118053
If \(\alpha\) and \(\beta\) are roots of the equation \(x^2+x+1\) \(=0\), then \(\alpha^2+\beta^2\) is
1 1
2 \(\frac{-1-\mathrm{i} \sqrt{3}}{2}\)
3 \(\frac{-1+\mathrm{i} \sqrt{3}}{2}\)
4 -1
Explanation:
D Given, \(\mathrm{x}^2+\mathrm{x}+1=0\) Sum of roots \(=\alpha+\beta=-1\) Product of roots \(=\alpha . \beta=1\) Now, \(\alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta\) \(=(-1)^2-2 \times 1=-1\)
Karnataka CET-2019
Complex Numbers and Quadratic Equation
118054
If \(\alpha, \beta, \gamma\) are the roots of the equation \(x^3+4 x+2=0\), then \(\alpha^3+\beta^3+\gamma^3=\)
118083
The number of roots of the equation \(|x|^2-7|x|+12=0\) is
1 1
2 2
3 3
4 4
Explanation:
D Given equation is- \(|\mathrm{x}|^2-7|\mathrm{x}|+12=0\) Let \(|\mathrm{x}|=\mathrm{p}\) Then, \(\mathrm{p}^2-7 \mathrm{p}+12=0\) \((\mathrm{p}-3)(\mathrm{p}-4)=0\) \(\mathrm{p}=3\) \& 4 So, \(|\mathrm{x}|=3\) \(\mathrm{x}= \pm 3\) \(|\mathrm{x}|=4\) \(\mathrm{x}= \pm 4\) Therefore total number of roots \(=4\)
UPSEE-2012
Complex Numbers and Quadratic Equation
118049
\(x+2\) is a factor of
1 : \((-2)^4+2=16+2=18 \neq 0\) \(\therefore(\mathrm{x}+2)\) is not the factor of \(\mathrm{x}^4+2\).
2 : \((-2)^4-(-2)^2+12=16-4+12=24 \neq 0\) \(\therefore(\mathrm{x}+2)\) is not the factor of \(\mathrm{x}^4-\mathrm{x}^2+12\)
3 : \((-2)^4-2(-2)^3-(-2)+2=16+16+2+2\) \(=36 \neq 0\) \(\therefore(\mathrm{x}+2)\) is not the factor of \(\mathrm{x}^4-2 \mathrm{x}^3-\mathrm{x}+2\)
4 : \((-2)^4+2(-2)^3-(-2)-2=16-16+2-2=0\) \(\therefore(x+2)\) is the factor of \(\mathrm{x}^4+2 \mathrm{x}^3-\mathrm{x}-2\)
Explanation:
D By option (a.) : \((-2)^4+2=16+2=18 \neq 0\) \(\therefore(\mathrm{x}+2)\) is not the factor of \(\mathrm{x}^4+2\). (b.) : \((-2)^4-(-2)^2+12=16-4+12=24 \neq 0\) \(\therefore(\mathrm{x}+2)\) is not the factor of \(\mathrm{x}^4-\mathrm{x}^2+12\) (c.) : \((-2)^4-2(-2)^3-(-2)+2=16+16+2+2\) \(=36 \neq 0\) \(\therefore(\mathrm{x}+2)\) is not the factor of \(\mathrm{x}^4-2 \mathrm{x}^3-\mathrm{x}+2\) (d.) : \((-2)^4+2(-2)^3-(-2)-2=16-16+2-2=0\) \(\therefore(x+2)\) is the factor of \(\mathrm{x}^4+2 \mathrm{x}^3-\mathrm{x}-2\)
SRM JEEE-2015
Complex Numbers and Quadratic Equation
118050
The number of solutions of \(\sqrt{4-x}+\sqrt{x+9}=5\) is
1 0
2 1
3 2
4 3
Explanation:
C Given that :- \(\sqrt{4-x}+\sqrt{x+9}=5\) \(\sqrt{4-x}=5-\sqrt{x+9}\) Squaring both sides, we get - \((4-x)=25+x+9-10 \sqrt{x+9}\) \(10 \sqrt{\mathrm{x}+9}=30+2 \mathrm{x}\) Again squaring both sides :- \(100(\mathrm{x}+9)=900+4 \mathrm{x}^2+120 \mathrm{x}\) \(\Rightarrow \mathrm{x}^2+5 \mathrm{x}=0\) \(\mathrm{x}(\mathrm{x}+5)=0\) \(\mathrm{x}=0,-5\) Hence, number of solution is 2 .
