Mark the correct alternative in the following question for the bi… - Vidyadip

MCQ

Mark the correct alternative in the following question for the binary operation $*$ on $Z$ defined by $a^ * b = a + b + 1,$ the identity element is:

A
$0$
✓
$-1$
C
$1$
D
$2$

✓

Answer

Correct option: B.

$-1$

We have,
$a ^* b = a + b + 1$
Let e be the identity element of $*.$ Then,
$a^ * e = a = e ^* a$
$a + e + 1 = a$
$e = a - a - 1$
$e = -1$

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 RELATIONS AND FUNCTIONS→RELATIONS AND FUNCTIONS — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Choose the correct answer from the given four options. Let $f : A \rightarrow B$ and $g : B \rightarrow C$ be the bijective functions. Then $(gof)^{-1}$ is:

$\int_{}^{} {{a^{3x + 3}}dx} = $

${d \over {dx}}(\sin 2{x^2})$ equals

If $|a|\,\, = 3,\,\,\,|b|\,\, = 4$ and $|a + b|\,\, = 5,$ then $|a - b|\,\, = $

Let $f (x) =$ $\left[ {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {\frac{{\begin{array}{*{20}{c}} {{2^x}}&{ + {2^{3 - x}}}&{ - 6} \end{array}}}{{\begin{array}{*{20}{c}} {\sqrt {{2^{ - x}}} }&{ - {2^{1 - x}}} \end{array}}}}&{if}&{x > 2} \end{array}} \\ {} \\ {} \\ {} \\ {\begin{array}{*{20}{c}} {\frac{{\begin{array}{*{20}{c}} {{x^2}}&{ - 4} \end{array}}}{{x - \sqrt {3x - 2} }}}&{if}&{x < 2} \end{array}} \end{array}} \right.$ then

The magnitude of the vector $\frac{1}{\sqrt{3}} \hat{i}+\frac{1}{\sqrt{3}} \hat{j}-\frac{1}{\sqrt{3}} \hat{k}$ is:

If $\vec{\text{a}}.\hat{\text{i}}=\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}\big)=\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)=1.$then $\vec{\text{a}}=$

Let f : R → R be a function defined by $\text{f(x)}=\frac{\text{e}^{|\text{x}|}-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}.$ Then,

The value of the integral $\int_{ - 1}^1 {\frac{d}{{dx}}\left( {{{\tan }^{ - 1}}\frac{1}{x}} \right)} \,dx$ is

$\int\tan^{-1}\sqrt{\text{x dx}}$ is equal to :