CBSE Class 9 Maths Exemplar Chapter 2: Polynomials NCERT Solutions

To determine which expression is a polynomial, we need to check if the exponents of the variables are non-negative integers.

1. $\backslash frac\{x^2\}\{2\}\; -\; \backslash frac\{2\}\{x^2\}$ can be rewritten as $\backslash frac\{1\}\{2\}x^2\; -\; 2x^\{-2\}$. The term $x^\{-2\}$ has a negative exponent (-2), so this is not a polynomial.

2. $\backslash sqrt\{2x\}\; -\; 1$ can be rewritten as $\backslash sqrt\{2\}x^\{\backslash frac\{1\}\{2\}\}\; -\; 1$. The term $x^\{\backslash frac\{1\}\{2\}\}$ has a fractional exponent (1/2), so this is not a polynomial.

3. $x^2\; +\; \backslash frac\{3x^\{\backslash frac\{3\}\{2\}\}\}\{\backslash sqrt\{x\}\}$ can be simplified. First, $\backslash sqrt\{x\}$ is $x^\{\backslash frac\{1\}\{2\}\}$. So, $\backslash frac\{3x^\{\backslash frac\{3\}\{2\}\}\}\{x^\{\backslash frac\{1\}\{2\}\}\}\; =\; 3x^\{\backslash frac\{3\}\{2\}\; -\; \backslash frac\{1\}\{2\}\}\; =\; 3x^\{\backslash frac\{2\}\{2\}\}\; =\; 3x^1$. The expression becomes $x^2\; +\; 3x$. Both terms have non-negative integer exponents (2 and 1), so this is a polynomial.

4. $\backslash frac\{x-1\}\{x+1\}$ involves division by a variable expression, which does not fit the standard form of a polynomial. It is a rational expression.

Therefore, the correct option is (C).

A constant term can be considered a polynomial. The term $\backslash sqrt\{2\}$ can be written in terms of a variable, say x, as $\backslash sqrt\{2\}x^0$. The degree of a polynomial is the highest power of the variable. In this case, the power of x is 0. Therefore, the degree of the constant polynomial $\backslash sqrt\{2\}$ is 0.

Hence, the correct option is (B).

The degree of a polynomial is defined as the highest power of the variable present in the polynomial. In the given polynomial $4x^4\; +\; 0\; \backslash times\; x^3\; +\; 0\; \backslash times\; x^5\; +\; 5x\; +\; 7$, the terms have powers of x as 4, 3, 5, 1, and 0 (for the constant term 7). The highest power among these is 5. Even though the coefficient of $x^5$ is 0, the term exists conceptually when determining the degree.

Therefore, the degree of the polynomial is 5.

Hence, the correct option is (B).

The zero polynomial is the polynomial where all the coefficients are zero (e.g., $0x^n\; +\; 0x^\{n-1\}\; +\; ...\; +\; 0$). Since there is no highest power of the variable that is uniquely defined (as all coefficients are zero), the degree of the zero polynomial is considered to be not defined.

Hence, the correct option is (D).

We are given the polynomial $p(x)\; =\; x^2\; -\; 2\backslash sqrt\{2\}x\; +\; 1$. To find the value of $p(2\backslash sqrt\{2\})$, we substitute $x\; =\; 2\backslash sqrt\{2\}$ into the polynomial:

$p(2\backslash sqrt\{2\})\; =\; (2\backslash sqrt\{2\})^2\; -\; 2\backslash sqrt\{2\}(2\backslash sqrt\{2\})\; +\; 1$

Now, we calculate the terms:

• $(2\backslash sqrt\{2\})^2\; =\; 2^2\; \backslash times\; (\backslash sqrt\{2\})^2\; =\; 4\; \backslash times\; 2\; =\; 8$

• $2\backslash sqrt\{2\}(2\backslash sqrt\{2\})\; =\; (2\; \backslash times\; 2)\; \backslash times\; (\backslash sqrt\{2\}\; \backslash times\; \backslash sqrt\{2\})\; =\; 4\; \backslash times\; 2\; =\; 8$

Substituting these values back into the expression for $p(2\backslash sqrt\{2\})$:

$p(2\backslash sqrt\{2\})\; =\; 8\; -\; 8\; +\; 1$

$p(2\backslash sqrt\{2\})\; =\; 1$

Hence, the correct option is (B).

We are given the polynomial $p(x)\; =\; 5x\; -\; 4x^2\; +\; 3$. We need to find the value of the polynomial when $x\; =\; -1$. Substitute $x\; =\; -1$ into the polynomial:

$p(-1)\; =\; 5(-1)\; -\; 4(-1)^2\; +\; 3$

Now, we evaluate the terms:

• $5(-1)\; =\; -5$

• $(-1)^2\; =\; 1$

• $4(-1)^2\; =\; 4(1)\; =\; 4$

Substitute these results back into the expression:

$p(-1)\; =\; -5\; -\; 4\; +\; 3$

$p(-1)\; =\; -9\; +\; 3$

$p(-1)\; =\; -6$

The value of the polynomial when $x\; =\; -1$ is -6.