118244
The sum of the real solutions of equation \(2|\mathbf{x}|^2+\mathbf{5 1}=|\mathbf{1}+\mathbf{2 0 x}| \text { is }\)
1 5
2 24
3 0
4 None of these
Explanation:
D Given, \(2|\mathrm{x}|^2+51=|1+20 \mathrm{x}|\) Then, following two possible cases arise : \(\text { Case I When } 1+20 \mathrm{x}>0 \Rightarrow \mathrm{x}>-\frac{1}{20}\) \(\therefore \quad 2 \mathrm{x}^2+51=(1+20 \mathrm{x})\) \(2 \mathrm{x}^2-20 \mathrm{x}+50=0\) \(\mathrm{x}^2-10 \mathrm{x}+25=0 \quad[\therefore 2 \neq 0]\) \((x-5)^2=0\) \(\mathrm{x}=5,5\) Case II When \(1+20 \mathrm{x}\lt 0 \Rightarrow \mathrm{x}\lt -\frac{1}{20}\) \(\therefore \quad 2 \mathrm{x}^2+51=-(1+20 \mathrm{x})\) \(2 \mathrm{x}^2+20 \mathrm{x}+52=0\) \(\mathrm{x}^2+10 \mathrm{x}+26=0\) Here, \(\mathrm{D}\lt 0\) Thus, roots are imaginary. Hence, sum of real roots \(=5+5=10\)
CG PET- 2015
Complex Numbers and Quadratic Equation
118245
How many real roots does the quadratic equation \(f(x)=x^2+3|x|+2=0\) have?
1 One
2 Two
3 Four
4 No real root
Explanation:
D We have, \(x^2+3|x|+2=0\) \(|x|^2+3|x|+2=0\) \((|x|+2)(|x|+1)=0\) \(|x|=-2,-1\)Hence, given quadratic equation has no real roots.
CG PET- 2017
Complex Numbers and Quadratic Equation
118246
All the fourth roots of unity are
1 \(1,1,-1,-1\)
2 i, i, -i, -i
3 \(1,-1\), i, -i
4 \(-\mathrm{i},-\mathrm{i},-\mathrm{i},-\mathrm{i}\) Where \(\mathbf{i}=\sqrt{-1}\).
Explanation:
C Let \(\mathrm{x}\) be in four fourth root of 1 , then \(\mathrm{x}=\sqrt[4]{1}\) \(\mathrm{x}=(1)^{3 / 4}\) \(\mathrm{x}^4=1\) \(\mathrm{x}^4-1=0\) \(\mathrm{x}^4-1^4=0 \quad\left\{\because 1^4=1\right\}\) \(\left(\mathrm{x}^2\right)^2-\left(1^2\right)^2=0\) \(\left(\mathrm{x}^2-1^2\right)\left(\mathrm{x}^2+1^2\right)=0\) \(x^2-1=0, \quad x^2+1=0\) \(\mathrm{x}^2=1 \text {, }\) \(\mathrm{x}^2=-1\) \(x= \pm 1 \text {, }\) \(\mathrm{x}= \pm \sqrt{-1}\) \(\mathrm{x}= \pm \mathrm{i} \quad\{\mathrm{i}=\sqrt{-1} \text { given }\}\)So, the four fourth root of \(1=\{1,-1, i,-i\}\)
SCRA-2015
Complex Numbers and Quadratic Equation
118247
If \(\alpha, \beta, \gamma, \delta\) are the roots of the equation \(x^4+x^3\) \(+\mathbf{x}^2+\mathbf{x}+1=0\), then \(\alpha^{2021]}+\beta^{2021]}+\gamma^{2021]}+\delta^{2021]}\) is equal to
1 -4
2 -1
3 1
4 4
Explanation:
B \(: \alpha, \beta, \gamma, \delta \text { root of the equation }\) \(x^4+x^3+x^2+x+1=0\) \(\text { which are } 5^{\text {th }} \text { roots of unity except } 1\) \(x^5=1\) \(\alpha^5=\beta^5=\gamma^5=\delta^5=1\) \(\text { then } \alpha^{2021}+\beta^{2021}+\gamma^{2021}+\delta^{2021}=\) \(\alpha+\beta+\gamma+\delta=-1\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Complex Numbers and Quadratic Equation
