118268
If one root of the cubic equation \(x^3+36=7 x^2\) is double of another, then the number of negative roots is
1 1
2 2
3 3
4 0
Explanation:
A \( Let \mathrm{a}, 2 \mathrm{a} \text { and } \mathrm{b} \text { be roots of the cubic }\) \(\text { equation } \mathrm{x}^3+36-7 \mathrm{x}^2=0\) \(\text { So, sum of roots }=-\frac{(-7)}{1}\) \(\therefore \mathrm{a}+2 \mathrm{a}+\mathrm{b}=7\) \(3 \mathrm{a}+\mathrm{b}=7\) \(\text { Now, } \mathrm{a}(2 \mathrm{a}) \mathrm{b}=36\) \(2 \mathrm{a} \mathrm{b}=36 \mathrm{and} \mathrm{ab}+\mathrm{a}(2 \mathrm{a})+2 \mathrm{a}(\mathrm{b})=0\) \(2 \mathrm{a}^2+3 \mathrm{ab}=0\) \(\mathrm{a}(2 \mathrm{a}+3 \mathrm{~b})=0\) \(\mathrm{~b}=-\frac{2 \mathrm{a}}{3}\) \(\text { From equation (i) }\) \(3 \mathrm{a}-\frac{2 \mathrm{a}}{3}=7\) \(9 \mathrm{a}-2 \mathrm{a}\) \(\frac{3}{3}=7\) \(9 \mathrm{a}-2 \mathrm{a}=21\) \(7 \mathrm{a}=21\) \(7 \mathrm{a}=21\) \(\mathrm{a}=3\) \(\text { From equation (i), }\) \(\text { and } \mathrm{b}=7-3 \mathrm{a}\) \(=7-3(3)=-2\) \(\text { Roots are } 3,6,-2\) \(\therefore \text { Number of negative roots is } 1 .\) From equation (i) From equation (i), and \(b=7-3 a\) Roots are \(3,6,-2\) \(\therefore\) Number of negative roots is 1 .
AP EAMCET-05.07.2022
Complex Numbers and Quadratic Equation
118269
If \(S=\left\{m \in R: x^2-2(1+3 m) x+7(3+2 m)=0\right.\) has distinct roots\}. Then the number of elements in \(\mathrm{S}\) is
1 2
2 3
3 4
4 Infinite
Explanation:
D Given, \(x^2-2(1+3 \mathrm{~m}) \mathrm{x}+7(3+2 \mathrm{~m})=0\) \(\mathrm{a}>0\) and \(\mathrm{D}=\mathrm{b}^2-4 \mathrm{ac}>0\) \(Expression is always positive, if roots are distinct\) then \(\mathrm{D}>0\) \({[-2(1+3 \mathrm{~m})]^2-4 \times 7(3+2 \mathrm{~m})>0}\) \(4\left[1+9 \mathrm{~m}^2+6 \mathrm{~m}\right]-84-56 \mathrm{~m}>0\) \(\begin{aligned} 36 m^2-32 m-80>0 \\ 9 m^2-8 m-20>0 \\ (m-2)\left(m+\frac{10}{9}\right)>0\end{aligned} right.\) \(\mathrm{m} \in\left(-\infty, \frac{-10}{9}\right) \cup(2, \infty)\)So, the number of element is infinite.
