118210
If \(\mathrm{z} \in \mathrm{C}\) and \(\mathrm{iz}^3+4 \mathrm{z}^2-\mathrm{z}+4 \mathrm{i}=0\), then a complex roots of this equation having minimum magnitude is
1 \(4 \mathrm{i}\)
2 \(\frac{1-\mathrm{i}}{\sqrt{2}}\)
3 \(\frac{\sqrt{3}+\mathrm{i}}{2}\)
4 \(\frac{1+\mathrm{i}}{\sqrt{2}}\)
Explanation:
B Given, equation \(=i z^3+4 z^2-z+4 i=0\) \(i z^3+4 z^2-z+4 i=0\) \(\mathrm{z}^2(\mathrm{iz}+4)+\mathrm{i}(\mathrm{iz}+4)=0\) \(\mathrm{z}=4 \mathrm{i}\) or \(\mathrm{z}^2=\mathrm{i}\) \(z=4 i\) or \(\pm\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}} \mathrm{i}\right)\) So, the complex root having minimum magnitude is \(\frac{1-\mathrm{i}}{\sqrt{2}} \text { or } \frac{-1+\mathrm{i}}{\sqrt{2}}\)
AP EAMCET-22.04.2018
Complex Numbers and Quadratic Equation
118211
If \(\alpha, \beta, \gamma\) are the roots of the equation \(x^3-a x^2+b x\) \(-\mathbf{c}=\mathbf{0}\), then \(\Sigma \boldsymbol{\alpha}^2(\boldsymbol{\beta}+\gamma)=\)
A Given equation : - \(\mathrm{x}^3-a \mathrm{x}^2+\mathrm{bx}-\mathrm{c}=0\) and its roots are, \(\alpha, \beta, \gamma\) \(x^3-a x^2+b x-c=0\) \(\alpha, \beta, \gamma\) are the roots of equation (given that) So, \(\alpha+\beta+\gamma=\mathrm{a}, \alpha \beta+\beta \gamma+\gamma \alpha=\mathrm{b}\) \(\alpha \beta \gamma=\mathrm{c}\) Now, \(\sum \alpha^2(\beta+\gamma)=\sum \alpha^2(a-\alpha) \quad[\because \alpha+\beta+\gamma=a]\)
AP EAMCET-20.04.2019
Complex Numbers and Quadratic Equation
118212
For \(x>2\), then equation \(\sqrt{x+2}-\sqrt{x-2}=\sqrt{4 x-2}\) has
1 one solution
2 two solutions
3 more than two solutions
4 No solution
Explanation:
D Given equation - \(\sqrt{\mathrm{x}+2}-\sqrt{\mathrm{x}-2}=\sqrt{4 \mathrm{x}-2}\) for, \(x>2\) (it is given that) \(\sqrt{\mathrm{x}+2}-\sqrt{\mathrm{x}-2}=\sqrt{4 \mathrm{x}-2}\) On squaring both side - we get \(x+2+x-2-2 \sqrt{x^2-4}=4 x-2\) \(1-x=\sqrt{x^2-4}\) As \(\mathrm{x}>2\) Then, \(1-\mathrm{x}\lt 0\) while \(\sqrt{\mathrm{x}^2-4}>0\) So, for the given equation there is no solution.
AP EAMCET-20.04.2019
Complex Numbers and Quadratic Equation
118099
What is the coefficient of \(x^2\) in \(p(x)\) ?
