NCERT Solutions for Class 11 Math Chapter 14 - Mathematical Reasoning

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The given sentence is false.
[8 is greater than 6]
Hence, it is a statement.

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The given sentence is false.
[There are sets, which are not finite] Hence, it is a statement.

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The given sentence is true.
[It is scientifically established fact ]
Hence, it is a statement.

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The given sentence is not always true.
[It may be fun for those who like Mathematics but it is not fun for others]
Hence, it is not a statement.

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This sentence is true.
[It is scientifically established natural phenomenon] Hence, it is a statement.

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This is not a statement.
[The sentence contains the word “here”]

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The given statement is :
p : Both the diagonals of a rectangle have same length.
The negation of the statment is :
~ p : It is false that both the diagonals of a rectangle have the same length.
Or ~ p : There is atleast one rectangle whose both diagonals do not have the same length.

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The component statements are :
p : The sky is blue.
q : The green is green.
The connecting word is ‘and’.

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The component statements are :
p : It is raining.
q : It is cold.
The connecting word is ‘and’.

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The component statements are :
p : All rational numbers are real.
q : All real numbers are complex.
The connecting word is ‘and’.

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The component statements are :
p : 0 is a positive number.
q : 0 is a negative number.
The connecting word is ‘or’.

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The component statements are :
p : A square is a quadrilateral.
q : A square has all its four sides equal.
Here p and q are both true.

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The component statements are :
p : All prime numbers are even.
q : All prime numbers are odd.
Here p and q are both false.

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The component statements are :
p : A person who has taken Mathematics can go for MCA.
q : A person who has taken Computer Science can go for MCA.
Here p and q are both true.

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The component statements are :
p : Chandigarh is the capital of Haryana.
q : Chandigarh is the capital of U.P.
Here p is true but q is false.

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The component statements are :
p : 24 is a multiple of 2.
q : 24 is a multiple of 4.
r : 24 is a multiple of 8.
Here p, q, r are all true.

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The component statements are :
p : A line is straight.
q : A line extends indefinitely in both directions.
Since p and q are both true,
$\therefore$ the compound statement (p $\wedge$ q) is true.

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The component statements are :
p : 0 is less than every positive integer.
q : 0 is less than every negative integer.
Since p is true and q is false,
$\therefore$ the compound statement (p $\wedge$ q) is false.

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The component statements are :
p : All living things have two legs.
q : All living things have two eyes.
Since p and q are both false,
$\therefore$ the compound statement (p $\wedge$ q) is false.

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p : To enter a country, you need a passport.
q : To enter a country, you need a voter registration card.
To enter a country, you need either a passport or a voter registration card as well as you have both.
Thus ‘Or’ is inclusive.

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p : The school is closed if it is a holiday. q : The school is closed if it is a Sunday.
The school is closed if it is either a holiday or it is Sunday as well as it is both.
Thus ‘Or’ is inclusive.

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p : Two lines intersect at a point.
q : Two lines are parallel.
Since two lines cannot intersect at a point as well as they are parallel,
$\therefore$‘Or’ is exclusive.

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p : Students can take French as their third language.
q : Students can take Sanskrit as their third language.
Since students cannot take both French and Sanskrit as their third language,
$\therefore$‘Or’ is exclusive.

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p : To get into a public library children need an identity card.
q : To get into a public library children need letter from the school authorities.
Children can enter the library if they have either identity card or the letter as well as when they have both.
Thus ‘Or’ is inclusive.
$\therefore$ The compound statement is also true when children have both card and the letter.

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p : A rectangle is a quadrilateral.
q : A rectangle is a 5-sided polygon.
Since p is true and q is false and ‘Or’ is exclusive,
$\therefore$ the compound statement is true.

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We have : antecedent : p : A number is divisible by 9.
consequent : q : The number is divisible by 3.
$\therefore$ ~ p : A number is not divisible by 9.
~ q : The number is not divisible by 3.
The contrapositive of the statement p $\Rightarrow$ q is ~ q $\Rightarrow$ ~ p.
Its statement is :
If a number is not divisible by 3, it is not divisible by 9.

