OP Malhotra Class 11 Maths Solutions Chapter 11 Inequalities Exercise 11 (C)

Question 1. \( x^2 - 4x + 3 < 0 \).
Answer: First, we need to factor the quadratic expression. The expression \( x^2 - 4x + 3 \) can be factored into \( (x - 1)(x - 3) \). So the inequality becomes \( (x - 1)(x - 3) < 0 \).
To find the critical points, we set the factors equal to zero: \( x - 1 = 0 \) gives \( x = 1 \), and \( x - 3 = 0 \) gives \( x = 3 \). These points divide the number line into three intervals: \( (-\infty, 1) \), \( (1, 3) \), and \( (3, \infty) \).
We check the sign of \( (x - 1)(x - 3) \) in each interval. For \( x < 1 \), both \( (x - 1) \) and \( (x - 3) \) are negative, so their product is positive. For \( 1 < x < 3 \), \( (x - 1) \) is positive and \( (x - 3) \) is negative, making their product negative. For \( x > 3 \), both factors are positive, so their product is positive.
Since we are looking for where \( (x - 1)(x - 3) < 0 \), the solution is the interval where the product is negative.
The required range is \( x \in (1, 3) \).
In simple words: We want to find the numbers \( x \) that make \( x^2 - 4x + 3 \) a negative number. By factoring it to \( (x-1)(x-3) \), we find that this happens for all numbers \( x \) that are between 1 and 3, but not including 1 or 3 themselves.

🎯 Exam Tip: When solving quadratic inequalities, always factor the quadratic expression first to find the critical points. Then, use a number line to test intervals and determine where the inequality holds true. Pay attention to whether the inequality is strict (<, >) or non-strict (≤, ≥) to decide if critical points are included.

Question 2. \( x^2 + 5x + 4 > 0 \).
Answer: We begin by factoring the quadratic expression \( x^2 + 5x + 4 \). This can be factored as \( (x + 1)(x + 4) \). So, the inequality is \( (x + 1)(x + 4) > 0 \).
To find the critical points, we set each factor to zero: \( x + 1 = 0 \) gives \( x = -1 \), and \( x + 4 = 0 \) gives \( x = -4 \). These critical points, -4 and -1, divide the number line into three intervals: \( (-\infty, -4) \), \( (-4, -1) \), and \( (-1, \infty) \).
Next, we test the sign of \( (x + 1)(x + 4) \) in each interval. If \( x < -4 \), both factors are negative, so their product is positive. If \( -4 < x < -1 \), \( (x + 4) \) is positive and \( (x + 1) \) is negative, making their product negative. If \( x > -1 \), both factors are positive, so their product is positive.
Since the inequality is \( (x + 1)(x + 4) > 0 \), we are looking for intervals where the product is positive.
The required range is \( x \in (-\infty, -4) \cup (-1, \infty) \), which means \( x < -4 \) or \( x > -1 \).
In simple words: We are looking for numbers \( x \) where the expression \( x^2 + 5x + 4 \) is positive. When we break it down, we find that this is true when \( x \) is smaller than -4 or when \( x \) is larger than -1.

🎯 Exam Tip: Remember that multiplying or dividing an inequality by a negative number reverses the inequality sign. Always check the sign of the factored expression in each interval defined by the critical points.

Question 3. \( x^2 + x - 6 \ge 0 \).
Answer: First, factor the quadratic expression \( x^2 + x - 6 \). It factors into \( (x - 2)(x + 3) \). So, the inequality we need to solve is \( (x - 2)(x + 3) \ge 0 \).
To find the critical points, set each factor to zero: \( x - 2 = 0 \) gives \( x = 2 \), and \( x + 3 = 0 \) gives \( x = -3 \). These points divide the number line into three intervals: \( (-\infty, -3) \), \( (-3, 2) \), and \( (2, \infty) \).
Let's check the sign of the product \( (x - 2)(x + 3) \) in each interval. For \( x < -3 \), both factors are negative, making their product positive. For \( -3 < x < 2 \), \( (x + 3) \) is positive and \( (x - 2) \) is negative, so their product is negative. For \( x > 2 \), both factors are positive, so their product is positive.
Since we are looking for where \( (x - 2)(x + 3) \ge 0 \), the solution includes intervals where the product is positive, and also the critical points because of the "equal to" part of the inequality.
The required range is \( x \in (-\infty, -3] \cup [2, \infty) \), which means \( x \le -3 \) or \( x \ge 2 \).
In simple words: We need to find the numbers \( x \) that make \( x^2 + x - 6 \) greater than or equal to zero. This happens when \( x \) is -3 or less, or when \( x \) is 2 or more. The values -3 and 2 are included in the answer.

🎯 Exam Tip: When the inequality includes "or equal to" (≤ or ≥), remember to include the critical points in your solution range, often shown with square brackets in interval notation or solid circles on a number line.

