KCET · Maths · Differentiation
For any two real numbers, an operation \( * \) defined by \( a * b=1+a b \) is
- A commutative but not associative
- B associative but not commutative
- C neither commutative nor associative
- D both commutative and associative
Answer & Solution
Correct Answer
(A) commutative but not associative
Step-by-step Solution
Detailed explanation
Given that, \(a * b=1+a b \rightarrow(1)\)
Now, \(b * a=1+b a=1+a b=a * b\)
\(\Rightarrow a * b=b^{*} a\)
Thus, \({ }^{*}\) is commutative.
Now, \((a * b) * c=(1+a b) * c\)
\(=1+(1+a b) c=1+c+a b c\)
But, \(a *(b * c)=a *(1+b c)\)
\(=1+a(1+b c)=1+a+a b c\)
So, \((a * b) * c \neq a *(b * c)\)
Thus, \({ }^{*}\) is not associative.
Hence, \({ }^{*}\) is commutative but not associative.
Now, \(b * a=1+b a=1+a b=a * b\)
\(\Rightarrow a * b=b^{*} a\)
Thus, \({ }^{*}\) is commutative.
Now, \((a * b) * c=(1+a b) * c\)
\(=1+(1+a b) c=1+c+a b c\)
But, \(a *(b * c)=a *(1+b c)\)
\(=1+a(1+b c)=1+a+a b c\)
So, \((a * b) * c \neq a *(b * c)\)
Thus, \({ }^{*}\) is not associative.
Hence, \({ }^{*}\) is commutative but not associative.
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