118111
If \(x^2-3 x+2\) is a factor of \(x^4-p x^2+q\), then the values \(p\) and \(q\) are
1 \(5,-4\)
2 5,4
3 \(-5,4\)
4 \(-5,-4\)
Explanation:
B \( x^2-3 x+2 \text { is a factor of } x^4-p x^2+q \text {. }\) \(\Rightarrow x^2-3 x+2\) \(x^2-2 x-x+2=0\) \(x(x-2)-1(x-2)=0\) \((x-1)(x-2)=0\) \(x_2=1,2\) \(\text { when } x=1 \text { then } x^4-p x^2+q=0\) \(1-p+q=0\) \(\text { And when } x=2 \text { then }\) \((2)^4-p(2)^2+q=0\) \(16-4 p+q=0\) \(\text { Subtracting (i) and (ii), we get }\) \(1-p+q=0\) \(16-4 p+q=0\) \(15-3 p=0\) \(p=5\) \(\text { Then, } 1-5+q=0\) \(q=4 \text { and } p=5\) Subtracting (i) and (ii), we get Then, \(1-5+q=0\)
AMU-2007
Complex Numbers and Quadratic Equation
118112
The sum of all the roots of the equation \(\left|x^2-\mathbf{8 x}+\mathbf{1 5}\right|-\mathbf{2 x}+\mathbf{7}=\mathbf{0} \text { is : }\)
1 \(9+\sqrt{3}\)
2 \(11+\sqrt{3}\)
3 \(9-\sqrt{3}\)
4 \(11-\sqrt{3}\)
Explanation:
A Given, the equation, \(\left|\mathrm{x}^2-8 \mathrm{x}+15\right|-2 \mathrm{x}+7=0\) \(\text { Case-I :- }\) \(\mathrm{x}^2-8 \mathrm{x}+15 \geq 0\) \(\Rightarrow \quad \mathrm{x}^2-5 \mathrm{x}-3 \mathrm{x}+15 \geq 0\) \(\text { or } \quad(x-5)(x-3) \geq 0\) \(\Rightarrow \quad \mathrm{x} \geq 5 \text {, then, in this case- }\) \(\mathrm{x}^2-8 \mathrm{x}+15-2 \mathrm{x}+7=0\) \(\mathrm{x}^2-10 \mathrm{x}+22=0\) \(\mathrm{x}=\frac{10 \pm \sqrt{100-88}}{2}\) \(\mathrm{x}=\frac{10 \pm \sqrt{12}}{2}=5 \pm \sqrt{3} \text {, But } \mathrm{x} \geq 5\) \(\mathrm{x}=5+\sqrt{2}\) \(\text { or }\) \(\mathrm{x}=\frac{10 \pm \sqrt{1}}{}\) \(\mathrm{x}=\frac{10 \pm \sqrt{12}}{2}\) \(\therefore \quad \mathrm{X}=5+\sqrt{3}\) \(\text { Case-II:- }\left(\mathrm{x}^2-8 \mathrm{x}+15\right) \leq 0\) \(\Rightarrow \quad(\mathrm{x}-5)(\mathrm{x}-3) \leq 0\) \(\Rightarrow \quad 3 \leq \mathrm{x} \leq 5\) \(\left|\mathrm{x}^2-8 \mathrm{x}+15\right|=-\left(\mathrm{x}^2-8 \mathrm{x}+15\right)\) \(\therefore \quad-x^2+8 \mathrm{x}-15-2 \mathrm{x}+7=0\) \(\text { Or } \quad x^2-6 x+8=0\) \(\mathrm{x}^2-4 \mathrm{x}-2 \mathrm{x}+8=0\) \(\mathrm{x}(\mathrm{x}-4)-2(\mathrm{x}-4)=0\) \((\mathrm{x}-2)(\mathrm{x}-4)=0\) \(\therefore \quad \mathrm{x}=2, \mathrm{x}=4\) \(\text { But } 3 \leq \mathrm{x} \leq 5\) \(\therefore \quad \mathrm{x}=4 \text { is the solution. }\) \(\therefore \quad\) The sum of the roots \((5+\sqrt{3}+4)\) is \((9+\sqrt{3})\)
