118380
For the equation \(\frac{1}{x+a}-\frac{1}{x+b}=\frac{1}{x+c}\), if the product of roots is zero, then the sum of roots is
1 0
2 \(\frac{2 a b}{b+c}\)
3 \(\frac{2 b c}{b+c}\)
4 \(\frac{-2 b c}{b+c}\)
Explanation:
D Given \(\frac{1}{\mathrm{x}+\mathrm{a}}-\frac{1}{\mathrm{x}+\mathrm{b}}=\frac{1}{\mathrm{x}+\mathrm{c}}\) \(\text { or } \mathrm{x}^2+(\mathrm{a}+\mathrm{b}) \mathrm{x}+\mathrm{ab}=(\mathrm{b}-\mathrm{a}) \mathrm{x}+(\mathrm{b}-\mathrm{a}) \mathrm{c}\) \(\text { or } \mathrm{x}^2+2 \mathrm{ax}+\mathrm{ab}+\mathrm{ca}-\mathrm{bc}=0\) \(\text { Since product of the roots }=0\) \(\mathrm{ab}+\mathrm{ca}-\mathrm{bc}=0 ; \mathrm{a}=\frac{\mathrm{bc}}{\mathrm{b}+\mathrm{c}}\) \(\text { Thus sum of roots }=-2 \mathrm{a}=\frac{-2 \mathrm{bc}}{\mathrm{b}+\mathrm{c}}\)
BITSAT-2009
Complex Numbers and Quadratic Equation
118381
If \(\alpha, \beta\) and \(\gamma\) are the roots of the equation \(x^3-\) \(3 x^2+3 x+7=0\), and \(\omega\) is cube root of unity, then the value of \(\frac{\alpha-1}{\beta-1}+\frac{\beta-1}{\gamma-1}+\frac{\gamma-1}{\alpha-1}\) is equal to
1 \(3 \omega^2\)
2 \(3 / \omega\)
3 \(2 \omega^2\)
4 None of these
Explanation:
A Given \(\alpha, \beta\) and \(\gamma\) are the roots of equation \(x^3-3 x^2+3 x+7=0\) \((\mathrm{x}+1)\left(\mathrm{x}^2-4 \mathrm{x}+7\right)=0\) \((x+1)=0\) \(\mathrm{x}=-1\) \(\left(x^2-4 x+7\right)=0\) \(x=\frac{+4 \pm \sqrt{(4)^2-4 \times 7}}{2}\) \(\mathrm{x}=\frac{4 \pm \sqrt{16-28}}{2}\) \(\mathrm{x}=2 \pm \sqrt{3} \mathrm{i}\) Given, \(\omega\) is cube root of unity Then, \(\quad x=2+\sqrt{3} i, 2-\sqrt{3} i\) Let, \(\alpha=-1, \beta=2+i \sqrt{3}, \gamma=2-i \sqrt{3}\) \(\alpha-1=-2, \beta-1=1+\sqrt{3} \mathrm{i}, \gamma-1=1-\sqrt{3} \mathrm{i}\) \(\alpha-1=-2, \beta-1=-2 \omega, \gamma-1=-2 \omega^2\) Now, \(\quad \frac{\alpha-1}{\beta-1}+\frac{\beta-1}{\gamma-1}+\frac{\gamma-1}{\alpha-1}\) \(=\frac{-2}{-2 \omega}+\left(\frac{-2 \omega}{-2 \omega^2}\right)+\left(\frac{-2 \omega^2}{-2}\right)\) \(=\frac{2}{\omega}+\omega^2=\frac{2+\omega^3}{\omega}=3 \omega^2\)
UPSEE-2018
Complex Numbers and Quadratic Equation
118383
One of the square roots of \(6+4 \sqrt{3}\) is
1 \(\sqrt{3}(\sqrt{3}+1)\)
2 \(-\sqrt{3}(\sqrt{3}-1)\)
3 \(\sqrt{3}(-\sqrt{3}+1)\)
4 None of these
Explanation:
D Square roots of \(6+4 \sqrt{3}\) Let, \(\quad \mathrm{x}=6+4 \sqrt{3}\) Then, square root is \(\sqrt{\mathrm{x}}=\sqrt{6+4 \sqrt{3}}\) \(\sqrt{\mathrm{x}}=\sqrt{2(3+2 \sqrt{3})}\) \(\sqrt{\mathrm{x}}=\sqrt{2} \sqrt{3+2 \sqrt{3}}\)So, answer is none of these.
