cls jeead-14-15 xi mat target-4 set-1 chapter-14

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SECTION - A

School/Board Exam. Type Questions

Very Short Answer Type Questions :

1. (i) p : Grass is green.

(ii) q : There are 366 days in a leap year.

(iii) r : Product of an even and odd number is an even number.

2. (i) p : He is a chemistry graduate

(ii) q : It x and y are two positive integer then x – y is always greater than zero.

(iii) r : How much old he is

3. Definition : If p is a statement, then the negation of p is also a statement and is denoted by ~p, and read as

not p.

Example : p : 4 32 is an irrational number

So, ~ : 4 32p is not an irrational number

4. Component statements are

q : Aeroplane flies in the air

r : Ships sails on the water

5. (i) p is false. In p basic connective “and” is used, and one component statement is false, so p is false.

(ii) q is true. In q basic connective “or” is used and in this case statement is true if any one of it’s component

statements are true.

6. (i) Inclusive OR

(ii) Exclusive OR

7. If a person cannot apply for post of P.O. then he has not a bachelor degree.

8. If x is complex number then it is not real.

9. Volume of a cone is one third of cylinder if and only if base radius and height of cone and cylinder are same.

10. n = 2, is a prime number and even number also.

Solutions

Chapter 14

Mathematical Reasoning

36 Mathematical Reasoning (School/Board Exams.) Solutions


Short Answer Type Questions :

11. (i) p : There are 33 days in a month. We know that a month cannot have 33 days, so above sentence is

false, so we can say this sentence is statement.

(ii) p : The product of 5 and 6 is 30. This sentence is true so it is statement.

(iii) p : Sum of two positive numbers is always positive. This sentence is always true, so it is statement.

12. (i) Call the police. It is an order, so it is not a statement.

(ii) If x and y are integers, then x

y is an integer. This sentence is sometimes true, sometimes false, so it

is not a statement.

(iii) Chandigarh is far from here. This sentence does not specify particular place, so it is not a statement.

13. (i) ~p : 0 is not a natural number. (True)

(ii) ~p : 70 is not a multiple of 20. (True)

(iii) ~p : Diameter is not the longent chord of circle.

14. Yes, they are negation of each other. Since negation of a statement p is also a statement, and we can form

negation of statement using phrases like “It is not the case” or “ It is false that” before p or, if possible by

inserting in p the word “not”.

15. ~p : It is false that square of an odd integer is of the form 8q + 1, for some positive integer q.

or

~p : It is not the case that square of an odd positive integer is of the form of 8q + 1, for some positive integer k.

or

~p : square of an odd positive integer is not of the form 8q + 1, for some positive integer q.

16. Component statements of p are

q : Taj Mahal is in India (True)

r : Niagara falls is in U.S.A. (True)


: 12q is a rational number (False)

: 12r is an irrational number (True)

: 12s is a complex number (False)

18. Compound statement using connective AND “rational number follows commutative property for addition and

multiplication”.

Compound statement using connective OR “rational number follows commutative property for addition or

multiplication”.

19. (i) By using basic connective “And”

All sides and all angles of two congruent triangles are equal

(ii) By using basic connective “OR”

All sides or all angles of two congruent triangles are equal

37(School/Board Exams.) Solutions Mathematical Reasoning



q : Chord of a circle lies within the circle (True)

r : Tangent of circle lies both outside and inside of circle (False)

In compound statement p basic connective “And” is used and it is true only when all component statement

are true.

So, compound statement p is false.


q : From a point outside the circle we can draw two tangents to a circle (True)

r : From a point inside the circle we can draw one tangent to circle (False)

In compound statement p basic connective “OR” is used, so p will be true if at least one component statement

is true, so compound statement is true.

22. Compound statements of p are

q : Solution of a quadratic equation ax2 + bx + c = 0 is real

r : Solution of a quadratic equation ax2 + bx + c = 0 is complex

We know that when q is true r is false and when r is true q is false, both cannot be true at a time. So Exclusive

OR is used.

23. (a) In p quantifier is “for every”.

