Let f: R - {2, 6} → R be real valued function defined as f(x) = (x² + 2x + 1)/(x² - 8x + 12). Then r

Q: Wait, why do we set the discriminant $D \ge 0$? I thought that was only for finding roots.

Great question! The discriminant condition $D \ge 0$ is our bridge between the world of $y$ and the world of $x$. By treating $y$ as a constant, we turn the function into a quadratic equation in $x$. If $x$ must be a real number, the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$ dictates that the term under the square root, $D$, cannot be negative. Thus, $D \ge 0$ ensures that for any chosen $y$, there is at least one real $x$ that produces it.

Q: What if the coefficient of $x^2$ becomes zero? Do we need to check $y=1$ separately?

You have a sharp eye! If $y=1$, the equation $(y-1)x^2 - (8y+2)x + (12y-1) = 0$ collapses into a linear equation: $-10x + 11 = 0$, which gives $x = 1.1$. Since $1.1$ is in the domain, $y=1$ is indeed a valid value in the range. Our discriminant method covers this implicitly, but it is excellent practice to verify that the linear case doesn't produce an $x$ value that is excluded from the domain.

Q: Why did we exclude $x=2$ and $x=6$ from the domain? Does it affect the range?

The values $x=2$ and $x=6$ are where the denominator vanishes, making the function undefined. These are vertical asymptotes. While they don't change the algebraic process of finding the range, they are crucial because if our discriminant method had yielded an $x$ value of $2$ or $6$ for a specific $y$, we would have had to exclude that $y$ from our range. Always keep an eye on those 'forbidden' $x$ values!

Q: Is the wavy curve method the only way to solve the final inequality $y(4y + 21) \ge 0$?

The wavy curve method (or sign scheme) is the most efficient and visual way to handle polynomial inequalities. You could also test intervals, but the wavy curve is your best friend in JEE Advanced because it prevents sign errors and helps you visualize the behavior of the function across the entire real line.

Q: Why is the range $(-\infty, -21/4] \cup [0, \infty)$ and not just the values between them?

The inequality $y(4y + 21) \ge 0$ tells us where the product is positive or zero. Since the parabola $4y^2 + 21y$ opens upwards, it is positive outside the roots and negative between them. Therefore, the range must include the 'outer' regions, leading to the union of two intervals.

Question

Let $f: \mathbb{R} - \{2, 6\} 	o \mathbb{R}$ be real valued function defined as $f(x) = \frac{x^2 + 2x + 1}{x^2 - 8x + 12}$. Then range of $f$ is

Options:

(A) $(-\infty, -21/4] \cup [0, \infty)$

(B) $(-\infty, -21/4) \cup (0, \infty)$

(C) $(-\infty, -21/4] \cup [21/4, \infty)$

(D) $(-\infty, -21/4] \cup [1, \infty)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving for a range; we are embarking on a journey to understand the very soul of a rational function.

When you look at the function:

$$f(x) = \frac{x^2 + 2x + 1}{x^2 - 8x + 12}$$

Do not see a mere collection of symbols. See a machine—a machine that takes an input $x$ and transforms it into an output $y$. Our goal is to map every possible destination this machine can reach.

### The Algebraic Bridge

To find the range, we must reverse the flow. We set:

$$y = \frac{x^2 + 2x + 1}{x^2 - 8x + 12}$$

This is our bridge. We want to know: for which values of $y$ does there exist at least one $x$ that satisfies this equation? Let us cross-multiply to bring the $x$ terms to the surface:

$$y(x^2 - 8x + 12) = x^2 + 2x + 1$$

Now, let us organize our thoughts. We rearrange this into a standard quadratic form in terms of $x$:

$$(y-1)x^2 - (8y+2)x + (12y-1) = 0$$

This is the heart of the problem. We have transformed a functional relationship into a quadratic equation where $y$ acts as a parameter. For $x$ to exist in the real world, this quadratic must have real roots.

### The Discriminant's Wisdom

How do we ensure $x$ is real? We invoke the power of the discriminant, $D = B^2 - 4AC$. For real roots, we demand $D \ge 0$.