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We have : antecedent : p : You are born in India.
consequent : q : You are a citizen of India.
$\therefore$ ~ p : You are not born in India.
~ q : You are not a citizen of India.
The contrapositive of the statement p $\Rightarrow$ q is ~ q $\Rightarrow$ ~ p.
Its statement is :
If you are not a citizen of India, then you are not born in India.

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We have :
antecedent : p : Triangle is equilateral.
consequent : q : Triangle is isosceles.
$\therefore$ ~ p : Triangle is not equilateral
~ q : Triangle is not isosceles.
The contrapositive of the statement p $\Rightarrow$ q is ~ q $\Rightarrow$ ~ p.
Its statement is :
If a triangle is not isosceles, then it is not equilateral.

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The converse of the given statements are :
If a number n² is even, then n is even.

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The converse of the given statements are :
If you get an A-grade in the class, then you have done all the exercises of the book.

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The converse of the given statements are :
If two integers a and b are such that a – b is always a positive integer, then a > b.

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The statement is of the from ‘if p, then q’.
(i) p : Triangle is equilateral
q : Triangle is isosceles.
Here p $\Rightarrow$ Triangle is equilateral
$\Rightarrow$ Triangle is isosceles [Equilateral triangle is isosceles]
$\Rightarrow$ q.
Hence, the compound statement is true.

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p : a, b are integers
q : ab is a rational number
p $\Rightarrow$ a, b are integers
$\Rightarrow$ (ab) is an integer
[Product of two integers is an integer]
$\Rightarrow$ ab is a rational number.
[Integer is a rational number]
$\Rightarrow$ q.
Hence, the compound statement is true.

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A rectangle is a square if and only if all its four sides are equal.

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A number is divisible by 3 if and only if the sum of its digits is divisible by 3.

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Let p : x, y $\in$ Z such that x and y are odd q : xy is odd.
(a) We can apply Rule–3, Case I (‘if – then’) :
i.e.if ‘p’ is true, then ‘q’ is true.
‘p’ is true $\Rightarrow$ x and y are odd integers.
Then x = 2l + 1 for l $\in$ Z
and y = 2m + 1 for m $\in$ Z.
Then xy = (2l + 1) (2m + 1)
= 2 (2lm + l + m) + 1
$\Rightarrow$ xy is odd.
Hence the given statement is true.
(b) We can also apply Rule–3, Case II (‘if – then) :
i.e.if ‘q’ is not true, then ‘p’ is not true.
Assume that ‘q’ is not true.
$\Rightarrow$ ~ q : xy is true.
This is possible only if either x or y is even
$\Rightarrow$ p is not true.
Thus $~$ q$\Rightarrow$$~$ p.
Hence the result.

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p : xy is odd
q : Both x and y are odd.
Here we check whether the statement p $\Rightarrow$ q is true or not by checking its contrapositive
i.e. $~$ q $\Rightarrow$ $~$ p.
Now $~$ q : It is false that both x and y are odd :
$\Rightarrow$ x (or y) is even.
Let x = 2n for some n $\in$ Z
$\therefore$ xy = 2ny for some n $\in$ Z
$\Rightarrow$ xy is even
$\Rightarrow$ $~$ p is true.
Thus $~$ q $\Rightarrow$ $~$ p.
Hence, the given statement is true.

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The given statement is of the type :
‘‘If p then q’’.
To show that this is false.
For this we need to show that ‘‘if p then ~ q”.
Take an odd integer, which is not prime.
Let n = 9.
With this, the given statement is false.

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‘‘Or’’ used in the given statement is inclusive.
[$\because$ It is possible that it rains and you are in the river]
The component statements of the given compound statement are :
p : You are wet when it rains.
q : You are wet when you are in a river.
Since ‘p’ and ‘q’ are both true,
$\therefore$ (p ∨ q), the compound statement, is also true.

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$~$ p : There exists a real number x such that x² < x.

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$~$ q : There does not exist a rational number x such that x² = 2.

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$~$ r : There exists a bird which has no wings.

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Let the statements be :
p : You drive over 80 km per hour
q : You will get a fine.
Now ‘if p, then q’ indicates that p is sufficient for q i.e. driving over 80 km per hour is sufficient to get a fine.
Hence the sufficient condition is :
“driving over 80 km per hour’’
Again ‘if p, then q’ indicates that q is necessary for p i.e. when you drive over 80 km per hour, you will necessarily get a fine.
Hence, the necessary condition is :
“getting a fine”.