UPSEE-2009
Complex Numbers and Quadratic Equation
118385
The equation \(x^3-3 x+4=0\) has only one real root. What is its first approximate value as obtained by the method of false position in \((-3,-2)\) ?
1 -2.125
2 2.125
3 -2.812
4 2.812
Explanation:
A Given, \(x^3-3 x+4=0\) We know that, According to false position method- Here, \(\mathrm{x}_{\mathrm{n}}=\mathrm{x}_1-\frac{\mathrm{x}_{\mathrm{n}-1}-\mathrm{x}_1}{\mathrm{f}\left(\mathrm{x}_{\mathrm{n}-1}\right)-\mathrm{f}\left(\mathrm{x}_1\right)} \mathrm{f}\left(\mathrm{x}_1\right)\) \(\mathrm{x}_{\mathrm{n}-1}=-2\) and \(\mathrm{x}_1=-3\) \(\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}(-3)=(-3)^3-3(-3)+4\) \(\mathrm{f}(-3)=-27+9+4=-14\) \(\mathrm{f}\left(\mathrm{x}_{\mathrm{n}-1}\right)=\mathrm{f}(-2)=(-2)^3-3(-2)+4\) \(=-8+6+4=2\) Now, \(x_n=x_1-\frac{x_{n-1}-x_1}{f\left(x_{n-1}\right)-f\left(x_1\right)} f\left(x_1\right)\) \(=-3-\frac{-2+3}{2-(-14)}(-14)\) \(=-3+\frac{14}{16}\) \(=-3+\frac{7}{8}=\frac{-24+7}{8}=-\frac{17}{8}\) \(=-2 \cdot 125\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Complex Numbers and Quadratic Equation
118380
For the equation \(\frac{1}{x+a}-\frac{1}{x+b}=\frac{1}{x+c}\), if the product of roots is zero, then the sum of roots is
1 0
2 \(\frac{2 a b}{b+c}\)
3 \(\frac{2 b c}{b+c}\)
4 \(\frac{-2 b c}{b+c}\)
Explanation:
D Given \(\frac{1}{\mathrm{x}+\mathrm{a}}-\frac{1}{\mathrm{x}+\mathrm{b}}=\frac{1}{\mathrm{x}+\mathrm{c}}\) \(\text { or } \mathrm{x}^2+(\mathrm{a}+\mathrm{b}) \mathrm{x}+\mathrm{ab}=(\mathrm{b}-\mathrm{a}) \mathrm{x}+(\mathrm{b}-\mathrm{a}) \mathrm{c}\) \(\text { or } \mathrm{x}^2+2 \mathrm{ax}+\mathrm{ab}+\mathrm{ca}-\mathrm{bc}=0\) \(\text { Since product of the roots }=0\) \(\mathrm{ab}+\mathrm{ca}-\mathrm{bc}=0 ; \mathrm{a}=\frac{\mathrm{bc}}{\mathrm{b}+\mathrm{c}}\) \(\text { Thus sum of roots }=-2 \mathrm{a}=\frac{-2 \mathrm{bc}}{\mathrm{b}+\mathrm{c}}\)
BITSAT-2009
Complex Numbers and Quadratic Equation
118381
If \(\alpha, \beta\) and \(\gamma\) are the roots of the equation \(x^3-\) \(3 x^2+3 x+7=0\), and \(\omega\) is cube root of unity, then the value of \(\frac{\alpha-1}{\beta-1}+\frac{\beta-1}{\gamma-1}+\frac{\gamma-1}{\alpha-1}\) is equal to
1 \(3 \omega^2\)
2 \(3 / \omega\)
3 \(2 \omega^2\)
4 None of these
Explanation:
