Question

Match the following (a > 0)
(1) | x| $\le$ a (P) -a $\le$ x $\le$ a
(2) | x| $\ge$ a (Q) x $\le$ -a or x $\ge$ a
(3) | x| - a = 0 (R) x = a or x = -a
(4) $|x-a|$ = 0 (S) x = a
(5) $x^2$ $\le$ $a^2$

• A. 1-Q;2-P;3-R;4-S;5-P
• B. 1-P;2-Q;3-R;4-S;5-P
• C. 1-P;2-Q;3-R;4-S;5-Q
• D. 1-Q;2-P;3-R;4-S;5-Q

Answer 1

The correct option is B

1-P;2-Q;3-R;4-S;5-P

The graph of y = |x| is

We have marked the point (0,a) on the y-axis.

The points where |x| becomes a are -a and a.

They are also marked on x-axis.

If |x| $\le$ a,all the values of x will lie between the points M and N.

$\Rightarrow$ x $\le$ a and x $\ge$ -a

If |x| $\ge$ a,then x will lie outside the line joining M and N.

$\Rightarrow$ x $\ge$ a and x $\le$ -a

(3) |x| - a = 0 $\Rightarrow$ |x| = a

For both a and -a, |x| = a

(4) |x-a| = 0 $\Rightarrow$ x - a = 0 or x = a

|x-a| is the distance of x from a.

The only point whose distance from a is zero, is a itself.

(5) $x^2$ $\le$ $a^2$ $\Rightarrow$ $\sqrt{x^2}$ $\le$ $\sqrt{a^2}$

$\Rightarrow$ |x| $\le$ |a| = a (a is +ve,given in question)

$\Rightarrow$ |x| $\le$ a

This is same case (1)