118138
If \(\alpha\) is a non-real root of \(x^6=1\), then \(\frac{\alpha^5+\alpha^3+\alpha+1}{\alpha^2+1}\) is equal to
1 \(\alpha^2\)
2 0
3 \(-\alpha^2\)
4 \(\alpha\)
Explanation:
C Given,\(\Rightarrow \alpha^3=1\) \(\frac{\alpha^5+\alpha^3+\alpha+1}{\alpha^2+1}=\frac{\alpha^2+1+\alpha+1}{1+\alpha^2}\) \(=\frac{(0+1)}{-\alpha}=\frac{-1}{\alpha}=\frac{-1 \cdot \alpha^2}{\alpha^3}\) \(\frac{\alpha^5+\alpha^3+\alpha+1}{\alpha^2+1}=-\alpha^2\) \(\because\left(1+\alpha+\alpha^2=-1\right)\) \(\alpha\) is a non real root of \(x^6=1\) \(\Rightarrow \alpha^6=1 \Rightarrow 1+\alpha+\alpha^2=0, \alpha^3=1\) i.e. Cube roots of units
AP EAMCET-2005
Complex Numbers and Quadratic Equation
118139
If \(\alpha, \beta, \gamma\) are the roots of \(x^3+2 x^2-3 x-1=0\) then \(\alpha^{-2}+\beta^{-2}+\gamma^{-2}=\)
1 12
2 13
3 14
4 15
Explanation:
B Given \(\alpha, \beta, \gamma\) are the roots of the equation, \(\mathrm{x}^3+2 \mathrm{x}^2-3 \mathrm{x}-1=0\) \(\therefore \alpha+\beta+\gamma=-2\) \(\alpha \beta+\beta \gamma+\alpha \gamma=-3\) \(\alpha \beta \gamma=+1\) \(\alpha^{-2}+\beta^{-2}+\gamma^{-2}=\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}\) \(=\frac{\beta^2 \gamma^2+\alpha^2 \gamma^2+\alpha^2 \beta^2}{(\alpha \beta \gamma)^2}\) \(=\frac{(\beta \gamma)^2+(\alpha \gamma)^2(\alpha \beta)^2}{(+1)^2}\) We know, \((\alpha \beta)^2+(\alpha \gamma)^2+(\beta \alpha)^2\) \(=(\alpha \beta+\beta \gamma+\alpha \gamma)^2-2\left(\alpha \beta^2 \gamma+\alpha \beta \gamma^2+\alpha^2 \beta \gamma\right)\) \(=[(\alpha \beta)+(\beta \gamma+\alpha \gamma)]^2-2 \alpha \beta \gamma(\alpha+\beta+\gamma)\) \(=(-3)^2-2 \times(+1)(-2)\) \(=9+4=13\) \(\alpha^{-2}+\beta^{-2}+\gamma^{-2}=13\)
AP EAMCET-2005
Complex Numbers and Quadratic Equation
118180
If the roots of the equation \(5 x^2-7 x+k=0\) are reciprocal of each other, then value of \(k\) is
1 5
2 2
3 2
4 1
Explanation:
A Given the equation - \(5 x^2-7 x+k=0\), Let the roots are \(\alpha, \beta\) Given \(\alpha=\frac{1}{\beta}\) Here, \(\alpha+\beta=\frac{7}{5}\) \(\alpha \cdot \beta=\frac{k}{5}\) Given, \(\alpha \beta=1 \Rightarrow \mathrm{k}=5\)
Jamia Millia Islamia-2012
Complex Numbers and Quadratic Equation
118140
If the roots of the equation \(4 x^3-12 x^2+11 x+k=\) 0 are in arithmetic progression, then \(k\) is equal to
1 -3
2 1
3 2
4 3
Explanation:
A Given the roots of the equation \(4 \mathrm{x}^3-12 \mathrm{x}^2+11 \mathrm{x}+\mathrm{k}=0 \text { is in } \mathrm{P} \text {. }\) \(\Rightarrow \mathrm{x}^3-3 \mathrm{x}^2+\frac{11}{4} \mathrm{x}+\frac{\mathrm{k}}{4}=0\) \(\text { Let the roots are } \alpha, \beta, \gamma \text {, We have }\) \(2 \beta=\alpha+\gamma\) \(\alpha+\beta+\gamma=3\) \(\alpha \beta+\beta \gamma+\alpha \gamma=11 / 4\) \(\alpha \beta \gamma=\frac{-\mathrm{k}}{4}\) \(\because \alpha+\beta+\gamma=3\) \(2 \beta+\beta=3\) \(\therefore \beta=1, \alpha+\gamma=2\) \(\beta(\alpha+\gamma)+\gamma \alpha=11 / 4\) \(1 \times 2+\alpha \gamma=11 / 4\) \(\alpha \gamma=\frac{11}{4}-2=\frac{3}{4}\) \(\therefore \frac{3}{4} \times 1=\frac{-\mathrm{k}}{4}\) \(\therefore \mathrm{k}=-3\) \(\therefore \text { Option (a) is correct. }\) Let the roots are \(\alpha, \beta, \gamma\), We have \(\therefore\) Option (a) is correct.