118053
If \(\alpha\) and \(\beta\) are roots of the equation \(x^2+x+1\) \(=0\), then \(\alpha^2+\beta^2\) is
1 1
2 \(\frac{-1-\mathrm{i} \sqrt{3}}{2}\)
3 \(\frac{-1+\mathrm{i} \sqrt{3}}{2}\)
4 -1
Explanation:
D Given, \(\mathrm{x}^2+\mathrm{x}+1=0\) Sum of roots \(=\alpha+\beta=-1\) Product of roots \(=\alpha . \beta=1\) Now, \(\alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta\) \(=(-1)^2-2 \times 1=-1\)
Karnataka CET-2019
Complex Numbers and Quadratic Equation
118054
If \(\alpha, \beta, \gamma\) are the roots of the equation \(x^3+4 x+2=0\), then \(\alpha^3+\beta^3+\gamma^3=\)
118083
The number of roots of the equation \(|x|^2-7|x|+12=0\) is
1 1
2 2
3 3
4 4
Explanation:
D Given equation is- \(|\mathrm{x}|^2-7|\mathrm{x}|+12=0\) Let \(|\mathrm{x}|=\mathrm{p}\) Then, \(\mathrm{p}^2-7 \mathrm{p}+12=0\) \((\mathrm{p}-3)(\mathrm{p}-4)=0\) \(\mathrm{p}=3\) \& 4 So, \(|\mathrm{x}|=3\) \(\mathrm{x}= \pm 3\) \(|\mathrm{x}|=4\) \(\mathrm{x}= \pm 4\) Therefore total number of roots \(=4\)
UPSEE-2012
Complex Numbers and Quadratic Equation
118049
\(x+2\) is a factor of
1 : \((-2)^4+2=16+2=18 \neq 0\) \(\therefore(\mathrm{x}+2)\) is not the factor of \(\mathrm{x}^4+2\).
2 : \((-2)^4-(-2)^2+12=16-4+12=24 \neq 0\) \(\therefore(\mathrm{x}+2)\) is not the factor of \(\mathrm{x}^4-\mathrm{x}^2+12\)
3 : \((-2)^4-2(-2)^3-(-2)+2=16+16+2+2\) \(=36 \neq 0\) \(\therefore(\mathrm{x}+2)\) is not the factor of \(\mathrm{x}^4-2 \mathrm{x}^3-\mathrm{x}+2\)
4 : \((-2)^4+2(-2)^3-(-2)-2=16-16+2-2=0\) \(\therefore(x+2)\) is the factor of \(\mathrm{x}^4+2 \mathrm{x}^3-\mathrm{x}-2\)
Explanation:
D By option (a.) : \((-2)^4+2=16+2=18 \neq 0\) \(\therefore(\mathrm{x}+2)\) is not the factor of \(\mathrm{x}^4+2\). (b.) : \((-2)^4-(-2)^2+12=16-4+12=24 \neq 0\) \(\therefore(\mathrm{x}+2)\) is not the factor of \(\mathrm{x}^4-\mathrm{x}^2+12\) (c.) : \((-2)^4-2(-2)^3-(-2)+2=16+16+2+2\) \(=36 \neq 0\) \(\therefore(\mathrm{x}+2)\) is not the factor of \(\mathrm{x}^4-2 \mathrm{x}^3-\mathrm{x}+2\) (d.) : \((-2)^4+2(-2)^3-(-2)-2=16-16+2-2=0\) \(\therefore(x+2)\) is the factor of \(\mathrm{x}^4+2 \mathrm{x}^3-\mathrm{x}-2\)
SRM JEEE-2015
Complex Numbers and Quadratic Equation
118050
The number of solutions of \(\sqrt{4-x}+\sqrt{x+9}=5\) is
1 0
2 1
3 2
4 3
Explanation:
C Given that :- \(\sqrt{4-x}+\sqrt{x+9}=5\) \(\sqrt{4-x}=5-\sqrt{x+9}\) Squaring both sides, we get - \((4-x)=25+x+9-10 \sqrt{x+9}\) \(10 \sqrt{\mathrm{x}+9}=30+2 \mathrm{x}\) Again squaring both sides :- \(100(\mathrm{x}+9)=900+4 \mathrm{x}^2+120 \mathrm{x}\) \(\Rightarrow \mathrm{x}^2+5 \mathrm{x}=0\) \(\mathrm{x}(\mathrm{x}+5)=0\) \(\mathrm{x}=0,-5\) Hence, number of solution is 2 .
