GSEB Class 11 Maths Solutions Chapter 14 Mathematical Reasoning Exercise 14.5

Question 1. Show that the statement p: “If x is a real number such that \( x^3 + 4x = 0 \), then \( x = 0 \) is true by
1. direct method
2. method of contradiction
3. method of contrapositive.

Answer:
1. Direct method:
We are given the equation \( x^3 + 4x = 0 \).
We can factor out \( x \) to get \( x(x^2 + 4) = 0 \).
For a product of two terms to be zero, at least one of the terms must be zero.
So, either \( x = 0 \) or \( x^2 + 4 = 0 \).
If \( x^2 + 4 = 0 \), then \( x^2 = -4 \). There is no real number \( x \) whose square is negative.
Therefore, for real numbers, the only possibility is \( x = 0 \).
In simple words: We factor the equation \( x^3 + 4x = 0 \) into \( x(x^2 + 4) = 0 \). Since \( x^2 + 4 \) can never be zero for real numbers, \( x \) must be zero.

Exam Tip: For direct proofs, show a clear logical progression from the given conditions to the conclusion using algebraic manipulation or definitions.

2. Method of contradiction:
We want to show that if \( x^3 + 4x = 0 \), then \( x = 0 \).
Assume, for the sake of contradiction, that \( x \neq 0 \).
Given \( x^3 + 4x = 0 \).
Factor out \( x \): \( x(x^2 + 4) = 0 \).
Since we assumed \( x \neq 0 \), for the product to be zero, the other term must be zero.
Thus, \( x^2 + 4 = 0 \).
This implies \( x^2 = -4 \).
However, the square of any real number cannot be negative. This creates a contradiction.
Therefore, our initial assumption that \( x \neq 0 \) must be false.
Hence, \( x = 0 \).
In simple words: We assume \( x \) is not zero. If \( x^3 + 4x = 0 \), this would lead to \( x^2 = -4 \), which is impossible for real numbers. This contradiction means our first assumption (that \( x \neq 0 \)) was wrong, so \( x \) must be zero.

Exam Tip: In a proof by contradiction, clearly state your assumption, derive a logical inconsistency, and then conclude that your initial assumption was incorrect.

3. Method of contrapositive:
The given statement is "If \( x^3 + 4x = 0 \), then \( x = 0 \)". Let this be \( P \rightarrow Q \).
The contrapositive statement is "If \( x \neq 0 \), then \( x^3 + 4x \neq 0 \)". This is \( \sim Q \rightarrow \sim P \).
We need to prove the contrapositive statement.
Assume \( x \neq 0 \).
Consider the expression \( x^3 + 4x \).
We can factor it as \( x(x^2 + 4) \).
Since we assumed \( x \neq 0 \).
Also, for any real number \( x \), \( x^2 \geq 0 \). So, \( x^2 + 4 \geq 4 \). This means \( x^2 + 4 \neq 0 \).
Since \( x \neq 0 \) and \( x^2 + 4 \neq 0 \), their product \( x(x^2 + 4) \) must also be non-zero.
Therefore, \( x^3 + 4x \neq 0 \).
This proves the contrapositive statement: if \( x \neq 0 \), then \( x^3 + 4x \neq 0 \).
Since the contrapositive is true, the original statement is also true.
In simple words: The contrapositive says: if \( x \) is not zero, then \( x^3 + 4x \) is also not zero. We show this by factoring \( x^3 + 4x \) into \( x(x^2 + 4) \). If \( x \) is not zero, and \( x^2 + 4 \) is also never zero for real numbers, then their product can't be zero. Since the contrapositive is true, the first statement must also be true.

Exam Tip: To prove a statement \( P \rightarrow Q \) by contrapositive, you must prove \( \sim Q \rightarrow \sim P \). Ensure you negate both the conclusion and the hypothesis correctly.

Question 2. Show that the statement "For any real numbers a and b, \( a^2 = b^2 \) implies that \( a = b \)" is not true by giving a counter example?
Answer: Let us consider a counterexample to show that the statement is false.
Let \( a = 1 \) and \( b = -1 \).
Then, \( a^2 = (1)^2 = 1 \).
And \( b^2 = (-1)^2 = 1 \).
So, \( a^2 = b^2 \) is true, as \( 1 = 1 \).
However, \( a = b \) is false, because \( 1 \neq -1 \).
Since we found an instance where \( a^2 = b^2 \) but \( a \neq b \), the original statement "For any real numbers a and b, \( a^2 = b^2 \) implies that \( a = b \)" is not true.
In simple words: If we take \( a = 1 \) and \( b = -1 \), their squares are equal \( (a^2 = 1, b^2 = 1) \), but \( a \) and \( b \) themselves are not equal. This shows the statement is false.

