RBSE Solutions Class 12 Maths Chapter 8 Application of Derivatives Exercise 8.2

Question 1. Prove that \( f(x) = x^2 \) is increasing in interval \( (0, \infty) \) and decreasing in interval \( (-\infty, 0) \).
Answer: Let's prove the function \( f(x) = x^2 \) is increasing in \( (0, \infty) \) and decreasing in \( (-\infty, 0) \).

For increasing function in \( [0, \infty) \):
Let \( X_1, X_2 \in [0, \infty) \) such that \( X_1 < X_2 \).
Since \( X_1 < X_2 \) and both are positive or zero, squaring them preserves the inequality:
\( X_1 < X_2 \implies X_1^2 < X_1X_2 \) .....(i)
Also, \( X_1 < X_2 \implies X_1X_2 < X_2^2 \) .....(ii)
From (i) and (ii), we get:
\( X_1 < X_2 \implies X_1^2 < X_2^2 \)
\( \implies f(X_1) < f(X_2) \)
This shows that \( f(x) = x^2 \) is an increasing function in the interval \( (0, \infty) \). For example, if we take \( X_1 = 2 \) and \( X_2 = 3 \), then \( f(2) = 4 \) and \( f(3) = 9 \), so \( 4 < 9 \).

For decreasing function in \( (-\infty, 0) \):
Let \( X_1, X_2 \in (-\infty, 0) \) such that \( X_1 < X_2 \).
This means both \( X_1 \) and \( X_2 \) are negative numbers.
Consider an example: let \( X_1 = -3 \) and \( X_2 = -2 \). Here \( X_1 < X_2 \).
\( f(X_1) = f(-3) = (-3)^2 = 9 \)
\( f(X_2) = f(-2) = (-2)^2 = 4 \)
We see that \( f(X_1) = 9 \) is greater than \( f(X_2) = 4 \), so \( f(X_1) > f(X_2) \).
Generally, if \( X_1 < X_2 \) and both are negative:
Multiplying by a negative number reverses the inequality:
\( X_1 < X_2 \implies -X_1 > -X_2 \)
Since both sides are now positive, squaring them preserves the inequality (or we can think of it as multiplying by positive values):
\( (-X_1)^2 > (-X_2)^2 \)
\( X_1^2 > X_2^2 \)
\( \implies f(X_1) > f(X_2) \)
Thus, \( f(x) = x^2 \) is a decreasing function in the interval \( (-\infty, 0) \).
In simple words: When numbers are positive, squaring them makes bigger numbers stay bigger. When numbers are negative, squaring them makes them positive, and a smaller negative number becomes a smaller positive number, showing a decreasing pattern.

🎯 Exam Tip: To prove increasing or decreasing nature without calculus, always pick two arbitrary points \( X_1 \) and \( X_2 \) in the given interval such that \( X_1 < X_2 \) and then compare \( f(X_1) \) and \( f(X_2) \).

Question 2. Prove that \( f(x) = a^x \), where \( 0 < a < 1 \), is decreasing in \( R \).
Answer: We need to prove that for \( f(x) = a^x \), where \( 0 < a < 1 \), the function is decreasing for all real numbers \( x \).
Let \( X_1, X_2 \in R \) (all real numbers) such that \( X_1 < X_2 \).
Since the base \( a \) is between 0 and 1 (e.g., \( a = 0.5 \)), raising it to a larger power will result in a smaller value.
For example, if \( a = 0.5 \), then \( (0.5)^1 = 0.5 \) and \( (0.5)^2 = 0.25 \). Here \( 1 < 2 \) but \( (0.5)^1 > (0.5)^2 \).
So, if \( X_1 < X_2 \) and \( 0 < a < 1 \), then \( a^{X_1} > a^{X_2} \).
\( \implies f(X_1) > f(X_2) \)
This means that if we pick two numbers \( X_1 \) and \( X_2 \) where \( X_1 \) is smaller than \( X_2 \), the value of the function at \( X_1 \) will be greater than the value of the function at \( X_2 \). This is the definition of a decreasing function.
In simple words: When the base number is between 0 and 1, like 0.5, raising it to a higher power actually makes the result smaller. This means the function is always going down as x gets bigger.

