GSEB Class 12 Maths Solutions Chapter 6 Application of Derivatives Exercise 6.2

Question 1. Show that the function given by \( f(x) = 3x + 17 \) is strictly increasing on R.
Answer: We have the function \( f(x) = 3x + 17 \).
To check if it is increasing, we find its derivative: \( f'(x) = 3 \).
Since \( f'(x) = 3 \), which is a positive number, it means the function always grows for any real value of x.
Therefore, \( f \) is strictly increasing on R.
In simple words: The function \( f(x) = 3x + 17 \) is always going up because its slope is a fixed positive number, 3.

Exam Tip: To show a function is strictly increasing, calculate its first derivative. If \( f'(x) > 0 \) for all \( x \) in the domain, the function is strictly increasing.

Question 2. Show that the function given by \( f(x) = e^{2x} \) is strictly increasing on R.
Answer: We have the function \( f(x) = e^{2x} \).
To determine if it is increasing, we first find its derivative:
\( f'(x) = \frac{d}{dx} (e^{2x}) = e^{2x} \cdot \frac{d}{dx} (2x) = 2e^{2x} \).

Now, we examine the sign of \( f'(x) \) for different cases of \( x \):
Case I: When \( x > 0 \).
In this case, \( e^{2x} \) is always positive, and multiplying by 2 keeps it positive. So, \( f'(x) = 2e^{2x} > 0 \).
Case II: When \( x = 0 \).
\( f'(0) = 2e^{2 \cdot 0} = 2e^0 = 2 \cdot 1 = 2 \).
Since \( f'(0) = 2 > 0 \), it is also positive.
Case III: When \( x < 0 \).
Let \( x = -y \), where \( y \) is a positive number. So, \( x \) is negative.
\( f'(x) = 2e^{2x} = 2e^{2(-y)} = 2e^{-2y} = \frac{2}{e^{2y}} \).
Since \( y > 0 \), \( e^{2y} \) is a positive number, so \( \frac{2}{e^{2y}} \) is also positive.
Thus, \( f'(x) > 0 \) for \( x < 0 \).

Since \( f'(x) > 0 \) for all \( x \in R \) (for positive, zero, and negative values of \( x \)), the function \( f(x) = e^{2x} \) is strictly increasing on R.
In simple words: The function \( f(x) = e^{2x} \) is always increasing. When we find its rate of change (derivative), we get \( 2e^{2x} \), which is always a positive number for any value of \( x \). This shows the function continuously goes up.

Exam Tip: Remember that the exponential function \( e^u \) is always positive for any real number \( u \). This property is key when analyzing the sign of derivatives involving exponential terms.

Question 3. Show that the function given by \( f(x) = \sin x \) is
(a) strictly increasing in \( (0, \frac{\pi}{2}) \)
(b) strictly decreasing in \( (\frac{\pi}{2}, \pi) \)
(c) neither increasing nor decreasing in \( (0, \pi) \)
Answer: We have the function \( f(x) = \sin x \).
First, find the derivative of \( f(x) \):
\( f'(x) = \frac{d}{dx} (\sin x) = \cos x \).

(a) For the interval \( (0, \frac{\pi}{2}) \):
In this interval, the value of \( \cos x \) is positive. For example, at \( x = \frac{\pi}{4} \), \( \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} > 0 \).
Since \( f'(x) = \cos x > 0 \) for \( x \in (0, \frac{\pi}{2}) \), \( f(x) \) is strictly increasing in this range.

(b) For the interval \( (\frac{\pi}{2}, \pi) \):
In this interval, the value of \( \cos x \) is negative. For example, at \( x = \frac{3\pi}{4} \), \( \cos(\frac{3\pi}{4}) = -\frac{\sqrt{2}}{2} < 0 \).
Since \( f'(x) = \cos x < 0 \) for \( x \in (\frac{\pi}{2}, \pi) \), \( f(x) \) is strictly decreasing in this range.

(c) For the interval \( (0, \pi) \):
We have seen that in \( (0, \frac{\pi}{2}) \), \( f'(x) \) is positive, and in \( (\frac{\pi}{2}, \pi) \), \( f'(x) \) is negative.
Since \( f'(x) \) does not maintain a single sign (it changes from positive to negative) across the entire interval \( (0, \pi) \), the function \( f(x) \) is neither strictly increasing nor strictly decreasing in \( (0, \pi) \). It increases then decreases.
In simple words: For \( f(x) = \sin x \), its slope is \( \cos x \). From 0 to 90 degrees (first quadrant), \( \cos x \) is positive, so \( \sin x \) goes up. From 90 to 180 degrees (second quadrant), \( \cos x \) is negative, so \( \sin x \) goes down. Because it goes up and then down in the full 0 to 180 degree range, it is not consistently increasing or decreasing.

Exam Tip: To analyze monotonicity for trigonometric functions, remember the signs of sine and cosine in different quadrants. The derivative's sign determines whether the function is increasing or decreasing in that interval.

Question 4. Find the intervals in which the function \( f \) given by, \( f(x) = 2x^2 - 3x \) is (a) strictly increasing, (b) strictly decreasing.
Answer: We have the function \( f(x) = 2x^2 - 3x \).
First, find the derivative of \( f(x) \):
\( f'(x) = \frac{d}{dx} (2x^2 - 3x) = 4x - 3 \).

To find the critical points, set \( f'(x) = 0 \):
\( 4x - 3 = 0 \)
\( 4x = 3 \)
\( x = \frac{3}{4} \).

This point \( x = \frac{3}{4} \) divides the real number line into two distinct intervals: \( (-\infty, \frac{3}{4}) \) and \( (\frac{3}{4}, \infty) \).

Now, we check the sign of \( f'(x) \) in each interval:
Interval 1: \( (-\infty, \frac{3}{4}) \)
Pick a test value, for example, \( x = 0 \).
\( f'(0) = 4(0) - 3 = -3 \).
Since \( f'(x) < 0 \) in this interval, \( f(x) \) is strictly decreasing in \( (-\infty, \frac{3}{4}) \).

Interval 2: \( (\frac{3}{4}, \infty) \)
Pick a test value, for example, \( x = 1 \).
\( f'(1) = 4(1) - 3 = 1 \).
Since \( f'(x) > 0 \) in this interval, \( f(x) \) is strictly increasing in \( (\frac{3}{4}, \infty) \).

Therefore:
(a) The function \( f(x) \) is strictly increasing in the interval \( (\frac{3}{4}, \infty) \).
(b) The function \( f(x) \) is strictly decreasing in the interval \( (-\infty, \frac{3}{4}) \).
In simple words: To see where \( f(x) = 2x^2 - 3x \) goes up or down, we first found its slope, which is \( 4x - 3 \). When the slope is zero, \( x = \frac{3}{4} \). This point splits the number line into two parts. Before \( \frac{3}{4} \), the slope is negative, so the function decreases. After \( \frac{3}{4} \), the slope is positive, so the function increases.

