*According to the latest update on the CBSE Syllabus 2023-24, this chapter has been removed.
The NCERT Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning contain solutions for all Exercise 14.5 questions. Students can download the NCERT Maths Solutions of Class 11 and practise offline to score more. By practising these questions, students can build their confidence level and boost their reasoning skills.
NCERT Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning Exercise 14.5
Access Other Exercise Solutions of Class 11 Maths Chapter 14 Mathematical Reasoning
Exercise 14.1 Solutions: 2 Questions
Exercise 14.2 Solutions: 3 Questions
Exercise 14.3 Solutions: 4 Questions
Exercise 14.4 Solutions: 4 Questions
Miscellaneous Exercise Solutions: 7 Questions
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NCERT Solutions for Class 11 Maths Chapter 14
Access Solutions for Class 11 Maths Chapter 14 Exercise 14.5
1. Show that the statement p: “If x is a real number such that x3 + 4x = 0, then x is 0” is true by
(i) direct method
(ii) method of contradiction
(iii) method of contrapositive
Solution:
Let p: ‘If x is a real number, such that x3 + 4x = 0, then x is 0.’
q: x is a real number, such that x3 + 4x = 0
r: x is 0
(i) We assume that q is true to show that statement p is true and then show that r is true.
Therefore, let statement q be true.
Hence, x3 + 4x = 0
x (x2 + 4) = 0
x = 0 or x2 + 4 = 0
Since x is real, it is 0.
So, statement r is true.
Hence, the given statement is true.
(ii) By contradiction, to show statement p to be true, we assume that p is not true.
Let x be a real number, such that x3 + 4x = 0 and let x ≠ 0.
Hence, x3 + 4x = 0
x (x2 + 4) = 0
x = 0 or x2 + 4 = 0
x = 0 or x2 = -4
However, x is real. Hence, x = 0, which is a contradiction since we have assumed that x ≠ 0.
Therefore, the given statement p is true.
(iii) By the contrapositive method, to prove statement p to be true, we assume that r is false and prove that q must be false.
∼r: x ≠ 0
Clearly, it can be seen that
(x2 + 4) will always be positive.
x ≠ 0 implies that the product of any positive real number with x is not zero.
Now, consider the product of x with (x2 + 4)
∴ x (x2 + 4) ≠ 0
x3 + 4x ≠ 0
This shows that statement q is not true.
Hence, it proved that
∼r ⇒ ∼q
Hence, the given statement p is true.
2. Show that the statement “For any real numbers a and b, a2 = b2 implies that a = b” is not true by giving a counter-example.
Solution:
The given statement can be written in the form of ‘if then’ is given below.
If a and b are real numbers, such that a2 = b2, then a = b.
Let p: a and b are real numbers, such that a2 = b2
q: a = b
The given statement has to be proved false. To show this, two real numbers, a and b, with a2 = b2 , are required, such that a ≠ b.
Let us consider a = 1 and b = – 1
a2 = (1)2
= 1 and
b2 = (-1)2
= 1
Hence, a2 = b2
However, a ≠ b
Therefore, it can be concluded that the given statement is false.
3. Show that the following statement is true by the method of contrapositive.
p: If x is an integer and x2 is even, then x is also even.
Solution:
Let p: If x is an integer and x2 is even, then x is also even.
Let q: x is an integer and x2 is even.
r: x is even
By the contrapositive method, to prove that p is true, we assume that r is false and prove that q is also false.
Let x is not even.
To prove that q is false, it has to be proved that x is not an integer or x2 is not even.
x does not even indicate that x2 is also not even.
Hence, statement q is false.
Therefore, the given statement p is true.
4. By giving a counter example, show that the following statements are not true.
(i) p: If all the angles of a triangle are equal, then the triangle is an obtuse angled triangle.
(ii) q: The equation x2 – 1 = 0 does not have a root lying between 0 and 2.
Solution:
(i) Let q: All the angles of a triangle are equal.
r: The triangle is an obtuse angled triangle.
The given statement p has to be proved false.
To show this, the required angles of a triangle should not be obtuse angles.
We know that the sum of all the angles of a triangle is 1800. Therefore, if all three angles are equal, then each angle measures 600, which is not obtuse.
In an equilateral triangle, all angles are equal. However, the triangle is not an obtuse-angled triangle.
Hence, it can be concluded that the given statement p is false.
(ii) The given statement is
q: The equation x2 – 1 = 0 does not have a root lying between 0 and 2.
This statement has to be proven false.
To show this, let us consider
x2 – 1 = 0
x2 = 1
x = ± 1
One root of equation x2 – 1 = 0, i.e., the root x = 1, lies between 0 and 2.
Therefore, the given statement is false.
5. Which of the following statements are true and which are false? In each case, give a valid reason for saying so.
(i) p: Each radius of a circle is a chord of the circle.
(ii) q: The centre of a circle bisects each chord of the circle.
(iii) r: Circle is a particular case of an ellipse.
(iv) s: If x and y are integers, such that x > y, then –x < –y.
(v) t: √11 is a rational number.
Solution:
(i) The given statement p is false.
By the definition of the chord, it should intersect the circle at two distinct points.
(ii) The given statement q is false.
The centre will not bisect that chord which is not the diameter of the circle.
In other words, the centre of a circle only bisects the diameter, which is the chord of the circle.
(iii) The equation of an ellipse is,
If we put a = b = 1, then we get
x2 + y2 = 1, which is an equation of a circle.
Hence, the circle is a particular case of an ellipse.
Therefore, statement r is true.
(iv) x > y
By a rule of inequality,
-x < – y
Hence, the given statement s is true.
(v) 11 is a prime number.
We know that the square root of any prime number is an irrational number.
Therefore, √11 is an irrational number.
Hence, the given statement t is false.
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