NEET Test Series from KOTA - 10 Papers In MS WORD
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Complex Numbers and Quadratic Equation
118068
If \(\alpha, \beta\) are the roots of \(x^2+p x+q=0\), and \(\omega\) is an imaginary cube root of unity, then value of \(\left(\omega \alpha+\omega^2 \beta\right)\left(\omega^2 \alpha+\omega \beta\right)\) is
1 \(\mathrm{p}^2\)
2 \(3 q\)
3 \(p^2-2 q\)
4 \(p^2-3 q\)
Explanation:
D \( \text { : Given equation: } \mathrm{x}^2+\mathrm{px}+\mathrm{q}=0\) \(\text { Sum of roots }(\alpha+\beta)=-\mathrm{p}\) \(\text { Product of roots }(\alpha \beta)=\mathrm{q}\) \(\text { Now, }\left(\omega \alpha+\omega^2 \beta\right)\left(\omega^2 \alpha+\omega \beta\right)\) \(=\alpha^2+\beta^2+\left(\omega+\omega^2\right) \alpha \beta=\alpha^2+\beta^2-\alpha \beta\left\{\because \omega+\omega^2=-1\right\}\) \(=(\alpha+\beta)^2-3 \alpha \beta=\mathrm{p}^2-3 \mathrm{q}\)
BITSAT-2006
Complex Numbers and Quadratic Equation
118069
The number of real roots of \(\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^3+\left(\mathrm{x}+\frac{\mathbf{1}}{\mathrm{x}}\right)=\mathbf{0}\)
1 0
2 2
3 4
4 6
Explanation:
A \( We have,\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^3+\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)=0\) \(\Rightarrow\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)\left[\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^2+1\right]=0\) \(\quad \Rightarrow \text { either } \mathrm{x}+\frac{1}{\mathrm{x}}=0 \Rightarrow \mathrm{x}^2=-1 \Rightarrow \mathrm{x}= \pm \mathrm{i}\) \(\text { or }\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^2+1=0\) \(\Rightarrow \mathrm{x}^2+\frac{1}{\mathrm{x}^2}+3=0 \Rightarrow \mathrm{x}^4+3 \mathrm{x}^2+1=0\) \(\Rightarrow \mathrm{x}^2=\frac{-3 \pm \sqrt{9-4}}{2}=\frac{-3 \pm \sqrt{5}}{2}\lt 0\) \(\therefore \text { There is no real root. }\)\(\therefore\) There is no real root.
BITSAT-2005
Complex Numbers and Quadratic Equation
118071
If \(x=\omega-\omega^2-2\), then the value of \(x^4+3 x^3+2 x^2-11 x-6\) is
1 1
2 -1
3 2
4 None of these
Explanation:
A We have \(x=\omega-\omega^2-2\) or \(x+2=\omega-\omega^2\) Squaring, \(x^2+4 x+4=\omega^2+\omega^4-2 \omega^3\) \(=\omega^2+\omega^3 \omega-2 \omega^3=\omega^2+\omega-2\) \(=-1-2=-3 \Rightarrow x^2+4 x+7=0\) Dividing \(x^4+3 x^3+2 x^2-11 x-6\) by \(x^2+4 x+7\) We get, \(x^4+3 x^3+2 x^2-11 x-6=\left(x^2+4 x+7\right)\left(x^2-x-1\right)+1\) \(=(0)\left(x^2-x-1\right)+1=0+1=1\)
BITSAT-2012
Complex Numbers and Quadratic Equation
118073
The roots of the equation \(x^2-2 \sqrt{2} x+1=0\) are
1 Real and different
2 Imaginary and different
3 Real and equal
4 Rational and different
Explanation:
A Given, equation \(x^2-2 \sqrt{2} x+1=0\) The discriminant of the equation \(\mathrm{D}>0\) \((-2 \sqrt{2})^2-4(1)(1)=8-4=4>0\) and a perfect square So, roots are real and different.
118068
If \(\alpha, \beta\) are the roots of \(x^2+p x+q=0\), and \(\omega\) is an imaginary cube root of unity, then value of \(\left(\omega \alpha+\omega^2 \beta\right)\left(\omega^2 \alpha+\omega \beta\right)\) is
1 \(\mathrm{p}^2\)
2 \(3 q\)
3 \(p^2-2 q\)
4 \(p^2-3 q\)
Explanation:
D \( \text { : Given equation: } \mathrm{x}^2+\mathrm{px}+\mathrm{q}=0\) \(\text { Sum of roots }(\alpha+\beta)=-\mathrm{p}\) \(\text { Product of roots }(\alpha \beta)=\mathrm{q}\) \(\text { Now, }\left(\omega \alpha+\omega^2 \beta\right)\left(\omega^2 \alpha+\omega \beta\right)\) \(=\alpha^2+\beta^2+\left(\omega+\omega^2\right) \alpha \beta=\alpha^2+\beta^2-\alpha \beta\left\{\because \omega+\omega^2=-1\right\}\) \(=(\alpha+\beta)^2-3 \alpha \beta=\mathrm{p}^2-3 \mathrm{q}\)
BITSAT-2006
Complex Numbers and Quadratic Equation
118069
The number of real roots of \(\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^3+\left(\mathrm{x}+\frac{\mathbf{1}}{\mathrm{x}}\right)=\mathbf{0}\)
1 0
2 2
3 4
4 6
Explanation:
A \( We have,\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^3+\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)=0\) \(\Rightarrow\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)\left[\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^2+1\right]=0\) \(\quad \Rightarrow \text { either } \mathrm{x}+\frac{1}{\mathrm{x}}=0 \Rightarrow \mathrm{x}^2=-1 \Rightarrow \mathrm{x}= \pm \mathrm{i}\) \(\text { or }\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^2+1=0\) \(\Rightarrow \mathrm{x}^2+\frac{1}{\mathrm{x}^2}+3=0 \Rightarrow \mathrm{x}^4+3 \mathrm{x}^2+1=0\) \(\Rightarrow \mathrm{x}^2=\frac{-3 \pm \sqrt{9-4}}{2}=\frac{-3 \pm \sqrt{5}}{2}\lt 0\) \(\therefore \text { There is no real root. }\)\(\therefore\) There is no real root.