Question 4. \( x^2 - 16 < 0 \).
Answer: We are given the inequality \( x^2 - 16 < 0 \).
First, we can rewrite this as \( x^2 < 16 \).
To solve for \( x \), we take the square root of both sides. When taking the square root of an inequality, we must consider both positive and negative roots, which leads to the absolute value form: \( |x| < \sqrt{16} \).
This simplifies to \( |x| < 4 \).
The property of absolute values states that if \( |x| < a \) where \( a > 0 \), then \( -a < x < a \).
Applying this, we get \( -4 < x < 4 \).
This means the value of \( x \) must be strictly between -4 and 4.
The required range is \( x \in (-4, 4) \).
In simple words: We want to find numbers \( x \) that, when squared, are less than 16. This means \( x \) must be any number between -4 and 4, but not -4 or 4 themselves.

🎯 Exam Tip: For inequalities of the form \( x^2 < a^2 \), the solution is always \( -a < x < a \). For \( x^2 > a^2 \), the solution is \( x < -a \) or \( x > a \). Always remember the absolute value interpretation.

Question 5. \( x^2 - 6x + 9 \ge 0 \).
Answer: The given inequality is \( x^2 - 6x + 9 \ge 0 \).
The expression \( x^2 - 6x + 9 \) is a perfect square trinomial, which can be factored as \( (x - 3)^2 \).
So, the inequality becomes \( (x - 3)^2 \ge 0 \).
When any real number is squared, the result is always greater than or equal to zero. This means that \( (x - 3)^2 \) will always be greater than or equal to zero, no matter what real value \( x \) takes. For example, if \( x=3 \), then \( (3-3)^2 = 0 \), which satisfies the inequality. If \( x=5 \), then \( (5-3)^2 = 2^2 = 4 \), which is also greater than zero.
Therefore, this inequality is true for all real numbers \( x \).
The required range is the set of all real numbers, denoted as \( R \) or \( (-\infty, \infty) \).
In simple words: We are looking for numbers \( x \) that make \( (x-3)^2 \) greater than or equal to zero. Since squaring any real number always gives a positive result or zero, this inequality is true for every single real number.

🎯 Exam Tip: Recognize perfect square trinomials like \( (ax \pm b)^2 \). Their value is always non-negative (\( \ge 0 \)), which can simplify solving inequalities significantly, sometimes meaning the solution is all real numbers.

Question 6. \( -x^2 + 2x + 3 < 0 \).
Answer: We are given the inequality \( -x^2 + 2x + 3 < 0 \).
To make the leading coefficient positive, we multiply the entire inequality by -1. Remember to reverse the inequality sign when multiplying or dividing by a negative number.
So, \( (-1)(-x^2 + 2x + 3) > (-1)(0) \).
This simplifies to \( x^2 - 2x - 3 > 0 \).
Now, factor the quadratic expression \( x^2 - 2x - 3 \). It factors into \( (x + 1)(x - 3) \).
The inequality becomes \( (x + 1)(x - 3) > 0 \).
The critical points are found by setting the factors to zero: \( x + 1 = 0 \) gives \( x = -1 \), and \( x - 3 = 0 \) gives \( x = 3 \). These points divide the number line into intervals: \( (-\infty, -1) \), \( (-1, 3) \), and \( (3, \infty) \).
We check the sign of \( (x + 1)(x - 3) \) in each interval. For \( x < -1 \), both factors are negative, product is positive. For \( -1 < x < 3 \), \( (x + 1) \) is positive and \( (x - 3) \) is negative, product is negative. For \( x > 3 \), both factors are positive, product is positive.
Since we need \( (x + 1)(x - 3) > 0 \), we select the intervals where the product is positive.
The required range is \( x \in (-\infty, -1) \cup (3, \infty) \), which means \( x < -1 \) or \( x > 3 \).
In simple words: First, we change the inequality so the \( x^2 \) term is positive. This flips the sign from \( < \) to \( > \). Then we find the numbers that make \( x^2 - 2x - 3 \) greater than zero. The answer is all numbers \( x \) that are smaller than -1 or larger than 3.

🎯 Exam Tip: Always make the leading coefficient of a quadratic inequality positive by multiplying by -1 (and flipping the inequality sign) before factoring and using the interval method. This helps prevent errors in determining the signs of the intervals.