AP EAMCET-05.07.2022
Complex Numbers and Quadratic Equation
118270
The equation \(x^{\left(\log _3 x\right)^2-\frac{9}{2} \log _3 x+5}=3 \sqrt{3}\) has
118271
Let \(z_1\) and \(z_2\) be two imaginary roots of \(z^2+p z\) \(+q=0\), where \(p\) and \(q\) are real. The points \(z_1\), \(z_2\) and origin form an equilateral triangle if
1 \(\mathrm{p}^2>3 \mathrm{q}\)
2 \(\mathrm{p}^2\lt 3 \mathrm{q}\)
3 \(\mathrm{p}^2=3 \mathrm{q}\)
4 \(\mathrm{p}^2=\mathrm{q}\)
Explanation:
C Given, \(z_1\) and \(z_2\) be two imaginary roots of \(\mathrm{z}^2+\mathrm{pz}+\mathrm{q}=0\) \(\therefore\) Sum of roots, \(\mathrm{z}_1+\mathrm{z}_2=-\mathrm{p}\) Product of roots, \(z_1 \cdot z_2=q\) \(\mathrm{z}_1, \mathrm{z}_2\) and origin form an equilateral triangle \(\therefore \mathrm{z}_1^2+\mathrm{z}_2^2=\mathrm{z}_1 \cdot \mathrm{z}_2\) \(\left(z_1+z_2\right)^2-2 z_1 z_2=z_1 \cdot z_2\) \(\mathrm{p}^2-2 \mathrm{q}=\mathrm{q}\) \(\mathrm{p}^2=3 \mathrm{q}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Complex Numbers and Quadratic Equation
118268
If one root of the cubic equation \(x^3+36=7 x^2\) is double of another, then the number of negative roots is
1 1
2 2
3 3
4 0
Explanation:
A \( Let \mathrm{a}, 2 \mathrm{a} \text { and } \mathrm{b} \text { be roots of the cubic }\) \(\text { equation } \mathrm{x}^3+36-7 \mathrm{x}^2=0\) \(\text { So, sum of roots }=-\frac{(-7)}{1}\) \(\therefore \mathrm{a}+2 \mathrm{a}+\mathrm{b}=7\) \(3 \mathrm{a}+\mathrm{b}=7\) \(\text { Now, } \mathrm{a}(2 \mathrm{a}) \mathrm{b}=36\) \(2 \mathrm{a} \mathrm{b}=36 \mathrm{and} \mathrm{ab}+\mathrm{a}(2 \mathrm{a})+2 \mathrm{a}(\mathrm{b})=0\) \(2 \mathrm{a}^2+3 \mathrm{ab}=0\) \(\mathrm{a}(2 \mathrm{a}+3 \mathrm{~b})=0\) \(\mathrm{~b}=-\frac{2 \mathrm{a}}{3}\) \(\text { From equation (i) }\) \(3 \mathrm{a}-\frac{2 \mathrm{a}}{3}=7\) \(9 \mathrm{a}-2 \mathrm{a}\) \(\frac{3}{3}=7\) \(9 \mathrm{a}-2 \mathrm{a}=21\) \(7 \mathrm{a}=21\) \(7 \mathrm{a}=21\) \(\mathrm{a}=3\) \(\text { From equation (i), }\) \(\text { and } \mathrm{b}=7-3 \mathrm{a}\) \(=7-3(3)=-2\) \(\text { Roots are } 3,6,-2\) \(\therefore \text { Number of negative roots is } 1 .\) From equation (i) From equation (i), and \(b=7-3 a\) Roots are \(3,6,-2\) \(\therefore\) Number of negative roots is 1 .
AP EAMCET-05.07.2022
Complex Numbers and Quadratic Equation
118269
If \(S=\left\{m \in R: x^2-2(1+3 m) x+7(3+2 m)=0\right.\) has distinct roots\}. Then the number of elements in \(\mathrm{S}\) is
1 2
2 3
3 4
4 Infinite
Explanation:
D Given, \(x^2-2(1+3 \mathrm{~m}) \mathrm{x}+7(3+2 \mathrm{~m})=0\) \(\mathrm{a}>0\) and \(\mathrm{D}=\mathrm{b}^2-4 \mathrm{ac}>0\) \(Expression is always positive, if roots are distinct\) then \(\mathrm{D}>0\) \({[-2(1+3 \mathrm{~m})]^2-4 \times 7(3+2 \mathrm{~m})>0}\) \(4\left[1+9 \mathrm{~m}^2+6 \mathrm{~m}\right]-84-56 \mathrm{~m}>0\) \(\begin{aligned} 36 m^2-32 m-80>0 \\ 9 m^2-8 m-20>0 \\ (m-2)\left(m+\frac{10}{9}\right)>0\end{aligned} right.\) \(\mathrm{m} \in\left(-\infty, \frac{-10}{9}\right) \cup(2, \infty)\)So, the number of element is infinite.
