Q51. - Algebra 2

Question

For which function does

$f(-\frac{1}{2})\ne -1?$

$\begin{array}{l}\left(A\right)f\left(x\right)=2x\\ \left(B\right)f\left(x\right)=|-2x|\\ \left(C\right)f\left(x\right)=\left[\text{}\left[x\right]\text{}\right]\\ \left(D\right)f\left(x\right)=\left[\text{}\left[2x\right]\text{}\right]\end{array}$

Step-by-Step Solution

Answer

Option (B) $f\left(x\right)=\left|-2x\right|$ is correct

Step-1 – Absolute value of the function.

An absolute value function is a function that contains an algebraic expression with in absolute value symbols

Step-2 – Substitute the value of x into the given function.

For option A) $f\left(x\right)=2x$

Put $x=-\frac{1}{2}$ into the given function

$\begin{array}{l}f\left(x\right)=2(-\frac{1}{2})\\ f\left(x\right)=-1\end{array}$

Hence, value of function is 

$f(-\frac{1}{2})=-1.$

Therefore, option (A) is not correct.

For option B)

Put $x=-\frac{1}{2}$ into$f\left(x\right)=\left|-2x\right|$

$\begin{array}{l}f\left(x\right)=|-2(-\frac{1}{2})|\\ f\left(x\right)=\left|1\right|\\ f\left(x\right)=1\end{array}$

Hence, value of function is$f\left(-\frac{1}{2}\right)\ne -1.$

Therefore, option (B) is correct.

Step - 3 – Solve for given function.

For (C)$f\left(x\right)=\left[\left[x\right]\right]$

Put  $x=-\frac{1}{2}$ into the given function

$\begin{array}{l}f\left(x\right)=\left[\text{}\left[x\right]\text{}\right]\\ f\left(x\right)=\left[\text{}[-\frac{1}{2}]\text{}\right]\\ f\left(x\right)=-1\end{array}$

Therefore, option (C) is incorrect.

 For (D) $f\left(x\right)=\left[\left[2x\right]\right]$

Put  $x=-\frac{1}{2}$ into the given function

$\begin{array}{l}f\left(x\right)=\left[\text{}\left[2x\right]\text{}\right]\\ f\left(x\right)=\left[\text{}\left[2(-\frac{1}{2})\right]\text{}\right]\\ f\left(x\right)=-1\end{array}$

Therefore, option (D) is incorrect.

Hence, value of function is $f\left(-\frac{1}{2}\right)=-1.$

Therefore, option (D) is not correct