BITSAT-2005
Complex Numbers and Quadratic Equation
118071
If \(x=\omega-\omega^2-2\), then the value of \(x^4+3 x^3+2 x^2-11 x-6\) is
1 1
2 -1
3 2
4 None of these
Explanation:
A We have \(x=\omega-\omega^2-2\) or \(x+2=\omega-\omega^2\) Squaring, \(x^2+4 x+4=\omega^2+\omega^4-2 \omega^3\) \(=\omega^2+\omega^3 \omega-2 \omega^3=\omega^2+\omega-2\) \(=-1-2=-3 \Rightarrow x^2+4 x+7=0\) Dividing \(x^4+3 x^3+2 x^2-11 x-6\) by \(x^2+4 x+7\) We get, \(x^4+3 x^3+2 x^2-11 x-6=\left(x^2+4 x+7\right)\left(x^2-x-1\right)+1\) \(=(0)\left(x^2-x-1\right)+1=0+1=1\)
BITSAT-2012
Complex Numbers and Quadratic Equation
118073
The roots of the equation \(x^2-2 \sqrt{2} x+1=0\) are
1 Real and different
2 Imaginary and different
3 Real and equal
4 Rational and different
Explanation:
A Given, equation \(x^2-2 \sqrt{2} x+1=0\) The discriminant of the equation \(\mathrm{D}>0\) \((-2 \sqrt{2})^2-4(1)(1)=8-4=4>0\) and a perfect square So, roots are real and different.
118068
If \(\alpha, \beta\) are the roots of \(x^2+p x+q=0\), and \(\omega\) is an imaginary cube root of unity, then value of \(\left(\omega \alpha+\omega^2 \beta\right)\left(\omega^2 \alpha+\omega \beta\right)\) is
1 \(\mathrm{p}^2\)
2 \(3 q\)
3 \(p^2-2 q\)
4 \(p^2-3 q\)
Explanation:
D \( \text { : Given equation: } \mathrm{x}^2+\mathrm{px}+\mathrm{q}=0\) \(\text { Sum of roots }(\alpha+\beta)=-\mathrm{p}\) \(\text { Product of roots }(\alpha \beta)=\mathrm{q}\) \(\text { Now, }\left(\omega \alpha+\omega^2 \beta\right)\left(\omega^2 \alpha+\omega \beta\right)\) \(=\alpha^2+\beta^2+\left(\omega+\omega^2\right) \alpha \beta=\alpha^2+\beta^2-\alpha \beta\left\{\because \omega+\omega^2=-1\right\}\) \(=(\alpha+\beta)^2-3 \alpha \beta=\mathrm{p}^2-3 \mathrm{q}\)
BITSAT-2006
Complex Numbers and Quadratic Equation
118069
The number of real roots of \(\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^3+\left(\mathrm{x}+\frac{\mathbf{1}}{\mathrm{x}}\right)=\mathbf{0}\)
1 0
2 2
3 4
4 6
Explanation:
A \( We have,\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^3+\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)=0\) \(\Rightarrow\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)\left[\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^2+1\right]=0\) \(\quad \Rightarrow \text { either } \mathrm{x}+\frac{1}{\mathrm{x}}=0 \Rightarrow \mathrm{x}^2=-1 \Rightarrow \mathrm{x}= \pm \mathrm{i}\) \(\text { or }\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^2+1=0\) \(\Rightarrow \mathrm{x}^2+\frac{1}{\mathrm{x}^2}+3=0 \Rightarrow \mathrm{x}^4+3 \mathrm{x}^2+1=0\) \(\Rightarrow \mathrm{x}^2=\frac{-3 \pm \sqrt{9-4}}{2}=\frac{-3 \pm \sqrt{5}}{2}\lt 0\) \(\therefore \text { There is no real root. }\)\(\therefore\) There is no real root.
