118150
The number of solutions for the equation \(x^2-5\) \(|x|+6=0\) is
1 4
2 3
3 2
4 1
Explanation:
A Given the equation. \(\mathrm{x}^2-5|\mathrm{x}|+6=0\) case \(1:-\mathrm{x}>0 \Rightarrow|\mathrm{x}|=\mathrm{x}\) \(\therefore \mathrm{x}^2-5 \mathrm{x}+6=0\) \(\mathrm{x}^2-3 \mathrm{x}-2 \mathrm{x}+6=0\) \(\mathrm{x}(\mathrm{x}-3)-2(\mathrm{x}-3)=0\) \((\mathrm{x}-2)(\mathrm{x}-3)=0\) \(\mathrm{x}=2,3\) \(\therefore\) case 2 : \(\mathrm{x}\lt 0 \Rightarrow|\mathrm{x}|=-\mathrm{x}\), then the given equation becomes. \(\mathrm{x}^2-5(-\mathrm{x})+6=0\) \(\Rightarrow \mathrm{x}^2+5 \mathrm{x}+6=0\) \(\Rightarrow \mathrm{x}^2+3 \mathrm{x}+2 \mathrm{x}+6=0\) \(\Rightarrow \mathrm{x}(\mathrm{x}+3)+2(\mathrm{x}+3)=0\) \((\mathrm{x}+2)(\mathrm{x}+3)=0\) \(\Rightarrow \mathrm{x}=-2,-3\) So, the number of solutions is 4
AP EAMCET-22.09.2020
Complex Numbers and Quadratic Equation
118151
The roots of the equation \(x^3-6 x^2+3 x+10=0\) are
1 in geometric progression
2 in arithmetic progression
3 in harmonic progression
4 Positive
Explanation:
B Given the equation \(\mathrm{x}^3-6 \mathrm{x}^2+3 \mathrm{x}+10=0\) has roots \(\alpha, \beta, \gamma\) where \(\alpha+\beta+\gamma=6\) \(\alpha \beta+\beta \gamma+\alpha \gamma=3\) \(\alpha \beta \gamma=-10\) \(\alpha, \beta, \gamma\) are in A. P if \(\beta=\frac{\alpha+\gamma}{2}\) From equation (i) \(\beta+2 \beta=6 \Rightarrow \beta=2\) From equation (iii) \(\alpha \beta . \gamma=-10\) \(\Rightarrow \alpha \gamma=\frac{-10}{2}=-5\) Now from equation to, \((\alpha+\gamma) \beta+\alpha \gamma=3\) \(4 \times 2+(-5)=3\), which satisfy equation (ii) \(\therefore \alpha, \beta, \gamma\) are in A.P
AP EAMCET-05.10.2021
Complex Numbers and Quadratic Equation
118152
The number of distinct real solutions for the equation \(\left|x^2+2 x-8\right|+x-2=0\) is
1 1
2 2
3 3
4 4
Explanation:
C Given the equation\(\left|x^2+2 x-8\right|+x-2=0\) \(\text { case } 1: x^2+2 x-8>0 \Rightarrow\left|x^2+2 x-8\right|=x^2+2 x-8\) \(x^2+2 x-8+x-2=0\) \(x^2+3 x-10=0\) \(\Rightarrow x^2+5 x-2 x-10=0\) \(\Rightarrow \mathrm{x}(\mathrm{x}+5)-2(\mathrm{x}+5)=0right.\)Ans: c Exp: (C) : Given the equation \(\therefore(\mathrm{x}-2)(\mathrm{x}+5)=0 \Rightarrow \mathrm{x}=(2,-5)\) \(\text { Now, } \mathrm{x}^2+2 \mathrm{x}-8>0\) \(\Rightarrow \mathrm{x}^2+4 \mathrm{x}-2 \mathrm{x}-8>0\) \(\Rightarrow \mathrm{x}(\mathrm{x}+4)-2(\mathrm{x}+4)>0 \Rightarrow(\mathrm{x}-2)(\mathrm{x}+4)>0\) \(\Rightarrow \mathrm{x}>2 \text { and } \mathrm{x}\lt -4\) \(\therefore 2,-5 \text { are the solutions }\) \(\text { case } 2:-\mathrm{x}^2+2 \mathrm{x}-8\lt 0\) \(\Rightarrow-4\lt \mathrm{x}\lt 2 \text { and }\left|\mathrm{x}^2+2 \mathrm{x}-8\right|=-\left(\mathrm{x}^2+2 \mathrm{x}-8\right)\) \(\therefore \text { The given equation becomes, }\) \(-\left(\mathrm{x}^2+2 \mathrm{x}-8\right)+\mathrm{x}-2=0\) \(0 \mathrm{r}-\mathrm{x}^2-2 \mathrm{x}+8+\mathrm{x}-2=0\) \(\mathrm{x}^2+\mathrm{x}-6=0\) \(\Rightarrow \mathrm{x}^2+3 \mathrm{x}-2 \mathrm{x}-6=0\) \(\Rightarrow \mathrm{x}(\mathrm{x}+3)-2(\mathrm{x}+3)=0\) \(\Rightarrow(\mathrm{x}+3)(\mathrm{x}-2)=0\) \(\therefore \mathrm{x}=2,-3\) \(\therefore 2,-3 \text { are the solutions. }\) \(\therefore \text { The solution are }\) \(2,-3,-5\) \(\therefore \text { The total No. of solutions are } 3 .\)\(\therefore\) The total No. of solutions are 3 .
AP EAMCET-24.04.2018
Complex Numbers and Quadratic Equation
118153
If \(x\) is real, then the sum of the maximum and the minimum values of the expression \(\frac{\mathrm{x}^2+4 \mathrm{x}+1}{\mathrm{x}^2+\mathrm{x}+1}\) is
1 -2
2 2
3 1
4 0
Explanation:
D Let, \(y=\frac{x^2+4 x+1}{x^2+x+1}\) \(\Rightarrow y x^2+y x+y=x^2+4 x+1\) \(\Rightarrow(y-1) x^2+(y-4) x+y-1=0\) \(\because \mathrm{x}\) is real, therefore. \((y-4)^2-4(y-1)(y-1) \geq 0\) \(\text { or } y^2+16-8 y-4\left(y^2+1-2 y\right) \geq 0\) \(-3 y^2+12 \geq 0\) \(3 y^2 \leq 12\) \(\text { or }-3 y^2 \leq 12\) \(\text { or } 3 y^2-12 \leq 0\) \(3\left(y^2-4\right) \leq 0 \Rightarrow(y-2)(y+2) \leq 0\) \(\Rightarrow-2 \leq y \leq 2\) \(\therefore\) The minimum value is -2 and max value is 2 Sum of minimum and maximum, \(-2+2=0\)So, option (d) is correct.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Complex Numbers and Quadratic Equation
118150
The number of solutions for the equation \(x^2-5\) \(|x|+6=0\) is
1 4
2 3
3 2
4 1
Explanation:
A Given the equation. \(\mathrm{x}^2-5|\mathrm{x}|+6=0\) case \(1:-\mathrm{x}>0 \Rightarrow|\mathrm{x}|=\mathrm{x}\) \(\therefore \mathrm{x}^2-5 \mathrm{x}+6=0\) \(\mathrm{x}^2-3 \mathrm{x}-2 \mathrm{x}+6=0\) \(\mathrm{x}(\mathrm{x}-3)-2(\mathrm{x}-3)=0\) \((\mathrm{x}-2)(\mathrm{x}-3)=0\) \(\mathrm{x}=2,3\) \(\therefore\) case 2 : \(\mathrm{x}\lt 0 \Rightarrow|\mathrm{x}|=-\mathrm{x}\), then the given equation becomes. \(\mathrm{x}^2-5(-\mathrm{x})+6=0\) \(\Rightarrow \mathrm{x}^2+5 \mathrm{x}+6=0\) \(\Rightarrow \mathrm{x}^2+3 \mathrm{x}+2 \mathrm{x}+6=0\) \(\Rightarrow \mathrm{x}(\mathrm{x}+3)+2(\mathrm{x}+3)=0\) \((\mathrm{x}+2)(\mathrm{x}+3)=0\) \(\Rightarrow \mathrm{x}=-2,-3\) So, the number of solutions is 4
