Q63 / 180 · KCET 2016 · Mathematics
Let \( { }^{*} \) be a binary operation defined on \( R \) by \( a * b=\frac{a+b}{4} \forall a, b \in R \) then the operation \( * \) is
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- ACommutative and Associative
- Correct answerBCommutative but not Associative
- CAssociative but not Commutative
- DNeither Associative nor Commutative
Solution
Given that, \( a * b=\frac{a+b}{4} \)
So, \( b * a=\frac{b+a}{4}=\frac{a+b}{4} \)
Then \( a^{*}=b^{*} a \)
Therefore, it is commutative
Now, \( a *(b * c)=a *\left(\frac{b+c}{4}\right) \)
\( =\frac{a+\frac{b+c}{4}}{4}=\frac{4 a+b+c}{16} \rightarrow(1) \)
and \( (a * b) * c=\left(\frac{a+b}{4}\right) * c \)
\( =\frac{\frac{a+b}{4}-c}{4}=\frac{a+b+4 c}{16} \rightarrow(2) \)
Since, \( a *(b * c) \neq(a * b) * c \)
Therefore, it is not associative.
So, \( b * a=\frac{b+a}{4}=\frac{a+b}{4} \)
Then \( a^{*}=b^{*} a \)
Therefore, it is commutative
Now, \( a *(b * c)=a *\left(\frac{b+c}{4}\right) \)
\( =\frac{a+\frac{b+c}{4}}{4}=\frac{4 a+b+c}{16} \rightarrow(1) \)
and \( (a * b) * c=\left(\frac{a+b}{4}\right) * c \)
\( =\frac{\frac{a+b}{4}-c}{4}=\frac{a+b+4 c}{16} \rightarrow(2) \)
Since, \( a *(b * c) \neq(a * b) * c \)
Therefore, it is not associative.
Hence, the correct answer is option: 'B'.