Samacheer Kalvi Class 12 Maths Solutions Chapter 7 Applications of Differential Calculus Exercise 7.6

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.6

Question 1. Find the absolute extrema of the following functions on the given closed interval.
(i) \( f(x) = x^2 – 12x + 10; [1, 2] \)
(ii) \( f(x) = 3x^4 – 4x^3; [-1, 2] \).
(iii) \( f(x) = 6x^{\frac{4}{3}} - 3x^{\frac{1}{3}}; [-1, 1] \)
(iv) \( f(x) = 2 \cos x + \sin 2x; [0, \frac{\pi}{2}] \)
Answer:
(i) Given function is \( f(x) = x^2 - 12x + 10 \).
First, we find the derivative of the function:
\( f'(x) = 2x - 12 \)
Next, we set the derivative to zero to find the critical points:
\( f'(x) = 0 \)
\( 2x - 12 = 0 \)
\( 2x = 12 \)
\( x = 6 \)
This critical point \( x = 6 \) is not within the given closed interval \( [1, 2] \). Therefore, we only need to check the function's value at the endpoints of the interval.
Evaluate \( f(x) \) at \( x = 1 \) and \( x = 2 \):
At \( x = 1 \): \( f(1) = (1)^2 - 12(1) + 10 = 1 - 12 + 10 = -1 \)
At \( x = 2 \): \( f(2) = (2)^2 - 12(2) + 10 = 4 - 24 + 10 = -10 \)
Comparing these values, the largest value is the absolute maximum, and the smallest is the absolute minimum.
Absolute maximum: \( f(1) = -1 \)
Absolute minimum: \( f(2) = -10 \)
In simple words: First, we find where the slope of the function is zero. This point is outside our chosen range. So, we check the function's value only at the start and end of the range. The biggest value found is the highest point, and the smallest value is the lowest point.

🎯 Exam Tip: Always check if critical points fall within the given closed interval. If they do, they must be included in the evaluation, along with the endpoints.

Answer:
(ii) Given function is \( f(x) = 3x^4 - 4x^3 \).
First, we find the derivative of the function:
\( f'(x) = 12x^3 - 12x^2 \)
Next, we set the derivative to zero to find the critical points:
\( f'(x) = 0 \)
\( 12x^3 - 12x^2 = 0 \)
\( 12x^2(x - 1) = 0 \)
This gives us two critical points: \( x = 0 \) or \( x = 1 \).
Both of these critical points, \( x = 0 \) and \( x = 1 \), are within the given closed interval \( [-1, 2] \).
Now, we evaluate the function \( f(x) \) at the critical points and the endpoints of the interval:
At \( x = -1 \) (endpoint): \( f(-1) = 3(-1)^4 - 4(-1)^3 = 3(1) - 4(-1) = 3 + 4 = 7 \)
At \( x = 0 \) (critical point): \( f(0) = 3(0)^4 - 4(0)^3 = 0 - 0 = 0 \)
At \( x = 1 \) (critical point): \( f(1) = 3(1)^4 - 4(1)^3 = 3 - 4 = -1 \)
At \( x = 2 \) (endpoint): \( f(2) = 3(2)^4 - 4(2)^3 = 3(16) - 4(8) = 48 - 32 = 16 \)
Comparing these values (\( 7, 0, -1, 16 \)), we find the largest and smallest values.
Absolute maximum: \( f(2) = 16 \)
Absolute minimum: \( f(1) = -1 \)
In simple words: We first find the derivative and set it to zero to get special points. These points are inside our chosen range. Then we check the function's value at these special points and at the range's start and end. The biggest value is the absolute highest, and the smallest is the absolute lowest.

🎯 Exam Tip: Ensure all critical points and interval endpoints are evaluated, as absolute extrema can occur at any of these locations.

