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# Which of the following is a linear polynomial?(a) $p\left( x \right) = x$ (b) $p\left( x \right) = y$ (c) $p\left( y \right) = x$ (d) $p\left( y \right) = 1$

Last updated date: 24th Jun 2024
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Hint: Here, we need to find which of the given options is a linear polynomial. We will use the concept of a linear polynomial to check whether the given options are a linear polynomial or not. A linear polynomial is a polynomial whose highest degree is 1. Degree is the power of the variable in the polynomial.

A linear polynomial in $x$ is of the form $p\left( x \right) = ax + b$, where $a$ is the coefficient of $x$, $b$ is the constant, and $a \ne 0$. Here, the highest degree of $x$ is 1.
Now, we will check which of the options is a linear polynomial.
The polynomial $p\left( x \right) = x$ is a polynomial in the variable $x$.
Rewriting the polynomial, we get
$\Rightarrow p\left( x \right) = {x^1}$
Here, the degree of $x$ is 1.
Therefore, $p\left( x \right) = x$ is a linear polynomial in $x$.
Let us now check the other options as well.
The polynomial $p\left( x \right) = y$ is a polynomial in the variable $x$.
Rewriting the polynomial, we get
$\begin{array}{l} \Rightarrow p\left( x \right) = 0 + y\\ \Rightarrow p\left( x \right) = 0 \times x + y\\ \Rightarrow p\left( x \right) = 0{x^1} + y\end{array}$
We can observe that $p\left( x \right) = y$ can be written in the form $p\left( x \right) = ax + b$ such that the highest degree of $x$ is 1.
However, since $a = 0$, this is not a linear polynomial.
Therefore, $p\left( x \right) = y$ is not a linear polynomial.
Thus, option (b) is incorrect.
The polynomial $p\left( y \right) = x$ is a polynomial in the variable $y$.
A linear polynomial in $y$ will be of the form $p\left( y \right) = ay + b$, where $a \ne 0$.
Rewriting the polynomial, we get
$\begin{array}{l} \Rightarrow p\left( y \right) = 0 + x\\ \Rightarrow p\left( y \right) = 0 \times y + x\\ \Rightarrow p\left( y \right) = 0{y^1} + x\end{array}$
We can observe that $p\left( y \right) = x$ can be written in the form $p\left( y \right) = ay + b$ such that the highest degree of $y$ is 1.
However, since $a = 0$, this is not a linear polynomial.
Therefore, $p\left( y \right) = x$ is not a linear polynomial.
Thus, option (c) is incorrect.
The polynomial $p\left( y \right) = 1$ is a polynomial in the variable $y$.
A linear polynomial in $y$ will be of the form $p\left( y \right) = ay + b$, where $a \ne 0$.
Rewriting the polynomial, we get
$\begin{array}{l} \Rightarrow p\left( y \right) = 0 + 1\\ \Rightarrow p\left( y \right) = 0 \times y + 1\\ \Rightarrow p\left( y \right) = 0{y^1} + 1\end{array}$
We can observe that $p\left( y \right) = 1$ can be written in the form $p\left( y \right) = ay + b$ such that the highest degree of $y$ is 1.
However, since $a = 0$, this is not a linear polynomial.
Therefore, $p\left( y \right) = 1$ is not a linear polynomial.
Thus, option (d) is incorrect.
Therefore, the only correct option is option (a) $p\left( x \right) = x$.

Note: Here we have asked to find linear polynomials. For this, first we need to understand the meaning of polynomial. A polynomial is an expression consisting of variables and constants, involving some operations between them, like addition, subtraction, multiplication, or division. Some examples of polynomials are $3x + 2$ and $3y + 6$.A polynomial in $x$ is usually denoted by $p\left( x \right)$.We have also used the property of exponents ${a^0} = 1$ to rewrite the given polynomials in the form $p\left( x \right) = ax + b$ or $p\left( y \right) = ay + b$. According to this property, any number raised to the power 0 is equal to 1.