🎯 Exam Tip: Remember the behavior of exponential functions: if the base is greater than 1, it's increasing; if the base is between 0 and 1, it's decreasing.

Question 3. \( f(x) = \log \sin x, x \in (0, \frac{\pi}{2}) \)
Answer: We have the function \( f(x) = \log \sin x \). We need to determine its behavior in the interval \( (0, \frac{\pi}{2}) \).
First, find the derivative of \( f(x) \) with respect to \( x \):
\( f'(x) = \frac{d}{dx} (\log \sin x) \)
\( f'(x) = \frac{1}{\sin x} \cdot \cos x \)
\( f'(x) = \cot x \)
Now, we check the sign of \( f'(x) \) in the given interval \( (0, \frac{\pi}{2}) \).
In the first quadrant, i.e., for \( x \in (0, \frac{\pi}{2}) \), the value of \( \cot x \) is always positive.
\( \implies \cot x > 0 \)
\( \implies f'(x) > 0 \)
Since the first derivative \( f'(x) \) is greater than 0 in the interval \( (0, \frac{\pi}{2}) \), the function \( f(x) \) is continuously increasing in this interval. This means as \( x \) increases from 0 to \( \frac{\pi}{2} \), the value of \( \sin x \) increases, and so does its logarithm.
In simple words: We find the slope of the function using derivatives. In the first part of the angle circle (0 to 90 degrees), the slope is positive. This means the function is always going up.

🎯 Exam Tip: For logarithmic functions, remember the chain rule for differentiation. Also, be familiar with the signs of trigonometric functions in different quadrants.

Question 4. \( f(x) = x^{100} + \sin x + 1, x \in (0, \frac{\pi}{2}) \)
Answer: We are given the function \( f(x) = x^{100} + \sin x + 1 \). We need to find if it is increasing or decreasing in the interval \( (0, \frac{\pi}{2}) \).
First, calculate the derivative of the function with respect to \( x \):
\( f'(x) = \frac{d}{dx} (x^{100} + \sin x + 1) \)
\( f'(x) = 100x^{99} + \cos x + 0 \)
\( f'(x) = 100x^{99} + \cos x \)
Now, let's examine the sign of \( f'(x) \) in the interval \( (0, \frac{\pi}{2}) \).
For any \( x \) in the interval \( (0, \frac{\pi}{2}) \):
\( x^{99} \) will always be positive because \( x \) is positive.
\( 100x^{99} \) will therefore also be positive.
\( \cos x \) will also be positive in the first quadrant \( (0, \frac{\pi}{2}) \).
Since both \( 100x^{99} \) and \( \cos x \) are positive values in this interval, their sum will definitely be positive.
\( \implies 100x^{99} + \cos x > 0 \)
\( \implies f'(x) > 0 \)
Because the derivative \( f'(x) \) is greater than 0 for all \( x \) in the given interval, the function \( f(x) \) is continuously increasing in \( (0, \frac{\pi}{2}) \). This makes sense as both \( x^{100} \) and \( \sin x \) are increasing in this interval.
In simple words: We find the slope of the function. For angles between 0 and 90 degrees, both parts of the slope calculation are positive. Adding two positive numbers always gives a positive result. So the slope is always positive, meaning the function is always going up.

🎯 Exam Tip: When dealing with sums of functions, check the sign of each term in the derivative for the given interval. If all terms have the same sign (or sum to a definite sign), the function's behavior is clear.

Question 5. \( f(x) = (x - 1)e^x + 1, x > 0. \)
Answer: We have the function \( f(x) = (x - 1)e^x + 1 \). We need to determine if it's increasing or decreasing for \( x > 0 \).
First, find the derivative of \( f(x) \) using the product rule for differentiation:
\( f'(x) = \frac{d}{dx} [(x - 1)e^x + 1] \)
Applying the product rule \( (uv)' = u'v + uv' \) to \( (x-1)e^x \):
Here, \( u = (x - 1) \) and \( v = e^x \). So \( u' = 1 \) and \( v' = e^x \).
\( f'(x) = (1)e^x + (x - 1)e^x + 0 \)
\( f'(x) = e^x + xe^x - e^x \)
\( f'(x) = xe^x \)
Now, let's check the sign of \( f'(x) \) for \( x > 0 \).
For \( x > 0 \), we know that \( x \) is positive.
Also, the exponential function \( e^x \) is always positive for any real \( x \), so \( e^x > 0 \).
Since \( x > 0 \) and \( e^x > 0 \), their product \( xe^x \) will also be positive.
\( \implies xe^x > 0 \)
\( \implies f'(x) > 0 \)
Since the derivative \( f'(x) \) is greater than 0 for all \( x > 0 \), the function \( f(x) \) is continuously increasing in the interval \( (0, \infty) \). This result is logical because both \( x \) and \( e^x \) contribute to growth when \( x \) is positive.
In simple words: After finding the slope of the function, we see it depends on \( x \) multiplied by \( e^x \). For any positive \( x \), both \( x \) and \( e^x \) are positive. A positive number times another positive number is always positive, so the slope is always positive. This means the function is always going up.