Answer:
(iii) Given function is \( f(x) = 6x^{\frac{4}{3}} - 3x^{\frac{1}{3}} \).
First, we find the derivative of the function:
\( f'(x) = 6 \cdot \frac{4}{3} x^{\frac{4}{3} - 1} - 3 \cdot \frac{1}{3} x^{\frac{1}{3} - 1} \)
\( f'(x) = 8x^{\frac{1}{3}} - x^{-\frac{2}{3}} \)
Next, we set the derivative to zero to find the critical points:
\( f'(x) = 0 \)
\( 8x^{\frac{1}{3}} - x^{-\frac{2}{3}} = 0 \)
We can factor out \( x^{-\frac{2}{3}} \):
\( x^{-\frac{2}{3}}(8x - 1) = 0 \)
This gives us \( x^{-\frac{2}{3}} = 0 \) (which has no solution) or \( 8x - 1 = 0 \).
\( 8x = 1 \)
\( x = \frac{1}{8} \)
The critical point \( x = \frac{1}{8} \) is within the given closed interval \( [-1, 1] \).
Now, we evaluate the function \( f(x) \) at the critical point and the endpoints of the interval:
At \( x = -1 \) (endpoint):
\( f(-1) = 6(-1)^{\frac{4}{3}} - 3(-1)^{\frac{1}{3}} \)
\( f(-1) = 6(1) - 3(-1) = 6 + 3 = 9 \)
At \( x = \frac{1}{8} \) (critical point):
\( f(\frac{1}{8}) = 6(\frac{1}{8})^{\frac{4}{3}} - 3(\frac{1}{8})^{\frac{1}{3}} \)
\( f(\frac{1}{8}) = 6((\frac{1}{2})^3)^{\frac{4}{3}} - 3((\frac{1}{2})^3)^{\frac{1}{3}} \)
\( f(\frac{1}{8}) = 6(\frac{1}{2})^4 - 3(\frac{1}{2})^1 \)
\( f(\frac{1}{8}) = 6(\frac{1}{16}) - 3(\frac{1}{2}) \)
\( f(\frac{1}{8}) = \frac{6}{16} - \frac{3}{2} = \frac{3}{8} - \frac{12}{8} = -\frac{9}{8} \)
At \( x = 1 \) (endpoint):
\( f(1) = 6(1)^{\frac{4}{3}} - 3(1)^{\frac{1}{3}} = 6(1) - 3(1) = 6 - 3 = 3 \)
Comparing these values (\( 9, -\frac{9}{8}, 3 \)), we find the largest and smallest values.
Absolute maximum: \( f(-1) = 9 \)
Absolute minimum: \( f(\frac{1}{8}) = -\frac{9}{8} \)
In simple words: We calculate the derivative of the function and find where it equals zero. This gives us a special point within our given range. We then check the function's value at this special point and at the two ends of the range. The largest of these values is the absolute maximum, and the smallest is the absolute minimum.

🎯 Exam Tip: Remember that terms with negative fractional exponents (like \( x^{-\frac{2}{3}} \)) can be written as fractions (e.g., \( \frac{1}{x^{\frac{2}{3}}} \)) to simplify finding common denominators and solving for \( x \).

