118130
If \(\alpha\) and \(\beta\) are the roots of \(x^2-2 x+4=0\), then the value of \(\alpha^6+\beta^6\) is
1 32
2 64
3 128
4 256
Explanation:
C \(\alpha^6+\beta^6=\left(\alpha^2\right)^3+\left(\beta^2\right)^3\) \(=\left(\alpha^2+\beta^2\right)\left(\left(\alpha^2+\beta^2\right)^2-3(\alpha \beta)^2\right)\) \(\text { We have the equation }\) \(x^2-2 x+4=0 \text { and } \alpha, \beta \text { are the roots of the eqn. }\) \(\therefore \alpha+\beta=2\) \(\alpha \beta=4\) \((\alpha+\beta)^2=\alpha^2+\beta^2+2 \alpha \beta\) \(\therefore \alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta=4-8=-4\) \(\therefore \alpha^6+\beta^6=-4\left((-4)^2-3\left(4^2\right)\right]\) \(=-4(16-48)\) \(=-4 \times(-32)\) \(=128\) \(\therefore \text { Option (c) is correct. }\)
AP EAMCET-2009
Complex Numbers and Quadratic Equation
118131
If \(\alpha\) and \(\beta\) are the roots of the equation \(a x^2+b x\) \(+\mathbf{c}=\mathbf{0}\) and, if \(\mathbf{p x}^2+\mathbf{q x}+\mathbf{r}=\mathbf{0}\) has roots \(\frac{1-\alpha}{\alpha}\) and \(\frac{1-\beta}{\beta}\), then \(r\) is equal to
1 \(a+2 b\)
2 \(a+b+c\)
3 \(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\)
4 abc
Explanation:
B \( \alpha \text { and } \beta \text { are the roots of }\) \(a x^2+b x+c=0\) \(\alpha+\beta=-b / a\) \(\alpha \beta=\mathrm{c} / \mathrm{a}\) \(\text { Given, } \mathrm{px}^2+\mathrm{qx}+\mathrm{r}=0 \text { has roots }\) \(\frac{1-\alpha}{\alpha} \text { and } \frac{1-\beta}{\beta}\) \(\Rightarrow \frac{1-\alpha}{\alpha}+\frac{1-\beta}{\beta}=\frac{-q}{p}\) \(\text { and }\left(\frac{1-\alpha}{\alpha}\right)\left(\frac{1-\beta}{\beta}\right)=\frac{\mathrm{r}}{\mathrm{p}}\) \(\text { or } \frac{1}{\alpha}+\frac{1}{\beta}=2-\frac{\mathrm{q}}{\mathrm{p}}\) \(\therefore \frac{(\alpha+\beta)}{\alpha \beta}=2-\frac{\mathrm{q}}{\mathrm{p}}=\frac{-\mathrm{b} / \mathrm{a}}{\mathrm{c} / \mathrm{a}}=\frac{-\mathrm{b}}{\mathrm{c}}\) \(\frac{(1-\alpha)(1-\beta)}{\alpha \beta}=\frac{1-\alpha-\beta+\alpha \beta}{\alpha \beta}\) \(=\frac{1-(\alpha+\beta)+\alpha \beta}{\alpha \beta}\) \(=\frac{1+\frac{\mathrm{b}}{\mathrm{a}}+\frac{\mathrm{c}}{\mathrm{a}}}{\frac{\mathrm{c}}{\mathrm{a}}}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{\frac{\mathrm{a}}{\mathrm{c}}}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{\mathrm{c}}\) \(\frac{r}{p}=\frac{a+b+c}{c}\) \(\mathrm{r}=\mathrm{a}+\mathrm{b}+\mathrm{c}\) From eq \({ }^{\text {n }}\) (i) \(\frac{1}{\alpha}-1+\frac{1}{\beta}-1=\frac{-q}{p}\) On comparing both side
AP EAMCET-2007
Complex Numbers and Quadratic Equation
118132
The set of values of \(x\) for which the inequalities \(\mathrm{x}^2-3 \mathrm{x}-10\lt 0, \quad 10 \mathrm{x}-\mathrm{x}^2-16>0\) hold simultaneously, is
118164
If one root of the equation \(x^2+p x+12=0\) is 4 while the equation \(x^2+\mathbf{p x}+\mathbf{q}=\mathbf{0}\) has equal roots, then the value of \(q\) is
1 \(\frac{49}{4}\)
2 12
3 3
4 4
Explanation:
A Given one root of the equation \(\mathrm{x}^2+\mathrm{px}+12=0\) is 4 \(\therefore \quad(4)^2+4 \mathrm{p}+12=0\) \(\Rightarrow \quad 4 \mathrm{p}=-12-16=-28\) \(\therefore \quad \mathrm{p}=\frac{-28}{4}=-7\) Also, given the equation \(\mathrm{x}^2+\mathrm{px}+\mathrm{q}=0\) has equal roots \(\Rightarrow \mathrm{D}=0,(\mathrm{p}\) is criminate \()=0\) i.e. \(\quad p^2-4 q=0\) \(\text { or }(-7)^2=4 q\) \(\therefore \quad \mathrm{q}=\frac{49}{4}\) \(\therefore \quad\) Option (a) is correct.
