JEE Advanced 2021

LEVELJEE Advanced

Animated Solution for Mathematics - Differentiation: Let $f : R \to R$ be defined by $f (x) = \frac{x ^{2} - 3 x - 6}{x ^{2} + 2 x + 4}$ . Then which of the following statements is (are) TRUE ?

Select Answer:

* Multiple Correct

Progressive Hints

Hint 1

Hint 1: Visualize the function as a ratio of two parabolas. The denominator

x^{2} + 2 x + 4

is always positive because its discriminant is negative. This means the function is continuous everywhere. To understand its behavior, don't just look at the values; look at how the slope changes.

Hint 2

Hint 2: To determine where the function increases or decreases, you must compute the derivative

f^{'} (x)

using the quotient rule. Remember, the sign of the derivative is all that matters for monotonicity. Focus on the numerator of

f^{'} (x)

after simplification.

Hint 3

Hint 3: To find the range, set

f (x) = y

and rearrange it into a quadratic equation in

x

:

(y - 1) x^{2} + (2 y + 3) x + (4 y + 6) = 0

. Since

x

must be a real number, the discriminant

D

of this quadratic must be greater than or equal to zero. Solve

D \geq 0

for

y

.

Visualized Solution

$f (x) = \frac{x ^{2} - 3 x - 6}{x ^{2} + 2 x + 4}$

$f (x) = \frac{x ^{2} - 3 x - 6}{x ^{2} + 2 x + 4}$
We need to check:
- Monotonicity (Increasing / Decreasing)
- Range and Surjectivity (Onto)

$f^{'} (x)$ Setup

Quotient Rule: $(\frac{u}{v})^{'} = \frac{u ^{'} v - u v ^{'}}{v ^{2}}$
Let $u = x^{2} - 3 x - 6$
Let $v = x^{2} + 2 x + 4$

$f^{'} (x) = \frac{u ^{'} v - u v ^{'}}{v ^{2}}$

$f^{'} (x) = \frac{( 2 x - 3 ) ( x ^{2} + 2 x + 4 ) - ( x ^{2} - 3 x - 6 ) ( 2 x + 2 )}{( x ^{2} + 2 x + 4 ) ^{2}}$

Simplifying Numerator

First part: $(2 x^{3} + 4 x^{2} + 8 x) - (3 x^{2} + 6 x + 12)$
$= 2 x^{3} + x^{2} + 2 x - 12$
Second part: $(2 x^{3} + 2 x^{2}) - (6 x^{2} + 6 x) - (12 x + 12)$
$= 2 x^{3} - 4 x^{2} - 18 x - 12$

Final Derivative

Subtracting: $(2 x^{3} + x^{2} + 2 x - 12) - (2 x^{3} - 4 x^{2} - 18 x - 12)$
$= 5 x^{2} + 20 x$
Factorizing: $5 x (x + 4)$
Final Derivative: $f^{'} (x) = \frac{5 x ( x + 4 )}{( x ^{2} + 2 x + 4 ) ^{2}}$

Critical Points

Denominator $(x^{2} + 2 x + 4)^{2} > 0$ for all $x \in R$
Sign of $f^{'} (x)$ depends entirely on the numerator: $x (x + 4)$
Set numerator to zero for critical points:
$x (x + 4) = 0 ⟹ x = 0, x = - 4$

Wavy Curve Method

For $x \in (- \infty, - 4)$ : $f^{'} (x) > 0 ⟹$ Increasing
For $x \in (- 4, 0)$ : $f^{'} (x) < 0 ⟹$ Decreasing
For $x \in (0, \infty)$ : $f^{'} (x) > 0 ⟹$ Increasing

Option A

Option A: $f$ is decreasing in $(- 2, - 1)$
We know $f$ is decreasing in $(- 4, 0)$
Since $(- 2, - 1) \subset (- 4, 0)$
The derivative $f^{'} (x) < 0$ in this interval.
$⟹$ Statement A is TRUE.

Option B

Option B: $f$ is increasing in $(1, 2)$
We know $f$ is increasing in $(0, \infty)$
Since $(1, 2) \subset (0, \infty)$
The derivative $f^{'} (x) > 0$ in this interval.
$⟹$ Statement B is TRUE.

Range Setup

To find the range, let $y = f (x)$
$y = \frac{x ^{2} - 3 x - 6}{x ^{2} + 2 x + 4}$
We will form a quadratic equation in $x$ .

Quadratic in $x$

Cross-multiplying: $y (x^{2} + 2 x + 4) = x^{2} - 3 x - 6$
$y x^{2} + 2 y x + 4 y - x^{2} + 3 x + 6 = 0$
Grouping terms by powers of $x$ :
$(y - 1) x^{2} + (2 y + 3) x + (4 y + 6) = 0$

Discriminant $D \geq 0$

For $x \in R$ , the Discriminant $D \geq 0$
$D = b^{2} - 4 a c$
Here, $a = (y - 1)$ , $b = (2 y + 3)$ , $c = (4 y + 6)$
$(2 y + 3)^{2} - 4 (y - 1) (4 y + 6) \geq 0$

Expanding $D$

Expanding the square: $(4 y^{2} + 12 y + 9) - 4 (4 y^{2} + 6 y - 4 y - 6) \geq 0$
Simplifying the second bracket: $(4 y^{2} + 12 y + 9) - 4 (4 y^{2} + 2 y - 6) \geq 0$
Multiplying by $- 4$ : $4 y^{2} + 12 y + 9 - 16 y^{2} - 8 y + 24 \geq 0$