JEE Main-06.04.2023
Complex Numbers and Quadratic Equation
118114
If \(\alpha, \beta, \gamma\) are the roots of the equation \(\mathrm{x}^3+\mathrm{ax}^2+\) \(b x+c=0\), then \(\alpha^{-1}+\beta^{-1}+\gamma^{-1}\) is equal to
1 \(\frac{\mathrm{a}}{\mathrm{c}}\)
2 \(\frac{\mathrm{c}}{\mathrm{a}}\)
3 \(-\frac{b}{c}\)
4 \(\frac{b}{a}\)
Explanation:
C Given the equation, Now, \(\mathrm{x}^3+\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0\) \(\alpha^{-1}+\beta^{-1}+\gamma^{-1} =\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=\frac{1}{\alpha \beta \gamma}(\alpha \beta+\beta \gamma+\alpha \gamma)\) \(=-\frac{b}{c}(\alpha \beta+\beta \gamma+\alpha \gamma=b, \alpha \beta \gamma=-c)\)
AP EAMCET-2002
Complex Numbers and Quadratic Equation
118115
If \(\alpha, \beta, \gamma\) are the roots of \(2 x^3-2 x-1=0\), then \((\Sigma \alpha \beta)^2\) is equal to
1 -1
2 1
3 2
4 3
Explanation:
B Given, \(\alpha, \beta, \gamma\) are the roots of- \(2 x^3-2 x-1=0\) \(\text { Or } \quad x^3-x-\frac{1}{2}=0\) We have, \(\quad \alpha \beta+\beta \gamma+\alpha \gamma=-1\) \((\Sigma \alpha \beta)^2=1\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Complex Numbers and Quadratic Equation
118111
If \(x^2-3 x+2\) is a factor of \(x^4-p x^2+q\), then the values \(p\) and \(q\) are
1 \(5,-4\)
2 5,4
3 \(-5,4\)
4 \(-5,-4\)
Explanation:
B \( x^2-3 x+2 \text { is a factor of } x^4-p x^2+q \text {. }\) \(\Rightarrow x^2-3 x+2\) \(x^2-2 x-x+2=0\) \(x(x-2)-1(x-2)=0\) \((x-1)(x-2)=0\) \(x_2=1,2\) \(\text { when } x=1 \text { then } x^4-p x^2+q=0\) \(1-p+q=0\) \(\text { And when } x=2 \text { then }\) \((2)^4-p(2)^2+q=0\) \(16-4 p+q=0\) \(\text { Subtracting (i) and (ii), we get }\) \(1-p+q=0\) \(16-4 p+q=0\) \(15-3 p=0\) \(p=5\) \(\text { Then, } 1-5+q=0\) \(q=4 \text { and } p=5\) Subtracting (i) and (ii), we get Then, \(1-5+q=0\)
AMU-2007
Complex Numbers and Quadratic Equation
118112
The sum of all the roots of the equation \(\left|x^2-\mathbf{8 x}+\mathbf{1 5}\right|-\mathbf{2 x}+\mathbf{7}=\mathbf{0} \text { is : }\)
1 \(9+\sqrt{3}\)
2 \(11+\sqrt{3}\)
3 \(9-\sqrt{3}\)
4 \(11-\sqrt{3}\)
Explanation:
A Given, the equation, \(\left|\mathrm{x}^2-8 \mathrm{x}+15\right|-2 \mathrm{x}+7=0\) \(\text { Case-I :- }\) \(\mathrm{x}^2-8 \mathrm{x}+15 \geq 0\) \(\Rightarrow \quad \mathrm{x}^2-5 \mathrm{x}-3 \mathrm{x}+15 \geq 0\) \(\text { or } \quad(x-5)(x-3) \geq 0\) \(\Rightarrow \quad \mathrm{x} \geq 5 \text {, then, in this case- }\) \(\mathrm{x}^2-8 \mathrm{x}+15-2 \mathrm{x}+7=0\) \(\mathrm{x}^2-10 \mathrm{x}+22=0\) \(\mathrm{x}=\frac{10 \pm \sqrt{100-88}}{2}\) \(\mathrm{x}=\frac{10 \pm \sqrt{12}}{2}=5 \pm \sqrt{3} \text {, But } \mathrm{x} \geq 5\) \(\mathrm{x}=5+\sqrt{2}\) \(\text { or }\) \(\mathrm{x}=\frac{10 \pm \sqrt{1}}{}\) \(\mathrm{x}=\frac{10 \pm \sqrt{12}}{2}\) \(\therefore \quad \mathrm{X}=5+\sqrt{3}\) \(\text { Case-II:- }\left(\mathrm{x}^2-8 \mathrm{x}+15\right) \leq 0\) \(\Rightarrow \quad(\mathrm{x}-5)(\mathrm{x}-3) \leq 0\) \(\Rightarrow \quad 3 \leq \mathrm{x} \leq 5\) \(\left|\mathrm{x}^2-8 \mathrm{x}+15\right|=-\left(\mathrm{x}^2-8 \mathrm{x}+15\right)\) \(\therefore \quad-x^2+8 \mathrm{x}-15-2 \mathrm{x}+7=0\) \(\text { Or } \quad x^2-6 x+8=0\) \(\mathrm{x}^2-4 \mathrm{x}-2 \mathrm{x}+8=0\) \(\mathrm{x}(\mathrm{x}-4)-2(\mathrm{x}-4)=0\) \((\mathrm{x}-2)(\mathrm{x}-4)=0\) \(\therefore \quad \mathrm{x}=2, \mathrm{x}=4\) \(\text { But } 3 \leq \mathrm{x} \leq 5\) \(\therefore \quad \mathrm{x}=4 \text { is the solution. }\) \(\therefore \quad\) The sum of the roots \((5+\sqrt{3}+4)\) is \((9+\sqrt{3})\)
JEE Main-06.04.2023
Complex Numbers and Quadratic Equation
118114
If \(\alpha, \beta, \gamma\) are the roots of the equation \(\mathrm{x}^3+\mathrm{ax}^2+\) \(b x+c=0\), then \(\alpha^{-1}+\beta^{-1}+\gamma^{-1}\) is equal to
1 \(\frac{\mathrm{a}}{\mathrm{c}}\)
2 \(\frac{\mathrm{c}}{\mathrm{a}}\)
3 \(-\frac{b}{c}\)
4 \(\frac{b}{a}\)
Explanation:
C Given the equation, Now, \(\mathrm{x}^3+\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0\) \(\alpha^{-1}+\beta^{-1}+\gamma^{-1} =\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=\frac{1}{\alpha \beta \gamma}(\alpha \beta+\beta \gamma+\alpha \gamma)\) \(=-\frac{b}{c}(\alpha \beta+\beta \gamma+\alpha \gamma=b, \alpha \beta \gamma=-c)\)
AP EAMCET-2002
Complex Numbers and Quadratic Equation
118115
If \(\alpha, \beta, \gamma\) are the roots of \(2 x^3-2 x-1=0\), then \((\Sigma \alpha \beta)^2\) is equal to
1 -1
2 1
3 2
4 3
Explanation:
B Given, \(\alpha, \beta, \gamma\) are the roots of- \(2 x^3-2 x-1=0\) \(\text { Or } \quad x^3-x-\frac{1}{2}=0\) We have, \(\quad \alpha \beta+\beta \gamma+\alpha \gamma=-1\) \((\Sigma \alpha \beta)^2=1\)