AIEEE-2004
Complex Numbers and Quadratic Equation
118133
If \(1,2,3\) and 4 are the roots of the equation \(x^4+\) \(\mathbf{a x}+\mathbf{b x}^2+\mathbf{c x}+\mathbf{d}=\mathbf{0}\), then \(a+2 b+c\) is equal to
1 -25
2 0
3 10
4 24
Explanation:
C Given the equation \(\mathrm{x}^4+a \mathrm{x}^3+b \mathrm{x}^2+\mathrm{cx}+\mathrm{d}=0\) If \(1,2,3\) and 4 are the roots of above Then, \(1+\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}=0\) \(16+8 \mathrm{a}+4 \mathrm{~b}+2 \mathrm{c}+\mathrm{d}=0\) \(81+27 a+a b+3 c+d=0\) \(256+64 a 16 b+4 c+d=0\) Subtracting (i) from (ii) we get \(15+7 a+3 b+c=0\) \(\text { Subtracting (ii) from (iii) }\) \(65+19 a+5 b+c=0\) Subtracting (iii) from (iv) \(175+37 \mathrm{a}+7 \mathrm{~b}+\mathrm{c}=0\) Subtracting (v) from (vi), we get \(50+12 a+2 b=0\) Subtracting (vi) from (vii), we get \(110+18 a+2 b=0\) Subtracting (viii) and (ix), we get \(60+6 a=0 \Rightarrow a=-10\) \(2 b=-110-18(-10)\) \(\text { or } 2 b=70\) \(\therefore b=35\) \(\therefore \text { From eqn (v) }\) \(15+7 x(-10)+3 \times 35+c=0\) \(\text { or } 15-70+105+c=0\) \(\text { or } c=-105-15+70\) \(=-120+70\) \(=-50\) \(\therefore a+2 b+c=-10+70-50\) \(=10\)\(\therefore\) Option (c) is correct.
118130
If \(\alpha\) and \(\beta\) are the roots of \(x^2-2 x+4=0\), then the value of \(\alpha^6+\beta^6\) is
1 32
2 64
3 128
4 256
Explanation:
C \(\alpha^6+\beta^6=\left(\alpha^2\right)^3+\left(\beta^2\right)^3\) \(=\left(\alpha^2+\beta^2\right)\left(\left(\alpha^2+\beta^2\right)^2-3(\alpha \beta)^2\right)\) \(\text { We have the equation }\) \(x^2-2 x+4=0 \text { and } \alpha, \beta \text { are the roots of the eqn. }\) \(\therefore \alpha+\beta=2\) \(\alpha \beta=4\) \((\alpha+\beta)^2=\alpha^2+\beta^2+2 \alpha \beta\) \(\therefore \alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta=4-8=-4\) \(\therefore \alpha^6+\beta^6=-4\left((-4)^2-3\left(4^2\right)\right]\) \(=-4(16-48)\) \(=-4 \times(-32)\) \(=128\) \(\therefore \text { Option (c) is correct. }\)
AP EAMCET-2009
Complex Numbers and Quadratic Equation
118131
If \(\alpha\) and \(\beta\) are the roots of the equation \(a x^2+b x\) \(+\mathbf{c}=\mathbf{0}\) and, if \(\mathbf{p x}^2+\mathbf{q x}+\mathbf{r}=\mathbf{0}\) has roots \(\frac{1-\alpha}{\alpha}\) and \(\frac{1-\beta}{\beta}\), then \(r\) is equal to
1 \(a+2 b\)
2 \(a+b+c\)
3 \(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\)
4 abc
Explanation:
B \( \alpha \text { and } \beta \text { are the roots of }\) \(a x^2+b x+c=0\) \(\alpha+\beta=-b / a\) \(\alpha \beta=\mathrm{c} / \mathrm{a}\) \(\text { Given, } \mathrm{px}^2+\mathrm{qx}+\mathrm{r}=0 \text { has roots }\) \(\frac{1-\alpha}{\alpha} \text { and } \frac{1-\beta}{\beta}\) \(\Rightarrow \frac{1-\alpha}{\alpha}+\frac{1-\beta}{\beta}=\frac{-q}{p}\) \(\text { and }\left(\frac{1-\alpha}{\alpha}\right)\left(\frac{1-\beta}{\beta}\right)=\frac{\mathrm{r}}{\mathrm{p}}\) \(\text { or } \frac{1}{\alpha}+\frac{1}{\beta}=2-\frac{\mathrm{q}}{\mathrm{p}}\) \(\therefore \frac{(\alpha+\beta)}{\alpha \beta}=2-\frac{\mathrm{q}}{\mathrm{p}}=\frac{-\mathrm{b} / \mathrm{a}}{\mathrm{c} / \mathrm{a}}=\frac{-\mathrm{b}}{\mathrm{c}}\) \(\frac{(1-\alpha)(1-\beta)}{\alpha \beta}=\frac{1-\alpha-\beta+\alpha \beta}{\alpha \beta}\) \(=\frac{1-(\alpha+\beta)+\alpha \beta}{\alpha \beta}\) \(=\frac{1+\frac{\mathrm{b}}{\mathrm{a}}+\frac{\mathrm{c}}{\mathrm{a}}}{\frac{\mathrm{c}}{\mathrm{a}}}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{\frac{\mathrm{a}}{\mathrm{c}}}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{\mathrm{c}}\) \(\frac{r}{p}=\frac{a+b+c}{c}\) \(\mathrm{r}=\mathrm{a}+\mathrm{b}+\mathrm{c}\) From eq \({ }^{\text {n }}\) (i) \(\frac{1}{\alpha}-1+\frac{1}{\beta}-1=\frac{-q}{p}\) On comparing both side
AP EAMCET-2007
Complex Numbers and Quadratic Equation
118132
The set of values of \(x\) for which the inequalities \(\mathrm{x}^2-3 \mathrm{x}-10\lt 0, \quad 10 \mathrm{x}-\mathrm{x}^2-16>0\) hold simultaneously, is
118164
If one root of the equation \(x^2+p x+12=0\) is 4 while the equation \(x^2+\mathbf{p x}+\mathbf{q}=\mathbf{0}\) has equal roots, then the value of \(q\) is
1 \(\frac{49}{4}\)
2 12
3 3
4 4
Explanation:
A Given one root of the equation \(\mathrm{x}^2+\mathrm{px}+12=0\) is 4 \(\therefore \quad(4)^2+4 \mathrm{p}+12=0\) \(\Rightarrow \quad 4 \mathrm{p}=-12-16=-28\) \(\therefore \quad \mathrm{p}=\frac{-28}{4}=-7\) Also, given the equation \(\mathrm{x}^2+\mathrm{px}+\mathrm{q}=0\) has equal roots \(\Rightarrow \mathrm{D}=0,(\mathrm{p}\) is criminate \()=0\) i.e. \(\quad p^2-4 q=0\) \(\text { or }(-7)^2=4 q\) \(\therefore \quad \mathrm{q}=\frac{49}{4}\) \(\therefore \quad\) Option (a) is correct.
AIEEE-2004
Complex Numbers and Quadratic Equation
118133
If \(1,2,3\) and 4 are the roots of the equation \(x^4+\) \(\mathbf{a x}+\mathbf{b x}^2+\mathbf{c x}+\mathbf{d}=\mathbf{0}\), then \(a+2 b+c\) is equal to
1 -25
2 0
3 10
4 24
Explanation:
C Given the equation \(\mathrm{x}^4+a \mathrm{x}^3+b \mathrm{x}^2+\mathrm{cx}+\mathrm{d}=0\) If \(1,2,3\) and 4 are the roots of above Then, \(1+\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}=0\) \(16+8 \mathrm{a}+4 \mathrm{~b}+2 \mathrm{c}+\mathrm{d}=0\) \(81+27 a+a b+3 c+d=0\) \(256+64 a 16 b+4 c+d=0\) Subtracting (i) from (ii) we get \(15+7 a+3 b+c=0\) \(\text { Subtracting (ii) from (iii) }\) \(65+19 a+5 b+c=0\) Subtracting (iii) from (iv) \(175+37 \mathrm{a}+7 \mathrm{~b}+\mathrm{c}=0\) Subtracting (v) from (vi), we get \(50+12 a+2 b=0\) Subtracting (vi) from (vii), we get \(110+18 a+2 b=0\) Subtracting (viii) and (ix), we get \(60+6 a=0 \Rightarrow a=-10\) \(2 b=-110-18(-10)\) \(\text { or } 2 b=70\) \(\therefore b=35\) \(\therefore \text { From eqn (v) }\) \(15+7 x(-10)+3 \times 35+c=0\) \(\text { or } 15-70+105+c=0\) \(\text { or } c=-105-15+70\) \(=-120+70\) \(=-50\) \(\therefore a+2 b+c=-10+70-50\) \(=10\)\(\therefore\) Option (c) is correct.