1 -1
2 1
3 0
4 None of the above
Explanation:
C Coefficient of \(\mathrm{x}^2\) in \(\mathrm{p}(\mathrm{x})\) is \(-(\alpha+\beta+\gamma)\). \(=-\left(\cos 75^{\circ}+\cos 45^{\circ}+\cos 165^{\circ}\right)\) \(=-\left[\frac{\sqrt{3}-1}{2 \sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{\sqrt{3}+1}{2 \sqrt{2}}\right]\) \(=-\left[\frac{\sqrt{3}-1+2-\sqrt{3}-1}{2 \sqrt{2}}\right]=0\)
SCRA-2012
Complex Numbers and Quadratic Equation
118056
If \(\alpha\) and \(\beta\) are the roots of \(x^2-a x+b^2=0\), then \(\alpha^2+\beta^2\) is equal to \(\qquad\)
118210
If \(\mathrm{z} \in \mathrm{C}\) and \(\mathrm{iz}^3+4 \mathrm{z}^2-\mathrm{z}+4 \mathrm{i}=0\), then a complex roots of this equation having minimum magnitude is
1 \(4 \mathrm{i}\)
2 \(\frac{1-\mathrm{i}}{\sqrt{2}}\)
3 \(\frac{\sqrt{3}+\mathrm{i}}{2}\)
4 \(\frac{1+\mathrm{i}}{\sqrt{2}}\)
Explanation:
B Given, equation \(=i z^3+4 z^2-z+4 i=0\) \(i z^3+4 z^2-z+4 i=0\) \(\mathrm{z}^2(\mathrm{iz}+4)+\mathrm{i}(\mathrm{iz}+4)=0\) \(\mathrm{z}=4 \mathrm{i}\) or \(\mathrm{z}^2=\mathrm{i}\) \(z=4 i\) or \(\pm\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}} \mathrm{i}\right)\) So, the complex root having minimum magnitude is \(\frac{1-\mathrm{i}}{\sqrt{2}} \text { or } \frac{-1+\mathrm{i}}{\sqrt{2}}\)
AP EAMCET-22.04.2018
Complex Numbers and Quadratic Equation
118211
If \(\alpha, \beta, \gamma\) are the roots of the equation \(x^3-a x^2+b x\) \(-\mathbf{c}=\mathbf{0}\), then \(\Sigma \boldsymbol{\alpha}^2(\boldsymbol{\beta}+\gamma)=\)
A Given equation : - \(\mathrm{x}^3-a \mathrm{x}^2+\mathrm{bx}-\mathrm{c}=0\) and its roots are, \(\alpha, \beta, \gamma\) \(x^3-a x^2+b x-c=0\) \(\alpha, \beta, \gamma\) are the roots of equation (given that) So, \(\alpha+\beta+\gamma=\mathrm{a}, \alpha \beta+\beta \gamma+\gamma \alpha=\mathrm{b}\) \(\alpha \beta \gamma=\mathrm{c}\) Now, \(\sum \alpha^2(\beta+\gamma)=\sum \alpha^2(a-\alpha) \quad[\because \alpha+\beta+\gamma=a]\)
AP EAMCET-20.04.2019
Complex Numbers and Quadratic Equation
118212
For \(x>2\), then equation \(\sqrt{x+2}-\sqrt{x-2}=\sqrt{4 x-2}\) has
1 one solution
2 two solutions
3 more than two solutions
4 No solution
Explanation:
D Given equation - \(\sqrt{\mathrm{x}+2}-\sqrt{\mathrm{x}-2}=\sqrt{4 \mathrm{x}-2}\) for, \(x>2\) (it is given that) \(\sqrt{\mathrm{x}+2}-\sqrt{\mathrm{x}-2}=\sqrt{4 \mathrm{x}-2}\) On squaring both side - we get \(x+2+x-2-2 \sqrt{x^2-4}=4 x-2\) \(1-x=\sqrt{x^2-4}\) As \(\mathrm{x}>2\) Then, \(1-\mathrm{x}\lt 0\) while \(\sqrt{\mathrm{x}^2-4}>0\) So, for the given equation there is no solution.
AP EAMCET-20.04.2019
Complex Numbers and Quadratic Equation
118099
What is the coefficient of \(x^2\) in \(p(x)\) ?