Question 7. \( 5x < 2 - 3x^2 \).
Answer: We have the inequality \( 5x < 2 - 3x^2 \).
First, rearrange the inequality to bring all terms to one side, aiming for a positive \( x^2 \) coefficient:
\( 3x^2 + 5x - 2 < 0 \).
Next, factor the quadratic expression \( 3x^2 + 5x - 2 \). This factors into \( (x + 2)(3x - 1) \).
So, the inequality becomes \( (x + 2)(3x - 1) < 0 \).
To find the critical points, set each factor to zero: \( x + 2 = 0 \) gives \( x = -2 \), and \( 3x - 1 = 0 \) gives \( x = \frac{1}{3} \). These critical points, -2 and \( \frac{1}{3} \), divide the number line into three intervals: \( (-\infty, -2) \), \( (-2, \frac{1}{3}) \), and \( (\frac{1}{3}, \infty) \).
Check the sign of \( (x + 2)(3x - 1) \) in each interval. For \( x < -2 \), both factors are negative, product is positive. For \( -2 < x < \frac{1}{3} \), \( (x + 2) \) is positive and \( (3x - 1) \) is negative, product is negative. For \( x > \frac{1}{3} \), both factors are positive, product is positive.
Since we need \( (x + 2)(3x - 1) < 0 \), we choose the interval where the product is negative.
The required range is \( x \in (-2, \frac{1}{3}) \).
In simple words: We want to solve for \( x \) where \( 5x \) is smaller than \( 2 - 3x^2 \). Rearranging and factoring leads us to find numbers \( x \) that are between -2 and \( \frac{1}{3} \). These are the numbers that make the expression negative.

🎯 Exam Tip: When factoring quadratic expressions like \( ax^2 + bx + c \), consider common methods such as splitting the middle term or using the quadratic formula to find roots, which then give factors.

Question 8. \( -x^2 - 4x - 5 < 0 \).
Answer: We are given the inequality \( -x^2 - 4x - 5 < 0 \).
First, multiply the entire inequality by -1 to make the leading coefficient positive. Remember to reverse the inequality sign.
\( (-1)(-x^2 - 4x - 5) > (-1)(0) \)
This becomes \( x^2 + 4x + 5 > 0 \).
Now, we try to factor this quadratic, or check its discriminant. The discriminant \( \Delta = b^2 - 4ac \) for \( x^2 + 4x + 5 \) is \( 4^2 - 4(1)(5) = 16 - 20 = -4 \).
Since the discriminant is negative and the leading coefficient (which is 1) is positive, the quadratic \( x^2 + 4x + 5 \) is always positive for all real values of \( x \). This means its graph is an upward-opening parabola that never crosses the x-axis.
Alternatively, we can complete the square: \( x^2 + 4x + 5 = (x^2 + 4x + 4) + 1 = (x + 2)^2 + 1 \).
Since \( (x + 2)^2 \) is always greater than or equal to 0 for any real \( x \), adding 1 means \( (x + 2)^2 + 1 \) will always be greater than or equal to 1. This means it is always strictly positive.
Thus, the inequality \( (x + 2)^2 + 1 > 0 \) is true for all real numbers \( x \).
The required range is the set of all real numbers, \( R \) or \( (-\infty, \infty) \).
In simple words: We want to find numbers \( x \) that make \( -x^2 - 4x - 5 \) negative. After changing the signs, we are looking for when \( x^2 + 4x + 5 \) is positive. Because this expression always stays positive for any number \( x \), the original inequality is true for all real numbers.

🎯 Exam Tip: For quadratic inequalities that don't easily factor, calculate the discriminant (\( b^2 - 4ac \)). If it's negative and the leading coefficient is positive, the quadratic is always positive. If it's negative and the leading coefficient is negative, the quadratic is always negative.

Question 9. \( 4x^2 + 1 > 4x \).
Answer: We are given the inequality \( 4x^2 + 1 > 4x \).
First, rearrange the inequality to bring all terms to one side, making the right side zero:
\( 4x^2 - 4x + 1 > 0 \).
The expression \( 4x^2 - 4x + 1 \) is a perfect square trinomial, which can be factored as \( (2x - 1)^2 \).
So, the inequality becomes \( (2x - 1)^2 > 0 \).
A squared real number is always greater than or equal to zero. For \( (2x - 1)^2 \) to be strictly greater than zero, it must not be equal to zero. The expression \( (2x - 1)^2 \) is equal to zero only when its base is zero: \( 2x - 1 = 0 \), which means \( 2x = 1 \), or \( x = \frac{1}{2} \).
So, the inequality \( (2x - 1)^2 > 0 \) is true for all real values of \( x \) except when \( x = \frac{1}{2} \). If \( x = \frac{1}{2} \), the expression becomes \( (2(\frac{1}{2}) - 1)^2 = (1 - 1)^2 = 0^2 = 0 \), which is not strictly greater than 0.
The required range is all real numbers except \( \frac{1}{2} \), which can be written as \( R - \{\frac{1}{2}\} \) or \( (-\infty, \frac{1}{2}) \cup (\frac{1}{2}, \infty) \).
In simple words: We want to find numbers \( x \) that make \( 4x^2 + 1 \) greater than \( 4x \). After moving all parts to one side, we get \( (2x-1)^2 > 0 \). Since any number squared is always positive or zero, this means the answer is all real numbers except for the one number that makes the expression zero, which is \( x = \frac{1}{2} \).

🎯 Exam Tip: When dealing with perfect square inequalities, remember that \( (ax+b)^2 > 0 \) means all real numbers except the root, while \( (ax+b)^2 \ge 0 \) means all real numbers. Pay close attention to the strictness of the inequality.