118111
If \(x^2-3 x+2\) is a factor of \(x^4-p x^2+q\), then the values \(p\) and \(q\) are
1 \(5,-4\)
2 5,4
3 \(-5,4\)
4 \(-5,-4\)
Explanation:
B \( x^2-3 x+2 \text { is a factor of } x^4-p x^2+q \text {. }\) \(\Rightarrow x^2-3 x+2\) \(x^2-2 x-x+2=0\) \(x(x-2)-1(x-2)=0\) \((x-1)(x-2)=0\) \(x_2=1,2\) \(\text { when } x=1 \text { then } x^4-p x^2+q=0\) \(1-p+q=0\) \(\text { And when } x=2 \text { then }\) \((2)^4-p(2)^2+q=0\) \(16-4 p+q=0\) \(\text { Subtracting (i) and (ii), we get }\) \(1-p+q=0\) \(16-4 p+q=0\) \(15-3 p=0\) \(p=5\) \(\text { Then, } 1-5+q=0\) \(q=4 \text { and } p=5\) Subtracting (i) and (ii), we get Then, \(1-5+q=0\)
AMU-2007
Complex Numbers and Quadratic Equation
118112
The sum of all the roots of the equation \(\left|x^2-\mathbf{8 x}+\mathbf{1 5}\right|-\mathbf{2 x}+\mathbf{7}=\mathbf{0} \text { is : }\)
1 \(9+\sqrt{3}\)
2 \(11+\sqrt{3}\)
3 \(9-\sqrt{3}\)
4 \(11-\sqrt{3}\)
Explanation:
A Given, the equation, \(\left|\mathrm{x}^2-8 \mathrm{x}+15\right|-2 \mathrm{x}+7=0\) \(\text { Case-I :- }\) \(\mathrm{x}^2-8 \mathrm{x}+15 \geq 0\) \(\Rightarrow \quad \mathrm{x}^2-5 \mathrm{x}-3 \mathrm{x}+15 \geq 0\) \(\text { or } \quad(x-5)(x-3) \geq 0\) \(\Rightarrow \quad \mathrm{x} \geq 5 \text {, then, in this case- }\) \(\mathrm{x}^2-8 \mathrm{x}+15-2 \mathrm{x}+7=0\) \(\mathrm{x}^2-10 \mathrm{x}+22=0\) \(\mathrm{x}=\frac{10 \pm \sqrt{100-88}}{2}\) \(\mathrm{x}=\frac{10 \pm \sqrt{12}}{2}=5 \pm \sqrt{3} \text {, But } \mathrm{x} \geq 5\) \(\mathrm{x}=5+\sqrt{2}\) \(\text { or }\) \(\mathrm{x}=\frac{10 \pm \sqrt{1}}{}\) \(\mathrm{x}=\frac{10 \pm \sqrt{12}}{2}\) \(\therefore \quad \mathrm{X}=5+\sqrt{3}\) \(\text { Case-II:- }\left(\mathrm{x}^2-8 \mathrm{x}+15\right) \leq 0\) \(\Rightarrow \quad(\mathrm{x}-5)(\mathrm{x}-3) \leq 0\) \(\Rightarrow \quad 3 \leq \mathrm{x} \leq 5\) \(\left|\mathrm{x}^2-8 \mathrm{x}+15\right|=-\left(\mathrm{x}^2-8 \mathrm{x}+15\right)\) \(\therefore \quad-x^2+8 \mathrm{x}-15-2 \mathrm{x}+7=0\) \(\text { Or } \quad x^2-6 x+8=0\) \(\mathrm{x}^2-4 \mathrm{x}-2 \mathrm{x}+8=0\) \(\mathrm{x}(\mathrm{x}-4)-2(\mathrm{x}-4)=0\) \((\mathrm{x}-2)(\mathrm{x}-4)=0\) \(\therefore \quad \mathrm{x}=2, \mathrm{x}=4\) \(\text { But } 3 \leq \mathrm{x} \leq 5\) \(\therefore \quad \mathrm{x}=4 \text { is the solution. }\) \(\therefore \quad\) The sum of the roots \((5+\sqrt{3}+4)\) is \((9+\sqrt{3})\)
