118200
\(\alpha\) is the maximum value of \(1-2 x-5 x^2\) and \(\beta\) is the minimum value of \(x^2-2 x+r\). If \(5 \alpha x^2+\beta x+6>0\) for all real values \(x\). then the interval in which \(r\) lies is
118201
Let \(\mathrm{S}\) be the set of all possible integral values of \(\lambda\) in the interval \((-3,7)\) for which the roots of the quadratic equation \(\lambda x^2+13 x+7=0\) are all rational numbers. Then the sum of the elements in \(\mathrm{S}\) is
1 4
2 2
3 3
4 1
Explanation:
A We have, \(\lambda \mathrm{x}^2+13 \mathrm{x}+7=0\) \(\mathrm{D}=(13)^2-4(\lambda)(7)=169-28 \lambda\) For rational roots, \(\mathrm{D}=\) perfect square So, \(\lambda \in(-3,7)\) has values \(-2,0,6\) that \(D\) become perfect square. \(\therefore\) Required sum \(=-2+0+6=4\)
TS EAMCET-10.09.2020
Complex Numbers and Quadratic Equation
118202
The equation \(x^2-4 x+[x]+3=x[x]\), where \([x]\) denotes the greatest integer function, has:
1 exactly two solutions in \((-\infty, \infty)\)
2 no solution
3 a unique solution in \((-\infty, 1)\)
4 a unique solution in \((-\infty, \infty)\)
Explanation:
Exp:(D): \(x^2-4 x+[x]+3=x[x]\) \(\Rightarrow x^2-4 x+3=x[x]-[x]\) \(\Rightarrow(x-1)(x-3)=[x] .(x-1)\) \(\Rightarrow x=1 \text { or } x-3=[x]\) \(\Rightarrow x-[x]=3\) \(\Rightarrow\{x\}=3 \text { (Not Possible) }\) \(\text { Only one solution } x=1 \text { in }(-\infty, \infty)\)Only one solution \(x=1\) in \((-\infty, \infty)\)
JEE Main-24.01.2023
Complex Numbers and Quadratic Equation
118203
The root of \(\mathrm{ax}^2+\mathrm{x}+1=0\), where \(\mathrm{a} \neq 0\) are in the ratio \(1: 1\). Then a is equal to
1 \(1 / 4\)
2 \(1 / 2\)
3 \(3 / 4\)
4 1
5 0
Explanation:
A Given the equation, \(a x^2+x+1=0, a \neq 0\) Let the roots of the above equation are \(\alpha\) and \(\beta\). \(\therefore \quad \alpha+\beta=\frac{-1}{\mathrm{a}}\) \(\alpha \beta=\frac{1}{\mathrm{a}}\) It is given that \(\alpha=\beta\) \(\Rightarrow 2 \alpha=\frac{-1}{a} \Rightarrow \alpha=\frac{-1}{2 a}\) \(\alpha^2=\frac{1}{a}\) Solving, \(\left(\frac{-1}{2 \mathrm{a}}\right)^2=\frac{1}{\mathrm{a}}\) Or \(\quad \frac{1}{4 \mathrm{a}^2}=\frac{1}{\mathrm{a}} \Rightarrow \mathrm{a}=\frac{1}{4}\)
118200
\(\alpha\) is the maximum value of \(1-2 x-5 x^2\) and \(\beta\) is the minimum value of \(x^2-2 x+r\). If \(5 \alpha x^2+\beta x+6>0\) for all real values \(x\). then the interval in which \(r\) lies is
118201
Let \(\mathrm{S}\) be the set of all possible integral values of \(\lambda\) in the interval \((-3,7)\) for which the roots of the quadratic equation \(\lambda x^2+13 x+7=0\) are all rational numbers. Then the sum of the elements in \(\mathrm{S}\) is
1 4
2 2
3 3
4 1
Explanation:
A We have, \(\lambda \mathrm{x}^2+13 \mathrm{x}+7=0\) \(\mathrm{D}=(13)^2-4(\lambda)(7)=169-28 \lambda\) For rational roots, \(\mathrm{D}=\) perfect square So, \(\lambda \in(-3,7)\) has values \(-2,0,6\) that \(D\) become perfect square. \(\therefore\) Required sum \(=-2+0+6=4\)