Answer:
(iv) Given function is \( f(x) = 2 \cos x + \sin 2x \), on the interval \( [0, \frac{\pi}{2}] \).
First, we find the derivative of the function:
\( f'(x) = -2 \sin x + 2 \cos 2x \)
Next, we set the derivative to zero to find the critical points:
\( f'(x) = 0 \)
\( -2 \sin x + 2 \cos 2x = 0 \)
We use the double angle identity \( \cos 2x = 1 - 2 \sin^2 x \):
\( -2 \sin x + 2 (1 - 2 \sin^2 x) = 0 \)
\( -2 \sin x + 2 - 4 \sin^2 x = 0 \)
Rearrange the terms into a quadratic equation in terms of \( \sin x \):
\( 4 \sin^2 x + 2 \sin x - 2 = 0 \)
Divide by 2:
\( 2 \sin^2 x + \sin x - 1 = 0 \)
Factor the quadratic:
\( (2 \sin x - 1)(\sin x + 1) = 0 \)
This gives two possibilities:
1) \( 2 \sin x - 1 = 0 \implies \sin x = \frac{1}{2} \)
In the interval \( [0, \frac{\pi}{2}] \), the solution is \( x = \frac{\pi}{6} \).
2) \( \sin x + 1 = 0 \implies \sin x = -1 \)
In the interval \( [0, 2\pi] \), \( x = \frac{3\pi}{2} \). This value is not in our given interval \( [0, \frac{\pi}{2}] \).
So, the only critical point in the interval is \( x = \frac{\pi}{6} \).
Now, we evaluate the function \( f(x) \) at this critical point and the endpoints of the interval \( [0, \frac{\pi}{2}] \):
At \( x = 0 \) (endpoint):
\( f(0) = 2 \cos 0 + \sin (2 \cdot 0) = 2(1) + \sin(0) = 2 + 0 = 2 \)
At \( x = \frac{\pi}{6} \) (critical point):
\( f(\frac{\pi}{6}) = 2 \cos (\frac{\pi}{6}) + \sin (2 \cdot \frac{\pi}{6}) \)
\( f(\frac{\pi}{6}) = 2(\frac{\sqrt{3}}{2}) + \sin (\frac{\pi}{3}) \)
\( f(\frac{\pi}{6}) = \sqrt{3} + \frac{\sqrt{3}}{2} = \frac{2\sqrt{3} + \sqrt{3}}{2} = \frac{3\sqrt{3}}{2} \)
(Approximately \( \frac{3 \times 1.732}{2} \approx 2.598 \))
At \( x = \frac{\pi}{2} \) (endpoint):
\( f(\frac{\pi}{2}) = 2 \cos (\frac{\pi}{2}) + \sin (2 \cdot \frac{\pi}{2}) \)
\( f(\frac{\pi}{2}) = 2(0) + \sin(\pi) = 0 + 0 = 0 \)
Comparing these values (\( 2, \frac{3\sqrt{3}}{2}, 0 \)), we find the largest and smallest values.
Absolute maximum: \( f(\frac{\pi}{6}) = \frac{3\sqrt{3}}{2} \)
Absolute minimum: \( f(\frac{\pi}{2}) = 0 \)
In simple words: We find the function's derivative and set it to zero to find special points. For this, we use a trigonometric identity. We only keep the special points that fall within our given range. Then, we check the function's value at these special points and at the ends of the range. The biggest value is the absolute highest point, and the smallest is the absolute lowest.

🎯 Exam Tip: When working with trigonometric functions, remember to simplify expressions using identities like \( \cos 2x = 1 - 2 \sin^2 x \) to make solving easier, and always check that critical points are in the specified interval.

Question 2. Find the intervals of monotonicities and hence find the local extremum for the following functions:
(i) \( f(x) = 2x^3 + 3x^2 – 12x \)
(ii) \( f(x) = \frac{x}{x-5} \)
(iii) \( f(x) = \frac{e^x}{1-e^x} \)
(iv) \( f(x) = \frac{x^3}{3} – \log x \)
(v) \( f(x) = \sin x \cos x + 5, x \in (0, 2\pi) \)
Answer:
(i) Given function: \( f(x) = 2x^3 + 3x^2 - 12x \).
First, we find the derivative of the function:
\( f'(x) = 6x^2 + 6x - 12 \)
Next, we set the derivative to zero to find the stationary points:
\( f'(x) = 0 \)
\( 6x^2 + 6x - 12 = 0 \)
Divide by 6:
\( x^2 + x - 2 = 0 \)
Factor the quadratic equation:
\( (x + 2)(x - 1) = 0 \)
This gives us two stationary points: \( x = -2 \) and \( x = 1 \).
These points divide the number line into three intervals: \( (-\infty, -2) \), \( (-2, 1) \), and \( (1, \infty) \).
Now, we check the sign of \( f'(x) \) in each interval:
1) In \( (-\infty, -2) \): Choose a test value, e.g., \( x = -3 \).
\( f'(-3) = 6(-3)^2 + 6(-3) - 12 = 6(9) - 18 - 12 = 54 - 30 = 24 > 0 \).
So, \( f(x) \) is strictly increasing in \( (-\infty, -2) \).
2) In \( (-2, 1) \): Choose a test value, e.g., \( x = 0 \).
\( f'(0) = 6(0)^2 + 6(0) - 12 = -12 < 0 \).
So, \( f(x) \) is strictly decreasing in \( (-2, 1) \).
3) In \( (1, \infty) \): Choose a test value, e.g., \( x = 2 \).
\( f'(2) = 6(2)^2 + 6(2) - 12 = 6(4) + 12 - 12 = 24 > 0 \).
So, \( f(x) \) is strictly increasing in \( (1, \infty) \).