118138
If \(\alpha\) is a non-real root of \(x^6=1\), then \(\frac{\alpha^5+\alpha^3+\alpha+1}{\alpha^2+1}\) is equal to
1 \(\alpha^2\)
2 0
3 \(-\alpha^2\)
4 \(\alpha\)
Explanation:
C Given,\(\Rightarrow \alpha^3=1\) \(\frac{\alpha^5+\alpha^3+\alpha+1}{\alpha^2+1}=\frac{\alpha^2+1+\alpha+1}{1+\alpha^2}\) \(=\frac{(0+1)}{-\alpha}=\frac{-1}{\alpha}=\frac{-1 \cdot \alpha^2}{\alpha^3}\) \(\frac{\alpha^5+\alpha^3+\alpha+1}{\alpha^2+1}=-\alpha^2\) \(\because\left(1+\alpha+\alpha^2=-1\right)\) \(\alpha\) is a non real root of \(x^6=1\) \(\Rightarrow \alpha^6=1 \Rightarrow 1+\alpha+\alpha^2=0, \alpha^3=1\) i.e. Cube roots of units
AP EAMCET-2005
Complex Numbers and Quadratic Equation
118139
If \(\alpha, \beta, \gamma\) are the roots of \(x^3+2 x^2-3 x-1=0\) then \(\alpha^{-2}+\beta^{-2}+\gamma^{-2}=\)
1 12
2 13
3 14
4 15
Explanation:
B Given \(\alpha, \beta, \gamma\) are the roots of the equation, \(\mathrm{x}^3+2 \mathrm{x}^2-3 \mathrm{x}-1=0\) \(\therefore \alpha+\beta+\gamma=-2\) \(\alpha \beta+\beta \gamma+\alpha \gamma=-3\) \(\alpha \beta \gamma=+1\) \(\alpha^{-2}+\beta^{-2}+\gamma^{-2}=\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}\) \(=\frac{\beta^2 \gamma^2+\alpha^2 \gamma^2+\alpha^2 \beta^2}{(\alpha \beta \gamma)^2}\) \(=\frac{(\beta \gamma)^2+(\alpha \gamma)^2(\alpha \beta)^2}{(+1)^2}\) We know, \((\alpha \beta)^2+(\alpha \gamma)^2+(\beta \alpha)^2\) \(=(\alpha \beta+\beta \gamma+\alpha \gamma)^2-2\left(\alpha \beta^2 \gamma+\alpha \beta \gamma^2+\alpha^2 \beta \gamma\right)\) \(=[(\alpha \beta)+(\beta \gamma+\alpha \gamma)]^2-2 \alpha \beta \gamma(\alpha+\beta+\gamma)\) \(=(-3)^2-2 \times(+1)(-2)\) \(=9+4=13\) \(\alpha^{-2}+\beta^{-2}+\gamma^{-2}=13\)
AP EAMCET-2005
Complex Numbers and Quadratic Equation
118180
If the roots of the equation \(5 x^2-7 x+k=0\) are reciprocal of each other, then value of \(k\) is
1 5
2 2
3 2
4 1
Explanation:
A Given the equation - \(5 x^2-7 x+k=0\), Let the roots are \(\alpha, \beta\) Given \(\alpha=\frac{1}{\beta}\) Here, \(\alpha+\beta=\frac{7}{5}\) \(\alpha \cdot \beta=\frac{k}{5}\) Given, \(\alpha \beta=1 \Rightarrow \mathrm{k}=5\)
Jamia Millia Islamia-2012
Complex Numbers and Quadratic Equation
