Samacheer Kalvi Class 11 Maths Solutions Chapter 2 Basic Algebra Exercise 2.8

Question 1. Find all values of x for which \( \frac{x^{3}(x-1)}{x-2} > 0 \)
Answer: The given inequality is \( f(x) = \frac{x^{3}(x-1)}{x-2} > 0 \).
To solve this, we first find the critical numbers, which are the values of x where \( f(x) = 0 \) or \( f(x) \) is undefined.
For \( f(x) = 0 \), we set the numerator to zero:
\( x^{3}(x-1) = 0 \)
\( \implies x = 0 \) or \( x = 1 \).
For \( f(x) \) to be undefined, we set the denominator to zero:
\( x-2 = 0 \)
\( \implies x = 2 \).
So, the critical numbers are 0, 1, and 2. These numbers divide the number line into four intervals: \( (-\infty, 0) \), \( (0, 1) \), \( (1, 2) \) and \( (2, \infty) \). We test a value from each interval to check the sign of \( f(x) \).

Let's examine each interval:

(1) In the interval \( (-\infty, 0) \), choose a test value, for example, \( x = -1 \).
Factor \( x^3 = (-1)^3 = -1 < 0 \)
Factor \( x-1 = -1-1 = -2 < 0 \)
Factor \( x-2 = -1-2 = -3 < 0 \)
Then, \( \frac{x^{3}(x-1)}{x-2} = \frac{(-)(-) }{(-) } = \frac{(+)}{(-)} = (-) \).
Since \( \frac{x^{3}(x-1)}{x-2} < 0 \) in this interval, it means \( \frac{x^{3}(x-1)}{x-2} > 0 \) is not true. So, there is no solution in \( (-\infty, 0) \).

(2) In the interval \( (0, 1) \), choose a test value, for example, \( x = 0.5 \).
Factor \( x^3 = (0.5)^3 > 0 \)
Factor \( x-1 = 0.5-1 = -0.5 < 0 \)
Factor \( x-2 = 0.5-2 = -1.5 < 0 \)
Then, \( \frac{x^{3}(x-1)}{x-2} = \frac{(+)(-) }{(-) } = \frac{(-)}{(-)} = (+) \).
Since \( \frac{x^{3}(x-1)}{x-2} > 0 \) in this interval, it means \( \frac{x^{3}(x-1)}{x-2} > 0 \) is true. So, this interval is a solution: \( (0, 1) \).

(3) In the interval \( (1, 2) \), choose a test value, for example, \( x = 1.5 \).
Factor \( x^3 = (1.5)^3 > 0 \)
Factor \( x-1 = 1.5-1 = 0.5 > 0 \)
Factor \( x-2 = 1.5-2 = -0.5 < 0 \)
Then, \( \frac{x^{3}(x-1)}{x-2} = \frac{(+)(+) }{(-) } = \frac{(+)}{(-)} = (-) \).
Since \( \frac{x^{3}(x-1)}{x-2} < 0 \) in this interval, it means \( \frac{x^{3}(x-1)}{x-2} > 0 \) is not true. So, there is no solution in \( (1, 2) \).

(4) In the interval \( (2, \infty) \), choose a test value, for example, \( x = 3 \).
Factor \( x^3 = (3)^3 > 0 \)
Factor \( x-1 = 3-1 = 2 > 0 \)
Factor \( x-2 = 3-2 = 1 > 0 \)
Then, \( \frac{x^{3}(x-1)}{x-2} = \frac{(+)(+) }{(+) } = \frac{(+)}{(+)} = (+) \).
Since \( \frac{x^{3}(x-1)}{x-2} > 0 \) in this interval, it means \( \frac{x^{3}(x-1)}{x-2} > 0 \) is true. So, this interval is a solution: \( (2, \infty) \).

Summary of signs:

The inequality \( \frac{x^{3}(x-1)}{x-2} > 0 \) is satisfied in the intervals where the sign is positive. These are \( (0, 1) \) and \( (2, \infty) \).
The solution set is the union of these intervals: \( (0, 1) \cup (2, \infty) \). This method helps visualize when the function is positive or negative.
In simple words: To find where the fraction is greater than zero, we first find the numbers that make the top or bottom zero. These "critical numbers" divide the number line into sections. Then, we test a number from each section to see if the whole fraction is positive. If it is positive, that section is part of the answer.

🎯 Exam Tip: Always remember to check for values that make the denominator zero, as these are points where the function is undefined and cannot be included in the solution set for strict inequalities.

 

Question 2. Find all values of x that satisfies the inequality \( \frac{2 x-3}{(x-2)(x-4)} < 0 \).
Answer: The given inequality is \( f(x) = \frac{2 x-3}{(x-2)(x-4)} < 0 \).
First, we find the critical numbers by setting the numerator and denominator to zero.
For the numerator: \( 2x-3 = 0 \implies 2x = 3 \implies x = \frac{3}{2} \).
For the denominator: \( (x-2)(x-4) = 0 \implies x-2 = 0 \) or \( x-4 = 0 \implies x = 2 \) or \( x = 4 \).
The critical numbers are \( \frac{3}{2}, 2, 4 \). These numbers divide the number line into four intervals: \( (-\infty, \frac{3}{2}) \), \( (\frac{3}{2}, 2) \), \( (2, 4) \), and \( (4, \infty) \). We test a value from each interval to determine the sign of \( f(x) \).