NEET Test Series from KOTA - 10 Papers In MS WORD
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Complex Numbers and Quadratic Equation
118130
If \(\alpha\) and \(\beta\) are the roots of \(x^2-2 x+4=0\), then the value of \(\alpha^6+\beta^6\) is
1 32
2 64
3 128
4 256
Explanation:
C \(\alpha^6+\beta^6=\left(\alpha^2\right)^3+\left(\beta^2\right)^3\) \(=\left(\alpha^2+\beta^2\right)\left(\left(\alpha^2+\beta^2\right)^2-3(\alpha \beta)^2\right)\) \(\text { We have the equation }\) \(x^2-2 x+4=0 \text { and } \alpha, \beta \text { are the roots of the eqn. }\) \(\therefore \alpha+\beta=2\) \(\alpha \beta=4\) \((\alpha+\beta)^2=\alpha^2+\beta^2+2 \alpha \beta\) \(\therefore \alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta=4-8=-4\) \(\therefore \alpha^6+\beta^6=-4\left((-4)^2-3\left(4^2\right)\right]\) \(=-4(16-48)\) \(=-4 \times(-32)\) \(=128\) \(\therefore \text { Option (c) is correct. }\)
AP EAMCET-2009
Complex Numbers and Quadratic Equation
118131
If \(\alpha\) and \(\beta\) are the roots of the equation \(a x^2+b x\) \(+\mathbf{c}=\mathbf{0}\) and, if \(\mathbf{p x}^2+\mathbf{q x}+\mathbf{r}=\mathbf{0}\) has roots \(\frac{1-\alpha}{\alpha}\) and \(\frac{1-\beta}{\beta}\), then \(r\) is equal to
1 \(a+2 b\)
2 \(a+b+c\)
3 \(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\)
4 abc
Explanation:
B \( \alpha \text { and } \beta \text { are the roots of }\) \(a x^2+b x+c=0\) \(\alpha+\beta=-b / a\) \(\alpha \beta=\mathrm{c} / \mathrm{a}\) \(\text { Given, } \mathrm{px}^2+\mathrm{qx}+\mathrm{r}=0 \text { has roots }\) \(\frac{1-\alpha}{\alpha} \text { and } \frac{1-\beta}{\beta}\) \(\Rightarrow \frac{1-\alpha}{\alpha}+\frac{1-\beta}{\beta}=\frac{-q}{p}\) \(\text { and }\left(\frac{1-\alpha}{\alpha}\right)\left(\frac{1-\beta}{\beta}\right)=\frac{\mathrm{r}}{\mathrm{p}}\) \(\text { or } \frac{1}{\alpha}+\frac{1}{\beta}=2-\frac{\mathrm{q}}{\mathrm{p}}\) \(\therefore \frac{(\alpha+\beta)}{\alpha \beta}=2-\frac{\mathrm{q}}{\mathrm{p}}=\frac{-\mathrm{b} / \mathrm{a}}{\mathrm{c} / \mathrm{a}}=\frac{-\mathrm{b}}{\mathrm{c}}\) \(\frac{(1-\alpha)(1-\beta)}{\alpha \beta}=\frac{1-\alpha-\beta+\alpha \beta}{\alpha \beta}\) \(=\frac{1-(\alpha+\beta)+\alpha \beta}{\alpha \beta}\) \(=\frac{1+\frac{\mathrm{b}}{\mathrm{a}}+\frac{\mathrm{c}}{\mathrm{a}}}{\frac{\mathrm{c}}{\mathrm{a}}}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{\frac{\mathrm{a}}{\mathrm{c}}}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{\mathrm{c}}\) \(\frac{r}{p}=\frac{a+b+c}{c}\) \(\mathrm{r}=\mathrm{a}+\mathrm{b}+\mathrm{c}\) From eq \({ }^{\text {n }}\) (i) \(\frac{1}{\alpha}-1+\frac{1}{\beta}-1=\frac{-q}{p}\) On comparing both side
