Embibe Experts Solutions for Chapter: Mathematical Reasoning, Exercise 4: Exercise-4

Consider statement $p$ : If two integers $a$ and $b$ are such that a divides $b$ then $b^3-a^2$ is a composite number. The converse of the statement $p$ is

Let $S$ be a non-empty subset of $R.$ Consider the following statement

$P$: There exist two irrational numbers $x\in S$ and $y\in S$ such that $x+y$ is a rational number.

Negation of the statement $P$ is

If the compound propositions $(q\leftrightarrow q)\wedge r$ is true and $p\to (~q)$ is false, then the truth values of $p, q$ and $r$ are respectively, 

The incorrect statement of the following is

$S_1: f\left(x\right)$ is not continuous in $\left[a, b\right]$

$S_2: f\left(x\right)$ is differentiable in $\left[a, b\right]$

${\mathrm{S}}_3: f^{\text{'}}\left(x\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$ for at least one $c\in (a, b)$

If a function $f\left(x\right)$ is continuous and differentiable in $[a, b]$ then $f^{\text{'}}\left(C\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$ for at least $c\in (a, b)$

which of the following is equivalent to the given statement

The statement $\left[\right(p\leftrightarrow ~q)\wedge ~p]\to q$ can be

Let $p$ and $q$ be two statements, then $~\left(~p\wedge q\right)\wedge \left(p\vee q\right)$ is logically equivalent to

Which of the following is False?