Question 10. \( -x^2 + x > 0 \).
Answer: We have the inequality \( -x^2 + x > 0 \).
To work with a positive leading coefficient, multiply the entire inequality by -1. This reverses the inequality sign.
\( (-1)(-x^2 + x) < (-1)(0) \)
This becomes \( x^2 - x < 0 \).
Now, factor out \( x \) from the expression: \( x(x - 1) < 0 \).
The critical points are found by setting each factor to zero: \( x = 0 \) and \( x - 1 = 0 \) gives \( x = 1 \). These points divide the number line into three intervals: \( (-\infty, 0) \), \( (0, 1) \), and \( (1, \infty) \).
Check the sign of \( x(x - 1) \) in each interval. For \( x < 0 \), \( x \) is negative and \( (x - 1) \) is negative, product is positive. For \( 0 < x < 1 \), \( x \) is positive and \( (x - 1) \) is negative, product is negative. For \( x > 1 \), both factors are positive, product is positive.
Since we need \( x(x - 1) < 0 \), we select the interval where the product is negative.
The required range is \( x \in (0, 1) \).
In simple words: We need to find numbers \( x \) for which \( -x^2 + x \) is positive. By changing the sign of the whole expression, we get \( x^2 - x < 0 \). This means we are looking for when \( x(x-1) \) is negative, which happens when \( x \) is between 0 and 1.

🎯 Exam Tip: Always factor out common terms first, if possible. For inequalities with only two factors like \( x(x-1) \), the critical points are simply the values of \( x \) that make each factor zero.

Question 11. \( 6 + x < 2x^2 \).
Answer: The given inequality is \( 6 + x < 2x^2 \).
Rearrange the terms to get all terms on one side, typically with \( x^2 \) being positive:
\( 0 < 2x^2 - x - 6 \)
So, \( 2x^2 - x - 6 > 0 \).
Now, factor the quadratic expression \( 2x^2 - x - 6 \). This factors into \( (x - 2)(2x + 3) \).
The inequality becomes \( (x - 2)(2x + 3) > 0 \).
To find the critical points, set each factor to zero: \( x - 2 = 0 \) gives \( x = 2 \), and \( 2x + 3 = 0 \) gives \( x = -\frac{3}{2} \). These critical points, \( -\frac{3}{2} \) and 2, divide the number line into three intervals: \( (-\infty, -\frac{3}{2}) \), \( (-\frac{3}{2}, 2) \), and \( (2, \infty) \).
Check the sign of \( (x - 2)(2x + 3) \) in each interval. For \( x < -\frac{3}{2} \), both factors are negative, product is positive. For \( -\frac{3}{2} < x < 2 \), \( (2x + 3) \) is positive and \( (x - 2) \) is negative, product is negative. For \( x > 2 \), both factors are positive, product is positive.
Since we need \( (x - 2)(2x + 3) > 0 \), we select the intervals where the product is positive.
The required range is \( x \in (-\infty, -\frac{3}{2}) \cup (2, \infty) \), which means \( x < -\frac{3}{2} \) or \( x > 2 \).
In simple words: We want to find numbers \( x \) where \( 6 + x \) is less than \( 2x^2 \). After moving terms around and factoring, we look for when \( (x-2)(2x+3) \) is positive. This happens when \( x \) is smaller than \( -\frac{3}{2} \) or larger than 2.

🎯 Exam Tip: Always make sure to factor the quadratic expression completely and correctly, as errors in factoring will lead to incorrect critical points and an incorrect solution set.

Question 12. Solve the following inequalities:
(i) \( (x - 4)(x + 6) > 0 \);
(ii) \( 3 - 2x^2 > 5x \)
Answer:
(i) We are given the inequality \( (x - 4)(x + 6) > 0 \).
The expression is already factored. The critical points are found by setting each factor to zero: \( x - 4 = 0 \) gives \( x = 4 \), and \( x + 6 = 0 \) gives \( x = -6 \). These points divide the number line into intervals: \( (-\infty, -6) \), \( (-6, 4) \), and \( (4, \infty) \).
Check the sign of \( (x - 4)(x + 6) \) in each interval. For \( x < -6 \), both factors are negative, product is positive. For \( -6 < x < 4 \), \( (x + 6) \) is positive and \( (x - 4) \) is negative, product is negative. For \( x > 4 \), both factors are positive, product is positive.
Since we need \( (x - 4)(x + 6) > 0 \), we choose the intervals where the product is positive.
The required range is \( x \in (-\infty, -6) \cup (4, \infty) \), which means \( x < -6 \) or \( x > 4 \).
(ii) We are given the inequality \( 3 - 2x^2 > 5x \).
Rearrange the terms to get all terms on one side, ensuring the \( x^2 \) coefficient is positive:
\( 0 > 2x^2 + 5x - 3 \)
So, \( 2x^2 + 5x - 3 < 0 \).
Now, factor the quadratic expression \( 2x^2 + 5x - 3 \). This factors into \( (x + 3)(2x - 1) \).
The inequality becomes \( (x + 3)(2x - 1) < 0 \).
To find the critical points, set each factor to zero: \( x + 3 = 0 \) gives \( x = -3 \), and \( 2x - 1 = 0 \) gives \( x = \frac{1}{2} \). These critical points, -3 and \( \frac{1}{2} \), divide the number line into three intervals: \( (-\infty, -3) \), \( (-3, \frac{1}{2}) \), and \( (\frac{1}{2}, \infty) \).
Check the sign of \( (x + 3)(2x - 1) \) in each interval. For \( x < -3 \), both factors are negative, product is positive. For \( -3 < x < \frac{1}{2} \), \( (x + 3) \) is positive and \( (2x - 1) \) is negative, product is negative. For \( x > \frac{1}{2} \), both factors are positive, product is positive.
Since we need \( (x + 3)(2x - 1) < 0 \), we select the interval where the product is negative.
The required range is \( x \in (-3, \frac{1}{2}) \).
In simple words: For part (i), we solve \( (x-4)(x+6) > 0 \) by finding the values of \( x \) that are smaller than -6 or larger than 4. For part (ii), we first rearrange \( 3 - 2x^2 > 5x \) to \( 2x^2 + 5x - 3 < 0 \). Then, we find that \( x \) must be between -3 and \( \frac{1}{2} \).

