Question
If $\text{AB}=\text{A}$ and $\text{BA = B}$ then $\text{B}^2 $ is equal to:
-
$\text{B}$
-
$\text{A}$
-
$\text{-B}$
-
$\text{B}^2$
If $\text{AB}=\text{A}$ and $\text{BA = B}$ then $\text{B}^2 $ is equal to:
$\text{B}$
$\text{A}$
$\text{-B}$
$\text{B}^2$
Solution:
We have, $\text{AB}=\text{A}$and $\text{BA = B}$
Since, $\text{B}^2=\text{B.B}$
$\text{B}^2=\text{(BA)}.\text{B}$
$\text{B}^2=\text{B}.\text{(AB)}$
$\text{B}^2=\text{B.A}$
$\text{B}^2=\text{B}$
Hence, this is the answer.
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($A$) $f$ has a local minimum at $x=2$
($B$) fhas a local maximum at $x=2$
($C$) $f^{\prime \prime}(2)>f(2)$
($D$) $f(x)-f^{\prime \prime}(x)=0$ for at least one $x \in \mathbb{R}$