AP EAMCET-22.09.2020
Complex Numbers and Quadratic Equation
118151
The roots of the equation \(x^3-6 x^2+3 x+10=0\) are
1 in geometric progression
2 in arithmetic progression
3 in harmonic progression
4 Positive
Explanation:
B Given the equation \(\mathrm{x}^3-6 \mathrm{x}^2+3 \mathrm{x}+10=0\) has roots \(\alpha, \beta, \gamma\) where \(\alpha+\beta+\gamma=6\) \(\alpha \beta+\beta \gamma+\alpha \gamma=3\) \(\alpha \beta \gamma=-10\) \(\alpha, \beta, \gamma\) are in A. P if \(\beta=\frac{\alpha+\gamma}{2}\) From equation (i) \(\beta+2 \beta=6 \Rightarrow \beta=2\) From equation (iii) \(\alpha \beta . \gamma=-10\) \(\Rightarrow \alpha \gamma=\frac{-10}{2}=-5\) Now from equation to, \((\alpha+\gamma) \beta+\alpha \gamma=3\) \(4 \times 2+(-5)=3\), which satisfy equation (ii) \(\therefore \alpha, \beta, \gamma\) are in A.P
AP EAMCET-05.10.2021
Complex Numbers and Quadratic Equation
118152
The number of distinct real solutions for the equation \(\left|x^2+2 x-8\right|+x-2=0\) is
1 1
2 2
3 3
4 4
Explanation:
C Given the equation\(\left|x^2+2 x-8\right|+x-2=0\) \(\text { case } 1: x^2+2 x-8>0 \Rightarrow\left|x^2+2 x-8\right|=x^2+2 x-8\) \(x^2+2 x-8+x-2=0\) \(x^2+3 x-10=0\) \(\Rightarrow x^2+5 x-2 x-10=0\) \(\Rightarrow \mathrm{x}(\mathrm{x}+5)-2(\mathrm{x}+5)=0right.\)Ans: c Exp: (C) : Given the equation \(\therefore(\mathrm{x}-2)(\mathrm{x}+5)=0 \Rightarrow \mathrm{x}=(2,-5)\) \(\text { Now, } \mathrm{x}^2+2 \mathrm{x}-8>0\) \(\Rightarrow \mathrm{x}^2+4 \mathrm{x}-2 \mathrm{x}-8>0\) \(\Rightarrow \mathrm{x}(\mathrm{x}+4)-2(\mathrm{x}+4)>0 \Rightarrow(\mathrm{x}-2)(\mathrm{x}+4)>0\) \(\Rightarrow \mathrm{x}>2 \text { and } \mathrm{x}\lt -4\) \(\therefore 2,-5 \text { are the solutions }\) \(\text { case } 2:-\mathrm{x}^2+2 \mathrm{x}-8\lt 0\) \(\Rightarrow-4\lt \mathrm{x}\lt 2 \text { and }\left|\mathrm{x}^2+2 \mathrm{x}-8\right|=-\left(\mathrm{x}^2+2 \mathrm{x}-8\right)\) \(\therefore \text { The given equation becomes, }\) \(-\left(\mathrm{x}^2+2 \mathrm{x}-8\right)+\mathrm{x}-2=0\) \(0 \mathrm{r}-\mathrm{x}^2-2 \mathrm{x}+8+\mathrm{x}-2=0\) \(\mathrm{x}^2+\mathrm{x}-6=0\) \(\Rightarrow \mathrm{x}^2+3 \mathrm{x}-2 \mathrm{x}-6=0\) \(\Rightarrow \mathrm{x}(\mathrm{x}+3)-2(\mathrm{x}+3)=0\) \(\Rightarrow(\mathrm{x}+3)(\mathrm{x}-2)=0\) \(\therefore \mathrm{x}=2,-3\) \(\therefore 2,-3 \text { are the solutions. }\) \(\therefore \text { The solution are }\) \(2,-3,-5\) \(\therefore \text { The total No. of solutions are } 3 .\)\(\therefore\) The total No. of solutions are 3 .
