Name: Real-Valued Functions
Uploaded: 2019-07-19
Duration: 4 min 7 s
Description: When the domain and range of the function belong to real numbers, then the function is called a real-valued function. Watch this video to learn more about it.

Question

Solve the following for $x,$ where $| x |$ is modulus function, $[x]$ is greatest integer function, ${x}$ is a fractional part function.

$|x^{2} - 9| + |x^{2} - 4| = 5$

Accepted Answer

Given that $|x^{2} - 9| + |x^{2} - 4| = 5$

Case $1 : x < - 3$

If $x < - 3$ then $x^{2} > 9$

$∴ |x^{2} - 9| = x^{2} - 9$ and $|x^{2} - 4| = x^{2} - 4$

equation becomes,

$x^{2} - 9 + x^{2} - 4 = 5$

$∴ 2 x^{2} = 18$
$∴ x^{2} = 9$
$\Rightarrow x = \pm 3$

But $x < - 3$ therefore $| x |$
Hence, there is no solution in this region.
Case $2 : - 3 \leq x < - 2$

If $- 3 \leq x < - 2$ then $4 < x^{2} \leq 9$

$∴ |x^{2} - 9| = 9 - x^{2} and |x^{2} - 4| = x^{2} - 4$

$∴$ equation becomes $9 - x^{2} + x^{2} - 4 = 5$

$∴ 5 = 5$ which is true

$∴ - 3 \leq x < - 2$ is the solution $. . . . . . (1)$

Case $3 : - 2 \leq x < 2$

If $- 2 \leq x < 2, then 0 \leq x^{2} < 4$

$∴ |x^{2} - 9| = 9 - x^{2} and |x^{2} - 4| = 4 - x^{2}$

$∴$ equation becomes,

$9 - x^{2} + 4 - x^{2} = 5$

$∴ 2 x^{2} = 8 ∴ x^{2} = 4$

$∴ x = \pm 2$

But $- 2 \leq x < 2 ∴ x \neq 2$

$∴ x = - 2$ is a solution ............ $(2)$

Case $4 : 2 \leq x < 3$

If $2 \leq x < 3$ then $4 \leq x^{2} < 9$

$∴ |x^{2} - 9| = 9 - x^{2} and |x^{2} - 4| = x^{2} - 4$

equation becomes,

$9 - x^{2} + x^{2} - 4 = 5$

$5 = 5$ which is true

$2 \leq x < 3$ is the solution $. . . . . . (3)$

Case $5 : x \geq 3$

If $x \geq 3$ then $x^{2} \geq 9$

$∴ |x^{2} - 9| = x^{2} - 9 and |x^{2} - 4| = x^{2} - 4$

$∴$ equation becomes

$x^{2} - 9 + x^{2} - 4 = 5$

$∴ 2 x^{2} = 18 ∴ x^{2} = 9$

$∴ x = \pm 3$

But $x \geq 3$ therefore $x \neq - 3$

$x = 3$ is a solution ............... $(4)$

From $(1), (2), (3) a n d (4)$

the solution set $= [- 3, - 2] \cup [2, 3]$

Hence, the solution for the given function $|x^{2} - 9| + |x^{2} - 4| = 5$ is $[- 3, - 2] \cup [2, 3]$

Solve the following for $x,$ where $| x |$ is modulus function, $[x]$ is greatest integer function, ${x}$ is a fractional part function.

$|x^{2} - 9| + |x^{2} - 4| = 5$

Important Questions on Functions

Learn and Prepare for any exam you want !

Solve the following for x, where |x| is modulus function, [x] is greatest integer function, {x} is a fractional part function. x2-9+x2-4=5

Important Questions on Functions

Learn and Prepare for any exam you want !

Solve the following for $x,$ where $| x |$ is modulus function, $[x]$ is greatest integer function, ${x}$ is a fractional part function.

$|x^{2} - 9| + |x^{2} - 4| = 5$