TS EAMCET-10.09.2020
Complex Numbers and Quadratic Equation
118202
The equation \(x^2-4 x+[x]+3=x[x]\), where \([x]\) denotes the greatest integer function, has:
1 exactly two solutions in \((-\infty, \infty)\)
2 no solution
3 a unique solution in \((-\infty, 1)\)
4 a unique solution in \((-\infty, \infty)\)
Explanation:
Exp:(D): \(x^2-4 x+[x]+3=x[x]\) \(\Rightarrow x^2-4 x+3=x[x]-[x]\) \(\Rightarrow(x-1)(x-3)=[x] .(x-1)\) \(\Rightarrow x=1 \text { or } x-3=[x]\) \(\Rightarrow x-[x]=3\) \(\Rightarrow\{x\}=3 \text { (Not Possible) }\) \(\text { Only one solution } x=1 \text { in }(-\infty, \infty)\)Only one solution \(x=1\) in \((-\infty, \infty)\)
JEE Main-24.01.2023
Complex Numbers and Quadratic Equation
118203
The root of \(\mathrm{ax}^2+\mathrm{x}+1=0\), where \(\mathrm{a} \neq 0\) are in the ratio \(1: 1\). Then a is equal to
1 \(1 / 4\)
2 \(1 / 2\)
3 \(3 / 4\)
4 1
5 0
Explanation:
A Given the equation, \(a x^2+x+1=0, a \neq 0\) Let the roots of the above equation are \(\alpha\) and \(\beta\). \(\therefore \quad \alpha+\beta=\frac{-1}{\mathrm{a}}\) \(\alpha \beta=\frac{1}{\mathrm{a}}\) It is given that \(\alpha=\beta\) \(\Rightarrow 2 \alpha=\frac{-1}{a} \Rightarrow \alpha=\frac{-1}{2 a}\) \(\alpha^2=\frac{1}{a}\) Solving, \(\left(\frac{-1}{2 \mathrm{a}}\right)^2=\frac{1}{\mathrm{a}}\) Or \(\quad \frac{1}{4 \mathrm{a}^2}=\frac{1}{\mathrm{a}} \Rightarrow \mathrm{a}=\frac{1}{4}\)
118200
\(\alpha\) is the maximum value of \(1-2 x-5 x^2\) and \(\beta\) is the minimum value of \(x^2-2 x+r\). If \(5 \alpha x^2+\beta x+6>0\) for all real values \(x\). then the interval in which \(r\) lies is
118201
Let \(\mathrm{S}\) be the set of all possible integral values of \(\lambda\) in the interval \((-3,7)\) for which the roots of the quadratic equation \(\lambda x^2+13 x+7=0\) are all rational numbers. Then the sum of the elements in \(\mathrm{S}\) is
1 4
2 2
3 3
4 1
Explanation:
A We have, \(\lambda \mathrm{x}^2+13 \mathrm{x}+7=0\) \(\mathrm{D}=(13)^2-4(\lambda)(7)=169-28 \lambda\) For rational roots, \(\mathrm{D}=\) perfect square So, \(\lambda \in(-3,7)\) has values \(-2,0,6\) that \(D\) become perfect square. \(\therefore\) Required sum \(=-2+0+6=4\)
TS EAMCET-10.09.2020
Complex Numbers and Quadratic Equation
118202
The equation \(x^2-4 x+[x]+3=x[x]\), where \([x]\) denotes the greatest integer function, has:
1 exactly two solutions in \((-\infty, \infty)\)
2 no solution
3 a unique solution in \((-\infty, 1)\)
4 a unique solution in \((-\infty, \infty)\)
Explanation:
Exp:(D): \(x^2-4 x+[x]+3=x[x]\) \(\Rightarrow x^2-4 x+3=x[x]-[x]\) \(\Rightarrow(x-1)(x-3)=[x] .(x-1)\) \(\Rightarrow x=1 \text { or } x-3=[x]\) \(\Rightarrow x-[x]=3\) \(\Rightarrow\{x\}=3 \text { (Not Possible) }\) \(\text { Only one solution } x=1 \text { in }(-\infty, \infty)\)Only one solution \(x=1\) in \((-\infty, \infty)\)
JEE Main-24.01.2023