Substituting our coefficients, we get:

$$[-(8y+2)]^2 - 4(y-1)(12y-1) \ge 0$$

Take a deep breath. This looks like a daunting expansion, but watch how the terms dance. We simplify this to:

$$4(4y+1)^2 - 4(12y^2 - 13y + 1) \ge 0$$

By dividing by $4$ and expanding the squares, we arrive at:

$$(16y^2 + 8y + 1) - (12y^2 - 13y + 1) \ge 0$$

The constants $1$ cancel out beautifully, leaving us with:

$$4y^2 + 21y \ge 0$$

### The Final Reveal

We are left with $y(4y + 21) \ge 0$. This is a simple quadratic inequality. The critical points are $y = 0$ and $y = -\frac{21}{4}$.

Using the wavy curve method, we see that the expression is positive or zero when $y$ is less than or equal to the smaller root or greater than or equal to the larger root.

Thus, the **range** is:

$$(-\infty, -\frac{21}{4}] \cup [0, \infty)$$

### Reflection

Look at what we have achieved. We didn't just calculate; we explored the constraints of the function. We identified the boundaries where the function exists and where it refuses to go.

This is the essence of **JEE Advanced mathematics**—taking a complex, intimidating expression and, through logical rigor, stripping it down to its fundamental behavior. You have mastered the range of this function. Carry this confidence forward; every function is just a story waiting for you to tell it.

Animated Solution for Mathematics - Quadratic Equations: Let $f : R - {2, 6} \to R$ be real valued function defined as $f (x) = \frac{x ^{2} + 2 x + 1}{x ^{2} - 8 x + 12}$ . Then range of $f$ is

Select Answer:

Visualized Solution

Define the Function $f (x)$

Set $y = f (x)$

Cross-Multiplication

Expand and Group Terms

Condition for Real $x$

Substitute into Discriminant

Simplify the Inequality

Expand and Subtract

Final Quadratic in $y$

Find Critical Points

Apply Wavy Curve Method

Final Range of $f$

The Sigma Insight: Maximum and Minimum Values of Quadratic Expressions

Analyzing the Setup

The Algebraic Bridge

The Discriminant's Wisdom

The Final Reveal

Reflection

Similar Questions

Let $f (x) = (1 + b^{2}) x^{2} + 2 b x + 1$ and let $m (b)$ be the minimum value of $f (x)$ . As $b$ varies, the range of $m (b)$ is

Find the range of values of $t$ for which $2 sin t = \frac{1 - 2 x + 5 x ^{2}}{3 x ^{2} - 2 x - 1}$ , $t \in [- π /2, π /2]$ .

For real $x$ , the function $\frac{( x - a ) ( x - b )}{x - c}$ will assume all real values provided

Let the sum of the maximum and the minimum values of the function $f (x) = \frac{2 x ^{2} - 3 x + 8}{2 x ^{2} + 3 x + 8}$ be $\frac{m}{n}$ , where $gcd (m, n) = 1$ . Then $m + n$ is equal to :

If $f (x) = x^{2} + 2 b x + 2 c^{2}$ and $g (x) = - x^{2} - 2 c x + b^{2}$ such that $min f (x) > max g (x)$ , then the relation between $b$ and $c$ , is

Let $f (x)$ be a quadratic expression which is positive for all the real values of $x$ . If $g (x) = f (x) + f^{'} (x) + f^{''} (x)$ , then for any real $x$ ,

If the roots of the equation $b x^{2} + c x + a = 0$ be imaginary, then for all real values of $x$ , the expression $3 b^{2} x^{2} + 6 b c x + 2 c^{2}$ is

For all 'x', $x^{2} + 2 a x + 10 - 3 a > 0$ , then the interval in which 'a' lies is

Let $f (x) = 2 x^{2} - x - 1$ and $S = {n \in Z : ∣ f (n) ∣ \leq 800}$ . Then, the value of $\sum_{n \in S} f (n)$ is equal to ________.

Animated Solution for Mathematics - Quadratic Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R−{2,6}→R be real valued function defined as f(x)=x2−8x+12x2+2x+1​. Then range of f is

Select Answer:

Visualized Solution

Define the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Set .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=f(x)