Let's examine each interval:

(1) In the interval \( (-\infty, \frac{3}{2}) \), choose a test value, for example, \( x = 0 \).
Factor \( 2x-3 = 2(0)-3 = -3 < 0 \)
Factor \( x-2 = 0-2 = -2 < 0 \)
Factor \( x-4 = 0-4 = -4 < 0 \)
Then, \( \frac{2x-3}{(x-2)(x-4)} = \frac{(-) }{(-)(-) } = \frac{(-)}{(+)} = (-) \).
Since \( \frac{2x-3}{(x-2)(x-4)} < 0 \) in this interval, it is true. So, \( (-\infty, \frac{3}{2}) \) is a solution.

(2) In the interval \( (\frac{3}{2}, 2) \), choose a test value, for example, \( x = 1.5 + 0.1 = 1.6 \) (or simply \( x = \frac{3+2}{2} = \frac{7}{4} = 1.75 \)). Let's use \( x = \frac{7}{4} \).
Factor \( 2x-3 = 2(\frac{7}{4})-3 = \frac{7}{2}-3 = \frac{7-6}{2} = \frac{1}{2} > 0 \)
Factor \( x-2 = \frac{7}{4}-2 = \frac{7-8}{4} = -\frac{1}{4} < 0 \)
Factor \( x-4 = \frac{7}{4}-4 = \frac{7-16}{4} = -\frac{9}{4} < 0 \)
Then, \( \frac{2x-3}{(x-2)(x-4)} = \frac{(+) }{(-)(-) } = \frac{(+)}{(+)} = (+) \).
Since \( \frac{2x-3}{(x-2)(x-4)} > 0 \) in this interval, it means \( \frac{2x-3}{(x-2)(x-4)} < 0 \) is not true. So, there is no solution in \( (\frac{3}{2}, 2) \).

(3) In the interval \( (2, 4) \), choose a test value, for example, \( x = 3 \).
Factor \( 2x-3 = 2(3)-3 = 6-3 = 3 > 0 \)
Factor \( x-2 = 3-2 = 1 > 0 \)
Factor \( x-4 = 3-4 = -1 < 0 \)
Then, \( \frac{2x-3}{(x-2)(x-4)} = \frac{(+) }{(+)(-) } = \frac{(+)}{(-)} = (-) \).
Since \( \frac{2x-3}{(x-2)(x-4)} < 0 \) in this interval, it is true. So, \( (2, 4) \) is a solution.

(4) In the interval \( (4, \infty) \), choose a test value, for example, \( x = 5 \).
Factor \( 2x-3 = 2(5)-3 = 10-3 = 7 > 0 \)
Factor \( x-2 = 5-2 = 3 > 0 \)
Factor \( x-4 = 5-4 = 1 > 0 \)
Then, \( \frac{2x-3}{(x-2)(x-4)} = \frac{(+) }{(+)(+) } = \frac{(+)}{(+)} = (+) \).
Since \( \frac{2x-3}{(x-2)(x-4)} > 0 \) in this interval, it means \( \frac{2x-3}{(x-2)(x-4)} < 0 \) is not true. So, there is no solution in \( (4, \infty) \).

Summary of signs:

The inequality \( \frac{2x-3}{(x-2)(x-4)} < 0 \) is satisfied in the intervals where the sign is negative. These are \( (-\infty, \frac{3}{2}) \) and \( (2, 4) \). It's important to remember that points where the denominator is zero (x=2, x=4) are never included in the solution for a strict inequality.
The solution set is the union of these intervals: \( (-\infty, \frac{3}{2}) \cup (2, 4) \).
In simple words: We want to find when this fraction is less than zero (negative). We list all numbers that make the top or bottom zero. These numbers split the number line into parts. We check each part to see if the fraction is negative there. The parts that make it negative are our answer.

🎯 Exam Tip: When dealing with strict inequalities like \( < 0 \) or \( > 0 \), always use open intervals (parentheses) around critical numbers, even if they come from the numerator, as the function itself cannot be zero for these inequalities.

 

Question 3. Solve: \( \frac{x^{2}-4}{x^{2}+4 x-15} \le 0 \)
Answer: The given inequality is \( f(x) = \frac{x^{2}-4}{x^{2}+4 x-15} \le 0 \).
First, we factorize the numerator and the denominator:
Numerator: \( x^2-4 = (x-2)(x+2) \)
Denominator: \( x^2+4x-15 = x^2+5x-3x-15 = x(x+5)-3(x+5) = (x+5)(x-3) \)
So, the inequality becomes \( f(x) = \frac{(x+2)(x-2)}{(x+5)(x-3)} \le 0 \).
Now, we find the critical numbers by setting the numerator and denominator to zero.
For the numerator: \( (x+2)(x-2) = 0 \implies x = -2 \) or \( x = 2 \).
For the denominator: \( (x+5)(x-3) = 0 \implies x = -5 \) or \( x = 3 \).
The critical numbers are -5, -2, 2, 3. These numbers divide the number line into five intervals: \( (-\infty, -5) \), \( (-5, -2) \), \( (-2, 2) \), \( (2, 3) \), and \( (3, \infty) \). We test a value from each interval to determine the sign of \( f(x) \).

