4 Marks Questions

Question 14 Marks

Show that the following statement is true by the method of contrapositive $p$ : If $x$ is an integer and $x^2$ is even, then $x$ is also even.

Answer

Let $q$ and $r$ be the statements given by $q$ : If $x$ is an integer and $x^2$ is even $r: x$ is an even integer.
Then $p$ : "If $q$, then $r$ "
If possible, let $r$ be true. Then, $r$ is false
$\Rightarrow x$ is not an even integer
$\Rightarrow x$ is an odd integer
$
\begin{array}{l}
\Rightarrow x=(2 n+1) \text { for some integer } n \\
\Rightarrow x^2=4 n^2+4 n+1 \\
\Rightarrow x^2=4 n(n+1)+1
\end{array}
$
$\Rightarrow x^2$ is an odd integer
$\Rightarrow q$ is false $\quad[\because 4 n(n+1)$ is even $]$
Thus, $r$ is false $\Rightarrow q$ is false
Hence, $p$ : "If $q$, then $r$ " is a true statement.

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Question 24 Marks

Check whether the following statement is true or not; "If $x$ and $y$ are odd integers, then $x y$ is an odd integer" by (i) Direct method (ii) Contrapositive method.

Answer

Let $p$ and $q$ be the statements given by
$p: x$ and $y$ are odd integers
$q: x y$ is an odd integer.
Then the given statement is
If $p$, then $q$
Direct Method : Let $p$ be true. Then $p$ is true
$\Rightarrow x$ and $y$ are odd integers
$
\begin{array}{ll}
\Rightarrow x=2 m+1, y=2 n+1, \text { for some integers } m, n \\
\Rightarrow x y=(2 m+1)(2 n+1) \\
\Rightarrow x y=2(m n+m+n)+1
\end{array}
$
$\Rightarrow x y$ is an odd integer
$\Rightarrow q$ is true
Thus, $p$ is true $\Rightarrow q$ is true
Hence, "If $p$, then $q$ " is true statement.
Contrapositive Method : Let $q$ be not true. Then $q$ is not true
$\Rightarrow x y$ is an even integer
$\Rightarrow$ either $x$ is even or $y$ is even or both $x$ and $y$ are even
$\Rightarrow p$ is not true
Thus, $q$ is false $\Rightarrow p$ is false.
Hence, "If $p$, then $q$ " is true statement.

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Question 34 Marks

Write the following statement in four different ways, conveying the same meaning.
$p$ : If a triangle is equiangular, then it is an obtuse angled triangle.

Answer

The given statement can be written in four different ways as follows :
(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.

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Question 44 Marks

Verify by method of contradiction $p: \sqrt{11}$ is irrational.

Answer

Let the given statement is false.
i.e., $\sim p: \sqrt{11}$ is rational
Therefore, $\sqrt{11}=\frac{p}{q}$ where $p$ and $q$ are coprime and $q \neq 0$
$\begin{array}{ll}\Rightarrow & 11=\frac{p^2}{q^2} \\ \Rightarrow & p^2=11 q^2 \\ \Rightarrow 11 \text { divides } p....(i)& \end{array}$
$\therefore r \in z$ such that: $p=11 r$
$\begin{array}{lrl}\Rightarrow \quad\quad\quad\quad\quad p^2 =121 r^2 \\ \Rightarrow\quad\quad\quad\quad 11 q^2 =121 r^2 \\ \Rightarrow 11 \text { divides } q....(ii)& \end{array}$
From eqs. (i) and (ii), we arrive at a contradiction, since $p$ and $q$ are co-prime
$\Rightarrow \sqrt{11}$ is irrational

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Applied Maths 4 Marks Questions

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