~p : For every natural number x, 7x is not greater than 7.

(b) In q quantifier is “There exists”

~q : There does not exist a tangent which is chord to the circle.

24. (a) p : If two circle touch other then we can draw three common tangents to both circle.

or

If three common tangent can be drawn to two circles then they touch each other.

(b) q : If chord of a circle subtends an angle 90º in alternate segment then it is diameter.

25. A convex polygon is pentagon if and only if it has 5 diagonals.

26. (i) If two chords are not diameter then they will not bisect each other.

(ii) If corresponding angles of two triangles are not equal then they are not similar.

27. (i) If radius of two circles are equal then they are congruent.

(ii) If two polygons overlap each other then they are congruent.

28. In order to prove that “p and q” is true, we have to follow following steps.

Step-1 : Show that the statement p is true.

Step-2 : Show that the statement q is true.

In order to show that “p or q” is true, one must consider the following

Case-1: By assuming that p is false, show that q must be true.

Case-2 : By assuming that q is false, show that p must be true.



29. Numerical value of area of a circle and area of a triangle can be equal but they are not congruent.

30. x3 – 6x2 + 11x – 6 = 0

(x – 1) (x – 2) (x – 3) = 0

So, given equation has three roots, so it will cut x-axis at three times.

Long Answer Type Questions :

31. Following are four types of sentences which are never considered as statement

Type-1 : Sentences involving variable time such as “Today”, tomorrow” or “Yesterday” are not statements,

because it is not known what time is referred here.

Example : Tomorrow is Friday.

Type-2 : Sentences which are an order are never considered as statements.

Example : Bring a glass of water

Type-3 : Sentences which are exclamation are not considered as statements.

Example : Too smooth.

Type-4 : Sentences with pronouns unless a particular person is referred to.

Example : He is a football player

32. (a) p contains pronoun “here” which refers to variable place, here it doesnot mention the place from which

we are measuring the distance of Hisar.

(b) q is true only on Sunday but not on other days.

(c) r is exclamation so it is not a sentence

33. (i) ~p : It is false that everyone in Europe plays football

or

~p : It is not the case that everyone in Europe plays football

or

~p : Everyone in Europe does not play football.

~p says that at least one person in Europe does not play football.

(ii) ~q : It is false that every quadratic equation has two real roots

or

~q : It is not the case that every quadratic equation has two real roots

or

~q : Every quadratic equation has not two real roots.

~q Says that at least one quadratic equation exists whose roots are not real

34. Definition : A compound statement is a statement which is made up of two or more statements. In this case,

each statement is called a component statement.

Examples are

(i) Ganga is a river and Mansarover is a lake.

(ii) 8 is a natural number of integer.

(iii) Product of a rational and an irrational number is real or irrational number.



35. We have the following rules regarding the connective “AND”.

Rule 1 : The compound statement with “And” is true if all its component statements are true.

Example : p : 110 is a multiple of 2, 5 and 55.

Component statements are

q : 110 is a multiple of 2 (True)

r : 110 is a multiple of 5 (True)

s : 110 is a multiple of 55 (True)

∵ All component statements are true so statement p is True.

Rule-2 : The compound statement with “And” is false if any of its component statements is false (This includes

the case that some of its component statements are false or all of it’s component statements are false).

Example, p : 110 is multiple of 2, 3 and 5


q : 110 is a multiple of 2 (True)

r : 110 is a multiple of 3 (Flase)

s : 110 is a multiple of 5 (True)

Since all the component statements are not true. So p is false.

36. We have the following rules regarding the connective “OR”

Rule-1 : A compound statement with an “OR” is true when one component statement is true or if both the

component statements are true.

Example, p : 15 is an odd or prime number.

Component statements of p are

q : 15 is an odd number (True)

r : 15 is a prime number (False)

One of the component statement is true, so p is true

Rule-2 : A compound statement with an “OR” is false when the component statement are false.

Example : p : 15 is an even or prime number.

Component statements of p are

q : 15 is an even number (False)

r : 15 is a prime number (False)

Since both component statements are false, so p is false.

