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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