AP EAMCET-2007
Complex Numbers and Quadratic Equation
118132
The set of values of \(x\) for which the inequalities \(\mathrm{x}^2-3 \mathrm{x}-10\lt 0, \quad 10 \mathrm{x}-\mathrm{x}^2-16>0\) hold simultaneously, is
118164
If one root of the equation \(x^2+p x+12=0\) is 4 while the equation \(x^2+\mathbf{p x}+\mathbf{q}=\mathbf{0}\) has equal roots, then the value of \(q\) is
1 \(\frac{49}{4}\)
2 12
3 3
4 4
Explanation:
A Given one root of the equation \(\mathrm{x}^2+\mathrm{px}+12=0\) is 4 \(\therefore \quad(4)^2+4 \mathrm{p}+12=0\) \(\Rightarrow \quad 4 \mathrm{p}=-12-16=-28\) \(\therefore \quad \mathrm{p}=\frac{-28}{4}=-7\) Also, given the equation \(\mathrm{x}^2+\mathrm{px}+\mathrm{q}=0\) has equal roots \(\Rightarrow \mathrm{D}=0,(\mathrm{p}\) is criminate \()=0\) i.e. \(\quad p^2-4 q=0\) \(\text { or }(-7)^2=4 q\) \(\therefore \quad \mathrm{q}=\frac{49}{4}\) \(\therefore \quad\) Option (a) is correct.
AIEEE-2004
Complex Numbers and Quadratic Equation
118133
If \(1,2,3\) and 4 are the roots of the equation \(x^4+\) \(\mathbf{a x}+\mathbf{b x}^2+\mathbf{c x}+\mathbf{d}=\mathbf{0}\), then \(a+2 b+c\) is equal to
1 -25
2 0
3 10
4 24
Explanation:
C Given the equation \(\mathrm{x}^4+a \mathrm{x}^3+b \mathrm{x}^2+\mathrm{cx}+\mathrm{d}=0\) If \(1,2,3\) and 4 are the roots of above Then, \(1+\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}=0\) \(16+8 \mathrm{a}+4 \mathrm{~b}+2 \mathrm{c}+\mathrm{d}=0\) \(81+27 a+a b+3 c+d=0\) \(256+64 a 16 b+4 c+d=0\) Subtracting (i) from (ii) we get \(15+7 a+3 b+c=0\) \(\text { Subtracting (ii) from (iii) }\) \(65+19 a+5 b+c=0\) Subtracting (iii) from (iv) \(175+37 \mathrm{a}+7 \mathrm{~b}+\mathrm{c}=0\) Subtracting (v) from (vi), we get \(50+12 a+2 b=0\) Subtracting (vi) from (vii), we get \(110+18 a+2 b=0\) Subtracting (viii) and (ix), we get \(60+6 a=0 \Rightarrow a=-10\) \(2 b=-110-18(-10)\) \(\text { or } 2 b=70\) \(\therefore b=35\) \(\therefore \text { From eqn (v) }\) \(15+7 x(-10)+3 \times 35+c=0\) \(\text { or } 15-70+105+c=0\) \(\text { or } c=-105-15+70\) \(=-120+70\) \(=-50\) \(\therefore a+2 b+c=-10+70-50\) \(=10\)\(\therefore\) Option (c) is correct.