37. When connective “OR” is used in the sense of statement “p or q or both, i.e., at least one of the two alternatives

we call it inclusive “OR”.

Example : A person who is graduate or has an experience of five years can apply for post of sales executive.

Here, in above example, A person who is graduate p and has an experience of five years can also apply, as

well as a person who is only graduate or only having a experience of five years can apply for sales executive.

So, this type of statements uses inclusive “OR”.



Now, when connective “OR” is used in the sense of statement “p or q but not both” i.e., exactly one of the

two alternative occurs is called exclusive “OR”

Example : p : All integers are even or odd.

Here all integers cannot be both even or odd.

If they are even then they are not odd and if they are odd then they are not even.

38. Following are the five different ways of writing p which convey the same meaning.

(i) Two circles are apart from each other implies we can draw four common tangents to them.

(ii) Knowing that two circles are apart from each other is sufficient to conclude that we can draw four common

tangents to them.

(iii) Two circles are apart from each other only if we can draw four common tangents to them.

(iv) For two circles to be apart from each other it is necessary condition that we can draw four common

tangents to them.

(v) If we cannot draw four common tangents to two circles then the two circles are not apart from each other.


q : A quadratic equation has real roots.

r : A quadratic equation has complex roots.

When statements q is true, r becomes false and when r is true q become false. So type of OR used in the

statement p is exclusive “OR”

So, at least one of the compound statement is true at a time

So, compound statement is true.


q : x3 is odd

r : x is odd.

We have to check whether the statement q r is true or not, that is, by checking its contrapositive

Statement i.e., ~q ~r.

Now, ~q : It is false that x3 is odd, then x3 is even

Then x3 = 2n for some integer n.

∵ x3 is even so x is also even.

It shows that ~q is true.

41. Suppose p is not true. Then there exist an 0,2

x⎡ ⎤ ⎢ ⎥

⎣ ⎦ for which sinx + cosx < 1.

Since 0,2

x⎡ ⎤ ⎢ ⎥

⎣ ⎦, neither sinx nor cosx is negative,

So, 0 sinx + cosx < 1

02 (sinx + cosx)2 < 12

02 sin2x + cos2x + 2sinx.cosx < 12

0 1 + 2sinx.cosx < 1



So, 1 + 2sinx cosx < 1

Subtracting 1 from both sides gives 2sinx.cosx < 0, but This contradicts the fact that neither sinx or cosx is

negative. So p is true

42. Let x be non-zero rational number and y be an irrational number. Then, we have to show that xy is an irrational

number. If possible let xy be a rational number. Since the quotient of two non-zero rational numbers is a rational

number.

So, xy is a rational number and x is a rational

xy

x

⎛ ⎞⎜ ⎟⎝ ⎠

is a rational number

y is a rational number

But, it contradicts the fact that y is an irrational number.

So, our assumption is wrong.

43. Let us assume that 3 is rational.

That is we can find integer a and b ( 0) such that 3a

b (where a and b are co-prime)

So, 3b a

Squaring on both side

3b2 = a2

Since a2 is divisible by 3, so a is also divisible by 3.

Now, we can write

a = 3c [for some integer c]

3 3b c

3b2 = 9c2

b2 = 3c2

So, b is also divisible by 3,

So, a and b both have a common factor 3, which is the contradiction of fact that a and b are co-prime.

So, our assumption is wrong and p is true.

44. Counter example is

a = 12, b = 3, c = 4

12|3*4, but neither 12|3 nor 12|4.

45. Let

p : You pass NTSE Test

q : You will get scholarship

So, given statement is in the form of If p then q. In such case q is necessary condition for p.

So, you will get scholarship is necessary condition.

Also, p is a sufficient condition for q.

To get scholarship it is sufficient condition that you pass NTSE Test.



SECTION - B

NCERT Questions

Exercise-14.1

1. (i) Given sentence is not correct since maximum number of days in a month is 31. So it is false and

according to definition of statement it is statement.

(ii) For some people mathematics is difficult, for some persons it is easy, so it is ambiguous. Hence it is

not statement.

