Let the sum of the maximum and the minimum values of the function $f(x) = \frac{2x^2-3x+8}{2x^2+3x+8}$ be $\frac{m}{n}$, where $\text{gcd}(m, n) = 1$. Then $m + n$ is equal to : Options: (A) 195 (B) 201 (C) 217 (D) 182

### Analyzing the Setup We are tasked with finding the range of the rational function: $$f(x) = \frac{2x^2-3x+8}{2x^2+3x+8}$$ To determine the range, we set $y = f(x)$ and identify all possible values of $y$ for which there exists a real $x$. ### The Quadratic Transformation By cross-multiplying the expression, we obtain: $$y(2x^2+3x+8) = 2x^2-3x+8$$ Rearranging this into a standard quadratic equation in terms of $x$, we get: $$(2y-2)x^2 + (3y+3)x + (8y-8) = 0$$ ### The Gatekeeper of Reality Since $x$ must be a real number, the discriminant $D = B^2 - 4AC$ of the quadratic equation must satisfy $D \ge 0$. Here, the coefficients are $A = 2y-2$, $B = 3y+3$, and $C = 8y-8$. Substituting these into the discriminant formula: $$(3y+3)^2 - 4(2y-2)(8y-8) \ge 0$$ ### Elegant Algebra We can simplify the expression by factoring out constants: $$9(y+1)^2 - 64(y-1)^2 \ge 0$$ Recognizing this as a difference of squares, $a^2 - b^2 = (a-b)(a+b)$, where $a = 3(y+1)$ and $b = 8(y-1)$, we have: $$[3(y+1) - 8(y-1)][3(y+1) + 8(y-1)] \ge 0$$ Simplifying the terms inside the brackets yields: $$(-5y+11)(11y-5) \ge 0$$ Multiplying by $-1$ to reverse the inequality, we obtain: $$(5y-11)(11y-5) \le 0$$ ### Final Calculation The critical points are $y = \frac{5}{11}$ and $y = \frac{11}{5}$. Thus, the range of the function is $y \in [\frac{5}{11}, \frac{11}{5}]$. The sum of the boundaries is: $$\frac{11}{5} + \frac{5}{11} = \frac{121+25}{55} = \frac{146}{55}$$ Given $m = 146$ and $n = 55$, where $\text{gcd}(146, 55) = 1$, the final result is: **$m + n = 146 + 55 = 201$**

JEE Main 2024 (04 Apr Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Quadratic Equations: 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 :

Q: Why do we set $D \ge 0$?

Because $x$ is a real number. If $D < 0$, the roots of the quadratic equation are complex (imaginary). Since we are dealing with a real-valued function, we must ensure that for every $y$ in the range, there exists a real $x$ that maps to it.

Select Answer:

Visualized Solution

Define the Function

Let $y = \frac{2 x ^{2} - 3 x + 8}{2 x ^{2} + 3 x + 8}$

Cross-Multiplication

$y (2 x^{2} + 3 x + 8) = 2 x^{2} - 3 x + 8$

Rearrange into Quadratic Form

$2 y x^{2} + 3 y x + 8 y = 2 x^{2} - 3 x + 8$
$(2 y - 2) x^{2} + (3 y + 3) x + (8 y - 8) = 0$

Condition for Real $x$

For $x \in R$ , Discriminant $D \geq 0$

Set up the Discriminant

$D = b^{2} - 4 a c \geq 0$
$(3 y + 3)^{2} - 4 (2 y - 2) (8 y - 8) \geq 0$

Factor Out Constants

$9 (y + 1)^{2} - 4 \cdot 2 (y - 1) \cdot 8 (y - 1) \geq 0$
$9 (y + 1)^{2} - 64 (y - 1)^{2} \geq 0$

Apply $a^{2} - b^{2}$ Identity

$(3 (y + 1))^{2} - (8 (y - 1))^{2} \geq 0$
$[3 (y + 1) - 8 (y - 1)] [3 (y + 1) + 8 (y - 1)] \geq 0$

Simplify the Factors

First factor: $3 y + 3 - 8 y + 8 = - 5 y + 11$
Second factor: $3 y + 3 + 8 y - 8 = 11 y - 5$
$(- 5 y + 11) (11 y - 5) \geq 0$

Solve the Inequality

Multiply by $- 1$ : $(5 y - 11) (11 y - 5) \leq 0$
Critical points: $y = \frac{11}{5}, y = \frac{5}{11}$
$y \in [\frac{5}{11}, \frac{11}{5}]$