118053
If \(\alpha\) and \(\beta\) are roots of the equation \(x^2+x+1\) \(=0\), then \(\alpha^2+\beta^2\) is
1 1
2 \(\frac{-1-\mathrm{i} \sqrt{3}}{2}\)
3 \(\frac{-1+\mathrm{i} \sqrt{3}}{2}\)
4 -1
Explanation:
D Given, \(\mathrm{x}^2+\mathrm{x}+1=0\) Sum of roots \(=\alpha+\beta=-1\) Product of roots \(=\alpha . \beta=1\) Now, \(\alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta\) \(=(-1)^2-2 \times 1=-1\)
Karnataka CET-2019
Complex Numbers and Quadratic Equation
118054
If \(\alpha, \beta, \gamma\) are the roots of the equation \(x^3+4 x+2=0\), then \(\alpha^3+\beta^3+\gamma^3=\)
118083
The number of roots of the equation \(|x|^2-7|x|+12=0\) is
1 1
2 2
3 3
4 4
Explanation:
D Given equation is- \(|\mathrm{x}|^2-7|\mathrm{x}|+12=0\) Let \(|\mathrm{x}|=\mathrm{p}\) Then, \(\mathrm{p}^2-7 \mathrm{p}+12=0\) \((\mathrm{p}-3)(\mathrm{p}-4)=0\) \(\mathrm{p}=3\) \& 4 So, \(|\mathrm{x}|=3\) \(\mathrm{x}= \pm 3\) \(|\mathrm{x}|=4\) \(\mathrm{x}= \pm 4\) Therefore total number of roots \(=4\)
UPSEE-2012
Complex Numbers and Quadratic Equation
118049
\(x+2\) is a factor of
1 : \((-2)^4+2=16+2=18 \neq 0\) \(\therefore(\mathrm{x}+2)\) is not the factor of \(\mathrm{x}^4+2\).
2 : \((-2)^4-(-2)^2+12=16-4+12=24 \neq 0\) \(\therefore(\mathrm{x}+2)\) is not the factor of \(\mathrm{x}^4-\mathrm{x}^2+12\)
3 : \((-2)^4-2(-2)^3-(-2)+2=16+16+2+2\) \(=36 \neq 0\) \(\therefore(\mathrm{x}+2)\) is not the factor of \(\mathrm{x}^4-2 \mathrm{x}^3-\mathrm{x}+2\)
4 : \((-2)^4+2(-2)^3-(-2)-2=16-16+2-2=0\) \(\therefore(x+2)\) is the factor of \(\mathrm{x}^4+2 \mathrm{x}^3-\mathrm{x}-2\)
Explanation:
D By option (a.) : \((-2)^4+2=16+2=18 \neq 0\) \(\therefore(\mathrm{x}+2)\) is not the factor of \(\mathrm{x}^4+2\). (b.) : \((-2)^4-(-2)^2+12=16-4+12=24 \neq 0\) \(\therefore(\mathrm{x}+2)\) is not the factor of \(\mathrm{x}^4-\mathrm{x}^2+12\) (c.) : \((-2)^4-2(-2)^3-(-2)+2=16+16+2+2\) \(=36 \neq 0\) \(\therefore(\mathrm{x}+2)\) is not the factor of \(\mathrm{x}^4-2 \mathrm{x}^3-\mathrm{x}+2\) (d.) : \((-2)^4+2(-2)^3-(-2)-2=16-16+2-2=0\) \(\therefore(x+2)\) is the factor of \(\mathrm{x}^4+2 \mathrm{x}^3-\mathrm{x}-2\)
SRM JEEE-2015
Complex Numbers and Quadratic Equation
118050
The number of solutions of \(\sqrt{4-x}+\sqrt{x+9}=5\) is
1 0
2 1
3 2
4 3
Explanation:
C Given that :- \(\sqrt{4-x}+\sqrt{x+9}=5\) \(\sqrt{4-x}=5-\sqrt{x+9}\) Squaring both sides, we get - \((4-x)=25+x+9-10 \sqrt{x+9}\) \(10 \sqrt{\mathrm{x}+9}=30+2 \mathrm{x}\) Again squaring both sides :- \(100(\mathrm{x}+9)=900+4 \mathrm{x}^2+120 \mathrm{x}\) \(\Rightarrow \mathrm{x}^2+5 \mathrm{x}=0\) \(\mathrm{x}(\mathrm{x}+5)=0\) \(\mathrm{x}=0,-5\) Hence, number of solution is 2 .