🎯 Exam Tip: When an inequality has multiple parts, solve each part separately to find its solution range. If solving a system of inequalities (like in Q13), find the intersection of these ranges. For simple quadratic inequalities, the "interval method" using critical points is very effective.

Question 13. Find all real values of \( x \) which satisfy \( x^2 - 3x + 2 > 0 \) and \( x^2 - 3x - 4 \le 0 \).
Answer: We need to solve two inequalities and find the values of \( x \) that satisfy both simultaneously.
**First Inequality:** \( x^2 - 3x + 2 > 0 \)
Factor the quadratic: \( (x - 1)(x - 2) > 0 \).
Critical points are \( x = 1 \) and \( x = 2 \).
Using the method of intervals, \( (x - 1)(x - 2) > 0 \) when \( x < 1 \) or \( x > 2 \). This is our first solution set (1).
**Second Inequality:** \( x^2 - 3x - 4 \le 0 \)
Factor the quadratic: \( (x + 1)(x - 4) \le 0 \).
Critical points are \( x = -1 \) and \( x = 4 \).
Using the method of intervals, \( (x + 1)(x - 4) \le 0 \) when \( -1 \le x \le 4 \). This is our second solution set (2).

Now, we need to find the intersection of the two solution sets:
\( (x < 1 \text{ or } x > 2) \) AND \( (-1 \le x \le 4) \)
Visually, on a single number line, we look for areas where both conditions are true.
\( \implies -1 \le x < 1 \) or \( 2 < x \le 4 \).
The final range is \( [-1, 1) \cup (2, 4] \).
In simple words: We have two math puzzles to solve at the same time. For the first one, \( x^2 - 3x + 2 > 0 \), the answer is \( x \) is less than 1 or greater than 2. For the second one, \( x^2 - 3x - 4 \le 0 \), the answer is \( x \) is between -1 and 4, including -1 and 4. When we put these two answers together, the numbers that work for both are those between -1 (inclusive) and 1 (exclusive), or those between 2 (exclusive) and 4 (inclusive).

🎯 Exam Tip: When solving a system of inequalities, solve each inequality separately. Then, represent each solution on a number line. The final solution is the intersection of these individual solutions, meaning the region where all conditions overlap.

Question 14. The set of values of \( x \) for which the inequalities \( x^2 - 3x - 10 < 0 \), \( 10x - x^2 - 16 > 0 \) hold simultaneously is
(a) \( (-2, 5) \)
(b) \( (2, 8) \)
(c) \( (-2,8) \)
(d) \( (2, 5) \)
Answer: (d) (2, 5)
Answer: We need to solve two inequalities and find their intersection.
**First Inequality:** \( x^2 - 3x - 10 < 0 \)
Factor the quadratic: \( (x + 2)(x - 5) < 0 \).
Critical points are \( x = -2 \) and \( x = 5 \).
Using the method of intervals, \( (x + 2)(x - 5) < 0 \) when \( -2 < x < 5 \). This is our first solution set (1).
**Second Inequality:** \( 10x - x^2 - 16 > 0 \)
Rearrange and multiply by -1 to get a positive \( x^2 \) coefficient, remembering to reverse the inequality sign:
\( -x^2 + 10x - 16 > 0 \)
\( x^2 - 10x + 16 < 0 \)
Factor the quadratic: \( (x - 2)(x - 8) < 0 \).
Critical points are \( x = 2 \) and \( x = 8 \).
Using the method of intervals, \( (x - 2)(x - 8) < 0 \) when \( 2 < x < 8 \). This is our second solution set (2).
Now, we find the intersection of the two solution sets: \( (-2 < x < 5) \) AND \( (2 < x < 8) \).
The common interval where both conditions hold is \( 2 < x < 5 \).
This corresponds to the interval \( (2, 5) \).
In simple words: We have two conditions that \( x \) must satisfy. The first one says \( x \) must be between -2 and 5. The second one says \( x \) must be between 2 and 8. For \( x \) to satisfy both at the same time, it must be in the overlap, which means \( x \) must be between 2 and 5. So, option (d) is correct.