Exam Tip: For polynomial functions, finding the critical points (where \( f'(x) = 0 \)) helps divide the number line into intervals. Then, test a value in each interval to determine the sign of the derivative and, consequently, the function's monotonicity.

Question 5. Find the intervals in which the function \( f \) given by \( f(x) = 2x^3 - 3x^2 - 36x + 7 \) is (a) strictly increasing, (b) strictly decreasing.
Answer: We have the function \( f(x) = 2x^3 - 3x^2 - 36x + 7 \).
First, find the derivative of \( f(x) \):
\( f'(x) = \frac{d}{dx} (2x^3 - 3x^2 - 36x + 7) = 6x^2 - 6x - 36 \).

To find the critical points, set \( f'(x) = 0 \):
\( 6x^2 - 6x - 36 = 0 \)
Divide by 6:
\( x^2 - x - 6 = 0 \)
Factor the quadratic equation:
\( (x - 3)(x + 2) = 0 \)
So, the critical points are \( x = 3 \) and \( x = -2 \).

These points \( x = -2 \) and \( x = 3 \) divide the real number line into three distinct intervals: \( (-\infty, -2) \), \( (-2, 3) \), and \( (3, \infty) \).

Now, check the sign of \( f'(x) \) in each interval:
Interval 1: \( (-\infty, -2) \)
Pick a test value, for example, \( x = -3 \).
\( f'(-3) = 6(-3 - 3)(-3 + 2) = 6(-6)(-1) = 36 \).
Since \( f'(x) > 0 \) in this interval, \( f(x) \) is strictly increasing in \( (-\infty, -2) \).

Interval 2: \( (-2, 3) \)
Pick a test value, for example, \( x = 0 \).
\( f'(0) = 6(0 - 3)(0 + 2) = 6(-3)(2) = -36 \).
Since \( f'(x) < 0 \) in this interval, \( f(x) \) is strictly decreasing in \( (-2, 3) \).

Interval 3: \( (3, \infty) \)
Pick a test value, for example, \( x = 4 \).
\( f'(4) = 6(4 - 3)(4 + 2) = 6(1)(6) = 36 \).
Since \( f'(x) > 0 \) in this interval, \( f(x) \) is strictly increasing in \( (3, \infty) \).

Therefore:
(a) The function \( f(x) \) is strictly increasing in the intervals \( (-\infty, -2) \cup (3, \infty) \).
(b) The function \( f(x) \) is strictly decreasing in the interval \( (-2, 3) \).
In simple words: To find where \( f(x) = 2x^3 - 3x^2 - 36x + 7 \) increases or decreases, we first found its slope, \( f'(x) = 6x^2 - 6x - 36 \). Setting the slope to zero gave us \( x = -2 \) and \( x = 3 \). These points create three sections on the number line. The function goes up before \( -2 \), then goes down between \( -2 \) and \( 3 \), and then goes up again after \( 3 \).

Exam Tip: For cubic functions, you typically get two critical points that divide the number line into three intervals. Always factorize the derivative correctly to find these points and then test each interval's sign.

Question 6. Find the intervals in which the following functions are strictly increasing or decreasing:
(b) \( 10 - 6x - 2x^2 \)
(c) \( -2x^3 - 9x^2 - 12x + 1 \)
(d) \( 6 - 9x - x^2 \)
(e) \( (x + 1)^3(x - 3)^3 \)
Answer:
(a) We have: \( f(x) = x^2 + 2x - 5 \).
The derivative is \( f'(x) = 2x + 2 = 2(x + 1) \).
For \( f(x) \) to be increasing, \( f'(x) > 0 \)
\( 2(x + 1) > 0 \)
\( x + 1 > 0 \)
\( x > -1 \).
For \( f(x) \) to be decreasing, \( f'(x) < 0 \)
\( 2(x + 1) < 0 \)
\( x + 1 < 0 \)
\( x < -1 \).
Thus, \( f(x) \) is strictly increasing in \( (-1, \infty) \) and strictly decreasing in \( (-\infty, -1) \).
In simple words: For \( f(x) = x^2 + 2x - 5 \), its slope is \( 2x + 2 \). When \( x \) is greater than -1, the slope is positive, meaning the function increases. When \( x \) is less than -1, the slope is negative, meaning the function decreases.

(b) We have: \( f(x) = 10 - 6x - 2x^2 \).
The derivative is \( f'(x) = -6 - 4x = -2(3 + 2x) \).
For \( f(x) \) to be increasing, \( f'(x) > 0 \)
\( -2(3 + 2x) > 0 \)
\( 3 + 2x < 0 \) (multiplying by a negative number reverses the inequality)
\( 2x < -3 \)
\( x < -\frac{3}{2} \).
For \( f(x) \) to be decreasing, \( f'(x) < 0 \)
\( -2(3 + 2x) < 0 \)
\( 3 + 2x > 0 \)
\( 2x > -3 \)
\( x > -\frac{3}{2} \).
Thus, \( f(x) \) is strictly increasing for \( x < -\frac{3}{2} \), i.e., in \( (-\infty, -\frac{3}{2}) \), and strictly decreasing for \( x > -\frac{3}{2} \), i.e., in \( (-\frac{3}{2}, \infty) \).
In simple words: For \( f(x) = 10 - 6x - 2x^2 \), its slope is \( -6 - 4x \). If \( x \) is smaller than \( -\frac{3}{2} \), the slope is positive, so the function increases. If \( x \) is larger than \( -\frac{3}{2} \), the slope is negative, so the function decreases.

(c) We have: \( f(x) = -2x^3 - 9x^2 - 12x + 1 \).
The derivative is \( f'(x) = -6x^2 - 18x - 12 = -6(x^2 + 3x + 2) \).
Set \( f'(x) = 0 \) to find critical points:
\( -6(x^2 + 3x + 2) = 0 \)
\( x^2 + 3x + 2 = 0 \)
\( (x + 1)(x + 2) = 0 \)
So, \( x = -1 \) or \( x = -2 \).
These points divide the real line into three intervals: \( (-\infty, -2) \), \( (-2, -1) \), and \( (-1, \infty) \).

Check the sign of \( f'(x) \) in each interval:
Interval 1: \( (-\infty, -2) \). Pick \( x = -3 \).
\( f'(-3) = -6(-3+1)(-3+2) = -6(-2)(-1) = -12 \).
Since \( f'(x) < 0 \), \( f(x) \) is strictly decreasing in \( (-\infty, -2) \).

