118107
The minimum value of the sum of the squares of the roots of \(x^2+(3-a) x+1=2 a\) is :
1 4
2 5
3 6
4 8
Explanation:
C Given, \(x^2+(3-a) x+1=2 a\) \(x^2+(3-a) x+1-2 a=0\) Let, \(\alpha, \beta\) are the roots of the above, Then, \(\alpha+\beta=(\mathrm{a}-3)\) \(\alpha \beta=1-2 \mathrm{a}\) \(\alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta\) \(=(a-3)^2-2(1-2 a)\) \(=a^2+9-6 a-2+4 a\) \(=\quad a^2-2 a+7\) \(=\quad(a-1)^2+7-1=(a-1)^2+6\) \(\therefore \quad \text { Minimum value of } \alpha^2+\beta^2=6\left[\because(a-1)^2\right]=0\)
JEE Main-26.07.2022
Complex Numbers and Quadratic Equation
118108
For \(x^2-(a+3)|x|+4=0\) to have real solutions, the range of a is
1 \((-\infty,-7] \cup[1, \infty)\)
2 \((-3, \infty)\)
3 \((-\infty,-7]\)
4 \([1, \infty)\)
Explanation:
A Given, \(x^2-(a+3)|x|+4=0\) Case I:- \(\mathrm{x}>0\), then \(|\mathrm{x}|=\mathrm{x}\) \(\therefore x^2-(a+3) x+4=0\) But \(\mathrm{x}\) is real i.e. real solution. \(D=(a+3)^2-16 \geq 0\) \(D=(a+3+4)(a+3-4) \geq 0\) \(=(a+7)(a-1) \geq 0\) \(= a \geq 1 \text { or } a \leq-7 \text { But } x>0\) \(\text { Case II }:- x\lt 0 \Rightarrow|x|=-x\) \(\therefore x^2-(a+3)(-x)+4=0\) \(\Rightarrow \quad x^2+(a+3) x+4=0\) \(\text { Discriminant }(D)=(a+3)^2-16 \geq 0\) \(\therefore \quad \text { We get, } a \geq 1 \text { or } a \leq-7\) \(\therefore \quad \mathrm{a} \in(-\infty,-7) \cup(1, \infty)\)
AMU-2019
Complex Numbers and Quadratic Equation
118109
The number of real solutions of equation \(3\left(x^2+\frac{1}{x^2}\right)-2\left(x+\frac{1}{x}\right)+5=0\) is
1 4
2 3
3 2
4 0
Explanation:
D Given, \(3\left(x^2+\frac{1}{x^2}\right)-2\left(x+\frac{1}{x}\right)+5=0\) \(3 \cdot\left[\left(x+\frac{1}{x}\right)^2-2\right]-2\left(x+\frac{1}{x}\right)+5=0\) Let, \(\left(x+\frac{1}{x}\right)=t\) we get- or \(3\left[\mathrm{t}^2-2\right]-2 \mathrm{t}+5=0\) \(3 \mathrm{t}^2-6-2 \mathrm{t}+5=0\) \(3 \mathrm{t}^2-2 \mathrm{t}-1=0\) \(3 \mathrm{t}^2-3 \mathrm{t}+\mathrm{t}-1=0\) \(3 \mathrm{t}(\mathrm{t}-1)+1(\mathrm{t}-1)=0\) \((3 \mathrm{t}+1)(\mathrm{t}-1)=0\) So, we get, \(3 t+1=0, t=-1 / 3\) \(x+\frac{1}{x}=1\) \(x^2+1=x\) or \(x^2-x+1=0\) \(p=(-1)^2-4=-3\lt 0\) \(\Rightarrow \quad\) No real solution. and \(3\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)+1=0\) \(\Rightarrow \quad 3\left(\mathrm{x}^2+1\right)+\mathrm{x}=0\) \(\Rightarrow \quad 3 \mathrm{x}^2+3+\mathrm{x}=0\) \(\therefore \quad \mathrm{D}=(1)^2-4 \times 3 \times 3=-35\) \(\Rightarrow \quad\) No real solution \(\therefore\) The number of real solutions is zero.