Local extrema:
At \( x = -2 \), \( f'(x) \) changes from positive to negative. This means there is a local maximum.
Local maximum value: \( f(-2) = 2(-2)^3 + 3(-2)^2 - 12(-2) = 2(-8) + 3(4) + 24 = -16 + 12 + 24 = 20 \).
At \( x = 1 \), \( f'(x) \) changes from negative to positive. This means there is a local minimum.
Local minimum value: \( f(1) = 2(1)^3 + 3(1)^2 - 12(1) = 2 + 3 - 12 = -7 \).
In simple words: We calculate the derivative and find points where it is zero. These points divide the number line into sections. We then check the sign of the derivative in each section to see if the function is going up or down. Where the function changes from up to down, it's a local peak (maximum). Where it changes from down to up, it's a local valley (minimum).

🎯 Exam Tip: Remember to use test points within each interval to determine the sign of the first derivative and confirm whether the function is increasing or decreasing there.

Answer:
(ii) Given function: \( f(x) = \frac{x}{x-5} \).
First, we find the derivative of the function using the quotient rule \( (\frac{u}{v})' = \frac{u'v - uv'}{v^2} \):
\( f'(x) = \frac{(1)(x-5) - (x)(1)}{(x-5)^2} \)
\( f'(x) = \frac{x-5-x}{(x-5)^2} \)
\( f'(x) = \frac{-5}{(x-5)^2} \)
Next, we try to find stationary points by setting the derivative to zero:
\( \frac{-5}{(x-5)^2} = 0 \)
This equation simplifies to \( -5 = 0 \), which is impossible. So, there are no critical points where \( f'(x) = 0 \).
However, the function \( f(x) \) is undefined when the denominator is zero, i.e., \( x - 5 = 0 \implies x = 5 \). This means there is a discontinuity at \( x=5 \).
The domain of the function is all real numbers except \( x=5 \). So, the intervals of monotonicity are \( (-\infty, 5) \) and \( (5, \infty) \).
Now, we check the sign of \( f'(x) \) in each interval:
For any \( x \neq 5 \), the term \( (x-5)^2 \) is always positive. Since the numerator is \( -5 \), which is negative, the entire derivative \( f'(x) \) will always be negative.
1) In \( (-\infty, 5) \): \( f'(x) = \frac{-5}{(x-5)^2} < 0 \). So, \( f(x) \) is strictly decreasing.
2) In \( (5, \infty) \): \( f'(x) = \frac{-5}{(x-5)^2} < 0 \). So, \( f(x) \) is strictly decreasing.

Local extrema:
Since the function is strictly decreasing in both intervals and there is a discontinuity at \( x=5 \), there are no local maximum or minimum points.
In simple words: We find the derivative and see if it can be zero. Here, it never is. But the function itself cannot exist at one point. This point splits the number line into two parts. In both parts, the derivative is negative, meaning the function is always going down. Because it always goes down and has a break, there are no local high or low points.

🎯 Exam Tip: Always identify points where the function itself or its derivative is undefined, as these points define the boundaries of the intervals of monotonicity.

Answer:
(iii) Given function: \( f(x) = \frac{e^x}{1-e^x} \).
First, we find the derivative of the function using the quotient rule \( (\frac{u}{v})' = \frac{u'v - uv'}{v^2} \):
\( f'(x) = \frac{(e^x)(1-e^x) - (e^x)(-e^x)}{(1-e^x)^2} \)
\( f'(x) = \frac{e^x - e^{2x} + e^{2x}}{(1-e^x)^2} \)
\( f'(x) = \frac{e^x}{(1-e^x)^2} \)
Next, we determine where the function \( f(x) \) is defined. The denominator cannot be zero:
\( 1 - e^x = 0 \implies e^x = 1 \implies x = 0 \)
So, \( x=0 \) is an excluded value from the domain, creating a discontinuity. The domain is \( (-\infty, 0) \cup (0, \infty) \).
Now, we look for stationary points by setting the derivative to zero:
\( \frac{e^x}{(1-e^x)^2} = 0 \)
This would imply \( e^x = 0 \), which has no solution (as \( e^x \) is always positive). Thus, there are no critical points where \( f'(x) = 0 \).
We check the sign of \( f'(x) \) in the intervals of its domain:
For any \( x \neq 0 \):
The numerator \( e^x \) is always positive.
The denominator \( (1-e^x)^2 \) is always positive (since it's a square of a non-zero real number).
Therefore, \( f'(x) = \frac{e^x}{(1-e^x)^2} \) is always positive for all \( x \) in its domain.
1) In \( (-\infty, 0) \): \( f'(x) > 0 \). So, \( f(x) \) is strictly increasing.
2) In \( (0, \infty) \): \( f'(x) > 0 \). So, \( f(x) \) is strictly increasing.