Interval 2: \( (-2, -1) \). Pick \( x = -1.5 \).
\( f'(-1.5) = -6(-1.5+1)(-1.5+2) = -6(-0.5)(0.5) = 1.5 \).
Since \( f'(x) > 0 \), \( f(x) \) is strictly increasing in \( (-2, -1) \).

Interval 3: \( (-1, \infty) \). Pick \( x = 0 \).
\( f'(0) = -6(0+1)(0+2) = -6(1)(2) = -12 \).
Since \( f'(x) < 0 \), \( f(x) \) is strictly decreasing in \( (-1, \infty) \).

Therefore: \( f(x) \) is strictly increasing in \( (-2, -1) \) and strictly decreasing in \( (-\infty, -2) \cup (-1, \infty) \).
In simple words: For \( f(x) = -2x^3 - 9x^2 - 12x + 1 \), its slope is \( -6(x+1)(x+2) \). The critical points are \( x = -2 \) and \( x = -1 \). The function goes down before \( -2 \), up between \( -2 \) and \( -1 \), and then down again after \( -1 \).

(d) We have: \( f(x) = 6 - 9x - x^2 \).
The derivative is \( f'(x) = -9 - 2x \).
For \( f(x) \) to be increasing, \( f'(x) > 0 \)
\( -9 - 2x > 0 \)
\( -2x > 9 \)
\( x < -\frac{9}{2} \).
For \( f(x) \) to be decreasing, \( f'(x) < 0 \)
\( -9 - 2x < 0 \)
\( -2x < 9 \)
\( x > -\frac{9}{2} \).
Thus, \( f(x) \) is strictly increasing in \( (-\infty, -\frac{9}{2}) \) and strictly decreasing in \( (-\frac{9}{2}, \infty) \).
In simple words: For \( f(x) = 6 - 9x - x^2 \), its slope is \( -9 - 2x \). The function increases when \( x \) is smaller than \( -\frac{9}{2} \) and decreases when \( x \) is larger than \( -\frac{9}{2} \).

(e) We have: \( f(x) = (x + 1)^3(x - 3)^3 \).
We can rewrite this as \( f(x) = ((x+1)(x-3))^3 = (x^2 - 2x - 3)^3 \).
The derivative is \( f'(x) = 3(x^2 - 2x - 3)^2 \cdot \frac{d}{dx}(x^2 - 2x - 3) \)
\( f'(x) = 3(x^2 - 2x - 3)^2 \cdot (2x - 2) \)
\( f'(x) = 3((x+1)(x-3))^2 \cdot 2(x - 1) \)
\( f'(x) = 6(x + 1)^2(x - 3)^2(x - 1) \).

To find critical points, set \( f'(x) = 0 \):
\( 6(x + 1)^2(x - 3)^2(x - 1) = 0 \)
This gives \( x = -1, x = 3, x = 1 \).
The points -1, 1, 3 divide the real line into four intervals: \( (-\infty, -1) \), \( (-1, 1) \), \( (1, 3) \), and \( (3, \infty) \).

Note that \( (x+1)^2 \) and \( (x-3)^2 \) are always \( \ge 0 \). So the sign of \( f'(x) \) depends only on the sign of \( (x-1) \).

For \( f(x) \) to be increasing, \( f'(x) > 0 \):
\( 6(x + 1)^2(x - 3)^2(x - 1) > 0 \)
Since \( 6(x + 1)^2(x - 3)^2 \ge 0 \), this inequality holds if \( (x - 1) > 0 \), which means \( x > 1 \).
However, at \( x = 3 \), \( f'(3) = 0 \). So, the function is strictly increasing for \( x \in (1, 3) \cup (3, \infty) \).

For \( f(x) \) to be decreasing, \( f'(x) < 0 \):
\( 6(x + 1)^2(x - 3)^2(x - 1) < 0 \)
This inequality holds if \( (x - 1) < 0 \), which means \( x < 1 \).
However, at \( x = -1 \), \( f'(-1) = 0 \). So, the function is strictly decreasing for \( x \in (-\infty, -1) \cup (-1, 1) \).
Therefore: \( f(x) \) is strictly increasing in \( (1, 3) \cup (3, \infty) \) and strictly decreasing in \( (-\infty, -1) \cup (-1, 1) \).
In simple words: For \( f(x) = (x + 1)^3(x - 3)^3 \), its slope is \( 6(x + 1)^2(x - 3)^2(x - 1) \). Because \( (x + 1)^2 \) and \( (x - 3)^2 \) are always positive or zero, the direction of the function only depends on \( (x - 1) \). The function increases when \( x > 1 \) (except at \( x=3 \) where it's flat for a moment) and decreases when \( x < 1 \) (except at \( x=-1 \) where it's also flat).

Exam Tip: When the derivative has squared terms like \( (x+1)^2 \), those terms do not change the sign of the derivative, but they can create points where \( f'(x) = 0 \) without changing the monotonicity around them (unless the factor is the only one left). Always analyze the sign contribution of each factor.

Question 7. Show that \( y = \log (1 + x) - \frac{2x}{2+x}, x > -1 \) is an increasing function of \( x \) throughout its domain.
Answer: Let \( f(x) = \log (1 + x) - \frac{2x}{2+x} \). The domain is \( x > -1 \).
First, find the derivative of \( f(x) \):
\( f'(x) = \frac{d}{dx} (\log(1+x)) - \frac{d}{dx} (\frac{2x}{2+x}) \)
\( = \frac{1}{1+x} - \frac{(2+x)(2) - (2x)(1)}{(2+x)^2} \)
\( = \frac{1}{1+x} - \frac{4 + 2x - 2x}{(2+x)^2} \)
\( = \frac{1}{1+x} - \frac{4}{(2+x)^2} \)
To combine these, find a common denominator:
\( = \frac{(2+x)^2 - 4(1+x)}{(1+x)(2+x)^2} \)
\( = \frac{(4 + 4x + x^2) - (4 + 4x)}{(1+x)(2+x)^2} \)
\( = \frac{x^2}{(1+x)(2+x)^2} \).

Now, analyze the sign of \( f'(x) \).
We know that for the domain \( x > -1 \):
1. \( x^2 \) is always greater than or equal to 0 (i.e., \( x^2 \ge 0 \)). It is 0 only when \( x=0 \).
2. \( (1+x) \) is positive because \( x > -1 \).
3. \( (2+x)^2 \) is always positive because \( x > -1 \) means \( 2+x > 1 \), so its square is positive.