Exam Tip: A single counterexample is enough to disprove a universal statement (a statement that claims something is true for "all" or "any" cases).

Question 3. Show that the following statement is true by the method of contrapositive: p: If x is an integer and \( x^2 \) is even, then x is also even.
Answer: We need to prove the statement "If \( x \) is an integer and \( x^2 \) is even, then \( x \) is also even" using the method of contrapositive.
Let \( P \) be the statement "\( x^2 \) is even" and \( Q \) be the statement "\( x \) is even". The original statement is \( P \rightarrow Q \).
The contrapositive statement is \( \sim Q \rightarrow \sim P \), which means "If \( x \) is not even (i.e., \( x \) is odd), then \( x^2 \) is not even (i.e., \( x^2 \) is odd)".
Let's assume \( x \) is an odd integer.
An odd integer can be represented as \( x = 2n + 1 \) for some integer \( n \).
Now, let's find \( x^2 \):
\( x^2 = (2n + 1)^2 \)
\( x^2 = (2n)^2 + 2(2n)(1) + 1^2 \)
\( x^2 = 4n^2 + 4n + 1 \)
\( x^2 = 4(n^2 + n) + 1 \)
Since \( 4(n^2 + n) \) is a multiple of 4, it is an even number. When 1 is added to an even number, the result is always an odd number.
So, \( x^2 \) is odd.
Thus, we have shown that if \( x \) is odd, then \( x^2 \) is odd. This means "If \( x \) is not even, then \( x^2 \) is not even" is true.
Since the contrapositive statement is true, the original statement "If \( x \) is an integer and \( x^2 \) is even, then \( x \) is also even" is also true.
In simple words: To prove the statement "if \( x^2 \) is even, then \( x \) is even" by contrapositive, we instead prove "if \( x \) is odd, then \( x^2 \) is odd". If \( x \) is odd, we can write it as \( 2n + 1 \). Squaring this gives \( 4n^2 + 4n + 1 \), which is an odd number. Since this is true, the original statement is also true.

Exam Tip: When proving a statement about even and odd numbers, represent an even number as \( 2n \) and an odd number as \( 2n + 1 \) for some integer \( n \).

Question 4. By giving counter examples show that the following statements are not true:
1. p: If all the angles of a triangles are equal, then the triangle is obtuse.
2. q : The equation \( x^2 - 1 = 0 \) does not have a root lying between 0 and 2.
Answer:
1. For statement p: If all the angles of a triangle are equal, then the triangle is obtuse.
A triangle where all angles are equal is called an equilateral triangle.
In an equilateral triangle, each angle measures \( 60^\circ \).
An obtuse angle is an angle greater than \( 90^\circ \).
Since \( 60^\circ \) is not greater than \( 90^\circ \), an equilateral triangle is not obtuse. It is an acute-angled triangle.
Therefore, the statement "If all the angles of a triangle are equal, then the triangle is obtuse" is not true.
In simple words: If all angles of a triangle are equal, each angle is \( 60^\circ \). An obtuse angle is larger than \( 90^\circ \). Since \( 60^\circ \) is not obtuse, the statement is false.

Exam Tip: Clearly define the terms used in the statement (e.g., equilateral, obtuse) before providing your counterexample.

2. For statement q : The equation \( x^2 - 1 = 0 \) does not have a root lying between 0 and 2.
First, let's find the roots of the equation \( x^2 - 1 = 0 \).
\( x^2 = 1 \)
\( \implies x = \sqrt{1} \)
\( \implies x = \pm 1 \)
So, the roots are \( x = 1 \) and \( x = -1 \).
Now, let's check if any of these roots lie between 0 and 2.
The root \( x = 1 \) is indeed a number that lies between 0 and 2 (since \( 0 < 1 < 2 \)).
The statement claims that there is no root lying between 0 and 2, which contradicts our finding.
Therefore, the statement "The equation \( x^2 - 1 = 0 \) does not have a root lying between 0 and 2" is not true.
In simple words: The equation \( x^2 - 1 = 0 \) has roots \( x = 1 \) and \( x = -1 \). Since \( x = 1 \) is clearly between 0 and 2, the statement claiming no roots are in that range is incorrect.