118244
The sum of the real solutions of equation \(2|\mathbf{x}|^2+\mathbf{5 1}=|\mathbf{1}+\mathbf{2 0 x}| \text { is }\)
1 5
2 24
3 0
4 None of these
Explanation:
D Given, \(2|\mathrm{x}|^2+51=|1+20 \mathrm{x}|\) Then, following two possible cases arise : \(\text { Case I When } 1+20 \mathrm{x}>0 \Rightarrow \mathrm{x}>-\frac{1}{20}\) \(\therefore \quad 2 \mathrm{x}^2+51=(1+20 \mathrm{x})\) \(2 \mathrm{x}^2-20 \mathrm{x}+50=0\) \(\mathrm{x}^2-10 \mathrm{x}+25=0 \quad[\therefore 2 \neq 0]\) \((x-5)^2=0\) \(\mathrm{x}=5,5\) Case II When \(1+20 \mathrm{x}\lt 0 \Rightarrow \mathrm{x}\lt -\frac{1}{20}\) \(\therefore \quad 2 \mathrm{x}^2+51=-(1+20 \mathrm{x})\) \(2 \mathrm{x}^2+20 \mathrm{x}+52=0\) \(\mathrm{x}^2+10 \mathrm{x}+26=0\) Here, \(\mathrm{D}\lt 0\) Thus, roots are imaginary. Hence, sum of real roots \(=5+5=10\)
CG PET- 2015
Complex Numbers and Quadratic Equation
118245
How many real roots does the quadratic equation \(f(x)=x^2+3|x|+2=0\) have?
1 One
2 Two
3 Four
4 No real root
Explanation:
D We have, \(x^2+3|x|+2=0\) \(|x|^2+3|x|+2=0\) \((|x|+2)(|x|+1)=0\) \(|x|=-2,-1\)Hence, given quadratic equation has no real roots.
CG PET- 2017
Complex Numbers and Quadratic Equation
118246
All the fourth roots of unity are
1 \(1,1,-1,-1\)
2 i, i, -i, -i
3 \(1,-1\), i, -i
4 \(-\mathrm{i},-\mathrm{i},-\mathrm{i},-\mathrm{i}\) Where \(\mathbf{i}=\sqrt{-1}\).
Explanation:
C Let \(\mathrm{x}\) be in four fourth root of 1 , then \(\mathrm{x}=\sqrt[4]{1}\) \(\mathrm{x}=(1)^{3 / 4}\) \(\mathrm{x}^4=1\) \(\mathrm{x}^4-1=0\) \(\mathrm{x}^4-1^4=0 \quad\left\{\because 1^4=1\right\}\) \(\left(\mathrm{x}^2\right)^2-\left(1^2\right)^2=0\) \(\left(\mathrm{x}^2-1^2\right)\left(\mathrm{x}^2+1^2\right)=0\) \(x^2-1=0, \quad x^2+1=0\) \(\mathrm{x}^2=1 \text {, }\) \(\mathrm{x}^2=-1\) \(x= \pm 1 \text {, }\) \(\mathrm{x}= \pm \sqrt{-1}\) \(\mathrm{x}= \pm \mathrm{i} \quad\{\mathrm{i}=\sqrt{-1} \text { given }\}\)So, the four fourth root of \(1=\{1,-1, i,-i\}\)
SCRA-2015
Complex Numbers and Quadratic Equation
118247
If \(\alpha, \beta, \gamma, \delta\) are the roots of the equation \(x^4+x^3\) \(+\mathbf{x}^2+\mathbf{x}+1=0\), then \(\alpha^{2021]}+\beta^{2021]}+\gamma^{2021]}+\delta^{2021]}\) is equal to
1 -4
2 -1
3 1
4 4
Explanation:
B \(: \alpha, \beta, \gamma, \delta \text { root of the equation }\) \(x^4+x^3+x^2+x+1=0\) \(\text { which are } 5^{\text {th }} \text { roots of unity except } 1\) \(x^5=1\) \(\alpha^5=\beta^5=\gamma^5=\delta^5=1\) \(\text { then } \alpha^{2021}+\beta^{2021}+\gamma^{2021}+\delta^{2021}=\) \(\alpha+\beta+\gamma+\delta=-1\)