(iii) This sentence is always true, since sum of 5 and 7 is 12, which is greater then 10., Hence, it is

statement.

(iv) In case of even number sentence is true but false when number is odd. So it is both true and false and

cannot be statement.

(v) In case of rhombus and square sentence is true but in other case it is false. So it is both true and false

and cannot be considered as statement.

(vi) It is an order, so it is not statement.

(vii) The product of (–1) and 8 is (–8). So given sentence is false. Hence it is statement.

(viii) The sentence is always true. So it is statement.

(ix) The day which is being referred is not evident from the sentence. Hence, it is not a statement.

(x) Set of complex number consist real and imaginary number. So given sentence is true. Hence it is

statement.

2. Following are three examples which are not sentences.

(i) She is a carpenter/

Reason : It is not evident from the sentence as to whom “she” is referred to . So it is not a statement.

(ii) Too large!

Sentence is exclamatory so it is not a sentence.

(iii) What is your name?

Sentence is a question so it is not a statement.

Exercise-14.2

1. (i) Chennai is not the capital of Tamil Nadu.

(ii) It is false that 2 is not a complex number.

(iii) It is not the case that all triangles are not equilateral triangle.

(iv) The number 2 is not greater than 7.

(v) It is false that every natural number is an integer.

2. (i) Negation of “The number x is not a rational number is “ It is false that number x is not rational number.

So given pair of statement is not negation of each other.

(ii) Negation of “The number x is a rational number” is “The number x is not a rational number. So, given

pair of statement is not negation of each other.

3. (i) Component statements are

p : Number 3 is a prime number (True)

q : Number 3 is odd (True)

Both component statements are true.



(ii) The component statements are as follows.

p : All integers are positive (False)

q : All integers are negative (False)

Both component statements are false

(iii) Component statements are

p : 100 is divisible by 3. (False)

q : 100 is divisible by 11. (False)

r : 100 is divisible by 5. (True)

Exercise-14.3

1. (i) Connecting word is “And”


p : All rational numbers are real.

q : All real numbers are not complex.

(ii) Connecting is “OR”


p : Square of an integer is positive.

q : Square of an integer is negative.

(iii) Connecting word is “And”.


p : The sand heats up quickly in the sun.

q : The sand does not cool down fast at night.

(iv) Connecting word is “AND”


p : x = 2 is a root of the equation 3x2 – x – 10 = 0

q : x = 3 is a root of the equation 3x2 – x – 10 = 0

2. (i) Quantifier is “There exists”.

It is false that there exists a number which is equal to its square.

(ii) Quantifier is “For every”.

For every real number x, x is not less than x + 1.

(iii) Quantifier is “There exists”.

It is false that there exists a capital for every state in India.

3. The negation of statement (i) is

There exists real number x and y such that x + y y + x. It is not the same as statement (ii).

So, the given statements are not the negation of each other.

4. (i) Here, “or” is exclusive since it is not possible for the Sun to rise and moon to set together.

(ii) Inclusive “or” since a person can have both a ration card and a passport to apply for a driving licence.

(iii) Exclusive “or” since all integers cannot be both positive and negative.



Exercise-14.4

1. Given statement is

“If a natural number is odd, then it’s square is also odd.

Following are the five different ways of conveying the some meaning/

(i) A natural number is odd implies that it’s square is odd.

(ii) Knowing that a natural number is odd is sufficient to conclude that it’s its square is odd.

(iii) A natural number is odd only if it’s square is odd.

(iv) When a natural number is odd then it’s square is necessarily odd.

(v) If square of a natural number is not odd then it is not a natural odd number.

2. (i) Contrapositive : If x is not prime number then it is not odd.

Converse : If x is an odd number then it is prime.

(ii) Contrapositive : If two lines intersect in the same plane then they are not parallel.

Converse : If two lines do not intersect in the same plane then they are parallel.

(iii) Given statement can be written as

If something is cold then it’s temperature is low

Contrapositive : If temperature of something is not low then it is not cold.

Converse : If something has low temperature then it is cold.