118140
If the roots of the equation \(4 x^3-12 x^2+11 x+k=\) 0 are in arithmetic progression, then \(k\) is equal to
1 -3
2 1
3 2
4 3
Explanation:
A Given the roots of the equation \(4 \mathrm{x}^3-12 \mathrm{x}^2+11 \mathrm{x}+\mathrm{k}=0 \text { is in } \mathrm{P} \text {. }\) \(\Rightarrow \mathrm{x}^3-3 \mathrm{x}^2+\frac{11}{4} \mathrm{x}+\frac{\mathrm{k}}{4}=0\) \(\text { Let the roots are } \alpha, \beta, \gamma \text {, We have }\) \(2 \beta=\alpha+\gamma\) \(\alpha+\beta+\gamma=3\) \(\alpha \beta+\beta \gamma+\alpha \gamma=11 / 4\) \(\alpha \beta \gamma=\frac{-\mathrm{k}}{4}\) \(\because \alpha+\beta+\gamma=3\) \(2 \beta+\beta=3\) \(\therefore \beta=1, \alpha+\gamma=2\) \(\beta(\alpha+\gamma)+\gamma \alpha=11 / 4\) \(1 \times 2+\alpha \gamma=11 / 4\) \(\alpha \gamma=\frac{11}{4}-2=\frac{3}{4}\) \(\therefore \frac{3}{4} \times 1=\frac{-\mathrm{k}}{4}\) \(\therefore \mathrm{k}=-3\) \(\therefore \text { Option (a) is correct. }\) Let the roots are \(\alpha, \beta, \gamma\), We have \(\therefore\) Option (a) is correct.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Complex Numbers and Quadratic Equation
118138
If \(\alpha\) is a non-real root of \(x^6=1\), then \(\frac{\alpha^5+\alpha^3+\alpha+1}{\alpha^2+1}\) is equal to
1 \(\alpha^2\)
2 0
3 \(-\alpha^2\)
4 \(\alpha\)
Explanation:
C Given,\(\Rightarrow \alpha^3=1\) \(\frac{\alpha^5+\alpha^3+\alpha+1}{\alpha^2+1}=\frac{\alpha^2+1+\alpha+1}{1+\alpha^2}\) \(=\frac{(0+1)}{-\alpha}=\frac{-1}{\alpha}=\frac{-1 \cdot \alpha^2}{\alpha^3}\) \(\frac{\alpha^5+\alpha^3+\alpha+1}{\alpha^2+1}=-\alpha^2\) \(\because\left(1+\alpha+\alpha^2=-1\right)\) \(\alpha\) is a non real root of \(x^6=1\) \(\Rightarrow \alpha^6=1 \Rightarrow 1+\alpha+\alpha^2=0, \alpha^3=1\) i.e. Cube roots of units
AP EAMCET-2005
Complex Numbers and Quadratic Equation
118139
If \(\alpha, \beta, \gamma\) are the roots of \(x^3+2 x^2-3 x-1=0\) then \(\alpha^{-2}+\beta^{-2}+\gamma^{-2}=\)
1 12
2 13
3 14
4 15
Explanation:
