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H Harsh Kankaria
The negation of the statement is: There exists a real number x such that neither x > 1 nor x < 1.

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H Harsh Kankaria
The statement is False. Since 11 is a prime number, therefore  is irrational.

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H Harsh Kankaria
The statement is True. Give, x>y Multiplying  -1 both sides (-1)x<(-1)y   -x < -y (When -1 is multiplied to both L.H.S & R.H.S, sign of inequality changes) By the rule of inequality.

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H Harsh Kankaria
The statement is True. In the equation of an ellipse if we put a = b, then it is a circle.  

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H Harsh Kankaria
The statement is False.  A chord is a line segment intersecting the circle in two points. But it is not necessary for a chord to pass through the centre.  

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H Harsh Kankaria
The statement is False. By definition, A chord is a line segment intersecting the circle in two points. But a radius is a line segment joining any point on circle to its centre.   

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H Harsh Kankaria
Given, The equation does not have a root lying between 0 and 2. Let x = 1 Hence 1 is a root of the equation . But 1 lies between 0 and 2. Hence the given statement is not true.  

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H Harsh Kankaria
We know, Sum of all the angles of a triangle = If all the three angles are equal, then each angle is  But  is not an obtuse angle, and hence none of the angles of the triangle is obtuse. Hence the triangle is not an obtuse-angled triangle. Hence the given statement is not true.

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H Harsh Kankaria
Given, If x is an integer and is even, then is also even.  Let, p : x is an integer and is even q:  is even In order to prove the statement “if p then q”  Contrapositive Method:  By assuming that q is false, prove that p must be false. So, q is false: x is not even  x is odd  x = 2n+1 (n is a natural number) Hence  is odd  is not even  Hence p is false. Hence the given statement is true.

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H Harsh Kankaria
Given, For any real numbers a and b,  implies that . Let a = 1 & b = -1 Now, = 1  = 1 But a  b Hence  does not imply that . Hence the given statement is not true.

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H Harsh Kankaria
If is a real number such that , then is 0 : (if p then q) p: x is a real number such that . q: x is 0. In order to prove the statement “if p then q”  Contrapositive Method:  By assuming that q is false, prove that p must be false. So, q is false:   x.(Positive number)  0.(Positive number) Therefore p is false.

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H Harsh Kankaria
If is a real number such that , then is 0 : (if p then q) p: x is a real number such that . q: x is 0. In order to prove the statement “if p then q”  Contradiction:  By assuming that p is true and q is false. So, p is true:  There exists a real number x such that  q is false:  Now,  Hence, x = 0 But we assumed . This contradicts our assumption. Therefore q is true.

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H Harsh Kankaria
If is a real number such that , then is 0 : (if p then q) p: x is a real number such that . q: x is 0. In order to prove the statement “if p then q”  Direct Method:  By assuming that p is true, prove that q must be true. So, p is true:There exists a real number x such that  Hence, x = 0 Therefore q is true.
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