118053
If \(\alpha\) and \(\beta\) are roots of the equation \(x^2+x+1\) \(=0\), then \(\alpha^2+\beta^2\) is
1 1
2 \(\frac{-1-\mathrm{i} \sqrt{3}}{2}\)
3 \(\frac{-1+\mathrm{i} \sqrt{3}}{2}\)
4 -1
Explanation:
D Given, \(\mathrm{x}^2+\mathrm{x}+1=0\) Sum of roots \(=\alpha+\beta=-1\) Product of roots \(=\alpha . \beta=1\) Now, \(\alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta\) \(=(-1)^2-2 \times 1=-1\)
Karnataka CET-2019
Complex Numbers and Quadratic Equation
118054
If \(\alpha, \beta, \gamma\) are the roots of the equation \(x^3+4 x+2=0\), then \(\alpha^3+\beta^3+\gamma^3=\)
118083
The number of roots of the equation \(|x|^2-7|x|+12=0\) is
1 1
2 2
3 3
4 4
Explanation:
D Given equation is- \(|\mathrm{x}|^2-7|\mathrm{x}|+12=0\) Let \(|\mathrm{x}|=\mathrm{p}\) Then, \(\mathrm{p}^2-7 \mathrm{p}+12=0\) \((\mathrm{p}-3)(\mathrm{p}-4)=0\) \(\mathrm{p}=3\) \& 4 So, \(|\mathrm{x}|=3\) \(\mathrm{x}= \pm 3\) \(|\mathrm{x}|=4\) \(\mathrm{x}= \pm 4\) Therefore total number of roots \(=4\)
UPSEE-2012
Complex Numbers and Quadratic Equation
118049
\(x+2\) is a factor of
1 : \((-2)^4+2=16+2=18 \neq 0\) \(\therefore(\mathrm{x}+2)\) is not the factor of \(\mathrm{x}^4+2\).
2 : \((-2)^4-(-2)^2+12=16-4+12=24 \neq 0\) \(\therefore(\mathrm{x}+2)\) is not the factor of \(\mathrm{x}^4-\mathrm{x}^2+12\)
3 : \((-2)^4-2(-2)^3-(-2)+2=16+16+2+2\) \(=36 \neq 0\) \(\therefore(\mathrm{x}+2)\) is not the factor of \(\mathrm{x}^4-2 \mathrm{x}^3-\mathrm{x}+2\)
4 : \((-2)^4+2(-2)^3-(-2)-2=16-16+2-2=0\) \(\therefore(x+2)\) is the factor of \(\mathrm{x}^4+2 \mathrm{x}^3-\mathrm{x}-2\)
Explanation:
D By option (a.) : \((-2)^4+2=16+2=18 \neq 0\) \(\therefore(\mathrm{x}+2)\) is not the factor of \(\mathrm{x}^4+2\). (b.) : \((-2)^4-(-2)^2+12=16-4+12=24 \neq 0\) \(\therefore(\mathrm{x}+2)\) is not the factor of \(\mathrm{x}^4-\mathrm{x}^2+12\) (c.) : \((-2)^4-2(-2)^3-(-2)+2=16+16+2+2\) \(=36 \neq 0\) \(\therefore(\mathrm{x}+2)\) is not the factor of \(\mathrm{x}^4-2 \mathrm{x}^3-\mathrm{x}+2\) (d.) : \((-2)^4+2(-2)^3-(-2)-2=16-16+2-2=0\) \(\therefore(x+2)\) is the factor of \(\mathrm{x}^4+2 \mathrm{x}^3-\mathrm{x}-2\)
SRM JEEE-2015
Complex Numbers and Quadratic Equation
118050
The number of solutions of \(\sqrt{4-x}+\sqrt{x+9}=5\) is
1 0
2 1
3 2
4 3
Explanation:
C Given that :- \(\sqrt{4-x}+\sqrt{x+9}=5\) \(\sqrt{4-x}=5-\sqrt{x+9}\) Squaring both sides, we get - \((4-x)=25+x+9-10 \sqrt{x+9}\) \(10 \sqrt{\mathrm{x}+9}=30+2 \mathrm{x}\) Again squaring both sides :- \(100(\mathrm{x}+9)=900+4 \mathrm{x}^2+120 \mathrm{x}\) \(\Rightarrow \mathrm{x}^2+5 \mathrm{x}=0\) \(\mathrm{x}(\mathrm{x}+5)=0\) \(\mathrm{x}=0,-5\) Hence, number of solution is 2 .