Let's examine each interval:

(1) In the interval \( (-\infty, -5) \), choose a test value, for example, \( x = -6 \).
Factor \( x+2 = -6+2 = -4 < 0 \)
Factor \( x-2 = -6-2 = -8 < 0 \)
Factor \( x+5 = -6+5 = -1 < 0 \)
Factor \( x-3 = -6-3 = -9 < 0 \)
Then, \( \frac{(x+2)(x-2)}{(x+5)(x-3)} = \frac{(-)(-) }{(-)(-) } = \frac{(+)}{(+)} = (+) \).
Since \( f(x) > 0 \) in this interval, \( f(x) \le 0 \) is not true. No solution here.

(2) In the interval \( (-5, -2) \), choose a test value, for example, \( x = -3 \).
Factor \( x+2 = -3+2 = -1 < 0 \)
Factor \( x-2 = -3-2 = -5 < 0 \)
Factor \( x+5 = -3+5 = 2 > 0 \)
Factor \( x-3 = -3-3 = -6 < 0 \)
Then, \( \frac{(x+2)(x-2)}{(x+5)(x-3)} = \frac{(-)(-) }{(+)(-) } = \frac{(+)}{(-)} = (-) \).
Since \( f(x) < 0 \) in this interval, \( f(x) \le 0 \) is true. So, \( (-5, -2] \) is a solution. (We include -2 because \( f(x) = 0 \) there, and the inequality is \( \le 0 \)). Note that -5 is excluded as it makes the denominator zero.

(3) In the interval \( (-2, 2) \), choose a test value, for example, \( x = 0 \).
Factor \( x+2 = 0+2 = 2 > 0 \)
Factor \( x-2 = 0-2 = -2 < 0 \)
Factor \( x+5 = 0+5 = 5 > 0 \)
Factor \( x-3 = 0-3 = -3 < 0 \)
Then, \( \frac{(x+2)(x-2)}{(x+5)(x-3)} = \frac{(+)(-) }{(+)(-) } = \frac{(-)}{(-)} = (+) \).
Since \( f(x) > 0 \) in this interval, \( f(x) \le 0 \) is not true. No solution here.

(4) In the interval \( (2, 3) \), choose a test value, for example, \( x = 2.5 \).
Factor \( x+2 = 2.5+2 = 4.5 > 0 \)
Factor \( x-2 = 2.5-2 = 0.5 > 0 \)
Factor \( x+5 = 2.5+5 = 7.5 > 0 \)
Factor \( x-3 = 2.5-3 = -0.5 < 0 \)
Then, \( \frac{(x+2)(x-2)}{(x+5)(x-3)} = \frac{(+)(+) }{(+)(-) } = \frac{(+)}{(-)} = (-) \).
Since \( f(x) < 0 \) in this interval, \( f(x) \le 0 \) is true. So, \( [2, 3) \) is a solution. (We include 2 because \( f(x) = 0 \) there, but exclude 3 as it makes the denominator zero).

(5) In the interval \( (3, \infty) \), choose a test value, for example, \( x = 5 \).
Factor \( x+2 = 5+2 = 7 > 0 \)
Factor \( x-2 = 5-2 = 3 > 0 \)
Factor \( x+5 = 5+5 = 10 > 0 \)
Factor \( x-3 = 5-3 = 2 > 0 \)
Then, \( \frac{(x+2)(x-2)}{(x+5)(x-3)} = \frac{(+)(+) }{(+)(+) } = \frac{(+)}{(+)} = (+) \).
Since \( f(x) > 0 \) in this interval, \( f(x) \le 0 \) is not true. No solution here.

Summary of signs:

The inequality \( \frac{x^{2}-4}{x^{2}+4 x-15} \le 0 \) is true where the sign is negative or zero. This happens in the intervals \( (-5, -2] \) and \( [2, 3) \). The critical numbers from the denominator (x=-5 and x=3) are never included because they make the expression undefined. The critical numbers from the numerator (x=-2 and x=2) are included because the inequality is "less than or equal to". Graphing these intervals on a number line helps to visualize the solution. The solution set is \( (-5, -2] \cup [2, 3) \).
In simple words: First, we change the top and bottom parts of the fraction into simpler multiplied terms. Then, we find all the numbers that make any of these terms zero. These numbers divide the number line into segments. We check each segment to see where the fraction is negative or exactly zero. Make sure to use square brackets for numbers that make the fraction zero (from the top part) and round brackets for numbers that make the bottom part zero (as these cannot be included).

🎯 Exam Tip: For inequalities that include "or equal to" (\( \le \) or \( \ge \)), always remember to include the roots of the numerator (values that make the expression zero) in the solution set by using square brackets, unless those values also make the denominator zero.