90110
If 2 is the zero of both the polynomials 3x\(^{1}\) + mx - 14 and 2x\(^{1}\) + nx\(^{1}\) + x - 2, then the value of m - 2n is:
1 5
2 -1
3 -9
4 9
Explanation:
9 According to the question, p(2) = 3x\(^{1}\) + mx - 14 = 0 ? 3(2)\(^{1}\) + m × 2 - 14 = 0 ? 12 + 2m - 14 = 0 ? m = 1
POLYNOMIALS
90111
If \(\sqrt{5}\) and \(-\sqrt{5}\) are two zeroes of the polynomial x\(^{1}\) + 3x\(^{1}\) - 5x - 15, then its third zero is:
1 3
2 -3
3 5
4 -5
Explanation:
-3 Let \(\alpha=\sqrt{5}\) and \(\beta=-\sqrt{5}\) be the given zeros and \(\gamma\) be the third zero of the polynomial x\(^{1}\) + 3x\(^{1}\) - 5x - 15 Then, By using \(\alpha+\beta+\gamma=\frac{-\text{Coefficient of x}^2}{\text{Coefficient of x}^3}\) \(\alpha+\beta+\gamma=\frac{-3}{1}\) \(\alpha+\beta+\gamma=-3\) Substituting \(\alpha=\sqrt{5}\) and \(\beta=-\sqrt{5}\) in \(\alpha+\beta+\gamma=-3\) We get \(\sqrt{5}-\sqrt{5}+\gamma=-3\) \(\gamma=-3\) Hence, the correct choice is (b)
POLYNOMIALS
90112
The zeros of the polynomial x\(^{1}\) - 2x - 3 are:
90113
If \(\alpha,\beta\) are the zeroes of the polynomial f(x) = x\(^{1}\) + x + 1, then \(\frac{1}{\alpha}+\frac{1}{\beta} = \)
1 0
2 -1
3 1
4 None of these
Explanation:
-1 Since \(\alpha\) and \(\beta\) are the zeroes of the quadratic polynomial f(x) = x\(^{1}\)+ x + 1 \(\alpha+\beta = \frac{-\text{Coefficient of }{\text{x}}}{\text{Coefficient of }\text{x}^{2}} = \frac{-1}{1} = {-1}\) \(\alpha\times\beta = \frac{\text{Constant term}}{\text{Coefficient of}{\text{x}^{2}}} = \frac{1}{1} = {1}\) Now \(\frac{1}{\alpha}+\frac{1}{\beta} = \frac{\beta+\alpha}{\alpha\beta}=\frac{-1}{1} = {-1}\) Thus the value of \(\frac{1}{\alpha}+\frac{1}{\beta}\) is -1
90110
If 2 is the zero of both the polynomials 3x\(^{1}\) + mx - 14 and 2x\(^{1}\) + nx\(^{1}\) + x - 2, then the value of m - 2n is:
1 5
2 -1
3 -9
4 9
Explanation:
9 According to the question, p(2) = 3x\(^{1}\) + mx - 14 = 0 ? 3(2)\(^{1}\) + m × 2 - 14 = 0 ? 12 + 2m - 14 = 0 ? m = 1
POLYNOMIALS
90111
If \(\sqrt{5}\) and \(-\sqrt{5}\) are two zeroes of the polynomial x\(^{1}\) + 3x\(^{1}\) - 5x - 15, then its third zero is:
1 3
2 -3
3 5
4 -5
Explanation:
-3 Let \(\alpha=\sqrt{5}\) and \(\beta=-\sqrt{5}\) be the given zeros and \(\gamma\) be the third zero of the polynomial x\(^{1}\) + 3x\(^{1}\) - 5x - 15 Then, By using \(\alpha+\beta+\gamma=\frac{-\text{Coefficient of x}^2}{\text{Coefficient of x}^3}\) \(\alpha+\beta+\gamma=\frac{-3}{1}\) \(\alpha+\beta+\gamma=-3\) Substituting \(\alpha=\sqrt{5}\) and \(\beta=-\sqrt{5}\) in \(\alpha+\beta+\gamma=-3\) We get \(\sqrt{5}-\sqrt{5}+\gamma=-3\) \(\gamma=-3\) Hence, the correct choice is (b)
POLYNOMIALS
90112
The zeros of the polynomial x\(^{1}\) - 2x - 3 are:
90113
If \(\alpha,\beta\) are the zeroes of the polynomial f(x) = x\(^{1}\) + x + 1, then \(\frac{1}{\alpha}+\frac{1}{\beta} = \)
1 0
2 -1
3 1
4 None of these
Explanation:
-1 Since \(\alpha\) and \(\beta\) are the zeroes of the quadratic polynomial f(x) = x\(^{1}\)+ x + 1 \(\alpha+\beta = \frac{-\text{Coefficient of }{\text{x}}}{\text{Coefficient of }\text{x}^{2}} = \frac{-1}{1} = {-1}\) \(\alpha\times\beta = \frac{\text{Constant term}}{\text{Coefficient of}{\text{x}^{2}}} = \frac{1}{1} = {1}\) Now \(\frac{1}{\alpha}+\frac{1}{\beta} = \frac{\beta+\alpha}{\alpha\beta}=\frac{-1}{1} = {-1}\) Thus the value of \(\frac{1}{\alpha}+\frac{1}{\beta}\) is -1
