Download App

Binary Operations — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSBinary Operations1 Mark
Question
The binary operation * defined on N by a * b = a + b + ab for all a, b ∈ N is:
  1. Commutative only.
  2. Associative only.
  3. Commutative and associative both.
  4. None of these.
✓

Answer

  1. Commutative and associative both.
Solution:
a * b = a + b + ab
b * a = b + a + ba
⇒ a * b = b * a
So * is commutative.
Now,
(a * b) * c
= (a + b + ab) * c
= a + b + ab + c + ca + cb + abc
a * (b * c)
= a * (b + c + bc)
= a + b + c + bc + ab + ac + abc
⇒ (a * b) * c = a * (b * c)
So * is associative.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from Binary Operations→Binary Operations — all question types→MATHS STD 12 Science question papers→Rajasthan Board STD 12 Science question papers→Rajasthan Board question paper generator→

Similar questions

If $\theta=\sin^{-1}\{\sin(-600^\circ)\},$ then one of the possible values of $\theta$ is:
  1. $\frac{\pi}{3}$
  2. $\frac{\pi}{2}$
  3. $\frac{2\pi}{3}$
  4. $-\frac{2\pi}{3}$
→
Total number of equivalence relations defined in the set S = {a, b, c} is:
  1. 5
  2. 3!
  3. 23
  4. 33
→
Find the particular solution for $2 x y+y^2-2 x^2 \frac{d y}{d x}=0 ; y=2$ when $x =1$
→
Matrix $X=\left[\begin{array}{lll}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{array}\right],\left(X^2-X\right)$
→
The area bounded by lines y=x and x = 2 in first quadrant is
→
In the set Z of all integers, which of the following relation R is not an equivalence relation?
  1. xRy : if $\text{x}\leq\text{y}$
  2. xRy : if x = y
  3. xRy : if x - y is an even integer
  4. xRy : if $\text{x}\equiv\text{y}\ (\text{mod 3})$
→
Differential coefficient of $\sec(\tan^{-1}\text{x})$ is:
  1. $\frac{\text{x}}{1+\text{x}^2}$
  2. $\text{x}\sqrt{1+\text{x}^2}$
  3. $\frac{\text{x}}{\sqrt{1+\text{x}^2}}$
  4. $\frac{\text{x}}{\sqrt{1+\text{x}^2}}$
→
In graphical solutions of linear inequalities, solution can be divided into.
  1. One subset
  2. Two subsets
  3. Three subsets
  4. Four subsets
→
If $A \subset B$, then the value of $P(A \cap B)$ will be-
→
If $\text{A}=\begin{bmatrix} \text{2 0 0 0 2 0 0 0 2}\end{bmatrix},$ then $A^5 =$
→