Final Inequality

Combining like terms: $- 12 y^{2} + 4 y + 33 \geq 0$
Multiply by $- 1$ (Inequality sign flips):
$12 y^{2} - 4 y - 33 \leq 0$

Range of $f (x)$

Roots of $12 y^{2} - 4 y - 33 = 0$ :
$y = \frac{4 \pm 16 - 4 ( 12 ) ( - 33 )}{24} = \frac{4 \pm 40}{24}$
$y_{1} = \frac{44}{24} = \frac{11}{6}$ , $y_{2} = \frac{- 36}{24} = - \frac{3}{2}$
Since $12 y^{2} - 4 y - 33 \leq 0$ :
$y \in [- \frac{3}{2}, \frac{11}{6}]$

Options C and D

Range of $f$ = $[- \frac{3}{2}, \frac{11}{6}]$
Option C: $f$ is onto $⟹$ False (Range $\neq = R$ )
Option D: Range is $[- \frac{3}{2}, 2]$ $⟹$ False (Max value is $\frac{11}{6} < 2$ )
Final Answer: Options A and B are TRUE.

00:00 / 00:00

The Sigma Insight: Monotonicity

Solution Diagram

Analyzing the Setup

We are tasked with analyzing the rational function:

f (x) = \frac{x ^{2} - 3 x - 6}{x ^{2} + 2 x + 4}

This function represents a ratio of two quadratics. To understand its behavior, we must examine its derivative and its range.

The Pulse of the Function

To determine the intervals of increase and decrease, we apply the Quotient Rule:

(\frac{u}{v})^{'} = \frac{u ^{'} v - u v ^{'}}{v ^{2}}

. Here,

u = x^{2} - 3 x - 6

and

v = x^{2} + 2 x + 4

.

After performing the differentiation and simplifying the numerator, we obtain:

f^{'} (x) = \frac{5 x ( x + 4 )}{( x ^{2} + 2 x + 4 ) ^{2}}

The denominator

(x^{2} + 2 x + 4)^{2}

is always positive because the discriminant of

x^{2} + 2 x + 4

is

2^{2} - 4 (4) = - 12 < 0

. Consequently, the sign of

f^{'} (x)

is determined entirely by the numerator

5 x (x + 4)

.

Using the Wavy Curve Method, we identify critical points at

x = 0

and

x = - 4

. The function increases on

(- \infty, - 4)

, decreases on

(- 4, 0)

, and increases on

(0, \infty)

.

The Quest for the Range

To find the range, we set

y = f (x)

and rearrange the equation into a quadratic form in terms of

x

:

(y - 1) x^{2} + (2 y + 3) x + (4 y + 6) = 0

For

y

to be in the range, this quadratic must have real roots for

x

. This requires the discriminant

D = b^{2} - 4 a c

to be non-negative:

D = (2 y + 3)^{2} - 4 (y - 1) (4 y + 6) \geq 0

Expanding the expression leads to the inequality:

- 12 y^{2} + 4 y + 33 \geq 0

Multiplying by

- 1

reverses the inequality sign:

12 y^{2} - 4 y - 33 \leq 0

Final Calculation

Solving the quadratic equation

12 y^{2} - 4 y - 33 = 0

using the quadratic formula, we find the roots:

y = \frac{4 \pm 16 - 4 ( 12 ) ( - 33 )}{24} = \frac{4 \pm 16 + 1584}{24} = \frac{4 \pm 1600}{24} = \frac{4 \pm 40}{24}

This yields the boundaries

y = - \frac{36}{24} = - \frac{3}{2}

and

y = \frac{44}{24} = \frac{11}{6}

.

Thus, the range of the function is $[- \frac{3}{2}, \frac{11}{6}]$ . Since this interval is not the set of all real numbers, the function is not onto, and the maximum value is strictly $\frac{11}{6}$ .

Common Traps & FAQs

Why can't we just plug in values to check if it's increasing or decreasing?

Plugging in values only tells you the behavior at specific points. To prove a function is increasing or decreasing on an entire interval, you must ensure the derivative maintains a constant sign throughout that interval. The derivative is the 'DNA' of the function's slope!

Wait, why do we set the discriminant

D \geq 0

to find the range?

Great question! By setting

f (x) = y

, we are asking: 'For which values of

y

does there exist at least one

x

such that

f (x) = y

?' Since

x

must be a real number, the quadratic equation in

x

must have real roots. The condition for real roots is exactly

D \geq 0

.

What happens if the coefficient of

x^{2}

in the quadratic equation becomes zero?

If

y - 1 = 0

, then

y = 1

. The equation becomes linear:

(2 (1) + 3) x + (4 (1) + 6) = 0

, which is

5 x + 10 = 0

, giving

x = - 2

. Since we found a valid

x

for

y = 1

,

y = 1

is definitely in the range. Always check the case where the leading coefficient vanishes!

Why is the function not 'onto'?

A function is 'onto' (surjective) if its range is equal to its codomain. Here, the codomain is

R

, but we calculated the range to be

[- \frac{3}{2}, \frac{11}{6}]

. Since the range is a small subset of

R

, the function misses infinitely many values, so it cannot be onto.

Is there a shortcut to finding the derivative of a rational function?

While the quotient rule is the standard path, you can sometimes simplify the expression using polynomial division if the degree of the numerator is greater than or equal to the denominator. However, here the degrees are equal, so the quotient rule is actually the most direct and reliable method.

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