Q.Find the value of the following: For all real values of $x$, the minimum value of $\frac{1-x+x^2}{1+x+x^2}$ is (A) $0$ (B) $1$ (C) $3$ (D) $\frac{1}{3}$

Concept: Rational Function Optimization — we treat the expression as a variable $y$, cross-multiply, and use the discriminant condition for real $x$.

Let

Since the denominator $1+x+x^2 > 0$ for all real $x$, we can cross-multiply:

For real $x$, the discriminant of this quadratic in $x$ must be $\ge 0$:

Simplify:

$(y+1)^2 - 4(y-1)^2 = (y^2+2y+1) - 4(y^2-2y+1) = -3y^2 + 10y - 3 \ge 0.$ …

Why This Approach Works

When you're asked for the minimum (or maximum) of a rational function like $\frac{1-x+x^2}{1+x+x^2}$, the standard calculus approach — differentiate, set to zero, check endpoints — works, but it's messy. There's a cleaner algebraic method that uses a powerful idea: if $y$ is a value the expression can take, then the equation $y = \frac{1-x+x^2}{1+x+x^2}$ must have a real solution for $x$. By rearranging this into a quadratic in $x$, we can use the discriminant condition ($\Delta \geq 0$) to find exactly which $y$ values are possible. The set of all such $y$ is the range, and the smallest $y$ in that set is the minimum.

This method is called Rational Function Optimization via Discriminant, and it's especially useful when the numerator and denominator are both quadratics with no common factors.

1. Set up the equation. Let $y = \frac{1-x+x^2}{1+x+x^2}$. Since the denominator $1+x+x^2$ is always positive (its discriminant $1-4 = -3 < 0$), the expression is defined for all real $x$. Multiply both sides by the denominator:

2. Rearrange into a quadratic in $x$. Bring all terms to one side:

Group powers of $x$:

This is a quadratic equation in $x$ (unless $y=1$, which we'll handle separately).

3. Apply the discriminant condition. For a real $x$ to exist, the discriminant must be non-negative. Here $a = y-1$, $b = y+1$, $c = y-1$. The discriminant is:

4. Simplify the discriminant. Expand both squares:

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

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

Factor the quadratic:

So the inequality becomes:

The product is $\leq 0$ when one factor is non-positive and the other non-negative. This happens for:

$\frac{1}{3} \leq y \leq 3$ …

Method: Finding the Range of a Ratio of Two Quadratics (the Discriminant Trick)

When a question asks for the maximum or minimum of a fraction whose numerator and denominator are both quadratics with no common real roots (so the expression is defined for every real $x$), it is often faster to find the whole range of the expression algebraically than to differentiate.

Steps

Step 1: Set the expression equal to a new variable $y$

where $p$ and $q$ are quadratics. First confirm $q(x)$ is never zero (e.g. by checking its own discriminant is negative), so the expression is defined for all real $x$.

Step 2: Cross-multiply to turn the equation into a quadratic in $x$

Collect terms so the equation reads $A(y)x^2 + B(y)x + C(y) = 0$.

Step 3: Apply the real-root condition (discriminant $\ge 0$)

A given value of $y$ is actually attained by the original expression only if the corresponding $x$ is real — which requires the discriminant of this quadratic-in-$x$ to be non-negative:

Step 4: Solve the resulting inequality in $y$ …

Common Mistakes

Mistake 1: Forgetting to flip the inequality when multiplying by $-1$

The discriminant condition often comes out as $-3y^2+10y-3 \ge 0$; multiplying through by $-1$ to get a positive leading coefficient must reverse the inequality to $3y^2-10y+3\le 0$. Skipping this flip gives the complement of the correct range — swapping which endpoint is the minimum and which is the maximum.

Mistake 2: Not checking the degenerate case where the $x^2$ coefficient vanishes

When $y$ equals the value that makes $A(y)=y-1=0$, the "quadratic in $x$" collapses to a linear equation, and the discriminant formula technically no longer applies. A student who never checks this value (here $y=1$) separately is skipping a genuine edge case, even though in this problem it turns out to still lie safely inside the valid range. …