Therefore, for \( x > -1 \):
\( f'(x) = \frac{x^2}{(1+x)(2+x)^2} \ge 0 \).
\( f'(x) \) is greater than 0 for all \( x > -1 \) except at \( x=0 \), where \( f'(0) = 0 \).
Since \( f'(x) \ge 0 \) throughout its domain \( x > -1 \), the function \( f(x) \) is an increasing function of \( x \).
In simple words: To show that \( y = \log (1 + x) - \frac{2x}{2+x} \) always goes up for \( x > -1 \), we took its derivative. After simplifying, the derivative came out to be \( \frac{x^2}{(1+x)(2+x)^2} \). Since all parts of this expression are positive (or zero at \( x=0 \)), the slope is always positive or zero. This tells us the function is always increasing.

Exam Tip: When proving a function is increasing or decreasing, after finding the derivative, carefully analyze the sign of each factor in the derivative across the given domain. Remember that a term like \( x^2 \) or \( (expression)^2 \) is always non-negative.

Question 8. Find the values of \( x \) for which \( y = [x(x - 2)]^2 \) is an increasing function.
Answer: We have the function \( y = [x(x - 2)]^2 \).
First, expand the function:
\( y = (x^2 - 2x)^2 = x^2(x^2 - 4x + 4) = x^4 - 4x^3 + 4x^2 \).
Now, find the derivative \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{d}{dx} (x^4 - 4x^3 + 4x^2) = 4x^3 - 12x^2 + 8x \).
Factor the derivative:
\( \frac{dy}{dx} = 4x(x^2 - 3x + 2) = 4x(x - 1)(x - 2) \).

For the function to be increasing, \( \frac{dy}{dx} > 0 \):
\( 4x(x - 1)(x - 2) > 0 \).

The critical points are \( x = 0, x = 1, x = 2 \). These points divide the real line into four intervals:
1. \( (-\infty, 0) \)
2. \( (0, 1) \)
3. \( (1, 2) \)
4. \( (2, \infty) \)

Let's check the sign of \( \frac{dy}{dx} \) in each interval using test values:
Interval 1: \( (-\infty, 0) \). Pick \( x = -1 \).
\( \frac{dy}{dx} = 4(-1)(-1-1)(-1-2) = 4(-1)(-2)(-3) = -24 \). (Negative)
So, \( y \) is decreasing in \( (-\infty, 0) \).

Interval 2: \( (0, 1) \). Pick \( x = 0.5 \).
\( \frac{dy}{dx} = 4(0.5)(0.5-1)(0.5-2) = 4(0.5)(-0.5)(-1.5) = 1.5 \). (Positive)
So, \( y \) is increasing in \( (0, 1) \).

Interval 3: \( (1, 2) \). Pick \( x = 1.5 \).
\( \frac{dy}{dx} = 4(1.5)(1.5-1)(1.5-2) = 4(1.5)(0.5)(-0.5) = -1.5 \). (Negative)
So, \( y \) is decreasing in \( (1, 2) \).

Interval 4: \( (2, \infty) \). Pick \( x = 3 \).
\( \frac{dy}{dx} = 4(3)(3-1)(3-2) = 4(3)(2)(1) = 24 \). (Positive)
So, \( y \) is increasing in \( (2, \infty) \).

Thus, the function \( y = [x(x - 2)]^2 \) is increasing when \( x \in (0, 1) \) or \( x \in (2, \infty) \).
The function is increasing for \( x \in (0, 1) \cup (2, \infty) \).
In simple words: To find where \( y = [x(x - 2)]^2 \) is increasing, we first wrote it out as \( x^4 - 4x^3 + 4x^2 \). Then we found its slope, \( \frac{dy}{dx} = 4x(x - 1)(x - 2) \). The slope is zero at \( x=0, 1, 2 \). These points create sections on the number line. We found that the function goes up when \( x \) is between 0 and 1, and also when \( x \) is greater than 2.

Exam Tip: When analyzing the sign of a factored polynomial derivative, it's helpful to list the roots (critical points) in order on a number line. Then, choose test points in each interval to quickly determine the derivative's sign. This method, often called a "sign chart," is very efficient.

Question 9. Prove that \( y = \frac{4 \sin \theta}{2+\cos \theta} - \theta \) is an increasing function of \( \theta \) in \( [0, \frac{\pi}{2}] \).
Answer: We have the function \( y = \frac{4 \sin \theta}{2+\cos \theta} - \theta \).
First, find the derivative \( \frac{dy}{d\theta} \):
\( \frac{dy}{d\theta} = \frac{d}{d\theta} \left( \frac{4 \sin \theta}{2+\cos \theta} \right) - \frac{d}{d\theta} (\theta) \)
Use the quotient rule for the first term: \( \frac{d}{d\theta} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \).
Here, \( u = 4 \sin \theta \implies u' = 4 \cos \theta \)
And \( v = 2 + \cos \theta \implies v' = -\sin \theta \)

So, \( \frac{d}{d\theta} \left( \frac{4 \sin \theta}{2+\cos \theta} \right) = \frac{(4 \cos \theta)(2+\cos \theta) - (4 \sin \theta)(-\sin \theta)}{(2+\cos \theta)^2} \)
\( = \frac{8 \cos \theta + 4 \cos^2 \theta + 4 \sin^2 \theta}{(2+\cos \theta)^2} \)
\( = \frac{8 \cos \theta + 4 (\cos^2 \theta + \sin^2 \theta)}{(2+\cos \theta)^2} \)
Since \( \cos^2 \theta + \sin^2 \theta = 1 \):
\( = \frac{8 \cos \theta + 4}{(2+\cos \theta)^2} \).

Now, substitute this back into the derivative of \( y \):
\( \frac{dy}{d\theta} = \frac{8 \cos \theta + 4}{(2+\cos \theta)^2} - 1 \)
To combine, find a common denominator:
\( = \frac{8 \cos \theta + 4 - (2+\cos \theta)^2}{(2+\cos \theta)^2} \)
\( = \frac{8 \cos \theta + 4 - (4 + 4 \cos \theta + \cos^2 \theta)}{(2+\cos \theta)^2} \)
\( = \frac{8 \cos \theta + 4 - 4 - 4 \cos \theta - \cos^2 \theta}{(2+\cos \theta)^2} \)
\( = \frac{4 \cos \theta - \cos^2 \theta}{(2+\cos \theta)^2} \)
Factor out \( \cos \theta \) from the numerator:
\( = \frac{\cos \theta (4 - \cos \theta)}{(2+\cos \theta)^2} \).