118130
If \(\alpha\) and \(\beta\) are the roots of \(x^2-2 x+4=0\), then the value of \(\alpha^6+\beta^6\) is
1 32
2 64
3 128
4 256
Explanation:
C \(\alpha^6+\beta^6=\left(\alpha^2\right)^3+\left(\beta^2\right)^3\) \(=\left(\alpha^2+\beta^2\right)\left(\left(\alpha^2+\beta^2\right)^2-3(\alpha \beta)^2\right)\) \(\text { We have the equation }\) \(x^2-2 x+4=0 \text { and } \alpha, \beta \text { are the roots of the eqn. }\) \(\therefore \alpha+\beta=2\) \(\alpha \beta=4\) \((\alpha+\beta)^2=\alpha^2+\beta^2+2 \alpha \beta\) \(\therefore \alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta=4-8=-4\) \(\therefore \alpha^6+\beta^6=-4\left((-4)^2-3\left(4^2\right)\right]\) \(=-4(16-48)\) \(=-4 \times(-32)\) \(=128\) \(\therefore \text { Option (c) is correct. }\)
AP EAMCET-2009
Complex Numbers and Quadratic Equation
118131
If \(\alpha\) and \(\beta\) are the roots of the equation \(a x^2+b x\) \(+\mathbf{c}=\mathbf{0}\) and, if \(\mathbf{p x}^2+\mathbf{q x}+\mathbf{r}=\mathbf{0}\) has roots \(\frac{1-\alpha}{\alpha}\) and \(\frac{1-\beta}{\beta}\), then \(r\) is equal to
1 \(a+2 b\)
2 \(a+b+c\)
3 \(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\)
4 abc
Explanation:
B \( \alpha \text { and } \beta \text { are the roots of }\) \(a x^2+b x+c=0\) \(\alpha+\beta=-b / a\) \(\alpha \beta=\mathrm{c} / \mathrm{a}\) \(\text { Given, } \mathrm{px}^2+\mathrm{qx}+\mathrm{r}=0 \text { has roots }\) \(\frac{1-\alpha}{\alpha} \text { and } \frac{1-\beta}{\beta}\) \(\Rightarrow \frac{1-\alpha}{\alpha}+\frac{1-\beta}{\beta}=\frac{-q}{p}\) \(\text { and }\left(\frac{1-\alpha}{\alpha}\right)\left(\frac{1-\beta}{\beta}\right)=\frac{\mathrm{r}}{\mathrm{p}}\) \(\text { or } \frac{1}{\alpha}+\frac{1}{\beta}=2-\frac{\mathrm{q}}{\mathrm{p}}\) \(\therefore \frac{(\alpha+\beta)}{\alpha \beta}=2-\frac{\mathrm{q}}{\mathrm{p}}=\frac{-\mathrm{b} / \mathrm{a}}{\mathrm{c} / \mathrm{a}}=\frac{-\mathrm{b}}{\mathrm{c}}\) \(\frac{(1-\alpha)(1-\beta)}{\alpha \beta}=\frac{1-\alpha-\beta+\alpha \beta}{\alpha \beta}\) \(=\frac{1-(\alpha+\beta)+\alpha \beta}{\alpha \beta}\) \(=\frac{1+\frac{\mathrm{b}}{\mathrm{a}}+\frac{\mathrm{c}}{\mathrm{a}}}{\frac{\mathrm{c}}{\mathrm{a}}}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{\frac{\mathrm{a}}{\mathrm{c}}}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{\mathrm{c}}\) \(\frac{r}{p}=\frac{a+b+c}{c}\) \(\mathrm{r}=\mathrm{a}+\mathrm{b}+\mathrm{c}\) From eq \({ }^{\text {n }}\) (i) \(\frac{1}{\alpha}-1+\frac{1}{\beta}-1=\frac{-q}{p}\) On comparing both side
AP EAMCET-2007
Complex Numbers and Quadratic Equation
118132
The set of values of \(x\) for which the inequalities \(\mathrm{x}^2-3 \mathrm{x}-10\lt 0, \quad 10 \mathrm{x}-\mathrm{x}^2-16>0\) hold simultaneously, is
118164
If one root of the equation \(x^2+p x+12=0\) is 4 while the equation \(x^2+\mathbf{p x}+\mathbf{q}=\mathbf{0}\) has equal roots, then the value of \(q\) is
1 \(\frac{49}{4}\)
2 12
3 3
4 4
Explanation:
A Given one root of the equation \(\mathrm{x}^2+\mathrm{px}+12=0\) is 4 \(\therefore \quad(4)^2+4 \mathrm{p}+12=0\) \(\Rightarrow \quad 4 \mathrm{p}=-12-16=-28\) \(\therefore \quad \mathrm{p}=\frac{-28}{4}=-7\) Also, given the equation \(\mathrm{x}^2+\mathrm{px}+\mathrm{q}=0\) has equal roots \(\Rightarrow \mathrm{D}=0,(\mathrm{p}\) is criminate \()=0\) i.e. \(\quad p^2-4 q=0\) \(\text { or }(-7)^2=4 q\) \(\therefore \quad \mathrm{q}=\frac{49}{4}\) \(\therefore \quad\) Option (a) is correct.