🎯 Exam Tip: For multiple-choice questions involving simultaneous inequalities, solving each inequality separately and then visually finding the intersection on a number line can quickly lead to the correct option. Pay attention to strict vs. non-strict inequalities.

Question 15. Solve the following inequalities : \( \frac{x+3}{x-1} > x \)
Answer: We are given the inequality \( \frac{x+3}{x-1} > x \).
First, bring all terms to one side to compare with zero:
\( \frac{x+3}{x-1} - x > 0 \)
Combine the terms using a common denominator:
\( \frac{x+3 - x(x-1)}{x-1} > 0 \)
Simplify the numerator:
\( \frac{x+3 - x^2 + x}{x-1} > 0 \)
\( \frac{-x^2 + 2x + 3}{x-1} > 0 \)
Multiply the numerator by -1 to make the \( x^2 \) term positive, and reverse the inequality sign:
\( \frac{x^2 - 2x - 3}{x-1} < 0 \)
Now, factor the numerator: \( \frac{(x + 1)(x - 3)}{x-1} < 0 \).
The critical points are \( x = -1 \) (from \( x+1=0 \)), \( x = 3 \) (from \( x-3=0 \)), and \( x = 1 \) (from \( x-1=0 \)). Note that \( x \ne 1 \) because the denominator cannot be zero.
These critical points divide the number line into four intervals: \( (-\infty, -1) \), \( (-1, 1) \), \( (1, 3) \), and \( (3, \infty) \).
We test a value from each interval to determine the sign of \( \frac{(x + 1)(x - 3)}{x-1} \):
* For \( x < -1 \) (e.g., \( x = -2 \)): \( \frac{(-1)(-5)}{-3} = -\frac{5}{3} < 0 \). This interval works.
* For \( -1 < x < 1 \) (e.g., \( x = 0 \)): \( \frac{(1)(-3)}{-1} = 3 > 0 \). This interval does not work.
* For \( 1 < x < 3 \) (e.g., \( x = 2 \)): \( \frac{(3)(-1)}{1} = -3 < 0 \). This interval works.
* For \( x > 3 \) (e.g., \( x = 4 \)): \( \frac{(5)(1)}{3} = \frac{5}{3} > 0 \). This interval does not work.
Since we need the expression to be less than 0, the solution is \( x \in (-\infty, -1) \cup (1, 3) \), which means \( x < -1 \) or \( 1 < x < 3 \).
In simple words: To solve \( \frac{x+3}{x-1} > x \), we first move \( x \) to the left side and combine everything into one fraction. After simplifying, we get \( \frac{x^2 - 2x - 3}{x-1} < 0 \). Then we find the special points (-1, 1, 3) and check the sign of the expression in between these points. The values of \( x \) that satisfy the inequality are those less than -1 or those between 1 and 3.

🎯 Exam Tip: When solving rational inequalities, always move all terms to one side to get a single fraction compared to zero. Identify critical points from both the numerator and the denominator, and remember that denominator roots are never included in the solution set.

Question 16. \( x + 4 < - \frac{2}{x+1} \)
Answer: We are given the inequality \( x + 4 < - \frac{2}{x+1} \).
First, bring all terms to one side to compare with zero:
\( x + 4 + \frac{2}{x+1} < 0 \)
Combine the terms using a common denominator \( (x+1) \):
\( \frac{(x+4)(x+1) + 2}{x+1} < 0 \)
Expand and simplify the numerator:
\( \frac{x^2 + x + 4x + 4 + 2}{x+1} < 0 \)
\( \frac{x^2 + 5x + 6}{x+1} < 0 \)
Factor the numerator:
\( \frac{(x+2)(x+3)}{x+1} < 0 \)
The critical points are found by setting each factor (from numerator and denominator) to zero: \( x = -2 \) (from \( x+2=0 \)), \( x = -3 \) (from \( x+3=0 \)), and \( x = -1 \) (from \( x+1=0 \)). Remember \( x \ne -1 \).
These points divide the number line into four intervals: \( (-\infty, -3) \), \( (-3, -2) \), \( (-2, -1) \), and \( (-1, \infty) \).
We test a value from each interval to determine the sign of \( \frac{(x+2)(x+3)}{x+1} \):
* For \( x < -3 \) (e.g., \( x = -4 \)): \( \frac{(-2)(-1)}{-3} = -\frac{2}{3} < 0 \). This interval works.
* For \( -3 < x < -2 \) (e.g., \( x = -2.5 \)): \( \frac{(-0.5)(0.5)}{-1.5} = \frac{-0.25}{-1.5} = \frac{1}{6} > 0 \). This interval does not work.
* For \( -2 < x < -1 \) (e.g., \( x = -1.5 \)): \( \frac{(0.5)(1.5)}{-0.5} = -1.5 < 0 \). This interval works.
* For \( x > -1 \) (e.g., \( x = 0 \)): \( \frac{(2)(3)}{1} = 6 > 0 \). This interval does not work.
Since we need the expression to be less than 0, the solution is \( x \in (-\infty, -3) \cup (-2, -1) \), which means \( x < -3 \) or \( -2 < x < -1 \).
In simple words: To solve this, we move all terms to one side to get \( \frac{(x+2)(x+3)}{x+1} < 0 \). We find the special numbers -3, -2, and -1. We then check what happens in the areas around these numbers. The solution is when \( x \) is smaller than -3, or when \( x \) is between -2 and -1. We must remember that \( x \) cannot be -1.