AP EAMCET-24.04.2018
Complex Numbers and Quadratic Equation
118153
If \(x\) is real, then the sum of the maximum and the minimum values of the expression \(\frac{\mathrm{x}^2+4 \mathrm{x}+1}{\mathrm{x}^2+\mathrm{x}+1}\) is
1 -2
2 2
3 1
4 0
Explanation:
D Let, \(y=\frac{x^2+4 x+1}{x^2+x+1}\) \(\Rightarrow y x^2+y x+y=x^2+4 x+1\) \(\Rightarrow(y-1) x^2+(y-4) x+y-1=0\) \(\because \mathrm{x}\) is real, therefore. \((y-4)^2-4(y-1)(y-1) \geq 0\) \(\text { or } y^2+16-8 y-4\left(y^2+1-2 y\right) \geq 0\) \(-3 y^2+12 \geq 0\) \(3 y^2 \leq 12\) \(\text { or }-3 y^2 \leq 12\) \(\text { or } 3 y^2-12 \leq 0\) \(3\left(y^2-4\right) \leq 0 \Rightarrow(y-2)(y+2) \leq 0\) \(\Rightarrow-2 \leq y \leq 2\) \(\therefore\) The minimum value is -2 and max value is 2 Sum of minimum and maximum, \(-2+2=0\)So, option (d) is correct.
118150
The number of solutions for the equation \(x^2-5\) \(|x|+6=0\) is
1 4
2 3
3 2
4 1
Explanation:
A Given the equation. \(\mathrm{x}^2-5|\mathrm{x}|+6=0\) case \(1:-\mathrm{x}>0 \Rightarrow|\mathrm{x}|=\mathrm{x}\) \(\therefore \mathrm{x}^2-5 \mathrm{x}+6=0\) \(\mathrm{x}^2-3 \mathrm{x}-2 \mathrm{x}+6=0\) \(\mathrm{x}(\mathrm{x}-3)-2(\mathrm{x}-3)=0\) \((\mathrm{x}-2)(\mathrm{x}-3)=0\) \(\mathrm{x}=2,3\) \(\therefore\) case 2 : \(\mathrm{x}\lt 0 \Rightarrow|\mathrm{x}|=-\mathrm{x}\), then the given equation becomes. \(\mathrm{x}^2-5(-\mathrm{x})+6=0\) \(\Rightarrow \mathrm{x}^2+5 \mathrm{x}+6=0\) \(\Rightarrow \mathrm{x}^2+3 \mathrm{x}+2 \mathrm{x}+6=0\) \(\Rightarrow \mathrm{x}(\mathrm{x}+3)+2(\mathrm{x}+3)=0\) \((\mathrm{x}+2)(\mathrm{x}+3)=0\) \(\Rightarrow \mathrm{x}=-2,-3\) So, the number of solutions is 4
AP EAMCET-22.09.2020
Complex Numbers and Quadratic Equation
118151
The roots of the equation \(x^3-6 x^2+3 x+10=0\) are
1 in geometric progression
2 in arithmetic progression
3 in harmonic progression
4 Positive
Explanation:
B Given the equation \(\mathrm{x}^3-6 \mathrm{x}^2+3 \mathrm{x}+10=0\) has roots \(\alpha, \beta, \gamma\) where \(\alpha+\beta+\gamma=6\) \(\alpha \beta+\beta \gamma+\alpha \gamma=3\) \(\alpha \beta \gamma=-10\) \(\alpha, \beta, \gamma\) are in A. P if \(\beta=\frac{\alpha+\gamma}{2}\) From equation (i) \(\beta+2 \beta=6 \Rightarrow \beta=2\) From equation (iii) \(\alpha \beta . \gamma=-10\) \(\Rightarrow \alpha \gamma=\frac{-10}{2}=-5\) Now from equation to, \((\alpha+\gamma) \beta+\alpha \gamma=3\) \(4 \times 2+(-5)=3\), which satisfy equation (ii) \(\therefore \alpha, \beta, \gamma\) are in A.P
