
x is an integer such that $ - x\left| x \right| > 4$. Which is greater:
A) $x$
B) 2
Answer
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Hint: As the modulus function will give two values of x. First, find the solution for the positive value of x. After that find the solution of the negative value of x. Then check whether the value of x is greater than 2 or not. It will give the desired result.
Complete step-by-step answer:
The expression in which a modulus can be defined is:
$f\left( x \right) = \left\{ \begin{gathered}
x & if\,x \geqslant 0 \\
- x & if\,x < 0 \\
\end{gathered} \right.$
Here, x represents any non-negative number, and the function generates a positive equivalent of x. For a negative number, \[x < 0\], the function generates (-x) where
$ - \left( { - x} \right) = $the positive value of x.
Case I: $x < 0$
Substitute $ - x$ in place of $\left| x \right|$ in the equation. Then,
$ \Rightarrow - x\left( { - x} \right) > 4$
Multiply the term on the left side and move the constant part on the left side,
$ \Rightarrow {x^2} - 4 > 0$
We know that,
${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$
Factorize the equation on the above formula,
$ \Rightarrow \left( {x - 2} \right)\left( {x + 2} \right) > 0$
As we know that when $\left( {x - a} \right)\left( {x - b} \right) > 0$ and $a > b$. Then,
$x > a$ or $x < b$
Use the above formula to find the value of $x$,
$ \Rightarrow x > 2$ or $x < - 2$
Since $x$ is negative.
$\therefore x < - 2$
Case II: $x > 0$
Substitute $x$ in place of $\left| x \right|$ in the equation. Then,
$ \Rightarrow - x\left( x \right) > 4$
Multiply the term on the left side,
$ \Rightarrow - {x^2} > 4$
Multiply both sides by -1,
$ \Rightarrow {x^2} < - 4$
As we know that the value of the square cannot be less than 0.
So neglect it.
Thus, $x < - 2$.
As $x < - 2$ means $x < 2$.
Hence, option (B) is greater.
Note: The modulus function generally refers to the function that gives the positive value of any variable or a number. Also known as the absolute value function, it can generate a non-negative value for any independent variable, irrespective of it being positive or negative. Commonly represented as: \[y = \left| x \right|\], where x represents a real number, and \[y = f\left( x \right)\], representing all positive real numbers, including 0, and \[f:R \to R\] and \[x \in R\].
Complete step-by-step answer:
The expression in which a modulus can be defined is:
$f\left( x \right) = \left\{ \begin{gathered}
x & if\,x \geqslant 0 \\
- x & if\,x < 0 \\
\end{gathered} \right.$
Here, x represents any non-negative number, and the function generates a positive equivalent of x. For a negative number, \[x < 0\], the function generates (-x) where
$ - \left( { - x} \right) = $the positive value of x.
Case I: $x < 0$
Substitute $ - x$ in place of $\left| x \right|$ in the equation. Then,
$ \Rightarrow - x\left( { - x} \right) > 4$
Multiply the term on the left side and move the constant part on the left side,
$ \Rightarrow {x^2} - 4 > 0$
We know that,
${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$
Factorize the equation on the above formula,
$ \Rightarrow \left( {x - 2} \right)\left( {x + 2} \right) > 0$
As we know that when $\left( {x - a} \right)\left( {x - b} \right) > 0$ and $a > b$. Then,
$x > a$ or $x < b$
Use the above formula to find the value of $x$,
$ \Rightarrow x > 2$ or $x < - 2$
Since $x$ is negative.
$\therefore x < - 2$
Case II: $x > 0$
Substitute $x$ in place of $\left| x \right|$ in the equation. Then,
$ \Rightarrow - x\left( x \right) > 4$
Multiply the term on the left side,
$ \Rightarrow - {x^2} > 4$
Multiply both sides by -1,
$ \Rightarrow {x^2} < - 4$
As we know that the value of the square cannot be less than 0.
So neglect it.
Thus, $x < - 2$.
As $x < - 2$ means $x < 2$.
Hence, option (B) is greater.
Note: The modulus function generally refers to the function that gives the positive value of any variable or a number. Also known as the absolute value function, it can generate a non-negative value for any independent variable, irrespective of it being positive or negative. Commonly represented as: \[y = \left| x \right|\], where x represents a real number, and \[y = f\left( x \right)\], representing all positive real numbers, including 0, and \[f:R \to R\] and \[x \in R\].
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