B Given \(\alpha, \beta, \gamma\) are the roots of the equation, \(\mathrm{x}^3+2 \mathrm{x}^2-3 \mathrm{x}-1=0\) \(\therefore \alpha+\beta+\gamma=-2\) \(\alpha \beta+\beta \gamma+\alpha \gamma=-3\) \(\alpha \beta \gamma=+1\) \(\alpha^{-2}+\beta^{-2}+\gamma^{-2}=\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}\) \(=\frac{\beta^2 \gamma^2+\alpha^2 \gamma^2+\alpha^2 \beta^2}{(\alpha \beta \gamma)^2}\) \(=\frac{(\beta \gamma)^2+(\alpha \gamma)^2(\alpha \beta)^2}{(+1)^2}\) We know, \((\alpha \beta)^2+(\alpha \gamma)^2+(\beta \alpha)^2\) \(=(\alpha \beta+\beta \gamma+\alpha \gamma)^2-2\left(\alpha \beta^2 \gamma+\alpha \beta \gamma^2+\alpha^2 \beta \gamma\right)\) \(=[(\alpha \beta)+(\beta \gamma+\alpha \gamma)]^2-2 \alpha \beta \gamma(\alpha+\beta+\gamma)\) \(=(-3)^2-2 \times(+1)(-2)\) \(=9+4=13\) \(\alpha^{-2}+\beta^{-2}+\gamma^{-2}=13\)
AP EAMCET-2005
Complex Numbers and Quadratic Equation
118180
If the roots of the equation \(5 x^2-7 x+k=0\) are reciprocal of each other, then value of \(k\) is
1 5
2 2
3 2
4 1
Explanation:
A Given the equation - \(5 x^2-7 x+k=0\), Let the roots are \(\alpha, \beta\) Given \(\alpha=\frac{1}{\beta}\) Here, \(\alpha+\beta=\frac{7}{5}\) \(\alpha \cdot \beta=\frac{k}{5}\) Given, \(\alpha \beta=1 \Rightarrow \mathrm{k}=5\)
Jamia Millia Islamia-2012
Complex Numbers and Quadratic Equation
118140
If the roots of the equation \(4 x^3-12 x^2+11 x+k=\) 0 are in arithmetic progression, then \(k\) is equal to
1 -3
2 1
3 2
4 3
Explanation:
A Given the roots of the equation \(4 \mathrm{x}^3-12 \mathrm{x}^2+11 \mathrm{x}+\mathrm{k}=0 \text { is in } \mathrm{P} \text {. }\) \(\Rightarrow \mathrm{x}^3-3 \mathrm{x}^2+\frac{11}{4} \mathrm{x}+\frac{\mathrm{k}}{4}=0\) \(\text { Let the roots are } \alpha, \beta, \gamma \text {, We have }\) \(2 \beta=\alpha+\gamma\) \(\alpha+\beta+\gamma=3\) \(\alpha \beta+\beta \gamma+\alpha \gamma=11 / 4\) \(\alpha \beta \gamma=\frac{-\mathrm{k}}{4}\) \(\because \alpha+\beta+\gamma=3\) \(2 \beta+\beta=3\) \(\therefore \beta=1, \alpha+\gamma=2\) \(\beta(\alpha+\gamma)+\gamma \alpha=11 / 4\) \(1 \times 2+\alpha \gamma=11 / 4\) \(\alpha \gamma=\frac{11}{4}-2=\frac{3}{4}\) \(\therefore \frac{3}{4} \times 1=\frac{-\mathrm{k}}{4}\) \(\therefore \mathrm{k}=-3\) \(\therefore \text { Option (a) is correct. }\) Let the roots are \(\alpha, \beta, \gamma\), We have \(\therefore\) Option (a) is correct.
