5 Marks Questions

Question 15 Marks

(a) Check the validity of the statements:
(i) $r: 100$ is multiple of 4 and 5
(ii) $p: 125$ is multiple of 5 and 7.
(b) Prove the following statement by contradiction method:
'The sum of an irrational and a rational number is irrational'.
(c) By giving the counter examples, show that the following statements are not true
(i) If all the angles of a triangle are equal, then triangle is an obtuse angled triangle.
(ii) If $n$ is an odd integer, then $n$ is prime.

Answer

(a) (i) Let $r=p \wedge q$
where, $p: 100$ is multiple of 4 , which is true $q: 100$ is multiple of 5 , which is true
As $p$ and $q$ are both true, so $p \wedge q$ is true. Hence, $r$ is true.
(ii) Let $p=q \wedge s$
where, $q: 125$ is divisible by 5 , which is true
$s: 125$ is divisible by 7 , which is false
Here, $q$ is true but $s$ is false, so $q \wedge s$ is false
Hence, $p$ is false.
(b) Let the given statement be not true i.e., the sum of an irrational number and a rational number is not irrational.
$\Rightarrow$ There exists an irrational number say $x$ and a rational number say $y$ such that $x+y$ is rational number, say $z$
$\begin{array}{rlrl}\Rightarrow x+y =z \\ \Rightarrow x =z-y,\end{array}$
but $z$ and $y$ are both rational number, so $z-y$ is also a rational number
$\Rightarrow x$ is a rational number
But $x$ is an irrational number, so we arrive at a contradiction. Hence, the given statement is true.
(c) (i) Let $A B C$ be an obtuse angled triangle and $\angle A$ be obtuse, then $\angle A$ is greater than $90^{\circ}$.
Since, all the angles of $\triangle A B C$ are equal, therefore $\angle A=\angle B=\angle C$
$\Rightarrow$ Sum of angles of $\triangle A B C=\angle A+\angle A+\angle A=$ $3 \angle A$
$\Rightarrow$ Sum of angles of $\triangle A B C>270^{\circ} \quad\left(\because A>90^{\circ}\right)$
Which is wrong (because sum of angles of a triangle is always $180^{\circ}$ ).
Hence, the given statement is not true.
(ii) We note that 15 is an odd integer but 15 is not prime. Hence, the given statement is not true

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Question 25 Marks

(a) Show that the following statement is true: Given a positive real number $p$, there exists a rational number $r$ such that $a<r<p$.
(b) Verify by the method of contradiction $p: \sqrt{7}$ is irrational

Answer

(a) To prove that 'there exists .......", basically we have to look for atleast one rational number satisfying the required property.
Since, $p$ is a positive real number, $\frac{1}{p}$ is also a positive real number. Also, we can find a positive integer $n$ such that $n > \frac{1}{p}$. Then $\frac{1}{n} < p$ (as both $n$ and $p$ are positive real).
Also, $\frac{1}{n}$ is a positive rational number. Letting $r=\frac{1}{n}$, we have $0 < r < p$. Thus, the statement is true.
(b)Let us assume that $p$ is false. This means that $\sqrt{7}$ is rational.
Then there exists positive integers $a$ and $b$ such that
$\sqrt{7}=\frac{a}{b}$
where $a, b$ have no common factors (except 1). Squaring both sides, we get
$\begin{aligned}7 & =\frac{a^2}{b^2} \\\Rightarrow \quad a^2 & =7 b^2\end{aligned}$
$\Rightarrow 7$ divides $a^2$
$\Rightarrow 7$ divides a (Since, 7 is a prime number)
Let $a=7 m$, then $a^2=7 b^2$
$\begin{array}{lr}\Rightarrow 49 m^2=7 b^2 \\\Rightarrow b^2=7 m^2 \\\Rightarrow 7 \text { divides } b^2 & \\\Rightarrow 7 \text { divides } b &\end{array}$
Thus, both $a, b$ are divisible by 7 , which contradicts our earlier assumption that $a, b$ have no common factors (except 1).
This shows that our assumption that ' $\sqrt{7}$ is rational' was wrong. Hence, the statement that ' $\sqrt{7}$ is irrational' is true.

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Question 35 Marks

Using the words "necessary and sufficient" rewrite the statement :
The integer $n$ is odd if $n^2$ is odd, and vice-versa Also, check whether the statement is true.