🎯 Exam Tip: Always pay attention to the value that makes the denominator zero. This value is a critical point but must be excluded from the final solution set, as division by zero is undefined.

Question 17. \( \frac{x^2-2x+3}{x^2-4x+3} > -3 \)
Answer: We are given the inequality \( \frac{x^2-2x+3}{x^2-4x+3} > -3 \).
First, move -3 to the left side and combine terms:
\( \frac{x^2-2x+3}{x^2-4x+3} + 3 > 0 \)
\( \frac{x^2-2x+3 + 3(x^2-4x+3)}{x^2-4x+3} > 0 \)
Simplify the numerator:
\( \frac{x^2-2x+3 + 3x^2-12x+9}{x^2-4x+3} > 0 \)
\( \frac{4x^2-14x+12}{x^2-4x+3} > 0 \)
Factor out 2 from the numerator and factor both quadratic expressions:
\( \frac{2(2x^2-7x+6)}{(x-1)(x-3)} > 0 \)
\( \frac{2(2x-3)(x-2)}{(x-1)(x-3)} > 0 \)
The constant factor 2 does not affect the sign, so we consider \( \frac{(2x-3)(x-2)}{(x-1)(x-3)} > 0 \).
Critical points are \( x = \frac{3}{2} \) (from \( 2x-3=0 \)), \( x = 2 \) (from \( x-2=0 \)), \( x = 1 \) (from \( x-1=0 \)), and \( x = 3 \) (from \( x-3=0 \)).
The domain restriction is \( x \ne 1 \) and \( x \ne 3 \).
These points divide the number line into five intervals: \( (-\infty, 1) \), \( (1, \frac{3}{2}) \), \( (\frac{3}{2}, 2) \), \( (2, 3) \), and \( (3, \infty) \).
We test a value from each interval to determine the sign of the expression:
* For \( x < 1 \) (e.g., \( x = 0 \)): \( \frac{2(-3)(-2)}{(-1)(-3)} = \frac{12}{3} = 4 > 0 \). This interval works.
* For \( 1 < x < \frac{3}{2} \) (e.g., \( x = 1.25 \)): \( \frac{(negative)(negative)}{(positive)(negative)} = \frac{positive}{negative} < 0 \). This interval does not work.
* For \( \frac{3}{2} < x < 2 \) (e.g., \( x = 1.75 \)): \( \frac{(positive)(negative)}{(positive)(negative)} = \frac{negative}{negative} > 0 \). This interval works.
* For \( 2 < x < 3 \) (e.g., \( x = 2.5 \)): \( \frac{(positive)(positive)}{(positive)(negative)} = \frac{positive}{negative} < 0 \). This interval does not work.
* For \( x > 3 \) (e.g., \( x = 4 \)): \( \frac{(positive)(positive)}{(positive)(positive)} = positive > 0 \). This interval works.
The solution is \( x \in (-\infty, 1) \cup (\frac{3}{2}, 2) \cup (3, \infty) \), which means \( x < 1 \) or \( \frac{3}{2} < x < 2 \) or \( x > 3 \).
In simple words: To solve this rational inequality, first we move -3 to the other side and combine the fractions. This simplifies to \( \frac{2(2x-3)(x-2)}{(x-1)(x-3)} > 0 \). We find the important numbers are 1, \( \frac{3}{2} \), 2, and 3. Then we check the signs in the areas around these numbers. The solution is \( x \) smaller than 1, or \( x \) between \( \frac{3}{2} \) and 2, or \( x \) larger than 3. Remember that \( x \) cannot be 1 or 3.

🎯 Exam Tip: When dealing with complex rational inequalities, always factor both numerator and denominator completely. This helps identify all critical points, which are essential for testing intervals accurately on the number line.