🎯 Exam Tip: Remember to use the product rule carefully when differentiating terms like \( (x-1)e^x \). The properties of \( e^x \) (always positive) and the given interval for \( x \) are key to determining the sign of the derivative.

Question 7. \( f(x) = \tan^{-1} x - x, x \in R. \)
Answer: We have the function \( f(x) = \tan^{-1} x - x \). We need to determine its behavior for all real numbers \( x \).
First, find the derivative of \( f(x) \) with respect to \( x \):
\( f'(x) = \frac{d}{dx} (\tan^{-1} x - x) \)
\( f'(x) = \frac{1}{1+x^2} - 1 \)
To find the sign of \( f'(x) \), let's combine the terms:
\( f'(x) = \frac{1 - (1+x^2)}{1+x^2} \)
\( f'(x) = \frac{1 - 1 - x^2}{1+x^2} \)
\( f'(x) = \frac{-x^2}{1+x^2} \)
Now, let's check the sign of \( f'(x) \) for all \( x \in R \).
For any real number \( x \):
\( x^2 \) will always be greater than or equal to 0. So, \( -x^2 \) will always be less than or equal to 0.
\( 1+x^2 \) will always be greater than or equal to 1, hence always positive.
So, \( \frac{-x^2}{1+x^2} \) will always be less than or equal to 0 (it is 0 only when \( x=0 \)).
\( \implies f'(x) \le 0 \)
Since the derivative \( f'(x) \) is less than or equal to 0 for all \( x \in R \), the function \( f(x) \) is continuously decreasing for all real numbers. This means the tangent inverse function grows slower than \( x \), leading to an overall decreasing trend.
In simple words: We calculate the slope of the function. The formula for the slope turns out to be \( -x^2 \) divided by \( 1+x^2 \). Since \( x^2 \) is always positive (or zero) and \( 1+x^2 \) is always positive, the fraction \( \frac{-x^2}{1+x^2} \) is always negative or zero. A negative slope means the function is always going down.

🎯 Exam Tip: Recall the derivative of \( \tan^{-1} x \). When determining the sign of a rational function like \( \frac{-x^2}{1+x^2} \), analyze the signs of the numerator and denominator separately.

Question 8. \( f(x) = \sin^4 x + \cos^4 x, x \in (0, \frac{\pi}{4}) \)
Answer: We are given the function \( f(x) = \sin^4 x + \cos^4 x \). We need to determine if it is increasing or decreasing in the interval \( (0, \frac{\pi}{4}) \).
First, find the derivative of \( f(x) \) with respect to \( x \):
\( f'(x) = \frac{d}{dx} (\sin^4 x + \cos^4 x) \)
Using the chain rule:
\( f'(x) = 4\sin^3 x (\cos x) + 4\cos^3 x (-\sin x) \)
\( f'(x) = 4\sin^3 x \cos x - 4\cos^3 x \sin x \)
Factor out common terms, \( 4\sin x \cos x \):
\( f'(x) = 4\sin x \cos x (\sin^2 x - \cos^2 x) \)
Now, use trigonometric identities: \( 2\sin x \cos x = \sin 2x \) and \( \sin^2 x - \cos^2 x = -(\cos^2 x - \sin^2 x) = -\cos 2x \).
So, \( f'(x) = (2)(2\sin x \cos x) (-\cos 2x) \)
\( f'(x) = 2\sin 2x (-\cos 2x) \)
\( f'(x) = -2\sin 2x \cos 2x \)
We can simplify this further using \( \sin 4x = 2\sin 2x \cos 2x \):
\( f'(x) = -\sin 4x \)
Now, we check the sign of \( f'(x) \) in the interval \( (0, \frac{\pi}{4}) \).
If \( x \in (0, \frac{\pi}{4}) \), then \( 2x \in (0, \frac{\pi}{2}) \) and \( 4x \in (0, \pi) \).
For \( 4x \in (0, \pi) \), the value of \( \sin 4x \) is always positive (since it's in the first or second quadrant).
Since \( \sin 4x > 0 \), then \( -\sin 4x < 0 \).
\( \implies f'(x) < 0 \)
Since the derivative \( f'(x) \) is less than 0 in the interval \( (0, \frac{\pi}{4}) \), the function \( f(x) \) is continuously decreasing in this interval. This means as \( x \) increases in this range, the function's value goes down.
In simple words: We find the slope of the function using trigonometry. It simplifies to \( -\sin 4x \). In the given range of angles, \( 4x \) is always between 0 and 180 degrees, where \( \sin \) is positive. So, \( -\sin 4x \) is always negative, meaning the function is always going down.