Answer

Given statement can be written as :
That integer $n$ is odd is necessary and sufficient condition for $n^2$ being odd.
Here, $p$ : Integer $n$ is odd
$q: n^2$ is odd
To check the validity of $p \leftrightarrow q$, we check 'sufficient' and 'necessary, parts'.
Step 1 : 'Sufficient part' To prove that 'If $p$ is true, then $q$ is true.
Assume that $p$ is true i.e., $n$ is odd
Let $n=2 k+1$, where k is an integer
$\therefore n^2=(2 k+1)^2=4 k^2+4 k+1=2\left(2 k^2+2 k\right)+1 \text {, }$
which is odd
Thus, whenever $p$ is true, $q$ is true.
Step 2: 'Necessary part' To prove that if $q$ is true then $p$ is also true.
Here, we have to prove that :
If $n^2$ is odd, then $n$ is odd
It is easy to prove this by contrapositive method i.e., by proving:
If $n$ is not odd then $n^2$ is not odd, which is same as :
If $n$ is even, then $n^2$ is even
Assume that $n$ is even, Let $n=2 k$ for some integer $k$ $\therefore n^2=(2 k)^2=4 k^2$, which is even.
As step 1 and step 2 are both proved, we have prove that the given biconditional statement is true.

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Question 45 Marks

Show that the statement $p$ : If $x$ is real number such 'that $x^3+4 x=0$, then $x$ is zero' is true by
(i) Direct method
(ii) Method of contradiction
(iii) Method of contropositive

Answer

Let $q$ and $r$ be the statements given by
$q: x$ is a real number such that $x^3+4 x=0$
$r: x$ is zero
Then, $p$ : If $q$, then $r$
(i) Direct Method: Let $q$ be true, then $q$ is true
$\Rightarrow x$ is a real number such that $x^3+4 x=0$
$\Rightarrow x$ is a real number such that $x\left(x^2+4\right)=0$
$\Rightarrow x=0$
$\Rightarrow r$ is true
Thus, $q$ is true $\Rightarrow r$ is true
Hence, $p$ is true.
(ii) Method of contradiction : If possible, let $p$ be not true, then $p$ is not true
$\Rightarrow \sim p$ is true
$\begin{array}{lr}\Rightarrow \sim(q \Rightarrow r) \text { is true } & {[\because p: q \Rightarrow r]} \\ \Rightarrow q \text { and } \sim r \text { is true } & {[\because \sim(q \Rightarrow r) \cong q \text { and } \sim r]}\end{array}$
$\Rightarrow x$ is a real number such that $x^3+4 x=0$ and $x \neq 0$
$\Rightarrow x=0 \text { and } x \neq 0$
This is a contradiction.
Hence, $p$ is true.
(iii) Method of contrapositive : Let $r$ be not true then, $r$ is not true
$\begin{array}{l}\Rightarrow x \neq 0, x \in R \\ \Rightarrow x\left(x^2+4\right) \neq 0, x \in R \\ \Rightarrow q \text { is not true }\end{array}$
Thus, $\sim r \Rightarrow \sim q$
Hence, $p: q \Rightarrow r$ is true

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Question 55 Marks

Define basic logical connectives and quantifiers with the help of examples.