(iv) Given statement as

“If you do not know how to reason deductively then you cannot comprehend geometry.

Contrapositive : If you can comprehend geometry then you know how to reason deductively.

Converse : If you cannot comprehend geometry then you do not know how to reason deductively.

(v) Girent statement can be written is

If x is an even number then it is divisible by 4.

Contrapositive : If a number x is not divisible by 4 then it is not an even number

Converse : If a number x is divisible by 4 then it is an even number.

3. (i) If you get a job, then your credentials are good.

(ii) If the Banana tree stays warm for a month, then it will bloom.

(iii) If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

(iv) If you want to get an A+ in the class, then you do all the exercises of the book.

4. (a) Let p : You live in Delhi

q : you have winter clothes

So, given statement is in the form if p then q

(i) It is in the form if ~q then ~p, so it is contrapositive.

(ii) It is in the form if q then p. So it is converse.

(b) Let p : A quadrilateral is parallelogram.

q : Diagonals of a quadrilateral bisect each other.

Given statement is in the form if p then q.



Now,

(i) If ~q then ~p form, so contrapositive.

(ii) If q then p form, so converse.

Exercise-14.5

1. Let q : x is a real number such that x3 + 4x = 0

r : x is 0.

So, the statement p is in the form “If q then r”

(i) Direct method : In this method we assume that q is true and prove that r is also true.

Let q be true

x3 + 4x = 0

x(x2 + 4) = 0

x = 0 or x2 + 4 = 0

However, since x is real, it is 0

So, statement r is true.

(ii) Method of contradiction : In this method, we assume that p is not true.

So, let x be a real number such that x. + 4x = 0 and let x is not 0

So, x3 + 4x = 0

x(x2 + 4) = 0

x = 0 or x2 = –4

∵ x is real, so x = 0, which is contradiction since we have assumed that x is not 0. So, the given

statement p is true.

(iii) Contrapositive method : In this method we assume that r is false and prove that q must be false.

So, we assume that x is not 0 (~r)

Now,

x3 + 4x = 0

x(x2 + 4x) = 0

(x2 + 4) is always positive, and x is not equal to zero,

So, x(x2 + 4) 0, so q is also not true.

Hence, statement p is valid.

2. Let two real numbers a = 3 and b = –3

then a2 = 9

and b2 = 9

So, a2 = b2

But a b [∵ a = 3 and b = –3]

So, given statement is false.



3. Contrapositive method :

Let x is not even i.e., x is odd

So, x = 2n + 1, where n is an integer

then x2 = (2n + 1)2

= 4n2 + 1 + 4n

= 4n(n + 1) + 1

= 2.2(n + 1) + 1

= Odd integer

So, x2 is even is false.

So, we proved that assuming that q is false we prove that p is also false.

4. (i) If all angles of a triangles are equal then it is equilateral triangle and equilateral triangles has all sides

equal, so it cannot be obtuse angle triangle.

Let q : All the angles of a triangle are equal.

r : The triangle is an obtuse angled triangle.

So, by above argument we proved that. If q then ~r.

So statement p is false.

(ii) Putting x = 1 in x2 – 1

(1)2 – 1 = 0

So, 1 is the root of equation and it is lying between 0 and 2.

So given statement q is false.

5. (i) p is false. Since chord is a line segment which joins any two distinct points liying on the circumference

of circle and radius is the line segment that joins centre of the circle to any point on the circumference

of the circle.

(ii) q is false. Centre of circle bisects only those chords which are diameter.

(iii) The equation of an ellipse is

2 2

2 21

x y

a b , if we put a = b = 1, then we obtain x2 + y2 = 1, which is an

equation of a circle. So circle is a particular case of an ellipse. So statement r is true.

(iv) s is true

∵ x > y

–x < –y (By a rule of inequality)

So, s in true

(v) t is false, 11 is not a perfect square, so 11 is an irrational number

Miscellaneous Exercise

1. (i) ~p : It is false that for every positive real number x, the number x – 1 is also positive.

(ii) ~q : It is not the case that all cats scratch.