1 -1
2 1
3 0
4 None of the above
Explanation:
C Coefficient of \(\mathrm{x}^2\) in \(\mathrm{p}(\mathrm{x})\) is \(-(\alpha+\beta+\gamma)\). \(=-\left(\cos 75^{\circ}+\cos 45^{\circ}+\cos 165^{\circ}\right)\) \(=-\left[\frac{\sqrt{3}-1}{2 \sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{\sqrt{3}+1}{2 \sqrt{2}}\right]\) \(=-\left[\frac{\sqrt{3}-1+2-\sqrt{3}-1}{2 \sqrt{2}}\right]=0\)
SCRA-2012
Complex Numbers and Quadratic Equation
118056
If \(\alpha\) and \(\beta\) are the roots of \(x^2-a x+b^2=0\), then \(\alpha^2+\beta^2\) is equal to \(\qquad\)
118210
If \(\mathrm{z} \in \mathrm{C}\) and \(\mathrm{iz}^3+4 \mathrm{z}^2-\mathrm{z}+4 \mathrm{i}=0\), then a complex roots of this equation having minimum magnitude is
1 \(4 \mathrm{i}\)
2 \(\frac{1-\mathrm{i}}{\sqrt{2}}\)
3 \(\frac{\sqrt{3}+\mathrm{i}}{2}\)
4 \(\frac{1+\mathrm{i}}{\sqrt{2}}\)
Explanation:
B Given, equation \(=i z^3+4 z^2-z+4 i=0\) \(i z^3+4 z^2-z+4 i=0\) \(\mathrm{z}^2(\mathrm{iz}+4)+\mathrm{i}(\mathrm{iz}+4)=0\) \(\mathrm{z}=4 \mathrm{i}\) or \(\mathrm{z}^2=\mathrm{i}\) \(z=4 i\) or \(\pm\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}} \mathrm{i}\right)\) So, the complex root having minimum magnitude is \(\frac{1-\mathrm{i}}{\sqrt{2}} \text { or } \frac{-1+\mathrm{i}}{\sqrt{2}}\)
AP EAMCET-22.04.2018
Complex Numbers and Quadratic Equation
118211
If \(\alpha, \beta, \gamma\) are the roots of the equation \(x^3-a x^2+b x\) \(-\mathbf{c}=\mathbf{0}\), then \(\Sigma \boldsymbol{\alpha}^2(\boldsymbol{\beta}+\gamma)=\)
A Given equation : - \(\mathrm{x}^3-a \mathrm{x}^2+\mathrm{bx}-\mathrm{c}=0\) and its roots are, \(\alpha, \beta, \gamma\) \(x^3-a x^2+b x-c=0\) \(\alpha, \beta, \gamma\) are the roots of equation (given that) So, \(\alpha+\beta+\gamma=\mathrm{a}, \alpha \beta+\beta \gamma+\gamma \alpha=\mathrm{b}\) \(\alpha \beta \gamma=\mathrm{c}\) Now, \(\sum \alpha^2(\beta+\gamma)=\sum \alpha^2(a-\alpha) \quad[\because \alpha+\beta+\gamma=a]\)
AP EAMCET-20.04.2019
Complex Numbers and Quadratic Equation
118212
For \(x>2\), then equation \(\sqrt{x+2}-\sqrt{x-2}=\sqrt{4 x-2}\) has
1 one solution
2 two solutions
3 more than two solutions
4 No solution
Explanation:
D Given equation - \(\sqrt{\mathrm{x}+2}-\sqrt{\mathrm{x}-2}=\sqrt{4 \mathrm{x}-2}\) for, \(x>2\) (it is given that) \(\sqrt{\mathrm{x}+2}-\sqrt{\mathrm{x}-2}=\sqrt{4 \mathrm{x}-2}\) On squaring both side - we get \(x+2+x-2-2 \sqrt{x^2-4}=4 x-2\) \(1-x=\sqrt{x^2-4}\) As \(\mathrm{x}>2\) Then, \(1-\mathrm{x}\lt 0\) while \(\sqrt{\mathrm{x}^2-4}>0\) So, for the given equation there is no solution.
AP EAMCET-20.04.2019
Complex Numbers and Quadratic Equation
118099
What is the coefficient of \(x^2\) in \(p(x)\) ?