A Given \(\alpha, \beta\) and \(\gamma\) are the roots of equation \(x^3-3 x^2+3 x+7=0\) \((\mathrm{x}+1)\left(\mathrm{x}^2-4 \mathrm{x}+7\right)=0\) \((x+1)=0\) \(\mathrm{x}=-1\) \(\left(x^2-4 x+7\right)=0\) \(x=\frac{+4 \pm \sqrt{(4)^2-4 \times 7}}{2}\) \(\mathrm{x}=\frac{4 \pm \sqrt{16-28}}{2}\) \(\mathrm{x}=2 \pm \sqrt{3} \mathrm{i}\) Given, \(\omega\) is cube root of unity Then, \(\quad x=2+\sqrt{3} i, 2-\sqrt{3} i\) Let, \(\alpha=-1, \beta=2+i \sqrt{3}, \gamma=2-i \sqrt{3}\) \(\alpha-1=-2, \beta-1=1+\sqrt{3} \mathrm{i}, \gamma-1=1-\sqrt{3} \mathrm{i}\) \(\alpha-1=-2, \beta-1=-2 \omega, \gamma-1=-2 \omega^2\) Now, \(\quad \frac{\alpha-1}{\beta-1}+\frac{\beta-1}{\gamma-1}+\frac{\gamma-1}{\alpha-1}\) \(=\frac{-2}{-2 \omega}+\left(\frac{-2 \omega}{-2 \omega^2}\right)+\left(\frac{-2 \omega^2}{-2}\right)\) \(=\frac{2}{\omega}+\omega^2=\frac{2+\omega^3}{\omega}=3 \omega^2\)
UPSEE-2018
Complex Numbers and Quadratic Equation
118383
One of the square roots of \(6+4 \sqrt{3}\) is
1 \(\sqrt{3}(\sqrt{3}+1)\)
2 \(-\sqrt{3}(\sqrt{3}-1)\)
3 \(\sqrt{3}(-\sqrt{3}+1)\)
4 None of these
Explanation:
D Square roots of \(6+4 \sqrt{3}\) Let, \(\quad \mathrm{x}=6+4 \sqrt{3}\) Then, square root is \(\sqrt{\mathrm{x}}=\sqrt{6+4 \sqrt{3}}\) \(\sqrt{\mathrm{x}}=\sqrt{2(3+2 \sqrt{3})}\) \(\sqrt{\mathrm{x}}=\sqrt{2} \sqrt{3+2 \sqrt{3}}\)So, answer is none of these.
UPSEE-2009
Complex Numbers and Quadratic Equation
118385
The equation \(x^3-3 x+4=0\) has only one real root. What is its first approximate value as obtained by the method of false position in \((-3,-2)\) ?
1 -2.125
2 2.125
3 -2.812
4 2.812
Explanation:
A Given, \(x^3-3 x+4=0\) We know that, According to false position method- Here, \(\mathrm{x}_{\mathrm{n}}=\mathrm{x}_1-\frac{\mathrm{x}_{\mathrm{n}-1}-\mathrm{x}_1}{\mathrm{f}\left(\mathrm{x}_{\mathrm{n}-1}\right)-\mathrm{f}\left(\mathrm{x}_1\right)} \mathrm{f}\left(\mathrm{x}_1\right)\) \(\mathrm{x}_{\mathrm{n}-1}=-2\) and \(\mathrm{x}_1=-3\) \(\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}(-3)=(-3)^3-3(-3)+4\) \(\mathrm{f}(-3)=-27+9+4=-14\) \(\mathrm{f}\left(\mathrm{x}_{\mathrm{n}-1}\right)=\mathrm{f}(-2)=(-2)^3-3(-2)+4\) \(=-8+6+4=2\) Now, \(x_n=x_1-\frac{x_{n-1}-x_1}{f\left(x_{n-1}\right)-f\left(x_1\right)} f\left(x_1\right)\) \(=-3-\frac{-2+3}{2-(-14)}(-14)\) \(=-3+\frac{14}{16}\) \(=-3+\frac{7}{8}=\frac{-24+7}{8}=-\frac{17}{8}\) \(=-2 \cdot 125\)
118380
For the equation \(\frac{1}{x+a}-\frac{1}{x+b}=\frac{1}{x+c}\), if the product of roots is zero, then the sum of roots is