118138
If \(\alpha\) is a non-real root of \(x^6=1\), then \(\frac{\alpha^5+\alpha^3+\alpha+1}{\alpha^2+1}\) is equal to
1 \(\alpha^2\)
2 0
3 \(-\alpha^2\)
4 \(\alpha\)
Explanation:
C Given,\(\Rightarrow \alpha^3=1\) \(\frac{\alpha^5+\alpha^3+\alpha+1}{\alpha^2+1}=\frac{\alpha^2+1+\alpha+1}{1+\alpha^2}\) \(=\frac{(0+1)}{-\alpha}=\frac{-1}{\alpha}=\frac{-1 \cdot \alpha^2}{\alpha^3}\) \(\frac{\alpha^5+\alpha^3+\alpha+1}{\alpha^2+1}=-\alpha^2\) \(\because\left(1+\alpha+\alpha^2=-1\right)\) \(\alpha\) is a non real root of \(x^6=1\) \(\Rightarrow \alpha^6=1 \Rightarrow 1+\alpha+\alpha^2=0, \alpha^3=1\) i.e. Cube roots of units
AP EAMCET-2005
Complex Numbers and Quadratic Equation
118139
If \(\alpha, \beta, \gamma\) are the roots of \(x^3+2 x^2-3 x-1=0\) then \(\alpha^{-2}+\beta^{-2}+\gamma^{-2}=\)
1 12
2 13
3 14
4 15
Explanation:
B Given \(\alpha, \beta, \gamma\) are the roots of the equation, \(\mathrm{x}^3+2 \mathrm{x}^2-3 \mathrm{x}-1=0\) \(\therefore \alpha+\beta+\gamma=-2\) \(\alpha \beta+\beta \gamma+\alpha \gamma=-3\) \(\alpha \beta \gamma=+1\) \(\alpha^{-2}+\beta^{-2}+\gamma^{-2}=\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}\) \(=\frac{\beta^2 \gamma^2+\alpha^2 \gamma^2+\alpha^2 \beta^2}{(\alpha \beta \gamma)^2}\) \(=\frac{(\beta \gamma)^2+(\alpha \gamma)^2(\alpha \beta)^2}{(+1)^2}\) We know, \((\alpha \beta)^2+(\alpha \gamma)^2+(\beta \alpha)^2\) \(=(\alpha \beta+\beta \gamma+\alpha \gamma)^2-2\left(\alpha \beta^2 \gamma+\alpha \beta \gamma^2+\alpha^2 \beta \gamma\right)\) \(=[(\alpha \beta)+(\beta \gamma+\alpha \gamma)]^2-2 \alpha \beta \gamma(\alpha+\beta+\gamma)\) \(=(-3)^2-2 \times(+1)(-2)\) \(=9+4=13\) \(\alpha^{-2}+\beta^{-2}+\gamma^{-2}=13\)
AP EAMCET-2005
Complex Numbers and Quadratic Equation
118180
If the roots of the equation \(5 x^2-7 x+k=0\) are reciprocal of each other, then value of \(k\) is
1 5
2 2
3 2
4 1
Explanation:
A Given the equation - \(5 x^2-7 x+k=0\), Let the roots are \(\alpha, \beta\) Given \(\alpha=\frac{1}{\beta}\) Here, \(\alpha+\beta=\frac{7}{5}\) \(\alpha \cdot \beta=\frac{k}{5}\) Given, \(\alpha \beta=1 \Rightarrow \mathrm{k}=5\)
Jamia Millia Islamia-2012
Complex Numbers and Quadratic Equation
118140
If the roots of the equation \(4 x^3-12 x^2+11 x+k=\) 0 are in arithmetic progression, then \(k\) is equal to
1 -3
2 1
3 2
4 3
Explanation:
A Given the roots of the equation \(4 \mathrm{x}^3-12 \mathrm{x}^2+11 \mathrm{x}+\mathrm{k}=0 \text { is in } \mathrm{P} \text {. }\) \(\Rightarrow \mathrm{x}^3-3 \mathrm{x}^2+\frac{11}{4} \mathrm{x}+\frac{\mathrm{k}}{4}=0\) \(\text { Let the roots are } \alpha, \beta, \gamma \text {, We have }\) \(2 \beta=\alpha+\gamma\) \(\alpha+\beta+\gamma=3\) \(\alpha \beta+\beta \gamma+\alpha \gamma=11 / 4\) \(\alpha \beta \gamma=\frac{-\mathrm{k}}{4}\) \(\because \alpha+\beta+\gamma=3\) \(2 \beta+\beta=3\) \(\therefore \beta=1, \alpha+\gamma=2\) \(\beta(\alpha+\gamma)+\gamma \alpha=11 / 4\) \(1 \times 2+\alpha \gamma=11 / 4\) \(\alpha \gamma=\frac{11}{4}-2=\frac{3}{4}\) \(\therefore \frac{3}{4} \times 1=\frac{-\mathrm{k}}{4}\) \(\therefore \mathrm{k}=-3\) \(\therefore \text { Option (a) is correct. }\) Let the roots are \(\alpha, \beta, \gamma\), We have \(\therefore\) Option (a) is correct.