JEE Main-06.04.2023
Complex Numbers and Quadratic Equation
118114
If \(\alpha, \beta, \gamma\) are the roots of the equation \(\mathrm{x}^3+\mathrm{ax}^2+\) \(b x+c=0\), then \(\alpha^{-1}+\beta^{-1}+\gamma^{-1}\) is equal to
1 \(\frac{\mathrm{a}}{\mathrm{c}}\)
2 \(\frac{\mathrm{c}}{\mathrm{a}}\)
3 \(-\frac{b}{c}\)
4 \(\frac{b}{a}\)
Explanation:
C Given the equation, Now, \(\mathrm{x}^3+\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0\) \(\alpha^{-1}+\beta^{-1}+\gamma^{-1} =\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=\frac{1}{\alpha \beta \gamma}(\alpha \beta+\beta \gamma+\alpha \gamma)\) \(=-\frac{b}{c}(\alpha \beta+\beta \gamma+\alpha \gamma=b, \alpha \beta \gamma=-c)\)
AP EAMCET-2002
Complex Numbers and Quadratic Equation
118115
If \(\alpha, \beta, \gamma\) are the roots of \(2 x^3-2 x-1=0\), then \((\Sigma \alpha \beta)^2\) is equal to
1 -1
2 1
3 2
4 3
Explanation:
B Given, \(\alpha, \beta, \gamma\) are the roots of- \(2 x^3-2 x-1=0\) \(\text { Or } \quad x^3-x-\frac{1}{2}=0\) We have, \(\quad \alpha \beta+\beta \gamma+\alpha \gamma=-1\) \((\Sigma \alpha \beta)^2=1\)
118111
If \(x^2-3 x+2\) is a factor of \(x^4-p x^2+q\), then the values \(p\) and \(q\) are
1 \(5,-4\)
2 5,4
3 \(-5,4\)
4 \(-5,-4\)
Explanation:
B \( x^2-3 x+2 \text { is a factor of } x^4-p x^2+q \text {. }\) \(\Rightarrow x^2-3 x+2\) \(x^2-2 x-x+2=0\) \(x(x-2)-1(x-2)=0\) \((x-1)(x-2)=0\) \(x_2=1,2\) \(\text { when } x=1 \text { then } x^4-p x^2+q=0\) \(1-p+q=0\) \(\text { And when } x=2 \text { then }\) \((2)^4-p(2)^2+q=0\) \(16-4 p+q=0\) \(\text { Subtracting (i) and (ii), we get }\) \(1-p+q=0\) \(16-4 p+q=0\) \(15-3 p=0\) \(p=5\) \(\text { Then, } 1-5+q=0\) \(q=4 \text { and } p=5\) Subtracting (i) and (ii), we get Then, \(1-5+q=0\)
AMU-2007
Complex Numbers and Quadratic Equation
118112
The sum of all the roots of the equation \(\left|x^2-\mathbf{8 x}+\mathbf{1 5}\right|-\mathbf{2 x}+\mathbf{7}=\mathbf{0} \text { is : }\)
1 \(9+\sqrt{3}\)
2 \(11+\sqrt{3}\)
3 \(9-\sqrt{3}\)
4 \(11-\sqrt{3}\)
Explanation:
A Given, the equation, \(\left|\mathrm{x}^2-8 \mathrm{x}+15\right|-2 \mathrm{x}+7=0\) \(\text { Case-I :- }\) \(\mathrm{x}^2-8 \mathrm{x}+15 \geq 0\) \(\Rightarrow \quad \mathrm{x}^2-5 \mathrm{x}-3 \mathrm{x}+15 \geq 0\) \(\text { or } \quad(x-5)(x-3) \geq 0\) \(\Rightarrow \quad \mathrm{x} \geq 5 \text {, then, in this case- }\) \(\mathrm{x}^2-8 \mathrm{x}+15-2 \mathrm{x}+7=0\) \(\mathrm{x}^2-10 \mathrm{x}+22=0\) \(\mathrm{x}=\frac{10 \pm \sqrt{100-88}}{2}\) \(\mathrm{x}=\frac{10 \pm \sqrt{12}}{2}=5 \pm \sqrt{3} \text {, But } \mathrm{x} \geq 5\) \(\mathrm{x}=5+\sqrt{2}\) \(\text { or }\) \(\mathrm{x}=\frac{10 \pm \sqrt{1}}{}\) \(\mathrm{x}=\frac{10 \pm \sqrt{12}}{2}\) \(\therefore \quad \mathrm{X}=5+\sqrt{3}\) \(\text { Case-II:- }\left(\mathrm{x}^2-8 \mathrm{x}+15\right) \leq 0\) \(\Rightarrow \quad(\mathrm{x}-5)(\mathrm{x}-3) \leq 0\) \(\Rightarrow \quad 3 \leq \mathrm{x} \leq 5\) \(\left|\mathrm{x}^2-8 \mathrm{x}+15\right|=-\left(\mathrm{x}^2-8 \mathrm{x}+15\right)\) \(\therefore \quad-x^2+8 \mathrm{x}-15-2 \mathrm{x}+7=0\) \(\text { Or } \quad x^2-6 x+8=0\) \(\mathrm{x}^2-4 \mathrm{x}-2 \mathrm{x}+8=0\) \(\mathrm{x}(\mathrm{x}-4)-2(\mathrm{x}-4)=0\) \((\mathrm{x}-2)(\mathrm{x}-4)=0\) \(\therefore \quad \mathrm{x}=2, \mathrm{x}=4\) \(\text { But } 3 \leq \mathrm{x} \leq 5\) \(\therefore \quad \mathrm{x}=4 \text { is the solution. }\) \(\therefore \quad\) The sum of the roots \((5+\sqrt{3}+4)\) is \((9+\sqrt{3})\)
JEE Main-06.04.2023
Complex Numbers and Quadratic Equation
118114
If \(\alpha, \beta, \gamma\) are the roots of the equation \(\mathrm{x}^3+\mathrm{ax}^2+\) \(b x+c=0\), then \(\alpha^{-1}+\beta^{-1}+\gamma^{-1}\) is equal to
1 \(\frac{\mathrm{a}}{\mathrm{c}}\)
2 \(\frac{\mathrm{c}}{\mathrm{a}}\)
3 \(-\frac{b}{c}\)
4 \(\frac{b}{a}\)
Explanation:
C Given the equation, Now, \(\mathrm{x}^3+\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0\) \(\alpha^{-1}+\beta^{-1}+\gamma^{-1} =\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=\frac{1}{\alpha \beta \gamma}(\alpha \beta+\beta \gamma+\alpha \gamma)\) \(=-\frac{b}{c}(\alpha \beta+\beta \gamma+\alpha \gamma=b, \alpha \beta \gamma=-c)\)
AP EAMCET-2002
Complex Numbers and Quadratic Equation
118115
If \(\alpha, \beta, \gamma\) are the roots of \(2 x^3-2 x-1=0\), then \((\Sigma \alpha \beta)^2\) is equal to
1 -1
2 1
3 2
4 3
Explanation:
B Given, \(\alpha, \beta, \gamma\) are the roots of- \(2 x^3-2 x-1=0\) \(\text { Or } \quad x^3-x-\frac{1}{2}=0\) We have, \(\quad \alpha \beta+\beta \gamma+\alpha \gamma=-1\) \((\Sigma \alpha \beta)^2=1\)