1 -1
2 1
3 0
4 None of the above
Explanation:
C Coefficient of \(\mathrm{x}^2\) in \(\mathrm{p}(\mathrm{x})\) is \(-(\alpha+\beta+\gamma)\). \(=-\left(\cos 75^{\circ}+\cos 45^{\circ}+\cos 165^{\circ}\right)\) \(=-\left[\frac{\sqrt{3}-1}{2 \sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{\sqrt{3}+1}{2 \sqrt{2}}\right]\) \(=-\left[\frac{\sqrt{3}-1+2-\sqrt{3}-1}{2 \sqrt{2}}\right]=0\)
SCRA-2012
Complex Numbers and Quadratic Equation
118056
If \(\alpha\) and \(\beta\) are the roots of \(x^2-a x+b^2=0\), then \(\alpha^2+\beta^2\) is equal to \(\qquad\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Complex Numbers and Quadratic Equation
118210
If \(\mathrm{z} \in \mathrm{C}\) and \(\mathrm{iz}^3+4 \mathrm{z}^2-\mathrm{z}+4 \mathrm{i}=0\), then a complex roots of this equation having minimum magnitude is
1 \(4 \mathrm{i}\)
2 \(\frac{1-\mathrm{i}}{\sqrt{2}}\)
3 \(\frac{\sqrt{3}+\mathrm{i}}{2}\)
4 \(\frac{1+\mathrm{i}}{\sqrt{2}}\)
Explanation:
B Given, equation \(=i z^3+4 z^2-z+4 i=0\) \(i z^3+4 z^2-z+4 i=0\) \(\mathrm{z}^2(\mathrm{iz}+4)+\mathrm{i}(\mathrm{iz}+4)=0\) \(\mathrm{z}=4 \mathrm{i}\) or \(\mathrm{z}^2=\mathrm{i}\) \(z=4 i\) or \(\pm\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}} \mathrm{i}\right)\) So, the complex root having minimum magnitude is \(\frac{1-\mathrm{i}}{\sqrt{2}} \text { or } \frac{-1+\mathrm{i}}{\sqrt{2}}\)
AP EAMCET-22.04.2018
Complex Numbers and Quadratic Equation
118211
If \(\alpha, \beta, \gamma\) are the roots of the equation \(x^3-a x^2+b x\) \(-\mathbf{c}=\mathbf{0}\), then \(\Sigma \boldsymbol{\alpha}^2(\boldsymbol{\beta}+\gamma)=\)
A Given equation : - \(\mathrm{x}^3-a \mathrm{x}^2+\mathrm{bx}-\mathrm{c}=0\) and its roots are, \(\alpha, \beta, \gamma\) \(x^3-a x^2+b x-c=0\) \(\alpha, \beta, \gamma\) are the roots of equation (given that) So, \(\alpha+\beta+\gamma=\mathrm{a}, \alpha \beta+\beta \gamma+\gamma \alpha=\mathrm{b}\) \(\alpha \beta \gamma=\mathrm{c}\) Now, \(\sum \alpha^2(\beta+\gamma)=\sum \alpha^2(a-\alpha) \quad[\because \alpha+\beta+\gamma=a]\)
AP EAMCET-20.04.2019
Complex Numbers and Quadratic Equation
118212
For \(x>2\), then equation \(\sqrt{x+2}-\sqrt{x-2}=\sqrt{4 x-2}\) has
1 one solution
2 two solutions
3 more than two solutions
4 No solution
Explanation:
D Given equation - \(\sqrt{\mathrm{x}+2}-\sqrt{\mathrm{x}-2}=\sqrt{4 \mathrm{x}-2}\) for, \(x>2\) (it is given that) \(\sqrt{\mathrm{x}+2}-\sqrt{\mathrm{x}-2}=\sqrt{4 \mathrm{x}-2}\) On squaring both side - we get \(x+2+x-2-2 \sqrt{x^2-4}=4 x-2\) \(1-x=\sqrt{x^2-4}\) As \(\mathrm{x}>2\) Then, \(1-\mathrm{x}\lt 0\) while \(\sqrt{\mathrm{x}^2-4}>0\) So, for the given equation there is no solution.
AP EAMCET-20.04.2019
Complex Numbers and Quadratic Equation
118099
What is the coefficient of \(x^2\) in \(p(x)\) ?