Local extrema:
Since \( f(x) \) is strictly increasing throughout its domain and has a discontinuity at \( x=0 \), there are no local maximum or minimum points.
In simple words: We find the derivative. We also find that the function cannot exist at a certain point. The derivative is always positive, which means the function is always going up. Because it always increases and has a break, it doesn't have any local high or low points.

🎯 Exam Tip: Exponential functions like \( e^x \) are always positive, which simplifies determining the sign of derivatives involving them.

Answer:
(iv) Given function: \( f(x) = \frac{x^3}{3} - \log x \).
First, determine the domain of the function. Since \( \log x \) is only defined for positive values, the domain is \( x > 0 \), or \( (0, \infty) \).
Next, we find the derivative of the function:
\( f'(x) = \frac{d}{dx}(\frac{x^3}{3}) - \frac{d}{dx}(\log x) \)
\( f'(x) = \frac{3x^2}{3} - \frac{1}{x} \)
\( f'(x) = x^2 - \frac{1}{x} \)
Now, we set the derivative to zero to find the critical points:
\( f'(x) = 0 \)
\( x^2 - \frac{1}{x} = 0 \)
Multiply by \( x \) (since \( x > 0 \)):
\( x^3 - 1 = 0 \)
\( x^3 = 1 \)
\( x = 1 \)
This critical point \( x = 1 \) is within the domain \( (0, \infty) \). It divides the domain into two intervals: \( (0, 1) \) and \( (1, \infty) \).
Now, we check the sign of \( f'(x) \) in each interval:
1) In \( (0, 1) \): Choose a test value, e.g., \( x = 0.5 \).
\( f'(0.5) = (0.5)^2 - \frac{1}{0.5} = 0.25 - 2 = -1.75 < 0 \).
So, \( f(x) \) is strictly decreasing in \( (0, 1) \).
2) In \( (1, \infty) \): Choose a test value, e.g., \( x = 2 \).
\( f'(2) = (2)^2 - \frac{1}{2} = 4 - 0.5 = 3.5 > 0 \).
So, \( f(x) \) is strictly increasing in \( (1, \infty) \).

Local extrema:
At \( x = 1 \), \( f'(x) \) changes from negative to positive. This means there is a local minimum.
Local minimum value: \( f(1) = \frac{1^3}{3} - \log 1 = \frac{1}{3} - 0 = \frac{1}{3} \).
In simple words: First, we find the derivative and set it to zero to find a special point. This point splits the function's allowed values into two sections. In the first section, the function is going down. In the second section, it is going up. Where it changes from going down to going up, there is a local lowest point.

🎯 Exam Tip: Always remember to consider the domain of functions involving logarithms or square roots before finding derivatives or critical points.