1 0
2 \(\frac{2 a b}{b+c}\)
3 \(\frac{2 b c}{b+c}\)
4 \(\frac{-2 b c}{b+c}\)
Explanation:
D Given \(\frac{1}{\mathrm{x}+\mathrm{a}}-\frac{1}{\mathrm{x}+\mathrm{b}}=\frac{1}{\mathrm{x}+\mathrm{c}}\) \(\text { or } \mathrm{x}^2+(\mathrm{a}+\mathrm{b}) \mathrm{x}+\mathrm{ab}=(\mathrm{b}-\mathrm{a}) \mathrm{x}+(\mathrm{b}-\mathrm{a}) \mathrm{c}\) \(\text { or } \mathrm{x}^2+2 \mathrm{ax}+\mathrm{ab}+\mathrm{ca}-\mathrm{bc}=0\) \(\text { Since product of the roots }=0\) \(\mathrm{ab}+\mathrm{ca}-\mathrm{bc}=0 ; \mathrm{a}=\frac{\mathrm{bc}}{\mathrm{b}+\mathrm{c}}\) \(\text { Thus sum of roots }=-2 \mathrm{a}=\frac{-2 \mathrm{bc}}{\mathrm{b}+\mathrm{c}}\)
BITSAT-2009
Complex Numbers and Quadratic Equation
118381
If \(\alpha, \beta\) and \(\gamma\) are the roots of the equation \(x^3-\) \(3 x^2+3 x+7=0\), and \(\omega\) is cube root of unity, then the value of \(\frac{\alpha-1}{\beta-1}+\frac{\beta-1}{\gamma-1}+\frac{\gamma-1}{\alpha-1}\) is equal to
1 \(3 \omega^2\)
2 \(3 / \omega\)
3 \(2 \omega^2\)
4 None of these
Explanation:
A Given \(\alpha, \beta\) and \(\gamma\) are the roots of equation \(x^3-3 x^2+3 x+7=0\) \((\mathrm{x}+1)\left(\mathrm{x}^2-4 \mathrm{x}+7\right)=0\) \((x+1)=0\) \(\mathrm{x}=-1\) \(\left(x^2-4 x+7\right)=0\) \(x=\frac{+4 \pm \sqrt{(4)^2-4 \times 7}}{2}\) \(\mathrm{x}=\frac{4 \pm \sqrt{16-28}}{2}\) \(\mathrm{x}=2 \pm \sqrt{3} \mathrm{i}\) Given, \(\omega\) is cube root of unity Then, \(\quad x=2+\sqrt{3} i, 2-\sqrt{3} i\) Let, \(\alpha=-1, \beta=2+i \sqrt{3}, \gamma=2-i \sqrt{3}\) \(\alpha-1=-2, \beta-1=1+\sqrt{3} \mathrm{i}, \gamma-1=1-\sqrt{3} \mathrm{i}\) \(\alpha-1=-2, \beta-1=-2 \omega, \gamma-1=-2 \omega^2\) Now, \(\quad \frac{\alpha-1}{\beta-1}+\frac{\beta-1}{\gamma-1}+\frac{\gamma-1}{\alpha-1}\) \(=\frac{-2}{-2 \omega}+\left(\frac{-2 \omega}{-2 \omega^2}\right)+\left(\frac{-2 \omega^2}{-2}\right)\) \(=\frac{2}{\omega}+\omega^2=\frac{2+\omega^3}{\omega}=3 \omega^2\)
UPSEE-2018
Complex Numbers and Quadratic Equation
118383
One of the square roots of \(6+4 \sqrt{3}\) is
1 \(\sqrt{3}(\sqrt{3}+1)\)
2 \(-\sqrt{3}(\sqrt{3}-1)\)
3 \(\sqrt{3}(-\sqrt{3}+1)\)
4 None of these
Explanation:
D Square roots of \(6+4 \sqrt{3}\) Let, \(\quad \mathrm{x}=6+4 \sqrt{3}\) Then, square root is \(\sqrt{\mathrm{x}}=\sqrt{6+4 \sqrt{3}}\) \(\sqrt{\mathrm{x}}=\sqrt{2(3+2 \sqrt{3})}\) \(\sqrt{\mathrm{x}}=\sqrt{2} \sqrt{3+2 \sqrt{3}}\)So, answer is none of these.