AIEEE-2004
Complex Numbers and Quadratic Equation
118133
If \(1,2,3\) and 4 are the roots of the equation \(x^4+\) \(\mathbf{a x}+\mathbf{b x}^2+\mathbf{c x}+\mathbf{d}=\mathbf{0}\), then \(a+2 b+c\) is equal to
1 -25
2 0
3 10
4 24
Explanation:
C Given the equation \(\mathrm{x}^4+a \mathrm{x}^3+b \mathrm{x}^2+\mathrm{cx}+\mathrm{d}=0\) If \(1,2,3\) and 4 are the roots of above Then, \(1+\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}=0\) \(16+8 \mathrm{a}+4 \mathrm{~b}+2 \mathrm{c}+\mathrm{d}=0\) \(81+27 a+a b+3 c+d=0\) \(256+64 a 16 b+4 c+d=0\) Subtracting (i) from (ii) we get \(15+7 a+3 b+c=0\) \(\text { Subtracting (ii) from (iii) }\) \(65+19 a+5 b+c=0\) Subtracting (iii) from (iv) \(175+37 \mathrm{a}+7 \mathrm{~b}+\mathrm{c}=0\) Subtracting (v) from (vi), we get \(50+12 a+2 b=0\) Subtracting (vi) from (vii), we get \(110+18 a+2 b=0\) Subtracting (viii) and (ix), we get \(60+6 a=0 \Rightarrow a=-10\) \(2 b=-110-18(-10)\) \(\text { or } 2 b=70\) \(\therefore b=35\) \(\therefore \text { From eqn (v) }\) \(15+7 x(-10)+3 \times 35+c=0\) \(\text { or } 15-70+105+c=0\) \(\text { or } c=-105-15+70\) \(=-120+70\) \(=-50\) \(\therefore a+2 b+c=-10+70-50\) \(=10\)\(\therefore\) Option (c) is correct.
118130
If \(\alpha\) and \(\beta\) are the roots of \(x^2-2 x+4=0\), then the value of \(\alpha^6+\beta^6\) is
1 32
2 64
3 128
4 256
Explanation:
C \(\alpha^6+\beta^6=\left(\alpha^2\right)^3+\left(\beta^2\right)^3\) \(=\left(\alpha^2+\beta^2\right)\left(\left(\alpha^2+\beta^2\right)^2-3(\alpha \beta)^2\right)\) \(\text { We have the equation }\) \(x^2-2 x+4=0 \text { and } \alpha, \beta \text { are the roots of the eqn. }\) \(\therefore \alpha+\beta=2\) \(\alpha \beta=4\) \((\alpha+\beta)^2=\alpha^2+\beta^2+2 \alpha \beta\) \(\therefore \alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta=4-8=-4\) \(\therefore \alpha^6+\beta^6=-4\left((-4)^2-3\left(4^2\right)\right]\) \(=-4(16-48)\) \(=-4 \times(-32)\) \(=128\) \(\therefore \text { Option (c) is correct. }\)
AP EAMCET-2009
Complex Numbers and Quadratic Equation
118131
If \(\alpha\) and \(\beta\) are the roots of the equation \(a x^2+b x\) \(+\mathbf{c}=\mathbf{0}\) and, if \(\mathbf{p x}^2+\mathbf{q x}+\mathbf{r}=\mathbf{0}\) has roots \(\frac{1-\alpha}{\alpha}\) and \(\frac{1-\beta}{\beta}\), then \(r\) is equal to
1 \(a+2 b\)
2 \(a+b+c\)
3 \(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\)
4 abc
Explanation:
B \( \alpha \text { and } \beta \text { are the roots of }\) \(a x^2+b x+c=0\) \(\alpha+\beta=-b / a\) \(\alpha \beta=\mathrm{c} / \mathrm{a}\) \(\text { Given, } \mathrm{px}^2+\mathrm{qx}+\mathrm{r}=0 \text { has roots }\) \(\frac{1-\alpha}{\alpha} \text { and } \frac{1-\beta}{\beta}\) \(\Rightarrow \frac{1-\alpha}{\alpha}+\frac{1-\beta}{\beta}=\frac{-q}{p}\) \(\text { and }\left(\frac{1-\alpha}{\alpha}\right)\left(\frac{1-\beta}{\beta}\right)=\frac{\mathrm{r}}{\mathrm{p}}\) \(\text { or } \frac{1}{\alpha}+\frac{1}{\beta}=2-\frac{\mathrm{q}}{\mathrm{p}}\) \(\therefore \frac{(\alpha+\beta)}{\alpha \beta}=2-\frac{\mathrm{q}}{\mathrm{p}}=\frac{-\mathrm{b} / \mathrm{a}}{\mathrm{c} / \mathrm{a}}=\frac{-\mathrm{b}}{\mathrm{c}}\) \(\frac{(1-\alpha)(1-\beta)}{\alpha \beta}=\frac{1-\alpha-\beta+\alpha \beta}{\alpha \beta}\) \(=\frac{1-(\alpha+\beta)+\alpha \beta}{\alpha \beta}\) \(=\frac{1+\frac{\mathrm{b}}{\mathrm{a}}+\frac{\mathrm{c}}{\mathrm{a}}}{\frac{\mathrm{c}}{\mathrm{a}}}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{\frac{\mathrm{a}}{\mathrm{c}}}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{\mathrm{c}}\) \(\frac{r}{p}=\frac{a+b+c}{c}\) \(\mathrm{r}=\mathrm{a}+\mathrm{b}+\mathrm{c}\) From eq \({ }^{\text {n }}\) (i) \(\frac{1}{\alpha}-1+\frac{1}{\beta}-1=\frac{-q}{p}\) On comparing both side
AP EAMCET-2007
Complex Numbers and Quadratic Equation
118132
The set of values of \(x\) for which the inequalities \(\mathrm{x}^2-3 \mathrm{x}-10\lt 0, \quad 10 \mathrm{x}-\mathrm{x}^2-16>0\) hold simultaneously, is
118164
If one root of the equation \(x^2+p x+12=0\) is 4 while the equation \(x^2+\mathbf{p x}+\mathbf{q}=\mathbf{0}\) has equal roots, then the value of \(q\) is
1 \(\frac{49}{4}\)
2 12
3 3
4 4
Explanation:
A Given one root of the equation \(\mathrm{x}^2+\mathrm{px}+12=0\) is 4 \(\therefore \quad(4)^2+4 \mathrm{p}+12=0\) \(\Rightarrow \quad 4 \mathrm{p}=-12-16=-28\) \(\therefore \quad \mathrm{p}=\frac{-28}{4}=-7\) Also, given the equation \(\mathrm{x}^2+\mathrm{px}+\mathrm{q}=0\) has equal roots \(\Rightarrow \mathrm{D}=0,(\mathrm{p}\) is criminate \()=0\) i.e. \(\quad p^2-4 q=0\) \(\text { or }(-7)^2=4 q\) \(\therefore \quad \mathrm{q}=\frac{49}{4}\) \(\therefore \quad\) Option (a) is correct.
AIEEE-2004
Complex Numbers and Quadratic Equation
118133
If \(1,2,3\) and 4 are the roots of the equation \(x^4+\) \(\mathbf{a x}+\mathbf{b x}^2+\mathbf{c x}+\mathbf{d}=\mathbf{0}\), then \(a+2 b+c\) is equal to
1 -25
2 0
3 10
4 24
Explanation:
C Given the equation \(\mathrm{x}^4+a \mathrm{x}^3+b \mathrm{x}^2+\mathrm{cx}+\mathrm{d}=0\) If \(1,2,3\) and 4 are the roots of above Then, \(1+\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}=0\) \(16+8 \mathrm{a}+4 \mathrm{~b}+2 \mathrm{c}+\mathrm{d}=0\) \(81+27 a+a b+3 c+d=0\) \(256+64 a 16 b+4 c+d=0\) Subtracting (i) from (ii) we get \(15+7 a+3 b+c=0\) \(\text { Subtracting (ii) from (iii) }\) \(65+19 a+5 b+c=0\) Subtracting (iii) from (iv) \(175+37 \mathrm{a}+7 \mathrm{~b}+\mathrm{c}=0\) Subtracting (v) from (vi), we get \(50+12 a+2 b=0\) Subtracting (vi) from (vii), we get \(110+18 a+2 b=0\) Subtracting (viii) and (ix), we get \(60+6 a=0 \Rightarrow a=-10\) \(2 b=-110-18(-10)\) \(\text { or } 2 b=70\) \(\therefore b=35\) \(\therefore \text { From eqn (v) }\) \(15+7 x(-10)+3 \times 35+c=0\) \(\text { or } 15-70+105+c=0\) \(\text { or } c=-105-15+70\) \(=-120+70\) \(=-50\) \(\therefore a+2 b+c=-10+70-50\) \(=10\)\(\therefore\) Option (c) is correct.