Answer
a) False
b) False
c) True
d) False
Work Step by Step
a) ∃!x(x > 1) means “There exists a unique x such that x > 1 is true.”
But x = 2 and x = 3 both satisfy P(x): x>1. Thus the statement is false.
b) For both x = 1 and x = -1, w have ${x}^2 =1$
Thus there does not exists a unique x such that ${x}^2 =1$ is true.
So, the statement is false.
c) Solving $x+3=2x$
$\equiv (3=2x-x)$
$\equiv \quad$ $ x=3$.
Thus there is a unique value of x for which $x+3=2x$.
So, the statement is true.
d) Let us solve $ x=x+1$
$\equiv\quad$ $x-x=1$
$\equiv\quad$ $0=1$
This equation can never be true. So, the given equation has no solution.
Since, the given equation has no solution, there does not exist any value of x for which $x=x+1$ is true.
Thus, the statement is false.