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Statement.
This is false.
[$\because$ A month cannot have 35 days]

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Not a statement.
Mathematics may be difficult to one but easy to the other.

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Statement.
This is true. [$\because$ 5 + 7 = 12 > 10]

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Not a statement.
The sentence is sometimes true and sometimes false.
For Ex. 22 = 4, which is even and 32 = 9, which is odd.

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Statement.
The sentence is sometimes true and sometimes false.
For Ex. Square and Rhombus have equal length whereas rectangle and trapezium have unequal lengths.

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Not a statement.
In fact it is an order.

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Statement.
This is false. [$\because$ (– 1) (8) = – 8]

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Statement.
This is always true.

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Not a statement.
From the context, it is not clear which day is referred.

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Statement.
This is true.
[$\because$ Each real number ‘a’ can be written as a + i0]

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The three examples are :
(i) It is a windy day.
Which day is a windy day ? We cannot say which day is windy.
(ii) She is very beautiful.
Who is beautiful ?

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Negation of the given statements are as given below :
Chennai is not the capital of Tamil Nadu.

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Negation of the given statements are as given below :

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Negation of the given statements are as given below :
All triangles are equilateral triangles.

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Negation of the given statements are as given below :
The number 2 is not greater than 7.

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Negation of the given statements are as given below :
Every natural number is not an integer.

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The negative of the statement
“The number x is not a rational number”
~ The number x is a rational number.
$\Rightarrow$The number x is not an irrational number.
Hence the statements are negation of each other.

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The negation of the statement
“The number is a rational number”
~ The number x is not a rational number.
$\Rightarrow$ x is an irrational number.
$\Rightarrow$ The statements are not nagation of each other.

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The component statements are :
p : Number 3 is prime.
q : Number 3 is odd.
Here p and q are both true.
The compound statement is true.

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The component statements are :
p : All integers are positive.
q : All integers are negative.
Here p and q are both false.
The compound statement is false.

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The component statements are :
p : 100 is divisible by 3.
q : 100 is divisible by 11.
and r : 100 is divisble by 5.
Here p, q are false and r is true.
$\therefore$ The compound statement is false.

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Connecting word is “And”.
The component statements are :
All rational numbers are real.
All real numbers are not complex.

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Connecting word is “Or”.
The component statements are :
Square of an integer is positive.
Square of an integer is negative.

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Connecting word is “And”. The component statements are : The sand heats up quickly in the sun. The sand does not cool down fast at night.

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Connecting word is “And”.
The component statements are :
x = 2 is root of the equation 3x² – x – 10 = 0.
x = 3 is root of the equation 3x² – x – 10 = 0.

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The given statement contains existential quantifier ‘there exists’.
Now the statement is :
p : A number is equal to its square.
The negation of the statement is obtained by replacing existential quantifier by a universal quantifier and then negate p.
$\therefore$ The negation of given statement is :
For all numbers they are not equal to their square.

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The given statement contains universal quantifier ‘for every’.
Now the statement is :
p : x is less than x + 1.
The negation of the statement is obtained by replacing universal quantifier by a existential quantifier and then negate p.
$\therefore$ The negation of given statement is :
“For some real number, x is not less than x + 1”.

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The given statement contains existential quantifier ‘there exists’.
Now the statement is :
p : A capital for every state in India.
The negation of the statement is obtained by replacing existential quantifier by a universal quantifier and then negate p.
$\therefore$ The negation of given statement is :
“There exists a state in India which does not have capital.”

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No.
Negation of statement is :
“There exist real numbers x and y for which x + y ≠ y + x”.

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Here “Or” is exclusive.
[$\because$ Sun rises and moon sets during day time]

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Here “Or” is inclusive.
[$\because$ A person can have both a ration card and a passport to apply for a driving licence]

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Here “Or” is exclusive.
[$\because$ All integers cannot be both positive as well as negative]

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The given statement can be rewritten in the following five different ways conveying the same meaning :
(i) A natural number is odd implies that its square is odd.
(ii) A natural number is odd only if its square is odd.
(iii) For a natural number to be odd, it is necessary that its square is odd.
(iv) For the square of a natural number to be odd, it is sufficient that the number is odd.
(v) If the square of a natural number is not odd, then the natural number is not odd.

