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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) \[p\left( y \right) = 1\]

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Last updated date: 24th Jun 2024
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Answer
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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.

Complete step-by-step answer:
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.