118233
If \(\alpha\) and \(\beta\) be the roots of \(x^2+b x+1=0\). then, the equation whose roots are \(-\left(\alpha+\frac{1}{\beta}\right)\) and \(-\left(\beta+\frac{1}{\alpha}\right)\) is
1 \(x^2=0\)
2 \(x^2+2 b x+4=0\)
3 \(x^2-2 b x+4=0\)
4 \(x^2-2 b x+1=0\)
Explanation:
C Given, \(x^2+b x+1=0\) and \(\alpha, \beta\) are roots Then, \(\alpha+\beta=-b\) \(\alpha \cdot \beta=1\) Now, the equation whose roots are \(-\left(\alpha+\frac{1}{\beta}\right) \text { and }-\left(\beta+\frac{1}{\alpha}\right)\) So, sum of roots \(=-\left(\alpha+\frac{1}{\beta}\right)-\left(\beta+\frac{1}{\alpha}\right)\) \(=-\left[(\alpha+\beta)+\frac{1}{\beta}+\frac{1}{\alpha}\right]\) \(=-\left[\alpha+\beta+\frac{\alpha+\beta}{\alpha \beta}\right]\) \(=-\left[-b+\frac{(-b)}{1}\right]\) \(=2 \mathrm{~b}\) Product of roots \(=-\left(\alpha+\frac{1}{\beta}\right)\left(-\left(\beta+\frac{1}{\alpha}\right)\right)\) \(=\alpha \cdot \beta+1+1+\frac{1}{\alpha \beta}\) \(=1+1+1+1=4\) Therefore equation is\(\mathrm{x}^2-(\) sum of roots) \(\mathrm{x}+\) (product of roots) \(=0\) \(x^2-(2 b) x+4=0\) \(x^2-2 b x+4=0\)
BCECE-2015
Complex Numbers and Quadratic Equation
118255
Let \(\mathbf{a}, \mathbf{b}, \mathbf{c}\) be three real numbers such that \(\mathbf{a}+\) \(2 b+4 c=0\). Then the equation \(a x^2+b x+c=0\)
1 has both the roots complex
2 has its roots lying within \(-1\lt x\lt 0\)
3 has one of the roots equal to \(\frac{1}{2}\)
4 has its roots lying within \(2\lt x\lt 6\)
Explanation:
C Given \(a^2+b x+c=0\) \(a+2 b+4 c=0\) \(\left(\frac{1}{4}\right) \mathrm{a}+\left(\frac{1}{2}\right) \mathrm{b}+\mathrm{c}=0\) \(\left(\frac{1}{2}\right)^2 \mathrm{a}+\left(\frac{1}{2}\right) \mathrm{b}+\mathrm{c}=0\) \(\text { on comparing with } \mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0\) \(\mathrm{x}=\frac{1}{2}\)on comparing with \(\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0\)
WB JEE-2011
Complex Numbers and Quadratic Equation
118234
If \(\alpha\) and \(\beta\) are the roots of \(x^2+5 x+4=0\), then equation whose roots are \(\frac{\alpha+2}{3}\) and \(\frac{\beta+2}{3}\), is
1 \(9 x^2+3 x+2=0\)
2 \(9 x^2-3 x+2=0\)
3 \(9 x^2+3 x-2=0\)
4 \(9 x^2-3 x-2=0\)
Explanation:
C Given, \(x^2+5 x+4=0 \text { and roots are } \alpha, \beta\) Then, sum of roots \((\alpha+\beta)\) \(\alpha+\beta=-5\) Product of roots \(=\alpha \cdot \beta\) \(\alpha \cdot \beta=4\) Now, the equation whose roots are \(\frac{\alpha+2}{3}, \frac{\beta+2}{3}\) So, sum of roots \(=\frac{\alpha+2}{3}+\frac{\beta+2}{3}\) \(=\frac{\alpha+\beta+4}{3}\) \(=\frac{-5+4}{3}=\frac{-1}{3}\) Product of roots \(=\frac{(\alpha+2)}{3} \times \frac{(\beta+2)}{3}\) \(=\frac{\alpha \beta+2(\alpha+\beta)+4}{9}\) \(=\frac{4+2(-5)+4}{9}=\frac{-2}{9}\) Therefore, equation is\(\mathrm{x}^2-(\) sum of roots) \(\mathrm{x}+\) product of roots \(=0\) \(x^2-\left(\frac{-1}{3}\right) x+\left(\frac{-2}{9}\right)=0\) \(9 x^2+3 x-2=0\)