UPSEE-2009
Complex Numbers and Quadratic Equation
118385
The equation \(x^3-3 x+4=0\) has only one real root. What is its first approximate value as obtained by the method of false position in \((-3,-2)\) ?
1 -2.125
2 2.125
3 -2.812
4 2.812
Explanation:
A Given, \(x^3-3 x+4=0\) We know that, According to false position method- Here, \(\mathrm{x}_{\mathrm{n}}=\mathrm{x}_1-\frac{\mathrm{x}_{\mathrm{n}-1}-\mathrm{x}_1}{\mathrm{f}\left(\mathrm{x}_{\mathrm{n}-1}\right)-\mathrm{f}\left(\mathrm{x}_1\right)} \mathrm{f}\left(\mathrm{x}_1\right)\) \(\mathrm{x}_{\mathrm{n}-1}=-2\) and \(\mathrm{x}_1=-3\) \(\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}(-3)=(-3)^3-3(-3)+4\) \(\mathrm{f}(-3)=-27+9+4=-14\) \(\mathrm{f}\left(\mathrm{x}_{\mathrm{n}-1}\right)=\mathrm{f}(-2)=(-2)^3-3(-2)+4\) \(=-8+6+4=2\) Now, \(x_n=x_1-\frac{x_{n-1}-x_1}{f\left(x_{n-1}\right)-f\left(x_1\right)} f\left(x_1\right)\) \(=-3-\frac{-2+3}{2-(-14)}(-14)\) \(=-3+\frac{14}{16}\) \(=-3+\frac{7}{8}=\frac{-24+7}{8}=-\frac{17}{8}\) \(=-2 \cdot 125\)
118380
For the equation \(\frac{1}{x+a}-\frac{1}{x+b}=\frac{1}{x+c}\), if the product of roots is zero, then the sum of roots is
1 0
2 \(\frac{2 a b}{b+c}\)
3 \(\frac{2 b c}{b+c}\)
4 \(\frac{-2 b c}{b+c}\)
Explanation:
D Given \(\frac{1}{\mathrm{x}+\mathrm{a}}-\frac{1}{\mathrm{x}+\mathrm{b}}=\frac{1}{\mathrm{x}+\mathrm{c}}\) \(\text { or } \mathrm{x}^2+(\mathrm{a}+\mathrm{b}) \mathrm{x}+\mathrm{ab}=(\mathrm{b}-\mathrm{a}) \mathrm{x}+(\mathrm{b}-\mathrm{a}) \mathrm{c}\) \(\text { or } \mathrm{x}^2+2 \mathrm{ax}+\mathrm{ab}+\mathrm{ca}-\mathrm{bc}=0\) \(\text { Since product of the roots }=0\) \(\mathrm{ab}+\mathrm{ca}-\mathrm{bc}=0 ; \mathrm{a}=\frac{\mathrm{bc}}{\mathrm{b}+\mathrm{c}}\) \(\text { Thus sum of roots }=-2 \mathrm{a}=\frac{-2 \mathrm{bc}}{\mathrm{b}+\mathrm{c}}\)
BITSAT-2009
Complex Numbers and Quadratic Equation
118381
If \(\alpha, \beta\) and \(\gamma\) are the roots of the equation \(x^3-\) \(3 x^2+3 x+7=0\), and \(\omega\) is cube root of unity, then the value of \(\frac{\alpha-1}{\beta-1}+\frac{\beta-1}{\gamma-1}+\frac{\gamma-1}{\alpha-1}\) is equal to
1 \(3 \omega^2\)
2 \(3 / \omega\)
3 \(2 \omega^2\)
4 None of these
Explanation:
A Given \(\alpha, \beta\) and \(\gamma\) are the roots of equation \(x^3-3 x^2+3 x+7=0\) \((\mathrm{x}+1)\left(\mathrm{x}^2-4 \mathrm{x}+7\right)=0\) \((x+1)=0\) \(\mathrm{x}=-1\) \(\left(x^2-4 x+7\right)=0\) \(x=\frac{+4 \pm \sqrt{(4)^2-4 \times 7}}{2}\) \(\mathrm{x}=\frac{4 \pm \sqrt{16-28}}{2}\) \(\mathrm{x}=2 \pm \sqrt{3} \mathrm{i}\) Given, \(\omega\) is cube root of unity Then, \(\quad x=2+\sqrt{3} i, 2-\sqrt{3} i\) Let, \(\alpha=-1, \beta=2+i \sqrt{3}, \gamma=2-i \sqrt{3}\) \(\alpha-1=-2, \beta-1=1+\sqrt{3} \mathrm{i}, \gamma-1=1-\sqrt{3} \mathrm{i}\) \(\alpha-1=-2, \beta-1=-2 \omega, \gamma-1=-2 \omega^2\) Now, \(\quad \frac{\alpha-1}{\beta-1}+\frac{\beta-1}{\gamma-1}+\frac{\gamma-1}{\alpha-1}\) \(=\frac{-2}{-2 \omega}+\left(\frac{-2 \omega}{-2 \omega^2}\right)+\left(\frac{-2 \omega^2}{-2}\right)\) \(=\frac{2}{\omega}+\omega^2=\frac{2+\omega^3}{\omega}=3 \omega^2\)
UPSEE-2018
Complex Numbers and Quadratic Equation
118383
One of the square roots of \(6+4 \sqrt{3}\) is
1 \(\sqrt{3}(\sqrt{3}+1)\)
2 \(-\sqrt{3}(\sqrt{3}-1)\)
3 \(\sqrt{3}(-\sqrt{3}+1)\)
4 None of these
Explanation:
D Square roots of \(6+4 \sqrt{3}\) Let, \(\quad \mathrm{x}=6+4 \sqrt{3}\) Then, square root is \(\sqrt{\mathrm{x}}=\sqrt{6+4 \sqrt{3}}\) \(\sqrt{\mathrm{x}}=\sqrt{2(3+2 \sqrt{3})}\) \(\sqrt{\mathrm{x}}=\sqrt{2} \sqrt{3+2 \sqrt{3}}\)So, answer is none of these.
UPSEE-2009
Complex Numbers and Quadratic Equation
118385
The equation \(x^3-3 x+4=0\) has only one real root. What is its first approximate value as obtained by the method of false position in \((-3,-2)\) ?
1 -2.125
2 2.125
3 -2.812
4 2.812
Explanation:
A Given, \(x^3-3 x+4=0\) We know that, According to false position method- Here, \(\mathrm{x}_{\mathrm{n}}=\mathrm{x}_1-\frac{\mathrm{x}_{\mathrm{n}-1}-\mathrm{x}_1}{\mathrm{f}\left(\mathrm{x}_{\mathrm{n}-1}\right)-\mathrm{f}\left(\mathrm{x}_1\right)} \mathrm{f}\left(\mathrm{x}_1\right)\) \(\mathrm{x}_{\mathrm{n}-1}=-2\) and \(\mathrm{x}_1=-3\) \(\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}(-3)=(-3)^3-3(-3)+4\) \(\mathrm{f}(-3)=-27+9+4=-14\) \(\mathrm{f}\left(\mathrm{x}_{\mathrm{n}-1}\right)=\mathrm{f}(-2)=(-2)^3-3(-2)+4\) \(=-8+6+4=2\) Now, \(x_n=x_1-\frac{x_{n-1}-x_1}{f\left(x_{n-1}\right)-f\left(x_1\right)} f\left(x_1\right)\) \(=-3-\frac{-2+3}{2-(-14)}(-14)\) \(=-3+\frac{14}{16}\) \(=-3+\frac{7}{8}=\frac{-24+7}{8}=-\frac{17}{8}\) \(=-2 \cdot 125\)