🎯 Exam Tip: Simplify trigonometric derivatives as much as possible using identities like \( \sin 2A = 2\sin A \cos A \) and \( \cos 2A = \cos^2 A - \sin^2 A \). This often makes it easier to determine the sign of the derivative.

Question 9. \( f(x) = \frac{3}{x} + 5, x \in R, x \ne 0. \)
Answer: We have the function \( f(x) = \frac{3}{x} + 5 \). We need to determine if it's increasing or decreasing for \( x \in R, x \ne 0 \).
First, find the derivative of \( f(x) \) with respect to \( x \):
\( f'(x) = \frac{d}{dx} (3x^{-1} + 5) \)
\( f'(x) = 3(-1)x^{-2} + 0 \)
\( f'(x) = -3x^{-2} \)
\( f'(x) = -\frac{3}{x^2} \)
Now, let's check the sign of \( f'(x) \) for all \( x \in R, x \ne 0 \).
For any non-zero real number \( x \), \( x^2 \) will always be positive.
For example, if \( x=2, x^2=4 \). If \( x=-2, x^2=4 \).
Since \( x^2 > 0 \), the term \( \frac{3}{x^2} \) will always be positive.
Therefore, \( -\frac{3}{x^2} \) will always be negative.
\( \implies f'(x) < 0 \)
Since the derivative \( f'(x) \) is always less than 0 for all \( x \in R, x \ne 0 \), the function \( f(x) \) is continuously decreasing throughout its domain. This means that as \( x \) increases (either in positive or negative numbers), the value of \( \frac{3}{x} \) changes in a way that makes the total function value go down.
In simple words: We calculate the slope of the function. The slope is \( -\frac{3}{x^2} \). Since \( x^2 \) is always positive, the whole fraction will always be negative. A negative slope means the function is always going down, or decreasing.

🎯 Exam Tip: When \( x \) is in the denominator, remember to rewrite it as a negative power before differentiating. The sign of \( x^2 \) is always positive (for non-zero \( x \)), which is crucial for determining the sign of the derivative.

Question 10. \( f(x) = x^2 - 2x + 3, x < 1. \)
Answer: We are given the function \( f(x) = x^2 - 2x + 3 \). We need to determine if it is increasing or decreasing for \( x < 1 \).
First, find the derivative of \( f(x) \) with respect to \( x \):
\( f'(x) = \frac{d}{dx} (x^2 - 2x + 3) \)
\( f'(x) = 2x - 2 + 0 \)
\( f'(x) = 2x - 2 \)
We can factor out a 2:
\( f'(x) = 2(x - 1) \)
Now, let's check the sign of \( f'(x) \) for the given interval \( x < 1 \).
If \( x < 1 \), then \( x - 1 \) will be a negative number.
For example, if \( x = 0 \), then \( x - 1 = -1 \). If \( x = 0.5 \), then \( x - 1 = -0.5 \).
Since \( x - 1 < 0 \), and we multiply it by a positive number (2), the product \( 2(x - 1) \) will also be negative.
\( \implies 2(x - 1) < 0 \)
\( \implies f'(x) < 0 \)
Since the derivative \( f'(x) \) is less than 0 for all \( x < 1 \), the function \( f(x) \) is continuously decreasing in the interval \( (-\infty, 1) \). This quadratic function has its vertex at \( x = 1 \), so it decreases before this point and increases after it.
In simple words: We find the slope of the function, which is \( 2(x-1) \). When \( x \) is smaller than 1, like 0 or -5, then \( x-1 \) will be a negative number. Multiplying a negative number by 2 keeps it negative. So the slope is always negative, meaning the function is always going down.

