Question
Which of the following statements is incorrect :
A.
$$x\,\operatorname{sgn} \,x = \left| x \right|$$
B.
$$\left| x \right|\operatorname{sgn} \,x = x$$
C.
$$x\left( {\operatorname{sgn} \,x} \right)\left( {\operatorname{sgn} \,x} \right) = x$$
D.
$$\left| x \right|{\left( {\operatorname{sgn} \,x} \right)^3} = \left| x \right|$$
Answer :
$$\left| x \right|{\left( {\operatorname{sgn} \,x} \right)^3} = \left| x \right|$$
Solution :
Checking the options
$$\eqalign{
& \left( {\bf{A}} \right){\text{ L}}{\text{.H}}{\text{.S}}{\text{.}} = x\,\operatorname{sgn} \,x = x \times \left( { + 1} \right) = x{\text{ if }}x > 0 \cr
& x \times 0 = 0{\text{ if }}x = 0 \cr
& x \times \left( { - 1} \right) = - x{\text{ if }}x < 0 \cr
& = \left| x \right| = {\text{R}}{\text{.H}}{\text{.S}}{\text{.}} \cr} $$
Therefore, statement is correct.
$$\eqalign{
& \left( {\bf{B}} \right){\text{ L}}{\text{.H}}{\text{.S}}{\text{.}} = \left| x \right|\operatorname{sgn} \,x = \left( { + x} \right) \times \left( { + 1} \right) = x{\text{ if }}x > 0 \cr
& 0 \times 0 = 0{\text{ if }}x = 0 \cr
& \left( { - x} \right) \times \left( { - 1} \right) = x{\text{ if }}x < 0 \cr} $$
Therefore, statement is correct.
$$\eqalign{
& \left( {\bf{C}} \right){\text{ L}}{\text{.H}}{\text{.S}}{\text{.}} = \left( {x\,\left( {\operatorname{sgn} \,x} \right)} \right) \times \operatorname{sgn} \,x \cr
& = \left| x \right|\operatorname{sgn} \,x\,\,\,\,\,\left( {{\text{from }}\left( {\text{A}} \right)} \right) \cr
& = x\,\,\,\,\,\left( {{\text{from }}\left( {\text{B}} \right)} \right) = {\text{R}}{\text{.H}}{\text{.S}}{\text{.}} \cr} $$
Therefore, statement is correct.
$$\eqalign{
& \left( {\bf{D}} \right){\text{ L}}{\text{.H}}{\text{.S}}{\text{.}} = \left( {\left| x \right|\operatorname{sgn} \,x} \right){\left( {\operatorname{sgn} \,x} \right)^2} \cr
& = x{\left( {\operatorname{sgn} \,x} \right)^2}\,\,\,\,\,\left( {{\text{from}}\left( {\text{B}} \right)} \right) \cr
& = \left| x \right| \times \left( {\operatorname{sgn} \,x} \right)\,\,\,\,\,\left( {{\text{from}}\left( {\text{A}} \right)} \right) \cr
& = x\,\,\,\,\,\left( {{\text{from}}\left( {\text{B}} \right)} \right) \ne {\text{R}}{\text{.H}}{\text{.S}}{\text{.}} \cr} $$
Therefore, statement is incorrect.