Identify Max and Min Values

Maximum value $y_{ma x} = \frac{11}{5}$
Minimum value $y_{min} = \frac{5}{11}$

Calculate the Sum

Sum $= \frac{11}{5} + \frac{5}{11}$
Sum $= \frac{121 + 25}{55} = \frac{146}{55}$

Find $m + n$

Given Sum $= \frac{m}{n}$
$\frac{146}{55} = \frac{m}{n}$
Since $gcd (146, 55) = 1$ , $m = 146, n = 55$
$m + n = 146 + 55 = 201$

00:00 / 00:00

The Sigma Insight: Maximum and Minimum Values of Quadratic Expressions

Solution Diagram

Analyzing the Setup

We are tasked with finding the range of the rational function:

f (x) = \frac{2 x ^{2} - 3 x + 8}{2 x ^{2} + 3 x + 8}

To determine the range, we set

y = f (x)

and identify all possible values of

y

for which there exists a real

x

The Quadratic Transformation

By cross-multiplying the expression, we obtain:

y (2 x^{2} + 3 x + 8) = 2 x^{2} - 3 x + 8

Rearranging this into a standard quadratic equation in terms of

x

, we get:

(2 y - 2) x^{2} + (3 y + 3) x + (8 y - 8) = 0

The Gatekeeper of Reality

Since

x

must be a real number, the discriminant

D = B^{2} - 4 A C

of the quadratic equation must satisfy

D \geq 0

. Here, the coefficients are

A = 2 y - 2

B = 3 y + 3

, and

C = 8 y - 8

Substituting these into the discriminant formula:

(3 y + 3)^{2} - 4 (2 y - 2) (8 y - 8) \geq 0

Elegant Algebra

We can simplify the expression by factoring out constants:

9 (y + 1)^{2} - 64 (y - 1)^{2} \geq 0

Recognizing this as a difference of squares,

a^{2} - b^{2} = (a - b) (a + b)

, where

a = 3 (y + 1)

and

b = 8 (y - 1)

, we have:

[3 (y + 1) - 8 (y - 1)] [3 (y + 1) + 8 (y - 1)] \geq 0

Simplifying the terms inside the brackets yields:

(- 5 y + 11) (11 y - 5) \geq 0

Multiplying by

- 1

to reverse the inequality, we obtain:

(5 y - 11) (11 y - 5) \leq 0

Final Calculation

The critical points are

y = \frac{5}{11}

and

y = \frac{11}{5}

. Thus, the range of the function is

y \in [\frac{5}{11}, \frac{11}{5}]

The sum of the boundaries is:

\frac{11}{5} + \frac{5}{11} = \frac{121 + 25}{55} = \frac{146}{55}

Given

m = 146

and

n = 55

, where

gcd (146, 55) = 1

, the final result is: $m + n = 146 + 55 = 201$

Common Traps & FAQs

Why can't we just use calculus and find

f^{'} (x) = 0

You absolutely can! Calculus is a beautiful tool. However, for rational functions of the form

\frac{q u a d r a t i c}{q u a d r a t i c}

, the discriminant method is often faster and less prone to algebraic errors during differentiation. It directly gives you the range without needing to evaluate the function at critical points.

Why do we set

D \geq 0

Because

x

is a real number. If

D < 0

, the roots of the quadratic equation are complex (imaginary). Since we are dealing with a real-valued function, we must ensure that for every

y

in the range, there exists a real

x

that maps to it.

What if the denominator becomes zero?

Great catch! Always check the domain. Here, the denominator is

2 x^{2} + 3 x + 8

. The discriminant of this quadratic is

3^{2} - 4 (2) (8) = 9 - 64 = - 55

. Since the discriminant is negative and the leading coefficient is positive, the denominator is always positive. It never touches zero, so the function is defined for all real

x

Animated Solution for Mathematics - Quadratic Equations: 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 :

Select Answer:

Visualized Solution

Define the Function

Cross-Multiplication

Rearrange into Quadratic Form

Condition for Real $x$

Set up the Discriminant

Factor Out Constants

Apply $a^{2} - b^{2}$ Identity

Simplify the Factors

Solve the Inequality

Identify Max and Min Values

Calculate the Sum

Find $m + n$

The Sigma Insight: Maximum and Minimum Values of Quadratic Expressions

Analyzing the Setup

The Quadratic Transformation

The Gatekeeper of Reality

Elegant Algebra

Final Calculation

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Visualized Solution

Define the Function

Cross-Multiplication

Rearrange into Quadratic Form

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Analyzing the Setup

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