(iii) ~r : It is false that for every real x, either x > 1 or x < 1.

(iv) ~s : There does not exist a number x such that 0 < x < 1.



2. (i) Statement p can be written as p : If a positive integer is prime, then it has no divisors other than 1 and

itself

Converse : If a positive integer has no divisors other then 1 and itself then it is prime.

Contrapositive : If a positive integer has divisors other than 1 and itself, then it is not prime.

(ii) The given statement can be written as follow, q : If it is a sunny day, then I go to a beach.

Converse : If I go to a beach then it is a sunny day.

Contrapositive : If I don’t go to a beach then it is not a sunny day.

(iii) Converse : If you feel thirsty then it is hot outside.

Contrapositive : If you do not feel thirsty then it is not hot outside.

3. (i) If you log on to the server, then you have a password

(ii) If it rains, then there is a traffic jam.

(iii) If you can access the website, then you pay a subscription fee.

4. (i) You watch television if and only if your mind is free.

(ii) You get an A grade if and only if you do all the homework regularly.

(iii) A quadrilateral is equiangular if and only if it is a rectangle.

5. p : 25 is a multiple of 5 (True)

q : 25 is a amultiple of 8 (False)

Compound statement using “And” is

r : 25 is a multiple of 5 and 8

All component statements of r is not true, so compound statement is not true.

Compound statement using “OR” is

s : 25 is a multiple of 5 or 8.

One component statement is true, so compound statement is true.

6. (i) Let x be a rational number and y be an irrational number.

Let the sum of (x + y) is rational.

Now, we know that difference of two rational number is always a rational number.

(x + y) is a rational number and x is a rational numbers

(x + y) – x is a rational number

y is a rational number

If contradicts the fact that y is an irrational number

So, (x + y) is an irrational number

So, given statement p is true.

(ii) Assuming that n is a real number with n > 3, but n2 > 9 is not true.

So, n2 < 9

Now,

∵ n > 3

Squaring both side

n2 > (3)2

n2 > 9, which is contradiction, since we have assumed that n2 < 9.

So the given statement is true.



7. Following are the five different way of writing statement p.

(i) A triangle is equiangular implies that it is an obtuse-angled triangle.

(ii) A triangle is equiangular only if it is an obtuse-angled triangle.

(iii) For a triangle to be equiangular, it is necessary that the triangle is an obtuse-angled triangle.

(iv) For a triangle to be an obtuse-angled triangle, it is sufficient that the triangle is equiangular.

(v) If a triangle is not an obtuse-angled triangle, then the triangle is not equiangular.

SECTION - C

Model Test Paper

Very Short Answer Type Questions :

1. (a) Is statement. Sum of all exterior angle of a convex polygon is 360º. So, sentence is false

(b) It is not sentence, in month of December, this statement is true, but for other months it is false, so it

is ambiguous and cannot be categorized as sentence.

2. Definition : A compound statement is a statement which is made up of two or more statements. In this case,

each statement is called a component statement.

Example : p : Bicycle has two wheels and a car has four wheels.

This statement is made up of two component statement

q : Cycle has two wheels

r : Car has four wheels.

3. Compound statements are

q : 185 is divisible by 5. (True)

r : 185 is divisible by 37. (True)

s : 18 is divisible by 19. (False)

Here connective “And” is used and all the component statements are not true, so it is not true.

4. Example of “exclusive OR” statement

p : A function is even or odd.

Example of “Inclusive OR” statement.

p : A student who has taken mathematics or computer science can apply for M.Sc. (IT).

5. If equal angles of a right angle triangle is 45º each then it is isosceles.

6. –5 is an integer and it is not a whole number

So, p is false.

Short Answer Type Questions :

7. ~p : It is false that all diagonals of a polygon lies inside the polygon.

or

~p : It is not the case that all diagonals of a polygon lies inside the polygon

or

~p : All diagonals of a polygon does not lie inside the polygon.

~p : Can be interpreated as there exist at least one polygon whose diagonals does not lie inside the triangle.


q : Irrational number between two irrational number a and b is ab (True)

r : Rational number between two rational number p and q is 2

p q (True)



Both component statements are true and basic connective used is “OR”

Compound statement p is also true.