AP EAMCET-05.07.2022
Complex Numbers and Quadratic Equation
118270
The equation \(x^{\left(\log _3 x\right)^2-\frac{9}{2} \log _3 x+5}=3 \sqrt{3}\) has
118271
Let \(z_1\) and \(z_2\) be two imaginary roots of \(z^2+p z\) \(+q=0\), where \(p\) and \(q\) are real. The points \(z_1\), \(z_2\) and origin form an equilateral triangle if
1 \(\mathrm{p}^2>3 \mathrm{q}\)
2 \(\mathrm{p}^2\lt 3 \mathrm{q}\)
3 \(\mathrm{p}^2=3 \mathrm{q}\)
4 \(\mathrm{p}^2=\mathrm{q}\)
Explanation:
C Given, \(z_1\) and \(z_2\) be two imaginary roots of \(\mathrm{z}^2+\mathrm{pz}+\mathrm{q}=0\) \(\therefore\) Sum of roots, \(\mathrm{z}_1+\mathrm{z}_2=-\mathrm{p}\) Product of roots, \(z_1 \cdot z_2=q\) \(\mathrm{z}_1, \mathrm{z}_2\) and origin form an equilateral triangle \(\therefore \mathrm{z}_1^2+\mathrm{z}_2^2=\mathrm{z}_1 \cdot \mathrm{z}_2\) \(\left(z_1+z_2\right)^2-2 z_1 z_2=z_1 \cdot z_2\) \(\mathrm{p}^2-2 \mathrm{q}=\mathrm{q}\) \(\mathrm{p}^2=3 \mathrm{q}\)
118268
If one root of the cubic equation \(x^3+36=7 x^2\) is double of another, then the number of negative roots is
1 1
2 2
3 3
4 0
Explanation:
A \( Let \mathrm{a}, 2 \mathrm{a} \text { and } \mathrm{b} \text { be roots of the cubic }\) \(\text { equation } \mathrm{x}^3+36-7 \mathrm{x}^2=0\) \(\text { So, sum of roots }=-\frac{(-7)}{1}\) \(\therefore \mathrm{a}+2 \mathrm{a}+\mathrm{b}=7\) \(3 \mathrm{a}+\mathrm{b}=7\) \(\text { Now, } \mathrm{a}(2 \mathrm{a}) \mathrm{b}=36\) \(2 \mathrm{a} \mathrm{b}=36 \mathrm{and} \mathrm{ab}+\mathrm{a}(2 \mathrm{a})+2 \mathrm{a}(\mathrm{b})=0\) \(2 \mathrm{a}^2+3 \mathrm{ab}=0\) \(\mathrm{a}(2 \mathrm{a}+3 \mathrm{~b})=0\) \(\mathrm{~b}=-\frac{2 \mathrm{a}}{3}\) \(\text { From equation (i) }\) \(3 \mathrm{a}-\frac{2 \mathrm{a}}{3}=7\) \(9 \mathrm{a}-2 \mathrm{a}\) \(\frac{3}{3}=7\) \(9 \mathrm{a}-2 \mathrm{a}=21\) \(7 \mathrm{a}=21\) \(7 \mathrm{a}=21\) \(\mathrm{a}=3\) \(\text { From equation (i), }\) \(\text { and } \mathrm{b}=7-3 \mathrm{a}\) \(=7-3(3)=-2\) \(\text { Roots are } 3,6,-2\) \(\therefore \text { Number of negative roots is } 1 .\) From equation (i) From equation (i), and \(b=7-3 a\) Roots are \(3,6,-2\) \(\therefore\) Number of negative roots is 1 .