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Contrapositive :
If a number x is not odd, then x is not a prime number.
Converse :
If a number is odd, then it a prime number.

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

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Contrapositive :
If something is not at low temperature, then it is not cold.
Converse :
If something is at low temperature, then it is cold.

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

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Contrapositive :
If x is not divisible by 4, then x is not an even number. Converse :
If x is divisible by 4, then x is an even number.

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If you get a job, then your credentials are good.

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If it stays for a month, then Banana tree will bloom.

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If a number is even, then it is a multiple of 2.

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

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The component statements are :
p : You live in Delhi.
q : You have winter clothes.
The given statements are of the type :
(i) p $\Rightarrow$ q
(ii) $~$ q $\Rightarrow$ $~$ p
(iii) q $\Rightarrow$ p.
Clearly (i) and (ii) are contrapositive of each other
and (i) and (iii) are converse of each other.

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The component statements are :
p : Quadrilateral is a parallelogram.
q : Diagonals bisect each other.
The given statements are of the type :
(i) p $\Rightarrow$ q
(ii) $~$ q $\Rightarrow$ $~$ p
(iii) q $\Rightarrow$ p.
Clearly (i) and (ii) one contrapositive of each other.
and (i) and (iii) are converse of each other.

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Let q : x is a real number such that x³ + 4x = 0 r : x is 0.
Then p : If q, then r.
Direct Method.
Let q be true.
Then x is a real number such that x³ + 4x = 0
$\Rightarrow$x is a real number such that x (x² + 4) = 0
$\Rightarrow$ x = 0 [$\because$ x² + 4 ≠ 0 for x $\in$ R]
$\Rightarrow$ r is true.
Thus q is true $\Rightarrow$ r is true.
Hence p is true.

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Let q : x is a real number such that x³ + 4x = 0 r : x is 0.
Then p : If q, then r.
Method of Contradiction.
Let us assume that ‘p’ is not true.
Then $~$ p is true
$\Rightarrow$ $~$ (q $\Rightarrow$ r) is true
$\Rightarrow$q and $~$ r are true [$\because$ $~$ (q $\Rightarrow$ r) ≡ q and $~$ r]
$\Rightarrow$x is a real number such that x³ + 4x = 0 and x ≠ 0
$\Rightarrow$ x = 0 and x ≠ 0.
This leads to contradiction.
Hence p is true.

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Let q : x is a real number such that x³ + 4x = 0 r : x is 0.
Then p : If q, then r.
Method of Contrapositive.
Let r be not true.
Then x ≠ 0; x $\in$ R
$\Rightarrow$ x³ + 4x ≠ 0; x $\in$ R
$\Rightarrow$ q is not true
$\Rightarrow$ $~$ q is true
Thus $~$ r $\Rightarrow$ $~$ q.
Hence p : q $\Rightarrow$ r is true.

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Counter Example :
Table a = 5 and b = – 5.
Here a² = b² [$\because$ each = 25]
but a ≠ b.

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Let the statements be :
q : If x is an integer and x2 is even r : x is even.
Thus p : “If q, then r”.
Let us suppose that r is false.
Then x is not an even integer
$\Rightarrow$x is an odd integer
$\Rightarrow$ x = 2m + 1 for some m $\in$ Z
$\Rightarrow$ x² = 4m² + 4m + 1
$\Rightarrow$ x² = 4m (m + 1) + 1
$\Rightarrow$x² is an odd integer
$\Rightarrow$ q is false.
Thus r is false
$\Rightarrow$ q is false.
Hence p : “If q, then r” is true.

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The given statement is in the form ‘if p then q’. To show that this is false, we are to show that if ‘p’ then $~$ q’.
We take a triangle whose all angles are less than 90$°$.
Let each angle be 65$°$.
Then sum of angles = 3 × 65$°$ = 195$°$ > 180$°$.
Thus p is true
$\Rightarrow$ q is false.
Hence the given statement is not true.

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Counter Example : The equation x² – 1 = 0 has a root – 1, which does not lie between 0 and 2.

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False.
By definition, the chord intersects the circle in two parts.

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False.
Counter Example. A chord, which is not a diameter.

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True.
By Rule of Inequality, x > y $\Rightarrow$ – x < – y.

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