9. We have the following rules regarding the connective “And”.

Rule-1 : The compound statement with “And” is true if all it’s component statements are true.

Rule-2 : The compound statement with “And” is false if any of it’s component statements is false, it includes

the case that some of its component statements are false or all of its component statements are false.

10. (a) “Exclusive OR” is used, since a person can avail either a discount of Rs 400 or a T-shirt not both on

purchase of two shirts from a shop.

(b) “Inclusive OR”. Since it’s possible that a number is both integer and natural number.

Example : 4

11. (a) Contrapositive : If a number is not a perfect number then sum of all factors of it is not twice the number

Converse : If a number is called perfect number then sum of all factors of it is twice the number.

(b) Contrapositive : If two numbers x and y will not follow commutative property then they are not rational

number

Converse : If two number x and y will follow commutative property then they are rational number.

12. (i) p is true. Since chord is a line segment which joins any two points on the circumference of circle and

diameter is also satisfy this condition and it is longest.

(ii) q is false. If a quadrilateral is cyclic only then sum of it’s opposite angles is 180º.

13. Check x = 2

(2)3 – 5(2)2 + 6(2) = 8 – 20 + 12

= 20 – 20

= 0

So, 2 is a root of equation. Hence p is not true.

Long Answer Type Questions :

14. Following are the five different ways in which we can write statement p.

(i) A quadrilateral is cyclic implies sum of its opposite angles is 180º.

(ii) Knowing that a quadrilateral is cyclic is sufficient to conclude that sum of its opposite angles is 180º.

(iii) A quadrilateral is cyclic only if sum of its opposite angles are 180º.

(iv) Sum of opposite angle of quadrilateral is 180º is necessary for quadrilateral to be cyclic.

(v) If sum of opposite angles of a quadrilateral is not 180º, then it is not cyclic.

15. We know that contrapositive of the statement “ if p, then q” is “if ~q, then ~p”.

Let both x, y z are not even i.e., they are odd

Then x = 2m + 1, where m is an integer.

y = 2n + 1, where n is an integer

Now,x × y = (2m + 1) (2n + 1)

= 4mn + 2m + 2n + 1

= 2.(2m + m + n) + 1

= 2� + 1 (where � = 2m + m + n is an integer)

So, product of xy is odd, it is not even.

So, given statement is true, since we proved that it ~q then ~p.



16. Let square root of n be rational number p

q such that p and q have no common factor and q 0, p , q z

Now, p

nq

2

2

2

pn

q

p2 = nq2

So, n is a factor of p2

n is a factor of p.

Let p = nm for some natural number m.

Then p = nm ...(i)

p2 = n2m2

nq2 = n2m2

q2 = nm2

n is factor of q2

n is factor q.

But n is a factor of p and n is a factor of q means n is a factor of both p and q. This contradicts our assumption

that p and q have no common factor. This means that our assumption is wrong.

Hence, n cannot be a rational number

i.e., n is irrational.

If n is not a perfect square.

17. (a) If discriminant of a quadratic equation is negative then it’s roots are non real.

(b) If nth term of a progression is linear than it will be in A.P.

18. Let x be an irrational number. Then we have to show that –x is also an irrational number. If possible, let –x

be a rational number.

We know that negative of a rational number is also a rational number.

So, –x is a rational number

–(–x) is a rational number

x is a rational number

It contradicts the fact that x is irrational.

Hence, –x is an irrational number and statement p is true.

19. The necessary and sufficient condition that roots of a quadratic equation are equal is that its discriminant must

be equal to zero.

20. Let statements

p : A function is bijective

q : A function is both one-one and onto.

So, the given statement r is in the form of “if p then q”. We know that in such case q is necessary condition

for p.

So, necessary condition is “Function is both one-one and onto”.

Also, p is sufficient for q.

So, “A function is bijective” is sufficient condition for it to be one-one and onto.

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cls jeead-14-15 xi mat target-4 set-1 chapter-14

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