Relations and Functions — MATHS STD 11 Science — Question

3. $\big[-3,-2\big]\cup\big[2,3\big] $

Solution:

$\text{f(x)}=\sqrt{5|\text{x}|-\text{x}^2-16}$

For f(x) to be defined, $5|\text{x}|-\text{x}^2-6\geq0$

$\Rightarrow5|\text{x}|-\text{x}^2-6\geq0$

$\Rightarrow\text{x}^2-5|\text{x}|+6\leq0$

For x > 0, |x| = x

$\Rightarrow\text{x}^2-5\text{x}+6\leq0$

$\Rightarrow(\text{x}-2)(\text{x}-3)\leq0$

$\Rightarrow\text{x}\in[2,3]\ ...(\text{i})$

For x < 0, |x| = -x

$\Rightarrow\text{x}^2+5\text{x}+6\leq0$

$\Rightarrow(\text{x}+2)(\text{x}+3)\leq0$

$\Rightarrow\text{x}\in\big[-3,-2\big]\ ...(\text{ii})$

From (i) and (ii),

$\text{x}\in\big[-3,-2\big]\cup\big[2,3\big]$

Or, $\text{domain(f)}=\big[-3,-2\big]\cup\big[2,3\big]$