TOPICS
Exercise - 14.5

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,
(ii) method of contradiction,
(iii) method of contrapositive

Solution :-
  p: “If x is a real number such that x3 + 4x = 0, then x is 0”. 
  Let q: x is a real number such that x3 + 4x = 0
  r: x is 0.
    
(i) To show that statement p is true, we assume that q 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 However, since x is real, it is 0.
  Thus, statement r is true.
  Therefore, the given statement is true.
    
(ii) To show statement p to be true by contradiction, we assume that p is not true. 
  Let x be a real number such that x3 + 4x = 0 and let x is not 0.
  Therefore, x3 + 4x = 0 x (x2 + 4) = 0
  x = 0 or x2 + 4 = 0 x = 0 or x2 = – 4
  However, x is real. Therefore, x = 0, which is a contradiction since we have assumed that x is not 0.
  Thus, the given statement p is true.
    
(iii) To prove statement p to be true by contrapositive method, we assume that r is false and prove that q must be false.
  Here, r is false implies that it is required to consider the negation of statement r. This obtains the following statement.
  ∼r: x is not 0.
  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. Let us consider the product of x with (x2 + 4).
  Therefore, x (x2 + 4) ≠ 0 ⇒ x3 + 4x ≠ 0
  This shows that statement q is not true.
  Thus, it has been proved that
  ∼r ⇒ ∼q
  Therefore, the given statement p is true.
   

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 statement can be written in the form of “if-then” as follows. 
  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. 
  For this purpose, it has to be proved that if p, then ∼q. 
  To show this, two real numbers, a and b, with a² = b² are required such that a ≠ b.
  Let a = 1 and b = –1
  a² = (1)² = 1 and b²= (– 1)² = 1
  Therefore, a² = b². However, a ≠ b.
  Thus, it can be concluded that the given statement is false.
   

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 :-
  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.
  To prove that p is true by contrapositive method, 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 is not even implies that x² is also not even.
  Therefore, statement q is false. 
  Thus, the given statement p is true.
    

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) The given statement is of the form “if q then r”. 
  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. 
  For this purpose, it has to be proved that if q, then ∼r.
  To show this, angles of a triangle are required such that none of them is an obtuse angle.
  It is known that the sum of all angles of a triangle is 180°. 
  Therefore, if all the three angles are equal, then each of them is of measure 60°, which is not an obtuse angle.
  In an equilateral triangle, the measure of all angles is equal. 
  However, the triangle is not an obtuse-angled triangle.
  Thus, it can be concluded that the given statement p is false. 
    
    
(ii) The given statement is as follows.
  q: The equation x² – 1 = 0 does not have a root lying between 0 and 2.
  This statement has to be proved false. 
  To show this, a counter example is required. 
  Consider x² – 1 = 0
  x² = 1
  x = ± 1
  One root of the equation x² – 1 = 0, i.e. the root x = 1, lies between 0 and 2. 
  Thus, the given statement is false.
   

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) The given statement p is false.
  According to the definition of chord, it should intersect the circle at two distinct points.
    
(ii) The given statement q is false.
  If the chord is not the diameter of the circle, then the centre will not bisect that chord.
  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, x²/a² + y²/b² = 1
  If we put a = b = 1, then we obtain x2 + y2 = 1, which is an equation of a circle 
  Therefore, circle is a particular case of an ellipse. 
  Thus, statement r is true.
    
(iv) x > y
  –x < –y (By a rule of inequality) 
  Thus, the given statement s is true.
    
(v) 11 is a prime number and we know that the square root of any prime number is anirrational number. 
  Therefore, √11 is an irrational number. 
  Thus, the given statement t is false.
   
CLASSES

Connect with us:

Copyright © 2015-17 by a1classes. All Rights Reserved.

www.000webhost.com