NCERT Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning Ex 14.5

# NCERT Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning Ex 14.5

## NCERT Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning Ex 14.5

NCERT Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning Ex 14.5 are the part of NCERT Solutions for Class 11 Maths. Here you can find the NCERT Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning Ex 14.5.

### Ex 14.5 Class 11 Maths Question 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,
(iii) method of contrapositive.

Solution:
The given compound statement is of the form “if p then q”
p: x Ïµ R such that x3 + 4x = 0
q: x = 0
(i) Direct method:
Let us assume that p is true, then
x Ïµ R such that x3 + 4x = 0
x Ïµ R such that x(x2 + 4) = 0
x Ïµ R such that x = 0 or x2 + 4 = 0
x = 0

q is true.
So, when p is true, q is true.
Thus, the given compound statement is true.

Let us assume that p is true and q is false, then
x Ïµ R such that x3 + 4x = 0
x Ïµ R such that x(x2 + 4) = 0
x Ïµ R such that x = 0 or x2 + 4 = 0
x = 0.
which is a contradiction. So, our assumption that x ≠ 0 is false. Thus, the given compound statement is true.

(iii) Method of contrapositive: Let us assume that q is false, then x ≠ 0
x Ïµ R such that x3 + 4x = 0
x Ïµ R such that x = 0 or x2 + 4 = 0
The statement q is false, so x ≠ 0. So, we have,
x Ïµ R such that x2 = -4
Which is not true for any x Ïµ R.
p is false.
So, when q is false, p is false.
Thus, the given compound statement is true.

### Ex 14.5 Class 11 Maths Question 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 compound statement is of the form “if p then q”
Let us assume that p is true, then a, b Ïµ R such that a2 = b2
Let us take a = -3 and b = 3
Now, a2 = b2, but a ≠ b
So, when p is true, q is false.
Thus, the given compound statement is not true.

### Ex 14.5 Class 11 Maths Question 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:
The given compound statement is of the form “if p then q”
p: x Ïµ Z and x2 is even.
q: x is an even integer.
Let us assume that q is false, then x is not an even integer.
x is an odd integer.
x2 is an odd integer.
p is false.
So, when q is false, p is false.
Thus, the given compound statement is true.

### Ex 14.5 Class 11 Maths Question 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) Since the triangle is an obtuse angled triangle then one angle measures greater than 90°.
Let one angle = 100°
Also, all the angles of the triangle are equal.
Sum of all angles of the triangle is 300°, which is not possible.
Thus, the given compound statement is not true.

(ii) We see that x = 1 is a root of the equation x2 – 1 = 0, which lies between 0 and 2. Thus, the given compound statement is not true.

### Ex 14.5 Class 11 Maths Question 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) A chord of a circle is a line whose two endpoints lie on the circle. So, the radius of a circle is not a chord of the circle. Thus, the given statement is false.
(ii) The centre of a circle bisects chord of circle when the chord is diameter of circle. When the chord is other than diameter, then the centre of the circle does not lie on the chord. Thus, the given statement is false.
(iii) In the equation of an ellipse if we put a = b, then we get an equation of a circle.
Thus, the given statement is true.
(iv) It is given that x, y Ïµ Z such that x > y. Multiplying both sides by negative sign, we get x, y Ïµ Z such that -x < -y.
Thus, the given statement is true.
(v) Since √11 cannot be expressed in the form a/b, where a and b are integers and b ≠ 0. Thus, the given statement is false.

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