118233
If \(\alpha\) and \(\beta\) be the roots of \(x^2+b x+1=0\). then, the equation whose roots are \(-\left(\alpha+\frac{1}{\beta}\right)\) and \(-\left(\beta+\frac{1}{\alpha}\right)\) is
1 \(x^2=0\)
2 \(x^2+2 b x+4=0\)
3 \(x^2-2 b x+4=0\)
4 \(x^2-2 b x+1=0\)
Explanation:
C Given, \(x^2+b x+1=0\) and \(\alpha, \beta\) are roots Then, \(\alpha+\beta=-b\) \(\alpha \cdot \beta=1\) Now, the equation whose roots are \(-\left(\alpha+\frac{1}{\beta}\right) \text { and }-\left(\beta+\frac{1}{\alpha}\right)\) So, sum of roots \(=-\left(\alpha+\frac{1}{\beta}\right)-\left(\beta+\frac{1}{\alpha}\right)\) \(=-\left[(\alpha+\beta)+\frac{1}{\beta}+\frac{1}{\alpha}\right]\) \(=-\left[\alpha+\beta+\frac{\alpha+\beta}{\alpha \beta}\right]\) \(=-\left[-b+\frac{(-b)}{1}\right]\) \(=2 \mathrm{~b}\) Product of roots \(=-\left(\alpha+\frac{1}{\beta}\right)\left(-\left(\beta+\frac{1}{\alpha}\right)\right)\) \(=\alpha \cdot \beta+1+1+\frac{1}{\alpha \beta}\) \(=1+1+1+1=4\) Therefore equation is\(\mathrm{x}^2-(\) sum of roots) \(\mathrm{x}+\) (product of roots) \(=0\) \(x^2-(2 b) x+4=0\) \(x^2-2 b x+4=0\)
BCECE-2015
Complex Numbers and Quadratic Equation
118255
Let \(\mathbf{a}, \mathbf{b}, \mathbf{c}\) be three real numbers such that \(\mathbf{a}+\) \(2 b+4 c=0\). Then the equation \(a x^2+b x+c=0\)
1 has both the roots complex
2 has its roots lying within \(-1\lt x\lt 0\)
3 has one of the roots equal to \(\frac{1}{2}\)
4 has its roots lying within \(2\lt x\lt 6\)
Explanation:
C Given \(a^2+b x+c=0\) \(a+2 b+4 c=0\) \(\left(\frac{1}{4}\right) \mathrm{a}+\left(\frac{1}{2}\right) \mathrm{b}+\mathrm{c}=0\) \(\left(\frac{1}{2}\right)^2 \mathrm{a}+\left(\frac{1}{2}\right) \mathrm{b}+\mathrm{c}=0\) \(\text { on comparing with } \mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0\) \(\mathrm{x}=\frac{1}{2}\)on comparing with \(\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0\)
WB JEE-2011
Complex Numbers and Quadratic Equation
118234
If \(\alpha\) and \(\beta\) are the roots of \(x^2+5 x+4=0\), then equation whose roots are \(\frac{\alpha+2}{3}\) and \(\frac{\beta+2}{3}\), is
1 \(9 x^2+3 x+2=0\)
2 \(9 x^2-3 x+2=0\)
3 \(9 x^2+3 x-2=0\)
4 \(9 x^2-3 x-2=0\)
Explanation:
C Given, \(x^2+5 x+4=0 \text { and roots are } \alpha, \beta\) Then, sum of roots \((\alpha+\beta)\) \(\alpha+\beta=-5\) Product of roots \(=\alpha \cdot \beta\) \(\alpha \cdot \beta=4\) Now, the equation whose roots are \(\frac{\alpha+2}{3}, \frac{\beta+2}{3}\) So, sum of roots \(=\frac{\alpha+2}{3}+\frac{\beta+2}{3}\) \(=\frac{\alpha+\beta+4}{3}\) \(=\frac{-5+4}{3}=\frac{-1}{3}\) Product of roots \(=\frac{(\alpha+2)}{3} \times \frac{(\beta+2)}{3}\) \(=\frac{\alpha \beta+2(\alpha+\beta)+4}{9}\) \(=\frac{4+2(-5)+4}{9}=\frac{-2}{9}\) Therefore, equation is\(\mathrm{x}^2-(\) sum of roots) \(\mathrm{x}+\) product of roots \(=0\) \(x^2-\left(\frac{-1}{3}\right) x+\left(\frac{-2}{9}\right)=0\) \(9 x^2+3 x-2=0\)