Complex Numbers and Quadratic Equation
118203
The root of \(\mathrm{ax}^2+\mathrm{x}+1=0\), where \(\mathrm{a} \neq 0\) are in the ratio \(1: 1\). Then a is equal to
1 \(1 / 4\)
2 \(1 / 2\)
3 \(3 / 4\)
4 1
5 0
Explanation:
A Given the equation, \(a x^2+x+1=0, a \neq 0\) Let the roots of the above equation are \(\alpha\) and \(\beta\). \(\therefore \quad \alpha+\beta=\frac{-1}{\mathrm{a}}\) \(\alpha \beta=\frac{1}{\mathrm{a}}\) It is given that \(\alpha=\beta\) \(\Rightarrow 2 \alpha=\frac{-1}{a} \Rightarrow \alpha=\frac{-1}{2 a}\) \(\alpha^2=\frac{1}{a}\) Solving, \(\left(\frac{-1}{2 \mathrm{a}}\right)^2=\frac{1}{\mathrm{a}}\) Or \(\quad \frac{1}{4 \mathrm{a}^2}=\frac{1}{\mathrm{a}} \Rightarrow \mathrm{a}=\frac{1}{4}\)
118200
\(\alpha\) is the maximum value of \(1-2 x-5 x^2\) and \(\beta\) is the minimum value of \(x^2-2 x+r\). If \(5 \alpha x^2+\beta x+6>0\) for all real values \(x\). then the interval in which \(r\) lies is
118201
Let \(\mathrm{S}\) be the set of all possible integral values of \(\lambda\) in the interval \((-3,7)\) for which the roots of the quadratic equation \(\lambda x^2+13 x+7=0\) are all rational numbers. Then the sum of the elements in \(\mathrm{S}\) is
1 4
2 2
3 3
4 1
Explanation:
A We have, \(\lambda \mathrm{x}^2+13 \mathrm{x}+7=0\) \(\mathrm{D}=(13)^2-4(\lambda)(7)=169-28 \lambda\) For rational roots, \(\mathrm{D}=\) perfect square So, \(\lambda \in(-3,7)\) has values \(-2,0,6\) that \(D\) become perfect square. \(\therefore\) Required sum \(=-2+0+6=4\)
TS EAMCET-10.09.2020
Complex Numbers and Quadratic Equation
118202
The equation \(x^2-4 x+[x]+3=x[x]\), where \([x]\) denotes the greatest integer function, has:
1 exactly two solutions in \((-\infty, \infty)\)
2 no solution
3 a unique solution in \((-\infty, 1)\)
4 a unique solution in \((-\infty, \infty)\)
Explanation:
Exp:(D): \(x^2-4 x+[x]+3=x[x]\) \(\Rightarrow x^2-4 x+3=x[x]-[x]\) \(\Rightarrow(x-1)(x-3)=[x] .(x-1)\) \(\Rightarrow x=1 \text { or } x-3=[x]\) \(\Rightarrow x-[x]=3\) \(\Rightarrow\{x\}=3 \text { (Not Possible) }\) \(\text { Only one solution } x=1 \text { in }(-\infty, \infty)\)Only one solution \(x=1\) in \((-\infty, \infty)\)
JEE Main-24.01.2023
Complex Numbers and Quadratic Equation
118203
The root of \(\mathrm{ax}^2+\mathrm{x}+1=0\), where \(\mathrm{a} \neq 0\) are in the ratio \(1: 1\). Then a is equal to
1 \(1 / 4\)
2 \(1 / 2\)
3 \(3 / 4\)
4 1
5 0
Explanation:
A Given the equation, \(a x^2+x+1=0, a \neq 0\) Let the roots of the above equation are \(\alpha\) and \(\beta\). \(\therefore \quad \alpha+\beta=\frac{-1}{\mathrm{a}}\) \(\alpha \beta=\frac{1}{\mathrm{a}}\) It is given that \(\alpha=\beta\) \(\Rightarrow 2 \alpha=\frac{-1}{a} \Rightarrow \alpha=\frac{-1}{2 a}\) \(\alpha^2=\frac{1}{a}\) Solving, \(\left(\frac{-1}{2 \mathrm{a}}\right)^2=\frac{1}{\mathrm{a}}\) Or \(\quad \frac{1}{4 \mathrm{a}^2}=\frac{1}{\mathrm{a}} \Rightarrow \mathrm{a}=\frac{1}{4}\)