Question 18. \( \frac{x^2+6x-11}{x+3} < -1 \).
Answer: We are given the inequality \( \frac{x^2+6x-11}{x+3} < -1 \).
First, bring -1 to the left side and combine terms:
\( \frac{x^2+6x-11}{x+3} + 1 < 0 \)
\( \frac{x^2+6x-11 + (x+3)}{x+3} < 0 \)
Simplify the numerator:
\( \frac{x^2+7x-8}{x+3} < 0 \)
Factor the numerator:
\( \frac{(x-1)(x+8)}{x+3} < 0 \)
The critical points are \( x = 1 \) (from \( x-1=0 \)), \( x = -8 \) (from \( x+8=0 \)), and \( x = -3 \) (from \( x+3=0 \)). Remember \( x \ne -3 \).
These points divide the number line into four intervals: \( (-\infty, -8) \), \( (-8, -3) \), \( (-3, 1) \), and \( (1, \infty) \).
We test a value from each interval to determine the sign of \( \frac{(x-1)(x+8)}{x+3} \):
* For \( x < -8 \) (e.g., \( x = -10 \)): \( \frac{(-11)(-2)}{-7} = -\frac{22}{7} < 0 \). This interval works.
* For \( -8 < x < -3 \) (e.g., \( x = -5 \)): \( \frac{(-6)(3)}{-2} = 9 > 0 \). This interval does not work.
* For \( -3 < x < 1 \) (e.g., \( x = 0 \)): \( \frac{(-1)(8)}{3} = -\frac{8}{3} < 0 \). This interval works.
* For \( x > 1 \) (e.g., \( x = 2 \)): \( \frac{(1)(10)}{5} = 2 > 0 \). This interval does not work.
The solution is \( x \in (-\infty, -8) \cup (-3, 1) \), which means \( x < -8 \) or \( -3 < x < 1 \).
In simple words: We want to solve \( \frac{x^2+6x-11}{x+3} < -1 \). First, we move -1 to the left side and combine the terms into a single fraction: \( \frac{(x-1)(x+8)}{x+3} < 0 \). Then we find the special numbers -8, -3, and 1. By checking the signs in the regions, the solution is when \( x \) is less than -8, or when \( x \) is between -3 and 1. Remember that \( x \) cannot be -3.

🎯 Exam Tip: When simplifying rational inequalities, ensure all factors in both the numerator and denominator are correctly identified to avoid missing critical points. Also, always exclude values that make the denominator zero from your solution set.

Question 19. \( \frac{x^2-3x+24}{x^2-3x+3} < 4 \)
Answer: We are given the inequality \( \frac{x^2-3x+24}{x^2-3x+3} < 4 \).
First, move 4 to the left side and combine terms:
\( \frac{x^2-3x+24}{x^2-3x+3} - 4 < 0 \)
\( \frac{x^2-3x+24 - 4(x^2-3x+3)}{x^2-3x+3} < 0 \)
Simplify the numerator:
\( \frac{x^2-3x+24 - 4x^2+12x-12}{x^2-3x+3} < 0 \)
\( \frac{-3x^2+9x+12}{x^2-3x+3} < 0 \)
Factor out -3 from the numerator and multiply the numerator by -1, reversing the inequality sign:
\( \frac{3x^2-9x-12}{x^2-3x+3} > 0 \)
\( \frac{3(x^2-3x-4)}{x^2-3x+3} > 0 \)
Factor the quadratic expressions in both numerator and denominator:
Numerator: \( x^2-3x-4 = (x+1)(x-4) \)
Denominator: \( x^2-3x+3 \). Let's check its discriminant: \( \Delta = (-3)^2 - 4(1)(3) = 9 - 12 = -3 \). Since the discriminant is negative and the leading coefficient (1) is positive, \( x^2-3x+3 \) is always positive for all real \( x \). This is an important detail.
So, the inequality simplifies to \( \frac{3(x+1)(x-4)}{always positive} > 0 \).
Since the denominator is always positive, the sign of the fraction depends only on the sign of the numerator \( 3(x+1)(x-4) \). The constant factor 3 does not affect the sign.
So, we need to solve \( (x+1)(x-4) > 0 \).
Critical points are \( x = -1 \) and \( x = 4 \).
Using the method of intervals: For \( x < -1 \), both factors are negative, product is positive. For \( -1 < x < 4 \), \( (x+1) \) is positive and \( (x-4) \) is negative, product is negative. For \( x > 4 \), both factors are positive, product is positive.
Since we need \( (x+1)(x-4) > 0 \), we select the intervals where the product is positive.
The solution is \( x \in (-\infty, -1) \cup (4, \infty) \), which means \( x < -1 \) or \( x > 4 \).
In simple words: We want to find numbers \( x \) that make the fraction \( \frac{x^2-3x+24}{x^2-3x+3} \) less than 4. After moving 4 to the other side and simplifying, we see that the bottom part of the fraction \( (x^2-3x+3) \) is always positive. This means the sign of the whole fraction depends only on the top part. We solve \( (x+1)(x-4) > 0 \), which gives us the answer that \( x \) is less than -1 or greater than 4.

🎯 Exam Tip: Before using the interval method for rational inequalities, always check the discriminant of the quadratic expressions. If a quadratic in the denominator is always positive (or always negative), it simplifies the analysis as it doesn't create new critical points or change signs based on \( x \).