🎯 Exam Tip: For polynomial functions, finding the derivative and factoring it helps easily identify the critical points (where \( f'(x)=0 \)) and the intervals where \( f'(x) \) changes sign.

Question 11. \( f(x) = 2x^3 - 3x^2 - 36x + 7 \)
Answer: We have the function \( f(x) = 2x^3 - 3x^2 - 36x + 7 \). We need to find the intervals where it is increasing or decreasing.
First, find the derivative of \( f(x) \) with respect to \( x \):
\( f'(x) = \frac{d}{dx} (2x^3 - 3x^2 - 36x + 7) \)
\( f'(x) = 6x^2 - 6x - 36 \)
To find the critical points, set \( f'(x) = 0 \):
\( 6x^2 - 6x - 36 = 0 \)
Divide the entire equation by 6:
\( x^2 - x - 6 = 0 \)
Factor the quadratic equation:
\( x^2 - 3x + 2x - 6 = 0 \)
\( x(x - 3) + 2(x - 3) = 0 \)
\( (x - 3)(x + 2) = 0 \)
This gives us two critical points: \( x - 3 = 0 \implies x = 3 \) and \( x + 2 = 0 \implies x = -2 \).
These points divide the real number line into three intervals: \( (-\infty, -2) \), \( (-2, 3) \), and \( (3, \infty) \).

(a) For the interval \( (-\infty, -2) \):
Let's pick a test value, for example, \( x = -3 \).
\( f'(-3) = 6(-3)^2 - 6(-3) - 36 \)
\( f'(-3) = 6(9) + 18 - 36 \)
\( f'(-3) = 54 + 18 - 36 \)
\( f'(-3) = 72 - 36 = 36 \)
Since \( f'(-3) = 36 > 0 \), the function \( f(x) \) is increasing in the interval \( (-\infty, -2) \).

(b) For the interval \( (-2, 3) \):
Let's pick a test value, for example, \( x = 0 \).
\( f'(0) = 6(0)^2 - 6(0) - 36 \)
\( f'(0) = 0 - 0 - 36 = -36 \)
Since \( f'(0) = -36 < 0 \), the function \( f(x) \) is decreasing in the interval \( (-2, 3) \).

(c) For the interval \( (3, \infty) \):
Let's pick a test value, for example, \( x = 4 \).
\( f'(4) = 6(4)^2 - 6(4) - 36 \)
\( f'(4) = 6(16) - 24 - 36 \)
\( f'(4) = 96 - 24 - 36 \)
\( f'(4) = 72 - 36 = 36 \)
Since \( f'(4) = 36 > 0 \), the function \( f(x) \) is increasing in the interval \( (3, \infty) \).

In summary, the function \( f(x) \) is increasing in intervals \( (-\infty, -2) \cup (3, \infty) \) and decreasing in the interval \( (-2, 3) \). Finding the derivative helps us understand where the function's slope is positive (increasing) or negative (decreasing).
In simple words: We found the "turn-around" points for the function by setting its slope to zero. These points divide the number line into sections. By testing numbers in each section, we found that the function goes up in the first and last sections, and goes down in the middle section.

🎯 Exam Tip: Always factor the derivative to find the critical points. Then, use a sign chart or test values in each interval to determine where the derivative is positive (increasing) or negative (decreasing).

