Step 1: Given Function
The given function is \( f(x) = \frac{x^2 - 3x - 6}{x^2 + 2x + 4} \), and we need to determine whether the function is increasing or decreasing in the specified intervals.
Step 2: First Derivative Calculation
To determine whether \( f(x) \) is increasing or decreasing in a given interval, we need to find the first derivative \( f'(x) \). Using the quotient rule for differentiation, we have:
\[
f'(x) = \frac{(x^2 + 2x + 4)(2x - 3) - (x^2 - 3x - 6)(2x + 2)}{(x^2 + 2x + 4)^2}
\]
Let's simplify the numerator:
\[
\text{Numerator} = (x^2 + 2x + 4)(2x - 3) - (x^2 - 3x - 6)(2x + 2)
\]
Expanding both terms:
\[
(x^2 + 2x + 4)(2x - 3) = 2x^3 - 3x^2 + 4x^2 - 6x + 8x - 12 = 2x^3 + x^2 + 2x - 12
\]
\[
(x^2 - 3x - 6)(2x + 2) = 2x^3 + 2x^2 - 6x^2 - 6x - 12x - 12 = 2x^3 - 4x^2 - 18x - 12
\]
Subtracting the two expressions:
\[
\text{Numerator} = (2x^3 + x^2 + 2x - 12) - (2x^3 - 4x^2 - 18x - 12) = 5x^2 + 20x
\]
Thus, the derivative becomes:
\[
f'(x) = \frac{5x^2 + 20x}{(x^2 + 2x + 4)^2}
\]
Step 3: Analyze the Sign of the Derivative
To determine whether the function is increasing or decreasing in an interval, we need to analyze the sign of \( f'(x) \). The denominator \( (x^2 + 2x + 4)^2 \) is always positive since the quadratic \( x^2 + 2x + 4 \) has no real roots and is always positive for all \( x \). Therefore, the sign of \( f'(x) \) depends on the numerator \( 5x^2 + 20x \), which can be factored as:
\[
f'(x) = \frac{5x(x + 4)}{(x^2 + 2x + 4)^2}
\]
The critical points are the values of \( x \) where \( f'(x) = 0 \), which occur when \( x = 0 \) or \( x = -4 \).
Step 4: Interval Analysis for Statement A
We are asked to check if \( f(x) \) is decreasing in the interval \( (-2, -1) \). To do this, we need to check the sign of \( f'(x) \) in this interval. Since the critical points are \( x = -4 \) and \( x = 0 \), we evaluate the sign of \( f'(x) \) in the interval \( (-2, -1) \). Choosing a test point, say \( x = -1.5 \), we find:
\[
f'(-1.5) = \frac{5(-1.5)(-1.5 + 4)}{( (-1.5)^2 + 2(-1.5) + 4 )^2}
\]
Simplifying:
\[
f'(-1.5) = \frac{5(-1.5)(2.5)}{(2.25 - 3 + 4)^2} = \frac{-18.75}{(3.25)^2} < 0
\]
Since \( f'(-1.5) < 0 \), the function is decreasing in the interval \( (-2, -1) \). Hence, statement (A) is true.
Step 5: Interval Analysis for Statement B
We are asked to check if \( f(x) \) is increasing in the interval \( (1, 2) \). Again, we check the sign of \( f'(x) \) in this interval. The critical points \( x = -4 \) and \( x = 0 \) are not in this interval, so we check the sign of \( f'(x) \) for \( x = 1.5 \):
\[
f'(1.5) = \frac{5(1.5)(1.5 + 4)}{( (1.5)^2 + 2(1.5) + 4 )^2}
\]
Simplifying:
\[
f'(1.5) = \frac{5(1.5)(5.5)}{(2.25 + 3 + 4)^2} = \frac{41.25}{9.25^2} > 0
\]
Since \( f'(1.5) > 0 \), the function is increasing in the interval \( (1, 2) \). Hence, statement (B) is true.
Final Answer:
Both statements (A) and (B) are correct:
- (A) \( f(x) \) is decreasing in the interval \( (-2, -1) \).
- (B) \( f(x) \) is increasing in the interval \( (1, 2) \).
The reaction sequence given below is carried out with 16 moles of X. The yield of the major product in each step is given below the product in parentheses. The amount (in grams) of S produced is ____. 
Use: Atomic mass (in amu): H = 1, C = 12, O = 16, Br = 80
Let $ \mathbb{R} $ denote the set of all real numbers. Then the area of the region $$ \left\{ (x, y) \in \mathbb{R} \times \mathbb{R} : x > 0, y > \frac{1}{x},\ 5x - 4y - 1 > 0,\ 4x + 4y - 17 < 0 \right\} $$ is
As shown in the figures, a uniform rod $ OO' $ of length $ l $ is hinged at the point $ O $ and held in place vertically between two walls using two massless springs of the same spring constant. The springs are connected at the midpoint and at the top-end $ (O') $ of the rod, as shown in Fig. 1, and the rod is made to oscillate by a small angular displacement. The frequency of oscillation of the rod is $ f_1 $. On the other hand, if both the springs are connected at the midpoint of the rod, as shown in Fig. 2, and the rod is made to oscillate by a small angular displacement, then the frequency of oscillation is $ f_2 $. Ignoring gravity and assuming motion only in the plane of the diagram, the value of $\frac{f_1}{f_2}$ is:
A relation R from a non-empty set B is a subset of the cartesian product A × B. The subset is derived by describing a relationship between the first element and the second element of the ordered pairs in A × B.
A relation f from a set A to a set B is said to be a function if every element of set A has one and only one image in set B. In other words, no two distinct elements of B have the same pre-image.
Relations and functions can be represented in different forms such as arrow representation, algebraic form, set-builder form, graphically, roster form, and tabular form. Define a function f: A = {1, 2, 3} → B = {1, 4, 9} such that f(1) = 1, f(2) = 4, f(3) = 9. Now, represent this function in different forms.