JEE Main-24.01.2023
Complex Numbers and Quadratic Equation
118110
Let \(\alpha\) be a root of the equation \(1+x^2+x^4=0\). Then the value of \(\alpha^{1011+} \alpha^{2022]}-\alpha^{3033}\) is equal to :
1 1
2 \(\alpha\)
3 \(1+\alpha\)
4 \(1+2 \alpha\)
Explanation:
A Given, \(1+\mathrm{x}^2+\mathrm{x}^4=0 .\) Let, \(\omega\) is root ( \(\omega\) is cube root of unity) Then, \(\omega^{1011}+\omega^{2022}-\omega^{3033}\) \(=\left(\omega^3\right)^{337}+\left(\omega^3\right)^{674}-\left(\omega^3\right)^{1011}\) \(=1+1-1=1\)
118107
The minimum value of the sum of the squares of the roots of \(x^2+(3-a) x+1=2 a\) is :
1 4
2 5
3 6
4 8
Explanation:
C Given, \(x^2+(3-a) x+1=2 a\) \(x^2+(3-a) x+1-2 a=0\) Let, \(\alpha, \beta\) are the roots of the above, Then, \(\alpha+\beta=(\mathrm{a}-3)\) \(\alpha \beta=1-2 \mathrm{a}\) \(\alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta\) \(=(a-3)^2-2(1-2 a)\) \(=a^2+9-6 a-2+4 a\) \(=\quad a^2-2 a+7\) \(=\quad(a-1)^2+7-1=(a-1)^2+6\) \(\therefore \quad \text { Minimum value of } \alpha^2+\beta^2=6\left[\because(a-1)^2\right]=0\)
JEE Main-26.07.2022
Complex Numbers and Quadratic Equation
118108
For \(x^2-(a+3)|x|+4=0\) to have real solutions, the range of a is
1 \((-\infty,-7] \cup[1, \infty)\)
2 \((-3, \infty)\)
3 \((-\infty,-7]\)
4 \([1, \infty)\)
Explanation:
A Given, \(x^2-(a+3)|x|+4=0\) Case I:- \(\mathrm{x}>0\), then \(|\mathrm{x}|=\mathrm{x}\) \(\therefore x^2-(a+3) x+4=0\) But \(\mathrm{x}\) is real i.e. real solution. \(D=(a+3)^2-16 \geq 0\) \(D=(a+3+4)(a+3-4) \geq 0\) \(=(a+7)(a-1) \geq 0\) \(= a \geq 1 \text { or } a \leq-7 \text { But } x>0\) \(\text { Case II }:- x\lt 0 \Rightarrow|x|=-x\) \(\therefore x^2-(a+3)(-x)+4=0\) \(\Rightarrow \quad x^2+(a+3) x+4=0\) \(\text { Discriminant }(D)=(a+3)^2-16 \geq 0\) \(\therefore \quad \text { We get, } a \geq 1 \text { or } a \leq-7\) \(\therefore \quad \mathrm{a} \in(-\infty,-7) \cup(1, \infty)\)
AMU-2019
Complex Numbers and Quadratic Equation
118109
The number of real solutions of equation \(3\left(x^2+\frac{1}{x^2}\right)-2\left(x+\frac{1}{x}\right)+5=0\) is
1 4
2 3
3 2
4 0
Explanation:
D Given, \(3\left(x^2+\frac{1}{x^2}\right)-2\left(x+\frac{1}{x}\right)+5=0\) \(3 \cdot\left[\left(x+\frac{1}{x}\right)^2-2\right]-2\left(x+\frac{1}{x}\right)+5=0\) Let, \(\left(x+\frac{1}{x}\right)=t\) we get- or \(3\left[\mathrm{t}^2-2\right]-2 \mathrm{t}+5=0\) \(3 \mathrm{t}^2-6-2 \mathrm{t}+5=0\) \(3 \mathrm{t}^2-2 \mathrm{t}-1=0\) \(3 \mathrm{t}^2-3 \mathrm{t}+\mathrm{t}-1=0\) \(3 \mathrm{t}(\mathrm{t}-1)+1(\mathrm{t}-1)=0\) \((3 \mathrm{t}+1)(\mathrm{t}-1)=0\) So, we get, \(3 t+1=0, t=-1 / 3\) \(x+\frac{1}{x}=1\) \(x^2+1=x\) or \(x^2-x+1=0\) \(p=(-1)^2-4=-3\lt 0\) \(\Rightarrow \quad\) No real solution. and \(3\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)+1=0\) \(\Rightarrow \quad 3\left(\mathrm{x}^2+1\right)+\mathrm{x}=0\) \(\Rightarrow \quad 3 \mathrm{x}^2+3+\mathrm{x}=0\) \(\therefore \quad \mathrm{D}=(1)^2-4 \times 3 \times 3=-35\) \(\Rightarrow \quad\) No real solution \(\therefore\) The number of real solutions is zero.