118244
The sum of the real solutions of equation \(2|\mathbf{x}|^2+\mathbf{5 1}=|\mathbf{1}+\mathbf{2 0 x}| \text { is }\)
1 5
2 24
3 0
4 None of these
Explanation:
D Given, \(2|\mathrm{x}|^2+51=|1+20 \mathrm{x}|\) Then, following two possible cases arise : \(\text { Case I When } 1+20 \mathrm{x}>0 \Rightarrow \mathrm{x}>-\frac{1}{20}\) \(\therefore \quad 2 \mathrm{x}^2+51=(1+20 \mathrm{x})\) \(2 \mathrm{x}^2-20 \mathrm{x}+50=0\) \(\mathrm{x}^2-10 \mathrm{x}+25=0 \quad[\therefore 2 \neq 0]\) \((x-5)^2=0\) \(\mathrm{x}=5,5\) Case II When \(1+20 \mathrm{x}\lt 0 \Rightarrow \mathrm{x}\lt -\frac{1}{20}\) \(\therefore \quad 2 \mathrm{x}^2+51=-(1+20 \mathrm{x})\) \(2 \mathrm{x}^2+20 \mathrm{x}+52=0\) \(\mathrm{x}^2+10 \mathrm{x}+26=0\) Here, \(\mathrm{D}\lt 0\) Thus, roots are imaginary. Hence, sum of real roots \(=5+5=10\)
CG PET- 2015
Complex Numbers and Quadratic Equation
118245
How many real roots does the quadratic equation \(f(x)=x^2+3|x|+2=0\) have?
1 One
2 Two
3 Four
4 No real root
Explanation:
D We have, \(x^2+3|x|+2=0\) \(|x|^2+3|x|+2=0\) \((|x|+2)(|x|+1)=0\) \(|x|=-2,-1\)Hence, given quadratic equation has no real roots.
CG PET- 2017
Complex Numbers and Quadratic Equation
118246
All the fourth roots of unity are
1 \(1,1,-1,-1\)
2 i, i, -i, -i
3 \(1,-1\), i, -i
4 \(-\mathrm{i},-\mathrm{i},-\mathrm{i},-\mathrm{i}\) Where \(\mathbf{i}=\sqrt{-1}\).
Explanation:
C Let \(\mathrm{x}\) be in four fourth root of 1 , then \(\mathrm{x}=\sqrt[4]{1}\) \(\mathrm{x}=(1)^{3 / 4}\) \(\mathrm{x}^4=1\) \(\mathrm{x}^4-1=0\) \(\mathrm{x}^4-1^4=0 \quad\left\{\because 1^4=1\right\}\) \(\left(\mathrm{x}^2\right)^2-\left(1^2\right)^2=0\) \(\left(\mathrm{x}^2-1^2\right)\left(\mathrm{x}^2+1^2\right)=0\) \(x^2-1=0, \quad x^2+1=0\) \(\mathrm{x}^2=1 \text {, }\) \(\mathrm{x}^2=-1\) \(x= \pm 1 \text {, }\) \(\mathrm{x}= \pm \sqrt{-1}\) \(\mathrm{x}= \pm \mathrm{i} \quad\{\mathrm{i}=\sqrt{-1} \text { given }\}\)So, the four fourth root of \(1=\{1,-1, i,-i\}\)
SCRA-2015
Complex Numbers and Quadratic Equation
118247
If \(\alpha, \beta, \gamma, \delta\) are the roots of the equation \(x^4+x^3\) \(+\mathbf{x}^2+\mathbf{x}+1=0\), then \(\alpha^{2021]}+\beta^{2021]}+\gamma^{2021]}+\delta^{2021]}\) is equal to
1 -4
2 -1
3 1
4 4
Explanation:
B \(: \alpha, \beta, \gamma, \delta \text { root of the equation }\) \(x^4+x^3+x^2+x+1=0\) \(\text { which are } 5^{\text {th }} \text { roots of unity except } 1\) \(x^5=1\) \(\alpha^5=\beta^5=\gamma^5=\delta^5=1\) \(\text { then } \alpha^{2021}+\beta^{2021}+\gamma^{2021}+\delta^{2021}=\) \(\alpha+\beta+\gamma+\delta=-1\)