Question 12. \( f(x) = x^4 - 2x^2. \)
Answer: We are given the function \( f(x) = x^4 - 2x^2 \). We need to find the intervals where it is increasing or decreasing.
First, find the derivative of \( f(x) \) with respect to \( x \):
\( f'(x) = \frac{d}{dx} (x^4 - 2x^2) \)
\( f'(x) = 4x^3 - 4x \)
To find the critical points, set \( f'(x) = 0 \):
\( 4x^3 - 4x = 0 \)
Factor out \( 4x \):
\( 4x(x^2 - 1) = 0 \)
Factor the difference of squares \( (x^2 - 1) = (x - 1)(x + 1) \):
\( 4x(x - 1)(x + 1) = 0 \)
This gives us three critical points: \( x = 0 \), \( x = 1 \), and \( x = -1 \).
These points divide the real number line into four intervals: \( (-\infty, -1) \), \( (-1, 0) \), \( (0, 1) \), and \( (1, \infty) \).

(a) For the interval \( (-\infty, -1) \):
Let's pick a test value, for example, \( x = -2 \).
\( f'(-2) = 4(-2)^3 - 4(-2) \)
\( f'(-2) = 4(-8) + 8 \)
\( f'(-2) = -32 + 8 = -24 \)
Since \( f'(-2) = -24 < 0 \), the function \( f(x) \) is decreasing in the interval \( (-\infty, -1) \).

(b) For the interval \( (-1, 0) \):
Let's pick a test value, for example, \( x = -0.5 \).
\( f'(-0.5) = 4(-0.5)^3 - 4(-0.5) \)
\( f'(-0.5) = 4(-0.125) + 2 \)
\( f'(-0.5) = -0.5 + 2 = 1.5 \)
Since \( f'(-0.5) = 1.5 > 0 \), the function \( f(x) \) is increasing in the interval \( (-1, 0) \).

(c) For the interval \( (0, 1) \):
Let's pick a test value, for example, \( x = 0.5 \).
\( f'(0.5) = 4(0.5)^3 - 4(0.5) \)
\( f'(0.5) = 4(0.125) - 2 \)
\( f'(0.5) = 0.5 - 2 = -1.5 \)
Since \( f'(0.5) = -1.5 < 0 \), the function \( f(x) \) is decreasing in the interval \( (0, 1) \).

(d) For the interval \( (1, \infty) \):
Let's pick a test value, for example, \( x = 2 \).
\( f'(2) = 4(2)^3 - 4(2) \)
\( f'(2) = 4(8) - 8 \)
\( f'(2) = 32 - 8 = 24 \)
Since \( f'(2) = 24 > 0 \), the function \( f(x) \) is increasing in the interval \( (1, \infty) \).

In summary, the function \( f(x) \) is increasing in intervals \( (-1, 0) \cup (1, \infty) \) and decreasing in intervals \( (-\infty, -1) \cup (0, 1) \). This function is symmetric about the y-axis, which is often seen in even-powered polynomial functions.
In simple words: We found the points where the function's slope is zero. These points divide the number line into four parts. By checking the slope in each part, we found the function goes down in the first and third parts, and goes up in the second and fourth parts.

🎯 Exam Tip: When dealing with more than two critical points, carefully manage the intervals. A sign chart for \( f'(x) \) with the critical points clearly marked can prevent errors in determining increasing/decreasing regions.

Question 13. \( f(x) = 2x^3 - 9x^2 + 12x + 5. \)
Answer: We have the function \( f(x) = 2x^3 - 9x^2 + 12x + 5 \). We need to find the intervals where it is increasing or decreasing.
First, find the derivative of \( f(x) \) with respect to \( x \):
\( f'(x) = \frac{d}{dx} (2x^3 - 9x^2 + 12x + 5) \)
\( f'(x) = 6x^2 - 18x + 12 \)
To find the critical points, set \( f'(x) = 0 \):
\( 6x^2 - 18x + 12 = 0 \)
Divide the entire equation by 6:
\( x^2 - 3x + 2 = 0 \)
Factor the quadratic equation:
\( (x - 1)(x - 2) = 0 \)
This gives us two critical points: \( x = 1 \) and \( x = 2 \).
These points divide the real number line into three intervals: \( (-\infty, 1) \), \( (1, 2) \), and \( (2, \infty) \).

(a) For the interval \( (-\infty, 1) \):
Let's pick a test value, for example, \( x = 0 \).
\( f'(0) = 6(0)^2 - 18(0) + 12 = 12 \)
Since \( f'(0) = 12 > 0 \), the function \( f(x) \) is increasing in the interval \( (-\infty, 1) \).