90110
If 2 is the zero of both the polynomials 3x\(^{1}\) + mx - 14 and 2x\(^{1}\) + nx\(^{1}\) + x - 2, then the value of m - 2n is:
1 5
2 -1
3 -9
4 9
Explanation:
9 According to the question, p(2) = 3x\(^{1}\) + mx - 14 = 0 ? 3(2)\(^{1}\) + m × 2 - 14 = 0 ? 12 + 2m - 14 = 0 ? m = 1
POLYNOMIALS
90111
If \(\sqrt{5}\) and \(-\sqrt{5}\) are two zeroes of the polynomial x\(^{1}\) + 3x\(^{1}\) - 5x - 15, then its third zero is:
1 3
2 -3
3 5
4 -5
Explanation:
-3 Let \(\alpha=\sqrt{5}\) and \(\beta=-\sqrt{5}\) be the given zeros and \(\gamma\) be the third zero of the polynomial x\(^{1}\) + 3x\(^{1}\) - 5x - 15 Then, By using \(\alpha+\beta+\gamma=\frac{-\text{Coefficient of x}^2}{\text{Coefficient of x}^3}\) \(\alpha+\beta+\gamma=\frac{-3}{1}\) \(\alpha+\beta+\gamma=-3\) Substituting \(\alpha=\sqrt{5}\) and \(\beta=-\sqrt{5}\) in \(\alpha+\beta+\gamma=-3\) We get \(\sqrt{5}-\sqrt{5}+\gamma=-3\) \(\gamma=-3\) Hence, the correct choice is (b)
POLYNOMIALS
90112
The zeros of the polynomial x\(^{1}\) - 2x - 3 are:
90113
If \(\alpha,\beta\) are the zeroes of the polynomial f(x) = x\(^{1}\) + x + 1, then \(\frac{1}{\alpha}+\frac{1}{\beta} = \)
1 0
2 -1
3 1
4 None of these
Explanation:
-1 Since \(\alpha\) and \(\beta\) are the zeroes of the quadratic polynomial f(x) = x\(^{1}\)+ x + 1 \(\alpha+\beta = \frac{-\text{Coefficient of }{\text{x}}}{\text{Coefficient of }\text{x}^{2}} = \frac{-1}{1} = {-1}\) \(\alpha\times\beta = \frac{\text{Constant term}}{\text{Coefficient of}{\text{x}^{2}}} = \frac{1}{1} = {1}\) Now \(\frac{1}{\alpha}+\frac{1}{\beta} = \frac{\beta+\alpha}{\alpha\beta}=\frac{-1}{1} = {-1}\) Thus the value of \(\frac{1}{\alpha}+\frac{1}{\beta}\) is -1
90110
If 2 is the zero of both the polynomials 3x\(^{1}\) + mx - 14 and 2x\(^{1}\) + nx\(^{1}\) + x - 2, then the value of m - 2n is:
1 5
2 -1
3 -9
4 9
Explanation:
9 According to the question, p(2) = 3x\(^{1}\) + mx - 14 = 0 ? 3(2)\(^{1}\) + m × 2 - 14 = 0 ? 12 + 2m - 14 = 0 ? m = 1
POLYNOMIALS
90111
If \(\sqrt{5}\) and \(-\sqrt{5}\) are two zeroes of the polynomial x\(^{1}\) + 3x\(^{1}\) - 5x - 15, then its third zero is:
1 3
2 -3
3 5
4 -5
Explanation:
-3 Let \(\alpha=\sqrt{5}\) and \(\beta=-\sqrt{5}\) be the given zeros and \(\gamma\) be the third zero of the polynomial x\(^{1}\) + 3x\(^{1}\) - 5x - 15 Then, By using \(\alpha+\beta+\gamma=\frac{-\text{Coefficient of x}^2}{\text{Coefficient of x}^3}\) \(\alpha+\beta+\gamma=\frac{-3}{1}\) \(\alpha+\beta+\gamma=-3\) Substituting \(\alpha=\sqrt{5}\) and \(\beta=-\sqrt{5}\) in \(\alpha+\beta+\gamma=-3\) We get \(\sqrt{5}-\sqrt{5}+\gamma=-3\) \(\gamma=-3\) Hence, the correct choice is (b)
POLYNOMIALS
90112
The zeros of the polynomial x\(^{1}\) - 2x - 3 are:
90113
If \(\alpha,\beta\) are the zeroes of the polynomial f(x) = x\(^{1}\) + x + 1, then \(\frac{1}{\alpha}+\frac{1}{\beta} = \)
1 0
2 -1
3 1
4 None of these
Explanation:
-1 Since \(\alpha\) and \(\beta\) are the zeroes of the quadratic polynomial f(x) = x\(^{1}\)+ x + 1 \(\alpha+\beta = \frac{-\text{Coefficient of }{\text{x}}}{\text{Coefficient of }\text{x}^{2}} = \frac{-1}{1} = {-1}\) \(\alpha\times\beta = \frac{\text{Constant term}}{\text{Coefficient of}{\text{x}^{2}}} = \frac{1}{1} = {1}\) Now \(\frac{1}{\alpha}+\frac{1}{\beta} = \frac{\beta+\alpha}{\alpha\beta}=\frac{-1}{1} = {-1}\) Thus the value of \(\frac{1}{\alpha}+\frac{1}{\beta}\) is -1