For the function to be increasing, \( \frac{dy}{d\theta} > 0 \). We need to show this holds for \( \theta \in [0, \frac{\pi}{2}] \).
In the interval \( [0, \frac{\pi}{2}] \):
1. \( \cos \theta \): For \( \theta \in [0, \frac{\pi}{2}] \), \( \cos \theta \) is always \( \ge 0 \). It is 0 only at \( \theta = \frac{\pi}{2} \).
2. \( (4 - \cos \theta) \): Since \( 0 \le \cos \theta \le 1 \) in this interval, \( 4 - \cos \theta \) will be between \( 4-1=3 \) and \( 4-0=4 \). So \( (4 - \cos \theta) \) is always positive.
3. \( (2+\cos \theta)^2 \): Since \( 0 \le \cos \theta \le 1 \), \( 2+\cos \theta \) will be between \( 2+0=2 \) and \( 2+1=3 \). So \( (2+\cos \theta)^2 \) is always positive.

Therefore, for \( \theta \in [0, \frac{\pi}{2}] \):
\( \frac{\cos \theta (4 - \cos \theta)}{(2+\cos \theta)^2} \ge 0 \).
The derivative \( \frac{dy}{d\theta} \) is \( \ge 0 \) for all \( \theta \in [0, \frac{\pi}{2}] \). It is 0 only at \( \theta = \frac{\pi}{2} \).
Thus, the function is an increasing function of \( \theta \) in \( [0, \frac{\pi}{2}] \).
In simple words: To prove that \( y = \frac{4 \sin \theta}{2+\cos \theta} - \theta \) always increases between 0 and 90 degrees, we found its slope, \( \frac{dy}{d\theta} \), which simplifies to \( \frac{\cos \theta (4 - \cos \theta)}{(2+\cos \theta)^2} \). In this range, \( \cos \theta \) is positive or zero, \( (4 - \cos \theta) \) is always positive, and \( (2+\cos \theta)^2 \) is always positive. This means the slope is always positive or zero, so the function is always going up.

Exam Tip: For trigonometric functions, pay close attention to the specified interval. The signs of sine and cosine (and related expressions) change across different quadrants, which directly impacts the sign of the derivative.

Question 10. Prove that the logarithmic function is strictly increasing in \( (0, \infty) \).
Answer: Let the logarithmic function be \( f(x) = \log x \). The domain of this function is \( (0, \infty) \), meaning \( x > 0 \).
To prove it is strictly increasing, we need to show that its derivative is always positive in this domain.
Find the derivative of \( f(x) \):
\( f'(x) = \frac{d}{dx} (\log x) = \frac{1}{x} \).

Now, we examine the sign of \( f'(x) \) for \( x \in (0, \infty) \).
Since \( x \) is in the interval \( (0, \infty) \), it means \( x \) is always a positive number.
Therefore, \( \frac{1}{x} \) will also always be a positive number.
So, \( f'(x) = \frac{1}{x} > 0 \) for all \( x \in (0, \infty) \).

Since the first derivative \( f'(x) \) is strictly positive throughout its entire domain \( (0, \infty) \), the logarithmic function \( f(x) = \log x \) is strictly increasing in \( (0, \infty) \).
In simple words: The logarithmic function, like \( \log x \), only works for numbers greater than zero. When we find its slope, we get \( \frac{1}{x} \). Since \( x \) must be positive, \( \frac{1}{x} \) is also always positive. A positive slope means the function is always going up, so it is strictly increasing.

Exam Tip: Remember the fundamental derivative of \( \log x \) as \( \frac{1}{x} \). For proofs of monotonicity, clearly state the derivative and analyze its sign across the function's specified domain.

Question 11. Prove that the function \( f \) given by \( f(x) = x^2 - x + 1 \) is neither strictly increasing nor strictly decreasing on \( (-1, 1) \).
Answer: We have the function \( f(x) = x^2 - x + 1 \).
First, find the derivative of \( f(x) \):
\( f'(x) = \frac{d}{dx} (x^2 - x + 1) = 2x - 1 \).

To find the critical point, set \( f'(x) = 0 \):
\( 2x - 1 = 0 \)
\( 2x = 1 \)
\( x = \frac{1}{2} \).

This critical point \( x = \frac{1}{2} \) lies within the given interval \( (-1, 1) \).
The point \( x = \frac{1}{2} \) divides the interval \( (-1, 1) \) into two sub-intervals: \( (-1, \frac{1}{2}) \) and \( (\frac{1}{2}, 1) \).

Now, let's check the sign of \( f'(x) \) in each sub-interval:
Interval 1: \( (-1, \frac{1}{2}) \). Pick a test value, for example, \( x = 0 \).
\( f'(0) = 2(0) - 1 = -1 \).
Since \( f'(x) < 0 \) in this interval, \( f(x) \) is strictly decreasing in \( (-1, \frac{1}{2}) \).

Interval 2: \( (\frac{1}{2}, 1) \). Pick a test value, for example, \( x = 0.75 \).
\( f'(0.75) = 2(0.75) - 1 = 1.5 - 1 = 0.5 \).
Since \( f'(x) > 0 \) in this interval, \( f(x) \) is strictly increasing in \( (\frac{1}{2}, 1) \).

Since the derivative \( f'(x) \) changes its sign (from negative to positive) within the interval \( (-1, 1) \), the function \( f(x) \) is neither strictly increasing nor strictly decreasing over the entire interval \( (-1, 1) \). It decreases first, then increases.
In simple words: For \( f(x) = x^2 - x + 1 \) in the range from -1 to 1, we found its slope, \( f'(x) = 2x - 1 \). The slope is zero at \( x = \frac{1}{2} \). Before \( \frac{1}{2} \), the slope is negative, so the function goes down. After \( \frac{1}{2} \), the slope is positive, so the function goes up. Because it changes direction, it's not consistently increasing or decreasing in the whole interval.

Exam Tip: To prove a function is "neither increasing nor decreasing" over an interval, you must show that its derivative changes sign (or becomes zero) at some point within that interval, indicating a change in monotonicity. Identifying the critical points is essential.

Question 12. Which of the following functions are strictly decreasing on \( (0, \frac{\pi}{2}) \)?
(A) \( \cos x \)
(B) \( \cos 2x \)
(C) \( \cos 3x \)
(D) \( \tan x \)
Answer: (A) \( \cos x \) and (B) \( \cos 2x \)
Let's analyze each option in the interval \( (0, \frac{\pi}{2}) \):

(A) For \( f(x) = \cos x \):
The derivative is \( f'(x) = -\sin x \).
In the interval \( (0, \frac{\pi}{2}) \), \( \sin x \) is positive. Therefore, \( f'(x) = -\sin x \) is negative.
So, \( f(x) = \cos x \) is strictly decreasing in \( (0, \frac{\pi}{2}) \).

(B) For \( f(x) = \cos 2x \):
The derivative is \( f'(x) = -2\sin 2x \).
In the interval \( (0, \frac{\pi}{2}) \), the argument \( 2x \) is in \( (0, \pi) \).
In the interval \( (0, \pi) \), \( \sin 2x \) is positive. Therefore, \( f'(x) = -2\sin 2x \) is negative.
So, \( f(x) = \cos 2x \) is strictly decreasing in \( (0, \frac{\pi}{2}) \).