1 -1
2 1
3 0
4 None of the above
Explanation:
C Coefficient of \(\mathrm{x}^2\) in \(\mathrm{p}(\mathrm{x})\) is \(-(\alpha+\beta+\gamma)\). \(=-\left(\cos 75^{\circ}+\cos 45^{\circ}+\cos 165^{\circ}\right)\) \(=-\left[\frac{\sqrt{3}-1}{2 \sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{\sqrt{3}+1}{2 \sqrt{2}}\right]\) \(=-\left[\frac{\sqrt{3}-1+2-\sqrt{3}-1}{2 \sqrt{2}}\right]=0\)
SCRA-2012
Complex Numbers and Quadratic Equation
118056
If \(\alpha\) and \(\beta\) are the roots of \(x^2-a x+b^2=0\), then \(\alpha^2+\beta^2\) is equal to \(\qquad\)
118210
If \(\mathrm{z} \in \mathrm{C}\) and \(\mathrm{iz}^3+4 \mathrm{z}^2-\mathrm{z}+4 \mathrm{i}=0\), then a complex roots of this equation having minimum magnitude is
1 \(4 \mathrm{i}\)
2 \(\frac{1-\mathrm{i}}{\sqrt{2}}\)
3 \(\frac{\sqrt{3}+\mathrm{i}}{2}\)
4 \(\frac{1+\mathrm{i}}{\sqrt{2}}\)
Explanation:
B Given, equation \(=i z^3+4 z^2-z+4 i=0\) \(i z^3+4 z^2-z+4 i=0\) \(\mathrm{z}^2(\mathrm{iz}+4)+\mathrm{i}(\mathrm{iz}+4)=0\) \(\mathrm{z}=4 \mathrm{i}\) or \(\mathrm{z}^2=\mathrm{i}\) \(z=4 i\) or \(\pm\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}} \mathrm{i}\right)\) So, the complex root having minimum magnitude is \(\frac{1-\mathrm{i}}{\sqrt{2}} \text { or } \frac{-1+\mathrm{i}}{\sqrt{2}}\)
AP EAMCET-22.04.2018
Complex Numbers and Quadratic Equation
118211
If \(\alpha, \beta, \gamma\) are the roots of the equation \(x^3-a x^2+b x\) \(-\mathbf{c}=\mathbf{0}\), then \(\Sigma \boldsymbol{\alpha}^2(\boldsymbol{\beta}+\gamma)=\)
A Given equation : - \(\mathrm{x}^3-a \mathrm{x}^2+\mathrm{bx}-\mathrm{c}=0\) and its roots are, \(\alpha, \beta, \gamma\) \(x^3-a x^2+b x-c=0\) \(\alpha, \beta, \gamma\) are the roots of equation (given that) So, \(\alpha+\beta+\gamma=\mathrm{a}, \alpha \beta+\beta \gamma+\gamma \alpha=\mathrm{b}\) \(\alpha \beta \gamma=\mathrm{c}\) Now, \(\sum \alpha^2(\beta+\gamma)=\sum \alpha^2(a-\alpha) \quad[\because \alpha+\beta+\gamma=a]\)
AP EAMCET-20.04.2019
Complex Numbers and Quadratic Equation
118212
For \(x>2\), then equation \(\sqrt{x+2}-\sqrt{x-2}=\sqrt{4 x-2}\) has
1 one solution
2 two solutions
3 more than two solutions
4 No solution
Explanation:
D Given equation - \(\sqrt{\mathrm{x}+2}-\sqrt{\mathrm{x}-2}=\sqrt{4 \mathrm{x}-2}\) for, \(x>2\) (it is given that) \(\sqrt{\mathrm{x}+2}-\sqrt{\mathrm{x}-2}=\sqrt{4 \mathrm{x}-2}\) On squaring both side - we get \(x+2+x-2-2 \sqrt{x^2-4}=4 x-2\) \(1-x=\sqrt{x^2-4}\) As \(\mathrm{x}>2\) Then, \(1-\mathrm{x}\lt 0\) while \(\sqrt{\mathrm{x}^2-4}>0\) So, for the given equation there is no solution.
AP EAMCET-20.04.2019
Complex Numbers and Quadratic Equation
118099
What is the coefficient of \(x^2\) in \(p(x)\) ?
1 -1
2 1
3 0
4 None of the above
Explanation:
C Coefficient of \(\mathrm{x}^2\) in \(\mathrm{p}(\mathrm{x})\) is \(-(\alpha+\beta+\gamma)\). \(=-\left(\cos 75^{\circ}+\cos 45^{\circ}+\cos 165^{\circ}\right)\) \(=-\left[\frac{\sqrt{3}-1}{2 \sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{\sqrt{3}+1}{2 \sqrt{2}}\right]\) \(=-\left[\frac{\sqrt{3}-1+2-\sqrt{3}-1}{2 \sqrt{2}}\right]=0\)
SCRA-2012
Complex Numbers and Quadratic Equation
118056
If \(\alpha\) and \(\beta\) are the roots of \(x^2-a x+b^2=0\), then \(\alpha^2+\beta^2\) is equal to \(\qquad\)