118233
If \(\alpha\) and \(\beta\) be the roots of \(x^2+b x+1=0\). then, the equation whose roots are \(-\left(\alpha+\frac{1}{\beta}\right)\) and \(-\left(\beta+\frac{1}{\alpha}\right)\) is
1 \(x^2=0\)
2 \(x^2+2 b x+4=0\)
3 \(x^2-2 b x+4=0\)
4 \(x^2-2 b x+1=0\)
Explanation:
C Given, \(x^2+b x+1=0\) and \(\alpha, \beta\) are roots Then, \(\alpha+\beta=-b\) \(\alpha \cdot \beta=1\) Now, the equation whose roots are \(-\left(\alpha+\frac{1}{\beta}\right) \text { and }-\left(\beta+\frac{1}{\alpha}\right)\) So, sum of roots \(=-\left(\alpha+\frac{1}{\beta}\right)-\left(\beta+\frac{1}{\alpha}\right)\) \(=-\left[(\alpha+\beta)+\frac{1}{\beta}+\frac{1}{\alpha}\right]\) \(=-\left[\alpha+\beta+\frac{\alpha+\beta}{\alpha \beta}\right]\) \(=-\left[-b+\frac{(-b)}{1}\right]\) \(=2 \mathrm{~b}\) Product of roots \(=-\left(\alpha+\frac{1}{\beta}\right)\left(-\left(\beta+\frac{1}{\alpha}\right)\right)\) \(=\alpha \cdot \beta+1+1+\frac{1}{\alpha \beta}\) \(=1+1+1+1=4\) Therefore equation is\(\mathrm{x}^2-(\) sum of roots) \(\mathrm{x}+\) (product of roots) \(=0\) \(x^2-(2 b) x+4=0\) \(x^2-2 b x+4=0\)
BCECE-2015
Complex Numbers and Quadratic Equation
118255
Let \(\mathbf{a}, \mathbf{b}, \mathbf{c}\) be three real numbers such that \(\mathbf{a}+\) \(2 b+4 c=0\). Then the equation \(a x^2+b x+c=0\)
1 has both the roots complex
2 has its roots lying within \(-1\lt x\lt 0\)
3 has one of the roots equal to \(\frac{1}{2}\)
4 has its roots lying within \(2\lt x\lt 6\)
Explanation:
C Given \(a^2+b x+c=0\) \(a+2 b+4 c=0\) \(\left(\frac{1}{4}\right) \mathrm{a}+\left(\frac{1}{2}\right) \mathrm{b}+\mathrm{c}=0\) \(\left(\frac{1}{2}\right)^2 \mathrm{a}+\left(\frac{1}{2}\right) \mathrm{b}+\mathrm{c}=0\) \(\text { on comparing with } \mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0\) \(\mathrm{x}=\frac{1}{2}\)on comparing with \(\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0\)
WB JEE-2011
Complex Numbers and Quadratic Equation
118234
If \(\alpha\) and \(\beta\) are the roots of \(x^2+5 x+4=0\), then equation whose roots are \(\frac{\alpha+2}{3}\) and \(\frac{\beta+2}{3}\), is
1 \(9 x^2+3 x+2=0\)
2 \(9 x^2-3 x+2=0\)
3 \(9 x^2+3 x-2=0\)
4 \(9 x^2-3 x-2=0\)
Explanation:
C Given, \(x^2+5 x+4=0 \text { and roots are } \alpha, \beta\) Then, sum of roots \((\alpha+\beta)\) \(\alpha+\beta=-5\) Product of roots \(=\alpha \cdot \beta\) \(\alpha \cdot \beta=4\) Now, the equation whose roots are \(\frac{\alpha+2}{3}, \frac{\beta+2}{3}\) So, sum of roots \(=\frac{\alpha+2}{3}+\frac{\beta+2}{3}\) \(=\frac{\alpha+\beta+4}{3}\) \(=\frac{-5+4}{3}=\frac{-1}{3}\) Product of roots \(=\frac{(\alpha+2)}{3} \times \frac{(\beta+2)}{3}\) \(=\frac{\alpha \beta+2(\alpha+\beta)+4}{9}\) \(=\frac{4+2(-5)+4}{9}=\frac{-2}{9}\) Therefore, equation is\(\mathrm{x}^2-(\) sum of roots) \(\mathrm{x}+\) product of roots \(=0\) \(x^2-\left(\frac{-1}{3}\right) x+\left(\frac{-2}{9}\right)=0\) \(9 x^2+3 x-2=0\)