(b) For the interval \( (1, 2) \):
Let's pick a test value, for example, \( x = 1.5 \).
\( f'(1.5) = 6(1.5)^2 - 18(1.5) + 12 \)
\( f'(1.5) = 6(2.25) - 27 + 12 \)
\( f'(1.5) = 13.5 - 27 + 12 = -1.5 \)
Since \( f'(1.5) = -1.5 < 0 \), the function \( f(x) \) is decreasing in the interval \( (1, 2) \).

(c) For the interval \( (2, \infty) \):
Let's pick a test value, for example, \( x = 3 \).
\( f'(3) = 6(3)^2 - 18(3) + 12 \)
\( f'(3) = 6(9) - 54 + 12 \)
\( f'(3) = 54 - 54 + 12 = 12 \)
Since \( f'(3) = 12 > 0 \), the function \( f(x) \) is increasing in the interval \( (2, \infty) \).

In summary, the function \( f(x) \) is increasing in intervals \( (-\infty, 1) \cup (2, \infty) \) and decreasing in the interval \( (1, 2) \). This method of using the first derivative helps determine the local behavior of a function.
In simple words: We found the points where the function's slope is zero, which are at \( x=1 \) and \( x=2 \). These points divide the number line into three sections. By testing numbers in each section, we found that the function goes up in the first and last sections, and goes down in the middle section.

🎯 Exam Tip: Always double-check your factorization of the quadratic derivative. Errors here will lead to incorrect critical points and thus incorrect intervals for increasing/decreasing behavior.

Question 14. \( f(x) = 2x^3 + 3x^2 + 12x + 5. \)
Answer: We are given the function \( f(x) = 2x^3 + 3x^2 + 12x + 5 \). We need to find the intervals where it is increasing or decreasing.
First, find the derivative of \( f(x) \) with respect to \( x \):
\( f'(x) = \frac{d}{dx} (2x^3 + 3x^2 + 12x + 5) \)
\( f'(x) = 6x^2 + 6x + 12 \)
To find the critical points, we can try to factor or use the discriminant. Let's check the discriminant \( D = b^2 - 4ac \) for the quadratic \( 6x^2 + 6x + 12 \):
Here, \( a = 6, b = 6, c = 12 \).
\( D = (6)^2 - 4(6)(12) \)
\( D = 36 - 288 \)
\( D = -252 \)
Since the discriminant \( D \) is negative (\( -252 < 0 \)) and the leading coefficient \( a = 6 \) is positive (\( a > 0 \)), the quadratic \( f'(x) = 6x^2 + 6x + 12 \) is always positive for all real values of \( x \).
\( \implies f'(x) > 0 \) for all \( x \in R \).
Since the derivative \( f'(x) \) is always greater than 0, the function \( f(x) \) is continuously increasing over its entire domain, \( R \). This means its slope is always upwards.
In simple words: We found the slope of the function. For this slope, we checked its special number called the discriminant. Because this number was negative and the main part of the slope was positive, it means the slope is always positive. A positive slope means the function is always going up.

🎯 Exam Tip: If a quadratic derivative \( ax^2+bx+c \) has \( a>0 \) and \( D<0 \), it's always positive. If \( a<0 \) and \( D<0 \), it's always negative. This shortcut avoids checking intervals.

Question 15. Find the minimum value of a, such that function \( f(x) = x^2 + ax + 5 \), is increasing in interval \( [1, 2] \)
Answer: We are given the function \( f(x) = x^2 + ax + 5 \), and it is stated to be increasing in the interval \( [1, 2] \).
For a function to be increasing in an interval, its derivative \( f'(x) \) must be greater than or equal to 0 in that interval.
First, find the derivative of \( f(x) \) with respect to \( x \):
\( f'(x) = \frac{d}{dx} (x^2 + ax + 5) \)
\( f'(x) = 2x + a \)
Now, we need \( f'(x) \ge 0 \) for all \( x \in [1, 2] \).
\( 2x + a \ge 0 \)
\( a \ge -2x \)
Since this condition must hold for all \( x \) in the interval \( [1, 2] \), we need \( a \) to be greater than or equal to the maximum value of \( -2x \) in that interval. The function \( -2x \) is a decreasing function.
The maximum value of \( -2x \) in the interval \( [1, 2] \) occurs at the smallest value of \( x \), which is \( x = 1 \).
When \( x = 1 \), \( -2x = -2(1) = -2 \).
So, to ensure \( a \ge -2x \) for all \( x \in [1, 2] \), \( a \) must be greater than or equal to \( -2 \).
\( \implies a \ge -2 \)
Therefore, the minimum value of \( a \) for which the function \( f(x) \) is increasing in the interval \( [1, 2] \) is \( -2 \). This value ensures that the slope is always non-negative in the given interval.
In simple words: For the function to always go up in the given range of \( x \) values, its slope must always be positive or zero. We found the formula for the slope is \( 2x+a \). To make sure \( 2x+a \) is always positive or zero when \( x \) is between 1 and 2, the smallest possible value for \( a \) is -2.