118053
If \(\alpha\) and \(\beta\) are roots of the equation \(x^2+x+1\) \(=0\), then \(\alpha^2+\beta^2\) is
1 1
2 \(\frac{-1-\mathrm{i} \sqrt{3}}{2}\)
3 \(\frac{-1+\mathrm{i} \sqrt{3}}{2}\)
4 -1
Explanation:
D Given, \(\mathrm{x}^2+\mathrm{x}+1=0\) Sum of roots \(=\alpha+\beta=-1\) Product of roots \(=\alpha . \beta=1\) Now, \(\alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta\) \(=(-1)^2-2 \times 1=-1\)
Karnataka CET-2019
Complex Numbers and Quadratic Equation
118054
If \(\alpha, \beta, \gamma\) are the roots of the equation \(x^3+4 x+2=0\), then \(\alpha^3+\beta^3+\gamma^3=\)
118083
The number of roots of the equation \(|x|^2-7|x|+12=0\) is
1 1
2 2
3 3
4 4
Explanation:
D Given equation is- \(|\mathrm{x}|^2-7|\mathrm{x}|+12=0\) Let \(|\mathrm{x}|=\mathrm{p}\) Then, \(\mathrm{p}^2-7 \mathrm{p}+12=0\) \((\mathrm{p}-3)(\mathrm{p}-4)=0\) \(\mathrm{p}=3\) \& 4 So, \(|\mathrm{x}|=3\) \(\mathrm{x}= \pm 3\) \(|\mathrm{x}|=4\) \(\mathrm{x}= \pm 4\) Therefore total number of roots \(=4\)
UPSEE-2012
Complex Numbers and Quadratic Equation
118049
\(x+2\) is a factor of
1 : \((-2)^4+2=16+2=18 \neq 0\) \(\therefore(\mathrm{x}+2)\) is not the factor of \(\mathrm{x}^4+2\).
2 : \((-2)^4-(-2)^2+12=16-4+12=24 \neq 0\) \(\therefore(\mathrm{x}+2)\) is not the factor of \(\mathrm{x}^4-\mathrm{x}^2+12\)
3 : \((-2)^4-2(-2)^3-(-2)+2=16+16+2+2\) \(=36 \neq 0\) \(\therefore(\mathrm{x}+2)\) is not the factor of \(\mathrm{x}^4-2 \mathrm{x}^3-\mathrm{x}+2\)
4 : \((-2)^4+2(-2)^3-(-2)-2=16-16+2-2=0\) \(\therefore(x+2)\) is the factor of \(\mathrm{x}^4+2 \mathrm{x}^3-\mathrm{x}-2\)
Explanation:
D By option (a.) : \((-2)^4+2=16+2=18 \neq 0\) \(\therefore(\mathrm{x}+2)\) is not the factor of \(\mathrm{x}^4+2\). (b.) : \((-2)^4-(-2)^2+12=16-4+12=24 \neq 0\) \(\therefore(\mathrm{x}+2)\) is not the factor of \(\mathrm{x}^4-\mathrm{x}^2+12\) (c.) : \((-2)^4-2(-2)^3-(-2)+2=16+16+2+2\) \(=36 \neq 0\) \(\therefore(\mathrm{x}+2)\) is not the factor of \(\mathrm{x}^4-2 \mathrm{x}^3-\mathrm{x}+2\) (d.) : \((-2)^4+2(-2)^3-(-2)-2=16-16+2-2=0\) \(\therefore(x+2)\) is the factor of \(\mathrm{x}^4+2 \mathrm{x}^3-\mathrm{x}-2\)
SRM JEEE-2015
Complex Numbers and Quadratic Equation
118050
The number of solutions of \(\sqrt{4-x}+\sqrt{x+9}=5\) is
1 0
2 1
3 2
4 3
Explanation:
C Given that :- \(\sqrt{4-x}+\sqrt{x+9}=5\) \(\sqrt{4-x}=5-\sqrt{x+9}\) Squaring both sides, we get - \((4-x)=25+x+9-10 \sqrt{x+9}\) \(10 \sqrt{\mathrm{x}+9}=30+2 \mathrm{x}\) Again squaring both sides :- \(100(\mathrm{x}+9)=900+4 \mathrm{x}^2+120 \mathrm{x}\) \(\Rightarrow \mathrm{x}^2+5 \mathrm{x}=0\) \(\mathrm{x}(\mathrm{x}+5)=0\) \(\mathrm{x}=0,-5\) Hence, number of solution is 2 .