(C) For \( f(x) = \cos 3x \):
The derivative is \( f'(x) = -3\sin 3x \).
In the interval \( (0, \frac{\pi}{2}) \), the argument \( 3x \) is in \( (0, \frac{3\pi}{2}) \).
In \( (0, \pi) \), \( \sin 3x \) is positive, making \( f'(x) \) negative (decreasing).
In \( (\pi, \frac{3\pi}{2}) \), \( \sin 3x \) is negative, making \( f'(x) \) positive (increasing).
Since \( f'(x) \) changes sign in \( (0, \frac{\pi}{2}) \) (specifically, within \( 3x \in (0, \frac{3\pi}{2}) \)), \( f(x) = \cos 3x \) is neither strictly increasing nor strictly decreasing in the entire interval \( (0, \frac{\pi}{2}) \).

(D) For \( f(x) = \tan x \):
The derivative is \( f'(x) = \sec^2 x \).
In the interval \( (0, \frac{\pi}{2}) \), \( \sec x \) is positive, so \( \sec^2 x \) is always positive.
So, \( f(x) = \tan x \) is strictly increasing in \( (0, \frac{\pi}{2}) \).

Based on the analysis, \( \cos x \) and \( \cos 2x \) are strictly decreasing functions on \( (0, \frac{\pi}{2}) \).
In simple words: We need functions that always go down between 0 and 90 degrees. For \( \cos x \), its slope is \( -\sin x \), which is negative in this range, so it decreases. For \( \cos 2x \), its slope is \( -2\sin 2x \), which is also negative in this range, so it decreases. For \( \cos 3x \), its slope changes, so it doesn't always decrease. For \( \tan x \), its slope is always positive, so it increases.

Exam Tip: For MCQs involving trigonometric functions, quickly recall the signs of the derivatives in the given interval. Pay attention to composite functions like \( \cos(kx) \), where the argument \( kx \) might span multiple quadrants, causing the derivative's sign to change.

Question 13. On which of the following intervals is the function \( f \) given by \( f(x) = x^{100} + \sin x - 1 \) strictly decreasing?
(A) \( (-1, 1) \)
(B) \( (0, 1) \)
(C) \( (\frac{\pi}{2}, \pi] \)
(D) \( (0, \frac{\pi}{2}) \)
Answer: (D) None of the options
We have the function \( f(x) = x^{100} + \sin x - 1 \).
First, find the derivative of \( f(x) \):
\( f'(x) = \frac{d}{dx} (x^{100} + \sin x - 1) = 100x^{99} + \cos x \).

Now, let's analyze the sign of \( f'(x) \) in each given interval:

(A) For the interval \( (-1, 1) \):
If \( x \in (-1, 0) \), then \( x^{99} \) is negative (e.g., \( (-0.5)^{99} \)), and \( \cos x \) is positive.
So, \( 100x^{99} \) is a negative number with a large magnitude, while \( \cos x \) is a positive number between 0 and 1. For example, if \( x = -0.5 \), \( 100(-0.5)^{99} \) is a very small negative number, while \( \cos(-0.5) \) is positive. This interval can have both positive and negative values for \( f'(x) \).
If \( x \in (0, 1) \), then \( x^{99} \) is positive and \( \cos x \) is positive. Thus, \( f'(x) = 100x^{99} + \cos x \) is positive.
Since \( f'(x) \) can be both positive and negative in \( (-1, 1) \), \( f(x) \) is neither strictly increasing nor strictly decreasing in \( (-1, 1) \).

(B) For the interval \( (0, 1) \):
If \( x \in (0, 1) \), then \( x^{99} \) is positive, and \( \cos x \) is also positive (since \( (0, 1) \) is approximately \( (0, \frac{\pi}{2}) \), and \( 1 \) radian is approx \( 57.3^\circ \)).
Thus, \( f'(x) = 100x^{99} + \cos x > 0 \) for \( x \in (0, 1) \).
So, \( f(x) \) is strictly increasing in \( (0, 1) \).

(C) For the interval \( (\frac{\pi}{2}, \pi] \):
If \( x \in (\frac{\pi}{2}, \pi] \), then \( x^{99} \) is positive (as \( x \) is positive). For example, \( (\frac{\pi}{2})^{99} \) is positive.
However, \( \cos x \) is negative in this interval. For example, \( \cos(\pi) = -1 \).
The term \( 100x^{99} \) is a very large positive number, while \( \cos x \) is between -1 and 0. So, \( 100x^{99} + \cos x \) will always be positive in this interval.
Thus, \( f'(x) = 100x^{99} + \cos x > 0 \) for \( x \in (\frac{\pi}{2}, \pi] \).
So, \( f(x) \) is strictly increasing in \( (\frac{\pi}{2}, \pi] \).

(D) For the interval \( (0, \frac{\pi}{2}) \):
If \( x \in (0, \frac{\pi}{2}) \), then \( x^{99} \) is positive, and \( \cos x \) is also positive.
Thus, \( f'(x) = 100x^{99} + \cos x > 0 \) for \( x \in (0, \frac{\pi}{2}) \).
So, \( f(x) \) is strictly increasing in \( (0, \frac{\pi}{2}) \).

Conclusion: In all the given options where \( f(x) \) is monotonic, it is strictly increasing. Therefore, none of the options show an interval where the function is strictly decreasing.
In simple words: To find where \( f(x) = x^{100} + \sin x - 1 \) decreases, we checked its slope, \( f'(x) = 100x^{99} + \cos x \). In most intervals, especially those including positive \( x \), both \( x^{99} \) and \( \cos x \) are positive or \( x^{99} \) is very large and positive, making the slope positive. So, the function is mostly increasing. No provided option shows an interval where it consistently decreases.

Exam Tip: Be careful with combined functions like \( x^{100} + \sin x \). While \( \sin x \) oscillates, \( x^{100} \) grows very rapidly for positive \( x \), often dominating the sign of the derivative for larger \( x \).

Question 14. Find the least value of \( a \) such that the function \( f \) given by \( f(x) = x^2 + ax + 1 \) is strictly increasing on \( (1, 2) \).
Answer: We have the function \( f(x) = x^2 + ax + 1 \).
First, find the derivative of \( f(x) \):
\( f'(x) = \frac{d}{dx} (x^2 + ax + 1) = 2x + a \).

For the function \( f(x) \) to be strictly increasing on the interval \( (1, 2) \), its derivative \( f'(x) \) must be strictly positive for all \( x \in (1, 2) \).
So, we require \( 2x + a > 0 \) for all \( x \in (1, 2) \).