🎯 Exam Tip: For a function to be increasing, \( f'(x) \ge 0 \). When solving for a constant, consider the range of \( x \) and how it affects the inequality for the derivative. For a linear expression like \( 2x+a \), the minimum or maximum occurs at the endpoints of the interval.

Question 16. Prove that, function \( f(x) = \tan^{-1} (\sin x + \cos x) \), is increasing function in interval \( (0, \frac{\pi}{4}) \)
Answer: We are given the function \( f(x) = \tan^{-1} (\sin x + \cos x) \). We need to prove that it is increasing in the interval \( (0, \frac{\pi}{4}) \).
First, find the derivative of \( f(x) \) with respect to \( x \). Recall that \( \frac{d}{dx} (\tan^{-1} u) = \frac{1}{1+u^2} \frac{du}{dx} \).
Here, \( u = \sin x + \cos x \).
So, \( \frac{du}{dx} = \frac{d}{dx} (\sin x + \cos x) = \cos x - \sin x \).
Now substitute these into the derivative formula:
\( f'(x) = \frac{1}{1 + (\sin x + \cos x)^2} (\cos x - \sin x) \)
Let's simplify the denominator: \( (\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + 2\sin x \cos x = 1 + \sin 2x \).
So, the denominator becomes \( 1 + (1 + \sin 2x) = 2 + \sin 2x \).
Therefore, the derivative is:
\( f'(x) = \frac{\cos x - \sin x}{2 + \sin 2x} \)
Now, we need to check the sign of \( f'(x) \) in the interval \( (0, \frac{\pi}{4}) \).
For \( x \in (0, \frac{\pi}{4}) \):
1. **Numerator \( (\cos x - \sin x) \):** In the interval \( (0, \frac{\pi}{4}) \), \( \cos x \) is greater than \( \sin x \) (e.g., at \( x=0 \), \( \cos x = 1, \sin x = 0 \); at \( x=\frac{\pi}{4} \), \( \cos x = \sin x = \frac{1}{\sqrt{2}} \)). Thus, \( \cos x - \sin x > 0 \) for \( x \in (0, \frac{\pi}{4}) \).
2. **Denominator \( (2 + \sin 2x) \):** In the interval \( (0, \frac{\pi}{4}) \), \( 2x \in (0, \frac{\pi}{2}) \). In this interval, \( \sin 2x \) is positive (specifically, \( 0 < \sin 2x < 1 \)). Therefore, \( 2 + \sin 2x \) will always be positive (it will be between 2 and 3).
Since the numerator is positive and the denominator is positive, the entire fraction \( f'(x) \) will be positive.
\( \implies f'(x) > 0 \)
As \( f'(x) > 0 \) for all \( x \in (0, \frac{\pi}{4}) \), the function \( f(x) \) is continuously increasing in this interval. This implies that as the angle increases from 0 to 45 degrees, the value of the function keeps rising.
In simple words: We found the slope of the function. For angles between 0 and 45 degrees, the top part of the slope calculation is positive because cosine is larger than sine. The bottom part is also positive because sine of an angle between 0 and 90 degrees is always positive. Since positive divided by positive is positive, the slope is always positive, meaning the function is always going up.

🎯 Exam Tip: When dealing with inverse trigonometric functions, remember their differentiation rules and apply the chain rule. Simplifying the derivative using trigonometric identities often makes determining its sign much easier.