AP EAMCET-05.07.2022
Complex Numbers and Quadratic Equation
118269
If \(S=\left\{m \in R: x^2-2(1+3 m) x+7(3+2 m)=0\right.\) has distinct roots\}. Then the number of elements in \(\mathrm{S}\) is
1 2
2 3
3 4
4 Infinite
Explanation:
D Given, \(x^2-2(1+3 \mathrm{~m}) \mathrm{x}+7(3+2 \mathrm{~m})=0\) \(\mathrm{a}>0\) and \(\mathrm{D}=\mathrm{b}^2-4 \mathrm{ac}>0\) \(Expression is always positive, if roots are distinct\) then \(\mathrm{D}>0\) \({[-2(1+3 \mathrm{~m})]^2-4 \times 7(3+2 \mathrm{~m})>0}\) \(4\left[1+9 \mathrm{~m}^2+6 \mathrm{~m}\right]-84-56 \mathrm{~m}>0\) \(\begin{aligned} 36 m^2-32 m-80>0 \\ 9 m^2-8 m-20>0 \\ (m-2)\left(m+\frac{10}{9}\right)>0\end{aligned} right.\) \(\mathrm{m} \in\left(-\infty, \frac{-10}{9}\right) \cup(2, \infty)\)So, the number of element is infinite.
AP EAMCET-05.07.2022
Complex Numbers and Quadratic Equation
118270
The equation \(x^{\left(\log _3 x\right)^2-\frac{9}{2} \log _3 x+5}=3 \sqrt{3}\) has
118271
Let \(z_1\) and \(z_2\) be two imaginary roots of \(z^2+p z\) \(+q=0\), where \(p\) and \(q\) are real. The points \(z_1\), \(z_2\) and origin form an equilateral triangle if
1 \(\mathrm{p}^2>3 \mathrm{q}\)
2 \(\mathrm{p}^2\lt 3 \mathrm{q}\)
3 \(\mathrm{p}^2=3 \mathrm{q}\)
4 \(\mathrm{p}^2=\mathrm{q}\)
Explanation:
C Given, \(z_1\) and \(z_2\) be two imaginary roots of \(\mathrm{z}^2+\mathrm{pz}+\mathrm{q}=0\) \(\therefore\) Sum of roots, \(\mathrm{z}_1+\mathrm{z}_2=-\mathrm{p}\) Product of roots, \(z_1 \cdot z_2=q\) \(\mathrm{z}_1, \mathrm{z}_2\) and origin form an equilateral triangle \(\therefore \mathrm{z}_1^2+\mathrm{z}_2^2=\mathrm{z}_1 \cdot \mathrm{z}_2\) \(\left(z_1+z_2\right)^2-2 z_1 z_2=z_1 \cdot z_2\) \(\mathrm{p}^2-2 \mathrm{q}=\mathrm{q}\) \(\mathrm{p}^2=3 \mathrm{q}\)
118268
If one root of the cubic equation \(x^3+36=7 x^2\) is double of another, then the number of negative roots is
1 1
2 2
3 3
4 0
Explanation:
A \( Let \mathrm{a}, 2 \mathrm{a} \text { and } \mathrm{b} \text { be roots of the cubic }\) \(\text { equation } \mathrm{x}^3+36-7 \mathrm{x}^2=0\) \(\text { So, sum of roots }=-\frac{(-7)}{1}\) \(\therefore \mathrm{a}+2 \mathrm{a}+\mathrm{b}=7\) \(3 \mathrm{a}+\mathrm{b}=7\) \(\text { Now, } \mathrm{a}(2 \mathrm{a}) \mathrm{b}=36\) \(2 \mathrm{a} \mathrm{b}=36 \mathrm{and} \mathrm{ab}+\mathrm{a}(2 \mathrm{a})+2 \mathrm{a}(\mathrm{b})=0\) \(2 \mathrm{a}^2+3 \mathrm{ab}=0\) \(\mathrm{a}(2 \mathrm{a}+3 \mathrm{~b})=0\) \(\mathrm{~b}=-\frac{2 \mathrm{a}}{3}\) \(\text { From equation (i) }\) \(3 \mathrm{a}-\frac{2 \mathrm{a}}{3}=7\) \(9 \mathrm{a}-2 \mathrm{a}\) \(\frac{3}{3}=7\) \(9 \mathrm{a}-2 \mathrm{a}=21\) \(7 \mathrm{a}=21\) \(7 \mathrm{a}=21\) \(\mathrm{a}=3\) \(\text { From equation (i), }\) \(\text { and } \mathrm{b}=7-3 \mathrm{a}\) \(=7-3(3)=-2\) \(\text { Roots are } 3,6,-2\) \(\therefore \text { Number of negative roots is } 1 .\) From equation (i) From equation (i), and \(b=7-3 a\) Roots are \(3,6,-2\) \(\therefore\) Number of negative roots is 1 .