Consider the expression \( g(x) = 2x + a \). This is a linear function, which means it's always increasing or decreasing. Since the coefficient of \( x \) is positive (2), \( g(x) \) is an increasing function.
For an increasing function to be greater than 0 over an interval, its smallest value in that interval must be greater than or equal to 0.
The smallest value of \( 2x + a \) on the interval \( (1, 2) \) occurs as \( x \) approaches 1.
So, we need \( \lim_{x \to 1^+} (2x + a) \ge 0 \).
\( 2(1) + a \ge 0 \)
\( 2 + a \ge 0 \)
\( a \ge -2 \).

Therefore, the least value of \( a \) for which \( f(x) \) is strictly increasing on \( (1, 2) \) is \( a = -2 \).
In simple words: We want the function \( f(x) = x^2 + ax + 1 \) to always go up in the range from 1 to 2. Its slope is \( 2x + a \). For the function to increase, its slope must always be positive. Since the slope itself is a rising line, its lowest point in the range (which is at \( x=1 \)) must be positive or zero. This gives us \( 2(1) + a \ge 0 \), which means \( a \ge -2 \). The smallest \( a \) can be is -2.

Exam Tip: When finding the least (or greatest) value of a parameter for monotonicity, remember that for a strictly increasing function, \( f'(x) \ge 0 \). If \( f'(x) \) is itself an increasing function, its minimum value in the interval must be non-negative.

Question 15. Let I be any interval disjoint from \( (-1, 1) \). Prove that the function \( f \) given by \( f(x) = x + \frac{1}{x} \) is strictly increasing on I.
Answer: We have the function \( f(x) = x + \frac{1}{x} \).
First, find the derivative of \( f(x) \):
\( f'(x) = \frac{d}{dx} (x + x^{-1}) = 1 - x^{-2} = 1 - \frac{1}{x^2} = \frac{x^2 - 1}{x^2} \).

For the function to be strictly increasing, \( f'(x) > 0 \).
\( \frac{x^2 - 1}{x^2} > 0 \).

We know that \( x^2 \) is always positive (since \( x \ne 0 \), as \( \frac{1}{x} \) is part of the function).
So, for \( \frac{x^2 - 1}{x^2} > 0 \) to hold, we need the numerator to be positive:
\( x^2 - 1 > 0 \)
\( x^2 > 1 \).

This inequality \( x^2 > 1 \) is true when \( x > 1 \) or \( x < -1 \).
The interval I is given as disjoint from \( (-1, 1) \). This means I is either \( (-\infty, -1) \) or \( (1, \infty) \) or a combination of intervals within these ranges, excluding \( x=0 \).
In both intervals \( (-\infty, -1) \) and \( (1, \infty) \), we have \( x^2 > 1 \).
Since \( x^2 > 1 \), it implies \( x^2 - 1 > 0 \).
And since \( x \ne 0 \), \( x^2 > 0 \).
Therefore, \( f'(x) = \frac{x^2 - 1}{x^2} > 0 \) for all \( x \in I \), where I is any interval disjoint from \( (-1, 1) \).
Thus, the function \( f(x) = x + \frac{1}{x} \) is strictly increasing on I.
In simple words: To show that \( f(x) = x + \frac{1}{x} \) always goes up in any range outside of -1 to 1, we found its slope, which is \( \frac{x^2 - 1}{x^2} \). For the function to increase, this slope must be positive. This happens when \( x^2 \) is greater than 1, meaning \( x \) is either greater than 1 or less than -1. Since the given interval I is exactly where \( x^2 > 1 \), the function's slope is always positive there, so it always increases.

Exam Tip: Pay close attention to the domain or intervals given. When dealing with fractions in the derivative, remember that the denominator's sign matters. For \( \frac{A}{B} > 0 \), both A and B must have the same sign.

Question 16. Prove that the function \( f \) given by \( f(x) = \log \sin x \) is strictly increasing on \( (0, \frac{\pi}{2}) \) and strictly decreasing on \( (\frac{\pi}{2}, \pi) \).
Answer: We have the function \( f(x) = \log \sin x \).
First, find the derivative of \( f(x) \):
\( f'(x) = \frac{d}{dx} (\log \sin x) = \frac{1}{\sin x} \cdot \frac{d}{dx} (\sin x) = \frac{1}{\sin x} \cdot \cos x = \frac{\cos x}{\sin x} = \cot x \).

Now, let's analyze the sign of \( f'(x) = \cot x \) in the given intervals:

Interval 1: \( (0, \frac{\pi}{2}) \).
In this interval, \( x \) is in the first quadrant. In the first quadrant, both \( \sin x \) and \( \cos x \) are positive.
Therefore, \( \cot x = \frac{\cos x}{\sin x} \) will be positive.
So, \( f'(x) = \cot x > 0 \) for \( x \in (0, \frac{\pi}{2}) \).
Thus, \( f(x) = \log \sin x \) is strictly increasing on \( (0, \frac{\pi}{2}) \).

Interval 2: \( (\frac{\pi}{2}, \pi) \).
In this interval, \( x \) is in the second quadrant. In the second quadrant, \( \sin x \) is positive, but \( \cos x \) is negative.
Therefore, \( \cot x = \frac{\cos x}{\sin x} \) will be negative.
So, \( f'(x) = \cot x < 0 \) for \( x \in (\frac{\pi}{2}, \pi) \).
Thus, \( f(x) = \log \sin x \) is strictly decreasing on \( (\frac{\pi}{2}, \pi) \).
In simple words: For \( f(x) = \log \sin x \), its slope is \( \cot x \). In the first quadrant (0 to 90 degrees), \( \cot x \) is positive, so the function goes up. In the second quadrant (90 to 180 degrees), \( \cot x \) is negative, so the function goes down. This matches what we needed to prove.

Exam Tip: When differentiating logarithmic functions of other functions (like \( \log(g(x)) \)), use the chain rule: \( \frac{d}{dx} \log(g(x)) = \frac{1}{g(x)} \cdot g'(x) \). Then, recall the signs of trigonometric ratios in various quadrants.

Question 17. Prove that the function \( f \) given by \( f(x) = \log \cos x \) is strictly decreasing on \( (0, \frac{\pi}{2}) \) and strictly increasing on \( (\frac{\pi}{2}, \pi) \).
Answer: We have the function \( f(x) = \log \cos x \).
First, find the derivative of \( f(x) \):
\( f'(x) = \frac{d}{dx} (\log \cos x) = \frac{1}{\cos x} \cdot \frac{d}{dx} (\cos x) = \frac{1}{\cos x} \cdot (-\sin x) = -\frac{\sin x}{\cos x} = -\tan x \).

