify the discriminant. Expand both squares:

So

For real $x$, we need $\Delta \geq 0$:

5. Solve the inequality. Multiply by $-1$ (reversing the inequality):

Factor the quadratic:

Check: $(3y - 1)(y - 3) = 3y^2 - 9y - y + 3 = 3y^2 - 10y + 3$. Correct.

The inequality $(3y - 1)(y - 3) \leq 0$ holds when $y$ lies between the roots:

6. Check the special case $y = 1$. …

We are asked: if $f(x) = \frac{x}{5} + \frac{5}{x}$ attains its relative maximum at $x = a$, find $\sqrt{a^2 + 2a - 6}$.

The first thing to notice: this function is of the form $f(x) = \frac{x}{5} + \frac{5}{x}$, which is a classic sum of a linear term and its reciprocal. For $x > 0$, by AM–GM, the minimum occurs when $\frac{x}{5} = \frac{5}{x}$, i.e., $x^2 = 25$, so $x = 5$. For $x < 0$, the function is negative and has a maximum (since both terms are negative, their sum is most negative at the same balance point $x = -5$). But the problem says "relative maximum" — and the function actually has no relative maximum for $x > 0$; it has only a minimum. This is a classic trick: students might blindly set derivative to zero and forget to check whether it's a max or min.

Let’s work through carefully.

1. Find critical points. Differentiate:

Set $f'(x) = 0$:

2. Classify each critical point. Second derivative:

• At $x = 5$: $f''(5) = \frac{10}{125} > 0$ → local minimum.
• At $x = -5$: $f''(-5) = \frac{10}{-125} < 0$ → local maximum.

So the only relative maximum is at $x = -5$, not at a positive $x$.

3. Interpret the problem’s wording. …

The problem asks for the equation of lowest degree with rational coefficients that has the given two roots. This is a classic trap: students often think that since the two given numbers are already conjugates (one has $+\sqrt{2}i$, the other $-\sqrt{2}i$), that's enough. But look carefully — the coefficients must be rational, not just real.

The numbers $\sqrt{3} + \sqrt{2}i$ and $\sqrt{3} - \sqrt{2}i$ involve $\sqrt{3}$, which is irrational. For a polynomial with rational coefficients, if a root contains an irrational square root, its conjugate under that square root must also be a root. Here, the irrational part is $\sqrt{3}$, so the conjugates $-\sqrt{3} + \sqrt{2}i$ and $-\sqrt{3} - \sqrt{2}i$ must also be roots. That gives us four roots in total.

Let's build the polynomial step by step.

1. Identify all four required roots.

   The given roots are $r_1 = \sqrt{3} + \sqrt{2}i$ and $r_2 = \sqrt{3} - \sqrt{2}i$.

   To make coefficients rational, we also need $r_3 = -\sqrt{3} + \sqrt{2}i$ and $r_4 = -\sqrt{3} - \sqrt{2}i$.

2. Group the roots into conjugate pairs that produce quadratic factors with rational coefficients.

   Pair $r_1$ and $r_2$: their sum is $2\sqrt{3}$ and product is $(\sqrt{3})^2 + (\sqrt{2})^2 = 3 + 2 = 5$.

   So the quadratic for this pair is $x^2 - (2\sqrt{3})x + 5 = 0$.

   But this has an irrational coefficient ($2\sqrt{3}$), so it's not yet a factor with rational coefficients. We need to pair differently.

3. Pair roots that eliminate the irrationality in the coefficients.

   The trick is to pair $\sqrt{3} + \sqrt{2}i$ with $-\sqrt{3} + \sqrt{2}i$ — these share the same imaginary part but opposite real parts.

   Their sum: $(\sqrt{3} - \sqrt{3}) + (2\sqrt{2}i) = 2\sqrt{2}i$ — still not rational.

   Instead, pair $\sqrt{3} + \sqrt{2}i$ with $-\sqrt{3} - \sqrt{2}i$:

   Sum = $0$, Product = $-(\sqrt{3})^2 - (\sqrt{2})^2 = -3 - 2 = -5$.

   That gives $x^2 + 0x - 5 = x^2 - 5$ — but this has irrational roots $\pm\sqrt{5}$, not our numbers. So that's wrong.

   The correct grouping: pair $\sqrt{3} + \sqrt{2}i$ with $\sqrt{3} - \sqrt{2}i$ to get $x^2 - 2\sqrt{3}x + 5$, and pair $-\sqrt{3} + \sqrt{2}i$ with $-\sqrt{3} - \sqrt{2}i$ to get $x^2 + 2\sqrt{3}x + 5$.

   Multiply these two quadratics:

   $(x^2 + 5 - 2\sqrt{3}x)(x^2 + 5 + 2\sqrt{3}x) = (x^2 + 5)^2 - (2\sqrt{3}x)^2$

   $= x^4 + 10x^2 + 25 - 12x^2 = x^4 - 2x^2 + 25$.

4. Now handle the other pair of conjugates.

   The remaining roots are $\sqrt{3} + \sqrt{2}i$ and $-\sqrt{3} + \sqrt{2}i$? No — we've used all four. Wait, we need to also form a factor from the other natural grouping: pair $\sqrt{3} + \sqrt{2}i$ with $-\sqrt{3} + \sqrt{2}i$? That gives $x^2 - 2\sqrt{2}i x - 5$, which has complex coefficients — not allowed.

   The correct second grouping: pair $\sqrt{3} + \sqrt{2}i$ with $-\sqrt{3} - \sqrt{2}i$? That gave $x^2 - 5$, which doesn't have our roots. So the only way to get rational coefficients is to take the product of the two quadratics we already found: $(x^2 - 2\sqrt{3}x + 5)(x^2 + 2\sqrt{3}x + 5) = x^4 - 2x^2 + 25$. That's one quartic factor.

5. But we need a polynomial that has all four roots. The quartic $x^4 - 2x^2 + 25$ has roots $\pm\sqrt{3} \pm \sqrt{2}i$? Let's check: if $x = \sqrt{3} + \sqrt{2}i$, then $x^2 = 3 + 2\sqrt{6}i - 2 = 1 + 2\sqrt{6}i$, so $x^4 = (1 + 2\sqrt{6}i)^2 = 1 + 4\sqrt{6}i - 24 = -23 + 4\sqrt{6}i$. Then $x^4 - 2x^2 + 25 = (-23 + 4\sqrt{6}i) - 2(1 + 2\sqrt{6}i) + 25 = (-23 - 2 + 25) + (4\sqrt{6}i - 4\sqrt{6}i) = 0$. Yes, it works. So $x^4 - 2x^2 + 25 = 0$ has all four roots $\pm\sqrt{3} \pm \sqrt{2}i$. …