118244
The sum of the real solutions of equation \(2|\mathbf{x}|^2+\mathbf{5 1}=|\mathbf{1}+\mathbf{2 0 x}| \text { is }\)
1 5
2 24
3 0
4 None of these
Explanation:
D Given, \(2|\mathrm{x}|^2+51=|1+20 \mathrm{x}|\) Then, following two possible cases arise : \(\text { Case I When } 1+20 \mathrm{x}>0 \Rightarrow \mathrm{x}>-\frac{1}{20}\) \(\therefore \quad 2 \mathrm{x}^2+51=(1+20 \mathrm{x})\) \(2 \mathrm{x}^2-20 \mathrm{x}+50=0\) \(\mathrm{x}^2-10 \mathrm{x}+25=0 \quad[\therefore 2 \neq 0]\) \((x-5)^2=0\) \(\mathrm{x}=5,5\) Case II When \(1+20 \mathrm{x}\lt 0 \Rightarrow \mathrm{x}\lt -\frac{1}{20}\) \(\therefore \quad 2 \mathrm{x}^2+51=-(1+20 \mathrm{x})\) \(2 \mathrm{x}^2+20 \mathrm{x}+52=0\) \(\mathrm{x}^2+10 \mathrm{x}+26=0\) Here, \(\mathrm{D}\lt 0\) Thus, roots are imaginary. Hence, sum of real roots \(=5+5=10\)
CG PET- 2015
Complex Numbers and Quadratic Equation
118245
How many real roots does the quadratic equation \(f(x)=x^2+3|x|+2=0\) have?
1 One
2 Two
3 Four
4 No real root
Explanation:
D We have, \(x^2+3|x|+2=0\) \(|x|^2+3|x|+2=0\) \((|x|+2)(|x|+1)=0\) \(|x|=-2,-1\)Hence, given quadratic equation has no real roots.
CG PET- 2017
Complex Numbers and Quadratic Equation
118246
All the fourth roots of unity are
1 \(1,1,-1,-1\)
2 i, i, -i, -i
3 \(1,-1\), i, -i
4 \(-\mathrm{i},-\mathrm{i},-\mathrm{i},-\mathrm{i}\) Where \(\mathbf{i}=\sqrt{-1}\).
Explanation:
C Let \(\mathrm{x}\) be in four fourth root of 1 , then \(\mathrm{x}=\sqrt[4]{1}\) \(\mathrm{x}=(1)^{3 / 4}\) \(\mathrm{x}^4=1\) \(\mathrm{x}^4-1=0\) \(\mathrm{x}^4-1^4=0 \quad\left\{\because 1^4=1\right\}\) \(\left(\mathrm{x}^2\right)^2-\left(1^2\right)^2=0\) \(\left(\mathrm{x}^2-1^2\right)\left(\mathrm{x}^2+1^2\right)=0\) \(x^2-1=0, \quad x^2+1=0\) \(\mathrm{x}^2=1 \text {, }\) \(\mathrm{x}^2=-1\) \(x= \pm 1 \text {, }\) \(\mathrm{x}= \pm \sqrt{-1}\) \(\mathrm{x}= \pm \mathrm{i} \quad\{\mathrm{i}=\sqrt{-1} \text { given }\}\)So, the four fourth root of \(1=\{1,-1, i,-i\}\)
SCRA-2015
Complex Numbers and Quadratic Equation
118247
If \(\alpha, \beta, \gamma, \delta\) are the roots of the equation \(x^4+x^3\) \(+\mathbf{x}^2+\mathbf{x}+1=0\), then \(\alpha^{2021]}+\beta^{2021]}+\gamma^{2021]}+\delta^{2021]}\) is equal to
1 -4
2 -1
3 1
4 4
Explanation:
B \(: \alpha, \beta, \gamma, \delta \text { root of the equation }\) \(x^4+x^3+x^2+x+1=0\) \(\text { which are } 5^{\text {th }} \text { roots of unity except } 1\) \(x^5=1\) \(\alpha^5=\beta^5=\gamma^5=\delta^5=1\) \(\text { then } \alpha^{2021}+\beta^{2021}+\gamma^{2021}+\delta^{2021}=\) \(\alpha+\beta+\gamma+\delta=-1\)