118233
If \(\alpha\) and \(\beta\) be the roots of \(x^2+b x+1=0\). then, the equation whose roots are \(-\left(\alpha+\frac{1}{\beta}\right)\) and \(-\left(\beta+\frac{1}{\alpha}\right)\) is
1 \(x^2=0\)
2 \(x^2+2 b x+4=0\)
3 \(x^2-2 b x+4=0\)
4 \(x^2-2 b x+1=0\)
Explanation:
C Given, \(x^2+b x+1=0\) and \(\alpha, \beta\) are roots Then, \(\alpha+\beta=-b\) \(\alpha \cdot \beta=1\) Now, the equation whose roots are \(-\left(\alpha+\frac{1}{\beta}\right) \text { and }-\left(\beta+\frac{1}{\alpha}\right)\) So, sum of roots \(=-\left(\alpha+\frac{1}{\beta}\right)-\left(\beta+\frac{1}{\alpha}\right)\) \(=-\left[(\alpha+\beta)+\frac{1}{\beta}+\frac{1}{\alpha}\right]\) \(=-\left[\alpha+\beta+\frac{\alpha+\beta}{\alpha \beta}\right]\) \(=-\left[-b+\frac{(-b)}{1}\right]\) \(=2 \mathrm{~b}\) Product of roots \(=-\left(\alpha+\frac{1}{\beta}\right)\left(-\left(\beta+\frac{1}{\alpha}\right)\right)\) \(=\alpha \cdot \beta+1+1+\frac{1}{\alpha \beta}\) \(=1+1+1+1=4\) Therefore equation is\(\mathrm{x}^2-(\) sum of roots) \(\mathrm{x}+\) (product of roots) \(=0\) \(x^2-(2 b) x+4=0\) \(x^2-2 b x+4=0\)
BCECE-2015
Complex Numbers and Quadratic Equation
118255
Let \(\mathbf{a}, \mathbf{b}, \mathbf{c}\) be three real numbers such that \(\mathbf{a}+\) \(2 b+4 c=0\). Then the equation \(a x^2+b x+c=0\)
1 has both the roots complex
2 has its roots lying within \(-1\lt x\lt 0\)
3 has one of the roots equal to \(\frac{1}{2}\)
4 has its roots lying within \(2\lt x\lt 6\)
Explanation:
C Given \(a^2+b x+c=0\) \(a+2 b+4 c=0\) \(\left(\frac{1}{4}\right) \mathrm{a}+\left(\frac{1}{2}\right) \mathrm{b}+\mathrm{c}=0\) \(\left(\frac{1}{2}\right)^2 \mathrm{a}+\left(\frac{1}{2}\right) \mathrm{b}+\mathrm{c}=0\) \(\text { on comparing with } \mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0\) \(\mathrm{x}=\frac{1}{2}\)on comparing with \(\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0\)
WB JEE-2011
Complex Numbers and Quadratic Equation
118234
If \(\alpha\) and \(\beta\) are the roots of \(x^2+5 x+4=0\), then equation whose roots are \(\frac{\alpha+2}{3}\) and \(\frac{\beta+2}{3}\), is
1 \(9 x^2+3 x+2=0\)
2 \(9 x^2-3 x+2=0\)
3 \(9 x^2+3 x-2=0\)
4 \(9 x^2-3 x-2=0\)
Explanation:
C Given, \(x^2+5 x+4=0 \text { and roots are } \alpha, \beta\) Then, sum of roots \((\alpha+\beta)\) \(\alpha+\beta=-5\) Product of roots \(=\alpha \cdot \beta\) \(\alpha \cdot \beta=4\) Now, the equation whose roots are \(\frac{\alpha+2}{3}, \frac{\beta+2}{3}\) So, sum of roots \(=\frac{\alpha+2}{3}+\frac{\beta+2}{3}\) \(=\frac{\alpha+\beta+4}{3}\) \(=\frac{-5+4}{3}=\frac{-1}{3}\) Product of roots \(=\frac{(\alpha+2)}{3} \times \frac{(\beta+2)}{3}\) \(=\frac{\alpha \beta+2(\alpha+\beta)+4}{9}\) \(=\frac{4+2(-5)+4}{9}=\frac{-2}{9}\) Therefore, equation is\(\mathrm{x}^2-(\) sum of roots) \(\mathrm{x}+\) product of roots \(=0\) \(x^2-\left(\frac{-1}{3}\right) x+\left(\frac{-2}{9}\right)=0\) \(9 x^2+3 x-2=0\)
