Exercise 3.1 Solutions – Class X Mathematics
These solutions are based on the Telangana State Class X Mathematics textbook, focusing on polynomials. Mathematical expressions are rendered using MathJax.
1. In \( p(x) = 5x^2 – 6x + 7x – 6 \), what is the
(i) coefficient of \( x^2 \)
First, simplify the polynomial: \( p(x) = 5x^2 – 6x + 7x – 6 \).
Combine like terms: \( -6x + 7x = x \), so \( p(x) = 5x^2 + x – 6 \).
The term with \( x^2 \) is \( 5x^2 \), so the coefficient of \( x^2 \) is 5.
Coefficient of \( x^2 \): 5
(ii) degree of \( p(x) \)
The simplified polynomial is \( p(x) = 5x^2 + x – 6 \).
The degree of a polynomial is the highest power of \( x \).
Here, the highest power is 2 (from \( 5x^2 \)).
Degree of \( p(x) \): 2
(iii) constant term
The simplified polynomial is \( p(x) = 5x^2 + x – 6 \).
The constant term is the term without \( x \), which is \( -6 \).
Constant term: -6
2. State which of the following statements are true and which are false? Give reasons for your choice.
(i) The degree of the polynomial \( \sqrt{2} x^2 – 3x + 1 \) is \( \sqrt{2} \).
The degree of a polynomial is the highest power of \( x \), not the coefficient.
In \( \sqrt{2} x^2 – 3x + 1 \), the highest power of \( x \) is 2 (from \( \sqrt{2} x^2 \)).
The degree is 2, not \( \sqrt{2} \), which is the coefficient of \( x^2 \).
Conclusion: False, Reason: Degree is 2, not \( \sqrt{2} \).
(ii) The coefficient of \( x^2 \) in the polynomial \( p(x) = 3x^3 – 4x^2 + 5x + 7 \) is 2.
The polynomial is \( p(x) = 3x^3 – 4x^2 + 5x + 7 \).
The term with \( x^2 \) is \( -4x^2 \), so the coefficient of \( x^2 \) is \( -4 \).
The statement claims the coefficient is 2, which is incorrect.
Conclusion: False, Reason: Coefficient of \( x^2 \) is \( -4 \), not 2.
(iii) The degree of a constant term is zero.
A constant term (e.g., 5) can be written as \( 5x^0 \), since \( x^0 = 1 \).
Thus, the degree of a constant term is 0.
Conclusion: True, Reason: A constant term has degree 0.
(iv) \( \frac{1}{x^2 – 5x + 6} \) is a quadratic polynomial.
A polynomial has non-negative integer powers of \( x \).
The expression \( \frac{1}{x^2 – 5x + 6} \) has \( x^2 – 5x + 6 \) in the denominator, making it a rational function.
The numerator is 1 (degree 0), and the denominator is a quadratic (degree 2), but the expression itself is not a polynomial.
Conclusion: False, Reason: The expression is a rational function, not a polynomial.
(v) The degree of a polynomial is one more than the number of terms in it.
Consider a polynomial like \( x^2 + x + 1 \): 3 terms, degree 2 (not 3 + 1).
Another example: \( 5x + 2 \), 2 terms, degree 1 (not 2 + 1).
The degree depends on the highest power, not the number of terms.
Conclusion: False, Reason: Degree is not related to the number of terms.
3. If \( p(t) = t^3 – 1 \), find the values of \( p(1), p(-1), p(0), p(2), p(-2) \).
The polynomial is \( p(t) = t^3 – 1 \).
For \( p(1) \): \( p(1) = 1^3 – 1 = 1 – 1 = 0 \).
For \( p(-1) \): \( p(-1) = (-1)^3 – 1 = -1 – 1 = -2 \).
For \( p(0) \): \( p(0) = 0^3 – 1 = 0 – 1 = -1 \).
For \( p(2) \): \( p(2) = 2^3 – 1 = 8 – 1 = 7 \).
For \( p(-2) \): \( p(-2) = (-2)^3 – 1 = -8 – 1 = -9 \).
Values: \( p(1) = 0, p(-1) = -2, p(0) = -1, p(2) = 7, p(-2) = -9 \)
4. Check whether -2 and 2 are the zeros of the polynomial \( x^4 – 16 \).
The polynomial is \( p(x) = x^4 – 16 \).
A number \( a \) is a zero if \( p(a) = 0 \).
For \( x = -2 \): \( p(-2) = (-2)^4 – 16 = 16 – 16 = 0 \).
For \( x = 2 \): \( p(2) = 2^4 – 16 = 16 – 16 = 0 \).
Since \( p(-2) = 0 \) and \( p(2) = 0 \), both -2 and 2 are zeros.
Conclusion: Yes, -2 and 2 are zeros of the polynomial.
5. Check whether 3 and -2 are the zeros of the polynomial \( p(x) \) when \( p(x) = x^2 – x – 6 \).
The polynomial is \( p(x) = x^2 – x – 6 \).
A number \( a \) is a zero if \( p(a) = 0 \).
For \( x = 3 \): \( p(3) = 3^2 – 3 – 6 = 9 – 3 – 6 = 0 \).
For \( x = -2 \): \( p(-2) = (-2)^2 – (-2) – 6 = 4 + 2 – 6 = 0 \).
Since \( p(3) = 0 \) and \( p(-2) = 0 \), both 3 and -2 are zeros.
Conclusion: Yes, 3 and -2 are zeros of the polynomial.