Answer:
(v) Given function: \( f(x) = \sin x \cos x + 5 \), on the interval \( x \in (0, 2\pi) \).
First, we can simplify the function using the double angle identity \( \sin 2x = 2 \sin x \cos x \), so \( \sin x \cos x = \frac{1}{2} \sin 2x \).
Thus, \( f(x) = \frac{1}{2} \sin 2x + 5 \).
Next, we find the derivative of the function:
\( f'(x) = \frac{d}{dx}(\frac{1}{2} \sin 2x + 5) \)
\( f'(x) = \frac{1}{2} (2 \cos 2x) + 0 \)
\( f'(x) = \cos 2x \)
Now, we set the derivative to zero to find the stationary points:
\( f'(x) = 0 \)
\( \cos 2x = 0 \)
Since \( x \in (0, 2\pi) \), the argument \( 2x \) will be in the interval \( (0, 4\pi) \).
In this interval, \( \cos (2x) = 0 \) when \( 2x \) takes values:
\( 2x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2} \)
Dividing by 2, we get the stationary points for \( x \):
\( x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \)
These stationary points divide the interval \( (0, 2\pi) \) into five sub-intervals:
\( (0, \frac{\pi}{4}), (\frac{\pi}{4}, \frac{3\pi}{4}), (\frac{3\pi}{4}, \frac{5\pi}{4}), (\frac{5\pi}{4}, \frac{7\pi}{4}), (\frac{7\pi}{4}, 2\pi) \).
Now, we check the sign of \( f'(x) = \cos 2x \) in each interval:
1) In \( (0, \frac{\pi}{4}) \): \( 2x \in (0, \frac{\pi}{2}) \). \( \cos 2x > 0 \). So, \( f(x) \) is strictly increasing.
2) In \( (\frac{\pi}{4}, \frac{3\pi}{4}) \): \( 2x \in (\frac{\pi}{2}, \frac{3\pi}{2}) \). \( \cos 2x < 0 \). So, \( f(x) \) is strictly decreasing.
3) In \( (\frac{3\pi}{4}, \frac{5\pi}{4}) \): \( 2x \in (\frac{3\pi}{2}, \frac{5\pi}{2}) \). \( \cos 2x > 0 \). So, \( f(x) \) is strictly increasing.
4) In \( (\frac{5\pi}{4}, \frac{7\pi}{4}) \): \( 2x \in (\frac{5\pi}{2}, \frac{7\pi}{2}) \). \( \cos 2x < 0 \). So, \( f(x) \) is strictly decreasing.
5) In \( (\frac{7\pi}{4}, 2\pi) \): \( 2x \in (\frac{7\pi}{2}, 4\pi) \). \( \cos 2x > 0 \). So, \( f(x) \) is strictly increasing.

Local extrema:
Local maximums occur where \( f'(x) \) changes from positive to negative. This happens at \( x = \frac{\pi}{4} \) and \( x = \frac{5\pi}{4} \).
At \( x = \frac{\pi}{4} \):
\( f(\frac{\pi}{4}) = \sin(\frac{\pi}{4}) \cos(\frac{\pi}{4}) + 5 = (\frac{\sqrt{2}}{2})(\frac{\sqrt{2}}{2}) + 5 = \frac{2}{4} + 5 = \frac{1}{2} + 5 = \frac{11}{2} \).
At \( x = \frac{5\pi}{4} \):
\( f(\frac{5\pi}{4}) = \sin(\frac{5\pi}{4}) \cos(\frac{5\pi}{4}) + 5 = (-\frac{\sqrt{2}}{2})(-\frac{\sqrt{2}}{2}) + 5 = \frac{2}{4} + 5 = \frac{1}{2} + 5 = \frac{11}{2} \).
Local minimums occur where \( f'(x) \) changes from negative to positive. This happens at \( x = \frac{3\pi}{4} \) and \( x = \frac{7\pi}{4} \).
At \( x = \frac{3\pi}{4} \):
\( f(\frac{3\pi}{4}) = \sin(\frac{3\pi}{4}) \cos(\frac{3\pi}{4}) + 5 = (\frac{\sqrt{2}}{2})(-\frac{\sqrt{2}}{2}) + 5 = -\frac{2}{4} + 5 = -\frac{1}{2} + 5 = \frac{9}{2} \).
At \( x = \frac{7\pi}{4} \):
\( f(\frac{7\pi}{4}) = \sin(\frac{7\pi}{4}) \cos(\frac{7\pi}{4}) + 5 = (-\frac{\sqrt{2}}{2})(\frac{\sqrt{2}}{2}) + 5 = -\frac{2}{4} + 5 = -\frac{1}{2} + 5 = \frac{9}{2} \).
In simple words: First, we change the function into a simpler form using a special math rule. Then, we find its derivative and find all the points where it is zero within the given range. These points break the range into many smaller parts. We check if the function is going up or down in each part. Where it changes from going up to down, it's a local peak. Where it changes from going down to up, it's a local valley.

🎯 Exam Tip: Simplifying trigonometric functions using identities like \( \sin x \cos x = \frac{1}{2} \sin 2x \) before differentiation can significantly ease calculations.