A) x sgn x = \[\left| x \right|\]
B) \[\left| x \right|\] sgn x = x
C) \[x(sgn\,x)(sgn\,x)=x\]
D) \[\left| x \right|\,{{(sgn\,x)}^{3}}=\left| x \right|\]
Correct Answer: D
Solution :
| [d] Checking the options |
| a. LHS \[=x\sgn x=x\times (+1)=x\,\,if\,\,x>0\] |
| \[x\times 0=0\,\,if\,\,x=0\] |
| \[x\times (-1)=-x\,\,if\,\,x<0\] |
| \[=\left| x \right|=RHS\] |
| b. LHS\[=\left| x \right|\operatorname{sgnx}=(+x)\times (+1)=x\,\,if\,\,x>0\] |
| \[0\times 0=0\,\,if\,\,\,x=0\] |
| \[(-x)\times (-1)=x\,\,if\,\,x<0\] |
| Therefore, statement is correct. |
| c. \[LHS=(x(sgn\,\,x))\times sgn\,x=\left| x \right|\sgn x\] (from (a))=x (from (b))= RHS |
| d. LHS \[(\left| x \right|sgn\,x){{(sgn\,x)}^{2}}=x{{(sgn\,x)}^{2}}(from(b))\] |
| \[=\left| x \right|\times (sgnx)\,\] (From (a)) |
| \[=x\] (From (b)) |
| \[\ne RHS\] |
| Therefore, statement is incorrect. |
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