BITSAT-2005
Complex Numbers and Quadratic Equation
118071
If \(x=\omega-\omega^2-2\), then the value of \(x^4+3 x^3+2 x^2-11 x-6\) is
1 1
2 -1
3 2
4 None of these
Explanation:
A We have \(x=\omega-\omega^2-2\) or \(x+2=\omega-\omega^2\) Squaring, \(x^2+4 x+4=\omega^2+\omega^4-2 \omega^3\) \(=\omega^2+\omega^3 \omega-2 \omega^3=\omega^2+\omega-2\) \(=-1-2=-3 \Rightarrow x^2+4 x+7=0\) Dividing \(x^4+3 x^3+2 x^2-11 x-6\) by \(x^2+4 x+7\) We get, \(x^4+3 x^3+2 x^2-11 x-6=\left(x^2+4 x+7\right)\left(x^2-x-1\right)+1\) \(=(0)\left(x^2-x-1\right)+1=0+1=1\)
BITSAT-2012
Complex Numbers and Quadratic Equation
118073
The roots of the equation \(x^2-2 \sqrt{2} x+1=0\) are
1 Real and different
2 Imaginary and different
3 Real and equal
4 Rational and different
Explanation:
A Given, equation \(x^2-2 \sqrt{2} x+1=0\) The discriminant of the equation \(\mathrm{D}>0\) \((-2 \sqrt{2})^2-4(1)(1)=8-4=4>0\) and a perfect square So, roots are real and different.
118068
If \(\alpha, \beta\) are the roots of \(x^2+p x+q=0\), and \(\omega\) is an imaginary cube root of unity, then value of \(\left(\omega \alpha+\omega^2 \beta\right)\left(\omega^2 \alpha+\omega \beta\right)\) is
1 \(\mathrm{p}^2\)
2 \(3 q\)
3 \(p^2-2 q\)
4 \(p^2-3 q\)
Explanation:
D \( \text { : Given equation: } \mathrm{x}^2+\mathrm{px}+\mathrm{q}=0\) \(\text { Sum of roots }(\alpha+\beta)=-\mathrm{p}\) \(\text { Product of roots }(\alpha \beta)=\mathrm{q}\) \(\text { Now, }\left(\omega \alpha+\omega^2 \beta\right)\left(\omega^2 \alpha+\omega \beta\right)\) \(=\alpha^2+\beta^2+\left(\omega+\omega^2\right) \alpha \beta=\alpha^2+\beta^2-\alpha \beta\left\{\because \omega+\omega^2=-1\right\}\) \(=(\alpha+\beta)^2-3 \alpha \beta=\mathrm{p}^2-3 \mathrm{q}\)
BITSAT-2006
Complex Numbers and Quadratic Equation
118069
The number of real roots of \(\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^3+\left(\mathrm{x}+\frac{\mathbf{1}}{\mathrm{x}}\right)=\mathbf{0}\)
1 0
2 2
3 4
4 6
Explanation:
A \( We have,\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^3+\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)=0\) \(\Rightarrow\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)\left[\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^2+1\right]=0\) \(\quad \Rightarrow \text { either } \mathrm{x}+\frac{1}{\mathrm{x}}=0 \Rightarrow \mathrm{x}^2=-1 \Rightarrow \mathrm{x}= \pm \mathrm{i}\) \(\text { or }\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^2+1=0\) \(\Rightarrow \mathrm{x}^2+\frac{1}{\mathrm{x}^2}+3=0 \Rightarrow \mathrm{x}^4+3 \mathrm{x}^2+1=0\) \(\Rightarrow \mathrm{x}^2=\frac{-3 \pm \sqrt{9-4}}{2}=\frac{-3 \pm \sqrt{5}}{2}\lt 0\) \(\therefore \text { There is no real root. }\)\(\therefore\) There is no real root.
BITSAT-2005
Complex Numbers and Quadratic Equation
118071
If \(x=\omega-\omega^2-2\), then the value of \(x^4+3 x^3+2 x^2-11 x-6\) is
1 1
2 -1
3 2
4 None of these
Explanation:
A We have \(x=\omega-\omega^2-2\) or \(x+2=\omega-\omega^2\) Squaring, \(x^2+4 x+4=\omega^2+\omega^4-2 \omega^3\) \(=\omega^2+\omega^3 \omega-2 \omega^3=\omega^2+\omega-2\) \(=-1-2=-3 \Rightarrow x^2+4 x+7=0\) Dividing \(x^4+3 x^3+2 x^2-11 x-6\) by \(x^2+4 x+7\) We get, \(x^4+3 x^3+2 x^2-11 x-6=\left(x^2+4 x+7\right)\left(x^2-x-1\right)+1\) \(=(0)\left(x^2-x-1\right)+1=0+1=1\)
BITSAT-2012
Complex Numbers and Quadratic Equation
118073
The roots of the equation \(x^2-2 \sqrt{2} x+1=0\) are
1 Real and different
2 Imaginary and different
3 Real and equal
4 Rational and different
Explanation:
A Given, equation \(x^2-2 \sqrt{2} x+1=0\) The discriminant of the equation \(\mathrm{D}>0\) \((-2 \sqrt{2})^2-4(1)(1)=8-4=4>0\) and a perfect square So, roots are real and different.