Now, let's analyze the sign of \( f'(x) = -\tan x \) in the given intervals:

Interval 1: \( (0, \frac{\pi}{2}) \).
In this interval, \( x \) is in the first quadrant. In the first quadrant, \( \tan x \) is positive.
Therefore, \( f'(x) = -\tan x \) will be negative.
So, \( f'(x) = -\tan x < 0 \) for \( x \in (0, \frac{\pi}{2}) \).
Thus, \( f(x) = \log \cos x \) is strictly decreasing on \( (0, \frac{\pi}{2}) \).

Interval 2: \( (\frac{\pi}{2}, \pi) \).
In this interval, \( x \) is in the second quadrant. In the second quadrant, \( \tan x \) is negative.
Therefore, \( f'(x) = -\tan x \) will be negative of a negative, which is positive.
So, \( f'(x) = -\tan x > 0 \) for \( x \in (\frac{\pi}{2}, \pi) \).
Thus, \( f(x) = \log \cos x \) is strictly increasing on \( (\frac{\pi}{2}, \pi) \).
In simple words: For \( f(x) = \log \cos x \), its slope is \( -\tan x \). In the first quadrant (0 to 90 degrees), \( \tan x \) is positive, so \( -\tan x \) is negative, meaning the function goes down. In the second quadrant (90 to 180 degrees), \( \tan x \) is negative, so \( -\tan x \) is positive, meaning the function goes up. This proves the required behavior.

Exam Tip: Similar to \( \log \sin x \), apply the chain rule. Be careful with the negative sign from differentiating \( \cos x \) and how it affects the sign of \( -\tan x \) in different quadrants.

Question 18. Prove that the function given by \( f(x) = x^3 - 3x^2 + 3x - 100 \) is increasing in R.
Answer: We have the function \( f(x) = x^3 - 3x^2 + 3x - 100 \).
First, find the derivative of \( f(x) \):
\( f'(x) = \frac{d}{dx} (x^3 - 3x^2 + 3x - 100) = 3x^2 - 6x + 3 \).
Factor the derivative:
\( f'(x) = 3(x^2 - 2x + 1) \).
Recognize the perfect square trinomial inside the parenthesis:
\( f'(x) = 3(x - 1)^2 \).

Now, analyze the sign of \( f'(x) \).
For any real number \( x \), \( (x - 1)^2 \) is always greater than or equal to 0 (i.e., \( (x - 1)^2 \ge 0 \)).
It is equal to 0 only when \( x = 1 \). Otherwise, it is positive.
Since \( (x - 1)^2 \ge 0 \), multiplying by 3 (a positive number) keeps the expression non-negative.
So, \( f'(x) = 3(x - 1)^2 \ge 0 \) for all \( x \in R \).

Since the first derivative \( f'(x) \) is always greater than or equal to 0 for all real numbers, the function \( f(x) \) is increasing throughout R.
(It is strictly increasing everywhere except at \( x=1 \), where its slope is momentarily zero. This still qualifies as increasing in R.)
In simple words: For \( f(x) = x^3 - 3x^2 + 3x - 100 \), we found its slope to be \( f'(x) = 3(x - 1)^2 \). Because a squared term like \( (x - 1)^2 \) is always positive or zero, and we multiply it by a positive 3, the slope is always positive or zero. This means the function is always going up or staying flat for an instant, so it is increasing across all real numbers.

Exam Tip: Be on the lookout for perfect square factorizations in the derivative. If \( f'(x) \) can be written as \( k(g(x))^2 \) where \( k>0 \), then \( f'(x) \ge 0 \), implying the function is increasing.

Question 19. The interval in which \( y = x^2 e^{-x} \) is increasing with respect to \( x \) is
(A) \( (-\infty, \infty) \)
(B) \( (-2, 0) \)
(C) \( (2, \infty) \)
(D) \( (0, 2) \)
Answer: (D) \( (0, 2) \)
We have the function \( f(x) = x^2 e^{-x} \).
First, find the derivative of \( f(x) \) using the product rule \( (uv)' = u'v + uv' \):
Let \( u = x^2 \implies u' = 2x \)
Let \( v = e^{-x} \implies v' = -e^{-x} \)
So, \( f'(x) = (2x)(e^{-x}) + (x^2)(-e^{-x}) \)
\( f'(x) = 2xe^{-x} - x^2e^{-x} \).
Factor out common terms, \( xe^{-x} \):
\( f'(x) = xe^{-x}(2 - x) \).

For the function to be increasing, \( f'(x) > 0 \):
\( xe^{-x}(2 - x) > 0 \).

Analyze the signs of the factors:
1. \( e^{-x} \): The exponential term \( e^{-x} \) is always positive for all real \( x \). So, it does not affect the sign of \( f'(x) \).
2. \( x \): This term is positive for \( x > 0 \) and negative for \( x < 0 \).
3. \( (2 - x) \): This term is positive for \( 2 - x > 0 \implies x < 2 \), and negative for \( 2 - x < 0 \implies x > 2 \).

The critical points where \( f'(x) = 0 \) are \( x = 0 \) and \( x = 2 \). These divide the real line into three intervals:
Interval 1: \( (-\infty, 0) \)
Pick \( x = -1 \). \( f'(-1) = (-1)e^{-(-1)}(2 - (-1)) = (-1)e^1(3) = -3e \). (Negative)
So, \( f(x) \) is decreasing in \( (-\infty, 0) \).

Interval 2: \( (0, 2) \)
Pick \( x = 1 \). \( f'(1) = (1)e^{-1}(2 - 1) = (1)e^{-1}(1) = e^{-1} = \frac{1}{e} \). (Positive)
So, \( f(x) \) is increasing in \( (0, 2) \).

Interval 3: \( (2, \infty) \)
Pick \( x = 3 \). \( f'(3) = (3)e^{-3}(2 - 3) = (3)e^{-3}(-1) = -3e^{-3} \). (Negative)
So, \( f(x) \) is decreasing in \( (2, \infty) \).

Therefore, the function \( y = x^2 e^{-x} \) is increasing in the interval \( (0, 2) \).
The correct option is (D).
In simple words: We want to know where \( y = x^2 e^{-x} \) goes up. We found its slope to be \( xe^{-x}(2 - x) \). The \( e^{-x} \) part is always positive, so we only need to look at \( x(2 - x) \). This expression is positive when \( x \) is between 0 and 2. So, the function increases in the interval \( (0, 2) \).

Exam Tip: When the derivative contains an exponential term like \( e^x \) or \( e^{-x} \), remember that these terms are always positive and won't affect the overall sign of the derivative. Focus on the polynomial factors to determine the intervals of increase or decrease.