Answer

Basic logical connectives : In mathematical reasoning, simple statements can be connected by using connectives like 'And', 'or', 'Implies' etc. 'And' and 'or' are basic logical connectives.
The connective 'And' : If any two simple statements are combined by the word "and" to form a compound statement, then the resulting statement is called the conjunction of the original statements. Symbolically the conjunction of two statements $p$ and $q$ is denoted by $p \wedge q$. The elements $p$ and $q$ of $p \wedge q$ are called its conjuncts (components).
e.g., Let us consider the statement
$p$ : The earth is round and the sun is hot.
This statement can be broken into two component statements, namely:
$q$ : The earth is round
$r$ : The sun is hot
Here, we observe that both statements $q$ and $r$ are true. So, the statement $p$ is also true.
The compound statement with 'And' is true if all the components are true. The compound statement with 'And' is false if any of the component statement is false.
The connective 'OR'
When two simple statements $p$ and $q$ are connected by the word 'or', then the new statement so formed is called disjunction of the original statements $p$ and q. Symbolically the disjuction of two statement $p$ and $q$ is denoted by $p \vee q$. The elements $p$ and $q$ are called conjucts (components).
e.g., Consider the following statements :
(i) He is very smart or he is very lucky. (He may be both smart and very lucky).
(ii) There is something wrong with the bulb or with the wiring (may be wrong with both bulb and wiring).
(iii) Rahul is dead or Rahul is alive (not both dead or alive)
(iv) I will stay home or I will go to see a movie (not both).
Note that in (i) and (ii) statements 'or' is used in the inclusive form and in (iii) and (iv) statements . 'or' is used in the exclusive form.
A compound statement with an 'or' is true when one component statement is true or both the components are true. A compound statement with an 'or' is false when both the components are flase.
Quantifiers: In the study of mathematics, we come across phrases like 'There exists', 'for every/ for all', 'for each'. Such pharses are called Quantifiers. Depending on the context the pharases can be replaced by equivalent phrases 'There is', 'There is atleast one' or 'It is possible to find' :
e.g. : Consider the following statements :
$p: x+3>2$, for all $x \in N$
$q$ : There exists an $x \in N$ such that $x+3<6$, or, for some $x, x+3<6$.
$r$ : There exists a rectangle whose all sides are equal.
$s$ : For every prime number $n, \sqrt{n}$ is an irrational number.
• The statement $p$ means that for every natural number $x, x+3>2$, because $x=1,2,3, \ldots \ldots$.
• The statement $q$ means that there is atleast one natural number $x$ such that $x+3<6$.
• The statement $r$ means that there is atleast one rectangle whose all sides are equal.
• The statement $S$ means that if $S$ be the set of all prime numbers, then for all the elements $n$ of the set $S, \sqrt{n}$ an irrational number.

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Question 65 Marks

(a) Give two examples of sentences which are not statements. Give reasons for the answer.
(b) Are the following pair of statements negation of each other :
(i) The number $x$ is a rational number. The number $x$ is an irrational number.
(ii) The number $x$ is not a rational number. The number $x$ is not an irrational number.

Answer

(a) Two examples are as follows:
(i) He is a student of commerce. This is not a statement because it is not clear that who is 'he'.
(ii) The value of $\cos \theta$ is always greather than $1 / 2$. Unless the angle $\theta$ is known, we cannot say whether the statement is true or false.
(b) We know that, if a number is not rational, then, it is irrational.
(i) Let $p$ : The number $x$ is a rational number. $\sim p$ : The number $x$ is not a rational number Here, the negation of first statement i.e., negation of $p$ is same as second statement.
The number $x$ is an irrational number.
Hence, the given pair is negation of each other.
(ii) Let $p$ : The number $x$ is not a rational number $\sim p$ : The number $x$ is a rational number Here, the negation of first statement i.e.,
negation of $p$ is same as second statement.
The number $x$ is not an irrational number.
Hence, the given pair is negation of each other.

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Question 75 Marks

Translate the following statements into symbolic form :
(i) Rahul passed in Hindi and English.
(ii) 2,3 and 6 are factors of $1 2$.
(iii) Either $x$ or $x+1$ is an odd integer.
(iv) $x$ and $y$ are even integers.
(v) A number is divisible by 2 or 3 .
(vi) Either $x=2$ or $x=3$ is a root of $3 x^2-x-10=0$

Answer

(i) $p:$ Rahul passed in Hindi.
$q$ : Rahul passed in English.
$\therefore p \wedge q$ : Rahul passed in Hindi and English.
(ii) $p: 2$ is a factor of 12
$q: 3$ is a factor of 12
$r: 6$ is a factor of 12
$\therefore p \wedge q \wedge r: 2,3$ and 6 are factors of 12 .
(iii) $p: x$ is an odd integer
$q: x+1$ is an odd integer
$\therefore p \vee q$ : either $x$ or $x+1$ is an odd integer.
(iv) $p: x$ is even integer.
$q: y$ is even integer.
$\therefore p \wedge q: x$ and $y$ are even integers.
(v) $p:$ A number is divisible by 2
$q$ : A number is divisible by 3
$\therefore p \vee q$ : A number is divisible by 2 or 3 .
(vi) $p: x=2$ is a root of $3 x^2-x-10=0$
$q: x=3$ is a root of $3 x^2-x-10=0$
$\therefore p \vee q$ : Either $x=2$ or $x=3$ is a root of $3 x^2-x-10=0$

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Applied Maths 5 Marks Questions

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