Exam Tip: When checking for roots within an interval, solve the equation completely and then test each root against the given interval boundaries.

Question 5. Which of the following statements are true and which are false? In each case, give a valid reason for saying so.
1. p: Each radius of a circle is a chord of the circle.
2. q : The centre of a circle bisects each chord of the circle.
3. r : Circle is a particular case of an ellipse.
4. s = If x and y are integers such that \( x > y \), then \( -x < -y \).
5. t: \( \sqrt{11} \) is a rational number.
Answer:
1. p: Each radius of a circle is a chord of the circle.
False.
A radius connects the center of the circle to a point on its circumference. A chord connects two distinct points on the circumference of the circle. A radius only has one endpoint on the circle; its other end is the center, which is not on the circle. Therefore, a radius does not fit the definition of a chord.
In simple words: A radius goes from the center to the edge. A chord connects two points on the edge. Since a radius only touches the edge at one point, it can't be a chord. So, false.

Exam Tip: Understand the precise definitions of geometric terms like 'radius' and 'chord' to differentiate between them correctly.

2. q : The centre of a circle bisects each chord of the circle.
False.
The center of a circle only bisects chords that pass through it. These chords are known as diameters. For any other chord that does not pass through the center, the center does not divide it into two equal parts. Only the perpendicular from the center to a chord bisects the chord.
In simple words: The center only cuts chords in half if those chords pass directly through the center (like a diameter). For other chords, the center does not bisect them. So, false.

Exam Tip: Remember that a diameter is a special type of chord, and only diameters are bisected by the circle's center.

3. r : Circle is a particular case of an ellipse.
True.
The general equation of an ellipse centered at the origin is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).
Here, \( a \) and \( b \) represent the lengths of the semi-major and semi-minor axes, respectively.
If we set \( a = b \) (meaning the lengths of both axes are equal), the equation becomes:
\( \frac{x^2}{a^2} + \frac{y^2}{a^2} = 1 \)
Multiplying by \( a^2 \), we get:
\( x^2 + y^2 = a^2 \)
This is the standard equation of a circle centered at the origin with radius \( a \). Thus, a circle is a specific form of an ellipse where both semi-axes are equal.
In simple words: An ellipse has two different "radii" (axes). If these two radii become equal, the ellipse turns into a perfect circle. So, a circle is just a special kind of ellipse. Thus, true.

Exam Tip: Familiarize yourself with the general equations of conic sections (circle, ellipse, parabola, hyperbola) to understand their relationships.

4. s = If x and y are integers such that \( x > y \), then \( -x < -y \).
True.
This statement relates to the properties of inequalities. When both sides of an inequality are multiplied or divided by a negative number, the direction of the inequality sign must be reversed.
Given \( x > y \).
Multiplying both sides by -1, we get \( -x < -y \).
For example, if \( x = 5 \) and \( y = 3 \), then \( 5 > 3 \) is true.
Multiplying by -1, we get \( -5 < -3 \), which is also true.
In simple words: When you multiply both sides of an inequality by a negative number, you must flip the greater than/less than sign. So, if \( x > y \), then \( -x \) will be less than \( -y \). Thus, true.

Exam Tip: Always remember to reverse the inequality sign when multiplying or dividing by a negative number; this is a common error point.

5. t: \( \sqrt{11} \) is a rational number.
False.
A rational number is a number that can be expressed as a simple fraction \( \frac{p}{q} \), where \( p \) and \( q \) are integers and \( q \neq 0 \).
An irrational number cannot be expressed in this way.
The number 11 is a prime number. The square root of any prime number is always an irrational number.
Therefore, \( \sqrt{11} \) cannot be written as a simple fraction and is an irrational number.
In simple words: A rational number can be written as a fraction. The number 11 is a prime number, and the square root of any prime number is always irrational. So, \( \sqrt{11} \) cannot be written as a fraction, meaning it's not rational. Thus, false.

Exam Tip: A key property to remember is that the square root of any non-square integer (especially a prime number) is always irrational.