Answer
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Hint: Here we will use the property of modulus function i.e.
\[\left| x \right| = \left[ {\begin{array}{*{20}{c}}
x&{x \geqslant 0} \\
{ - x}&{x < 0}
\end{array}} \right]\] and then solve for the value of x from each of the equations formed.
Complete step-by-step answer:
The given equation is:-
\[\left| {\dfrac{{x - 8}}{2}} \right| = 5\]
Splitting the modulus function we get:-
$ \Rightarrow$\[\dfrac{{\left| {x - 8} \right|}}{{\left| 2 \right|}} = 5\]
\[ \Rightarrow \dfrac{{\left| {x - 8} \right|}}{2} = 5\]
On cross multiplying we get:-
$ \Rightarrow$\[\left| {x - 8} \right| = 10\]……………………. (1)
Now we know that the property of modulus function is:-
$ \Rightarrow$\[\left| x \right| = \left[ {\begin{array}{*{20}{c}}
x&{x \geqslant 0} \\
{ - x}&{x < 0}
\end{array}} \right]\]
Applying this property we get:-
$ \Rightarrow$\[\left| {x - 8} \right| = \left[ {\begin{array}{*{20}{c}}
{x - 8}&{x \geqslant 8} \\
{ - \left( {x - 8} \right)}&{x < 8}
\end{array}} \right]\]
Simplifying it further we get:-
$ \Rightarrow$\[\left| {x - 8} \right| = \left[ {\begin{array}{*{20}{c}}
{x - 8}&{x \geqslant 8} \\
{8 - x}&{x < 8}
\end{array}} \right]\]
Putting this value in equation 1 we get:-
$ \Rightarrow$\[\left| {x - 8} \right| = \left[ {\begin{array}{*{20}{c}}
{x - 8}&{x \geqslant 8} \\
{8 - x}&{x < 8}
\end{array}} \right] = 10\]
Now let us consider the case \[x \geqslant 8\]we get:-
$ \Rightarrow$\[x - 8 = 10\]
Solving for the value of x we get:-
$ \Rightarrow$\[x = 18\]
Now let us consider the case \[x < 8\] we get:-
$ \Rightarrow$\[8 - x = 10\]
Solving for the value of x we get:-
$ \Rightarrow$\[x = - 2\]
Therefore, we get two values of x as 18 and -2.
Additional Information: The modulus function \[\left| x \right| = \left[ {\begin{array}{*{20}{c}}
x&{x \geqslant 0} \\
{ - x}&{x < 0}
\end{array}} \right]\]is discontinuous at x=0.
For any real number x, \[\sqrt {{x^2}} = \left| x \right|\]
The domain of modulus function is any real number i.e. R.
The range of the modulus function is all positive real numbers i.e. \[\left[ {0,\infty } \right)\]
Note: Students should take a note that the value of modulus is always positive i.e. even if the value inside the modulus function is negative then also, the resultant value of mod would be positive.
Also, we find the open modulus function at the critical point i.e. the point where the value of the quantity inside the mod becomes equal to zero.
\[\left| x \right| = \left[ {\begin{array}{*{20}{c}}
x&{x \geqslant 0} \\
{ - x}&{x < 0}
\end{array}} \right]\] and then solve for the value of x from each of the equations formed.
Complete step-by-step answer:
The given equation is:-
\[\left| {\dfrac{{x - 8}}{2}} \right| = 5\]
Splitting the modulus function we get:-
$ \Rightarrow$\[\dfrac{{\left| {x - 8} \right|}}{{\left| 2 \right|}} = 5\]
\[ \Rightarrow \dfrac{{\left| {x - 8} \right|}}{2} = 5\]
On cross multiplying we get:-
$ \Rightarrow$\[\left| {x - 8} \right| = 10\]……………………. (1)
Now we know that the property of modulus function is:-
$ \Rightarrow$\[\left| x \right| = \left[ {\begin{array}{*{20}{c}}
x&{x \geqslant 0} \\
{ - x}&{x < 0}
\end{array}} \right]\]
Applying this property we get:-
$ \Rightarrow$\[\left| {x - 8} \right| = \left[ {\begin{array}{*{20}{c}}
{x - 8}&{x \geqslant 8} \\
{ - \left( {x - 8} \right)}&{x < 8}
\end{array}} \right]\]
Simplifying it further we get:-
$ \Rightarrow$\[\left| {x - 8} \right| = \left[ {\begin{array}{*{20}{c}}
{x - 8}&{x \geqslant 8} \\
{8 - x}&{x < 8}
\end{array}} \right]\]
Putting this value in equation 1 we get:-
$ \Rightarrow$\[\left| {x - 8} \right| = \left[ {\begin{array}{*{20}{c}}
{x - 8}&{x \geqslant 8} \\
{8 - x}&{x < 8}
\end{array}} \right] = 10\]
Now let us consider the case \[x \geqslant 8\]we get:-
$ \Rightarrow$\[x - 8 = 10\]
Solving for the value of x we get:-
$ \Rightarrow$\[x = 18\]
Now let us consider the case \[x < 8\] we get:-
$ \Rightarrow$\[8 - x = 10\]
Solving for the value of x we get:-
$ \Rightarrow$\[x = - 2\]
Therefore, we get two values of x as 18 and -2.
Additional Information: The modulus function \[\left| x \right| = \left[ {\begin{array}{*{20}{c}}
x&{x \geqslant 0} \\
{ - x}&{x < 0}
\end{array}} \right]\]is discontinuous at x=0.
For any real number x, \[\sqrt {{x^2}} = \left| x \right|\]
The domain of modulus function is any real number i.e. R.
The range of the modulus function is all positive real numbers i.e. \[\left[ {0,\infty } \right)\]
Note: Students should take a note that the value of modulus is always positive i.e. even if the value inside the modulus function is negative then also, the resultant value of mod would be positive.
Also, we find the open modulus function at the critical point i.e. the point where the value of the quantity inside the mod becomes equal to zero.
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