JEE Main-24.01.2023
Complex Numbers and Quadratic Equation
118110
Let \(\alpha\) be a root of the equation \(1+x^2+x^4=0\). Then the value of \(\alpha^{1011+} \alpha^{2022]}-\alpha^{3033}\) is equal to :
1 1
2 \(\alpha\)
3 \(1+\alpha\)
4 \(1+2 \alpha\)
Explanation:
A Given, \(1+\mathrm{x}^2+\mathrm{x}^4=0 .\) Let, \(\omega\) is root ( \(\omega\) is cube root of unity) Then, \(\omega^{1011}+\omega^{2022}-\omega^{3033}\) \(=\left(\omega^3\right)^{337}+\left(\omega^3\right)^{674}-\left(\omega^3\right)^{1011}\) \(=1+1-1=1\)
118107
The minimum value of the sum of the squares of the roots of \(x^2+(3-a) x+1=2 a\) is :
1 4
2 5
3 6
4 8
Explanation:
C Given, \(x^2+(3-a) x+1=2 a\) \(x^2+(3-a) x+1-2 a=0\) Let, \(\alpha, \beta\) are the roots of the above, Then, \(\alpha+\beta=(\mathrm{a}-3)\) \(\alpha \beta=1-2 \mathrm{a}\) \(\alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta\) \(=(a-3)^2-2(1-2 a)\) \(=a^2+9-6 a-2+4 a\) \(=\quad a^2-2 a+7\) \(=\quad(a-1)^2+7-1=(a-1)^2+6\) \(\therefore \quad \text { Minimum value of } \alpha^2+\beta^2=6\left[\because(a-1)^2\right]=0\)
JEE Main-26.07.2022
Complex Numbers and Quadratic Equation
118108
For \(x^2-(a+3)|x|+4=0\) to have real solutions, the range of a is
1 \((-\infty,-7] \cup[1, \infty)\)
2 \((-3, \infty)\)
3 \((-\infty,-7]\)
4 \([1, \infty)\)
Explanation:
A Given, \(x^2-(a+3)|x|+4=0\) Case I:- \(\mathrm{x}>0\), then \(|\mathrm{x}|=\mathrm{x}\) \(\therefore x^2-(a+3) x+4=0\) But \(\mathrm{x}\) is real i.e. real solution. \(D=(a+3)^2-16 \geq 0\) \(D=(a+3+4)(a+3-4) \geq 0\) \(=(a+7)(a-1) \geq 0\) \(= a \geq 1 \text { or } a \leq-7 \text { But } x>0\) \(\text { Case II }:- x\lt 0 \Rightarrow|x|=-x\) \(\therefore x^2-(a+3)(-x)+4=0\) \(\Rightarrow \quad x^2+(a+3) x+4=0\) \(\text { Discriminant }(D)=(a+3)^2-16 \geq 0\) \(\therefore \quad \text { We get, } a \geq 1 \text { or } a \leq-7\) \(\therefore \quad \mathrm{a} \in(-\infty,-7) \cup(1, \infty)\)
AMU-2019
Complex Numbers and Quadratic Equation
118109
The number of real solutions of equation \(3\left(x^2+\frac{1}{x^2}\right)-2\left(x+\frac{1}{x}\right)+5=0\) is
1 4
2 3
3 2
4 0
Explanation:
D Given, \(3\left(x^2+\frac{1}{x^2}\right)-2\left(x+\frac{1}{x}\right)+5=0\) \(3 \cdot\left[\left(x+\frac{1}{x}\right)^2-2\right]-2\left(x+\frac{1}{x}\right)+5=0\) Let, \(\left(x+\frac{1}{x}\right)=t\) we get- or \(3\left[\mathrm{t}^2-2\right]-2 \mathrm{t}+5=0\) \(3 \mathrm{t}^2-6-2 \mathrm{t}+5=0\) \(3 \mathrm{t}^2-2 \mathrm{t}-1=0\) \(3 \mathrm{t}^2-3 \mathrm{t}+\mathrm{t}-1=0\) \(3 \mathrm{t}(\mathrm{t}-1)+1(\mathrm{t}-1)=0\) \((3 \mathrm{t}+1)(\mathrm{t}-1)=0\) So, we get, \(3 t+1=0, t=-1 / 3\) \(x+\frac{1}{x}=1\) \(x^2+1=x\) or \(x^2-x+1=0\) \(p=(-1)^2-4=-3\lt 0\) \(\Rightarrow \quad\) No real solution. and \(3\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)+1=0\) \(\Rightarrow \quad 3\left(\mathrm{x}^2+1\right)+\mathrm{x}=0\) \(\Rightarrow \quad 3 \mathrm{x}^2+3+\mathrm{x}=0\) \(\therefore \quad \mathrm{D}=(1)^2-4 \times 3 \times 3=-35\) \(\Rightarrow \quad\) No real solution \(\therefore\) The number of real solutions is zero.
JEE Main-24.01.2023
Complex Numbers and Quadratic Equation
118110
Let \(\alpha\) be a root of the equation \(1+x^2+x^4=0\). Then the value of \(\alpha^{1011+} \alpha^{2022]}-\alpha^{3033}\) is equal to :
1 1
2 \(\alpha\)
3 \(1+\alpha\)
4 \(1+2 \alpha\)
Explanation:
A Given, \(1+\mathrm{x}^2+\mathrm{x}^4=0 .\) Let, \(\omega\) is root ( \(\omega\) is cube root of unity) Then, \(\omega^{1011}+\omega^{2022}-\omega^{3033}\) \(=\left(\omega^3\right)^{337}+\left(\omega^3\right)^{674}-\left(\omega^3\right)^{1011}\) \(=1+1-1=1\)
118107
The minimum value of the sum of the squares of the roots of \(x^2+(3-a) x+1=2 a\) is :
1 4
2 5
3 6
4 8
Explanation:
C Given, \(x^2+(3-a) x+1=2 a\) \(x^2+(3-a) x+1-2 a=0\) Let, \(\alpha, \beta\) are the roots of the above, Then, \(\alpha+\beta=(\mathrm{a}-3)\) \(\alpha \beta=1-2 \mathrm{a}\) \(\alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta\) \(=(a-3)^2-2(1-2 a)\) \(=a^2+9-6 a-2+4 a\) \(=\quad a^2-2 a+7\) \(=\quad(a-1)^2+7-1=(a-1)^2+6\) \(\therefore \quad \text { Minimum value of } \alpha^2+\beta^2=6\left[\because(a-1)^2\right]=0\)
JEE Main-26.07.2022
Complex Numbers and Quadratic Equation
118108
For \(x^2-(a+3)|x|+4=0\) to have real solutions, the range of a is
1 \((-\infty,-7] \cup[1, \infty)\)
2 \((-3, \infty)\)
3 \((-\infty,-7]\)
4 \([1, \infty)\)
Explanation:
A Given, \(x^2-(a+3)|x|+4=0\) Case I:- \(\mathrm{x}>0\), then \(|\mathrm{x}|=\mathrm{x}\) \(\therefore x^2-(a+3) x+4=0\) But \(\mathrm{x}\) is real i.e. real solution. \(D=(a+3)^2-16 \geq 0\) \(D=(a+3+4)(a+3-4) \geq 0\) \(=(a+7)(a-1) \geq 0\) \(= a \geq 1 \text { or } a \leq-7 \text { But } x>0\) \(\text { Case II }:- x\lt 0 \Rightarrow|x|=-x\) \(\therefore x^2-(a+3)(-x)+4=0\) \(\Rightarrow \quad x^2+(a+3) x+4=0\) \(\text { Discriminant }(D)=(a+3)^2-16 \geq 0\) \(\therefore \quad \text { We get, } a \geq 1 \text { or } a \leq-7\) \(\therefore \quad \mathrm{a} \in(-\infty,-7) \cup(1, \infty)\)
AMU-2019
Complex Numbers and Quadratic Equation
118109
The number of real solutions of equation \(3\left(x^2+\frac{1}{x^2}\right)-2\left(x+\frac{1}{x}\right)+5=0\) is
1 4
2 3
3 2
4 0
Explanation:
D Given, \(3\left(x^2+\frac{1}{x^2}\right)-2\left(x+\frac{1}{x}\right)+5=0\) \(3 \cdot\left[\left(x+\frac{1}{x}\right)^2-2\right]-2\left(x+\frac{1}{x}\right)+5=0\) Let, \(\left(x+\frac{1}{x}\right)=t\) we get- or \(3\left[\mathrm{t}^2-2\right]-2 \mathrm{t}+5=0\) \(3 \mathrm{t}^2-6-2 \mathrm{t}+5=0\) \(3 \mathrm{t}^2-2 \mathrm{t}-1=0\) \(3 \mathrm{t}^2-3 \mathrm{t}+\mathrm{t}-1=0\) \(3 \mathrm{t}(\mathrm{t}-1)+1(\mathrm{t}-1)=0\) \((3 \mathrm{t}+1)(\mathrm{t}-1)=0\) So, we get, \(3 t+1=0, t=-1 / 3\) \(x+\frac{1}{x}=1\) \(x^2+1=x\) or \(x^2-x+1=0\) \(p=(-1)^2-4=-3\lt 0\) \(\Rightarrow \quad\) No real solution. and \(3\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)+1=0\) \(\Rightarrow \quad 3\left(\mathrm{x}^2+1\right)+\mathrm{x}=0\) \(\Rightarrow \quad 3 \mathrm{x}^2+3+\mathrm{x}=0\) \(\therefore \quad \mathrm{D}=(1)^2-4 \times 3 \times 3=-35\) \(\Rightarrow \quad\) No real solution \(\therefore\) The number of real solutions is zero.
JEE Main-24.01.2023
Complex Numbers and Quadratic Equation
118110
Let \(\alpha\) be a root of the equation \(1+x^2+x^4=0\). Then the value of \(\alpha^{1011+} \alpha^{2022]}-\alpha^{3033}\) is equal to :
1 1
2 \(\alpha\)
3 \(1+\alpha\)
4 \(1+2 \alpha\)
Explanation:
A Given, \(1+\mathrm{x}^2+\mathrm{x}^4=0 .\) Let, \(\omega\) is root ( \(\omega\) is cube root of unity) Then, \(\omega^{1011}+\omega^{2022}-\omega^{3033}\) \(=\left(\omega^3\right)^{337}+\left(\omega^3\right)^{674}-\left(\omega^3\right)^{1011}\) \(=1+1-1=1\)