AP EAMCET-05.07.2022
Complex Numbers and Quadratic Equation
118269
If \(S=\left\{m \in R: x^2-2(1+3 m) x+7(3+2 m)=0\right.\) has distinct roots\}. Then the number of elements in \(\mathrm{S}\) is
1 2
2 3
3 4
4 Infinite
Explanation:
D Given, \(x^2-2(1+3 \mathrm{~m}) \mathrm{x}+7(3+2 \mathrm{~m})=0\) \(\mathrm{a}>0\) and \(\mathrm{D}=\mathrm{b}^2-4 \mathrm{ac}>0\) \(Expression is always positive, if roots are distinct\) then \(\mathrm{D}>0\) \({[-2(1+3 \mathrm{~m})]^2-4 \times 7(3+2 \mathrm{~m})>0}\) \(4\left[1+9 \mathrm{~m}^2+6 \mathrm{~m}\right]-84-56 \mathrm{~m}>0\) \(\begin{aligned} 36 m^2-32 m-80>0 \\ 9 m^2-8 m-20>0 \\ (m-2)\left(m+\frac{10}{9}\right)>0\end{aligned} right.\) \(\mathrm{m} \in\left(-\infty, \frac{-10}{9}\right) \cup(2, \infty)\)So, the number of element is infinite.
AP EAMCET-05.07.2022
Complex Numbers and Quadratic Equation
118270
The equation \(x^{\left(\log _3 x\right)^2-\frac{9}{2} \log _3 x+5}=3 \sqrt{3}\) has
118271
Let \(z_1\) and \(z_2\) be two imaginary roots of \(z^2+p z\) \(+q=0\), where \(p\) and \(q\) are real. The points \(z_1\), \(z_2\) and origin form an equilateral triangle if
1 \(\mathrm{p}^2>3 \mathrm{q}\)
2 \(\mathrm{p}^2\lt 3 \mathrm{q}\)
3 \(\mathrm{p}^2=3 \mathrm{q}\)
4 \(\mathrm{p}^2=\mathrm{q}\)
Explanation:
C Given, \(z_1\) and \(z_2\) be two imaginary roots of \(\mathrm{z}^2+\mathrm{pz}+\mathrm{q}=0\) \(\therefore\) Sum of roots, \(\mathrm{z}_1+\mathrm{z}_2=-\mathrm{p}\) Product of roots, \(z_1 \cdot z_2=q\) \(\mathrm{z}_1, \mathrm{z}_2\) and origin form an equilateral triangle \(\therefore \mathrm{z}_1^2+\mathrm{z}_2^2=\mathrm{z}_1 \cdot \mathrm{z}_2\) \(\left(z_1+z_2\right)^2-2 z_1 z_2=z_1 \cdot z_2\) \(\mathrm{p}^2-2 \mathrm{q}=\mathrm{q}\) \(\mathrm{p}^2=3 \mathrm{q}\)