AP EAMCET-05.10.2021
Complex Numbers and Quadratic Equation
118152
The number of distinct real solutions for the equation \(\left|x^2+2 x-8\right|+x-2=0\) is
1 1
2 2
3 3
4 4
Explanation:
C Given the equation\(\left|x^2+2 x-8\right|+x-2=0\) \(\text { case } 1: x^2+2 x-8>0 \Rightarrow\left|x^2+2 x-8\right|=x^2+2 x-8\) \(x^2+2 x-8+x-2=0\) \(x^2+3 x-10=0\) \(\Rightarrow x^2+5 x-2 x-10=0\) \(\Rightarrow \mathrm{x}(\mathrm{x}+5)-2(\mathrm{x}+5)=0right.\)Ans: c Exp: (C) : Given the equation \(\therefore(\mathrm{x}-2)(\mathrm{x}+5)=0 \Rightarrow \mathrm{x}=(2,-5)\) \(\text { Now, } \mathrm{x}^2+2 \mathrm{x}-8>0\) \(\Rightarrow \mathrm{x}^2+4 \mathrm{x}-2 \mathrm{x}-8>0\) \(\Rightarrow \mathrm{x}(\mathrm{x}+4)-2(\mathrm{x}+4)>0 \Rightarrow(\mathrm{x}-2)(\mathrm{x}+4)>0\) \(\Rightarrow \mathrm{x}>2 \text { and } \mathrm{x}\lt -4\) \(\therefore 2,-5 \text { are the solutions }\) \(\text { case } 2:-\mathrm{x}^2+2 \mathrm{x}-8\lt 0\) \(\Rightarrow-4\lt \mathrm{x}\lt 2 \text { and }\left|\mathrm{x}^2+2 \mathrm{x}-8\right|=-\left(\mathrm{x}^2+2 \mathrm{x}-8\right)\) \(\therefore \text { The given equation becomes, }\) \(-\left(\mathrm{x}^2+2 \mathrm{x}-8\right)+\mathrm{x}-2=0\) \(0 \mathrm{r}-\mathrm{x}^2-2 \mathrm{x}+8+\mathrm{x}-2=0\) \(\mathrm{x}^2+\mathrm{x}-6=0\) \(\Rightarrow \mathrm{x}^2+3 \mathrm{x}-2 \mathrm{x}-6=0\) \(\Rightarrow \mathrm{x}(\mathrm{x}+3)-2(\mathrm{x}+3)=0\) \(\Rightarrow(\mathrm{x}+3)(\mathrm{x}-2)=0\) \(\therefore \mathrm{x}=2,-3\) \(\therefore 2,-3 \text { are the solutions. }\) \(\therefore \text { The solution are }\) \(2,-3,-5\) \(\therefore \text { The total No. of solutions are } 3 .\)\(\therefore\) The total No. of solutions are 3 .
AP EAMCET-24.04.2018
Complex Numbers and Quadratic Equation
118153
If \(x\) is real, then the sum of the maximum and the minimum values of the expression \(\frac{\mathrm{x}^2+4 \mathrm{x}+1}{\mathrm{x}^2+\mathrm{x}+1}\) is
1 -2
2 2
3 1
4 0
Explanation:
D Let, \(y=\frac{x^2+4 x+1}{x^2+x+1}\) \(\Rightarrow y x^2+y x+y=x^2+4 x+1\) \(\Rightarrow(y-1) x^2+(y-4) x+y-1=0\) \(\because \mathrm{x}\) is real, therefore. \((y-4)^2-4(y-1)(y-1) \geq 0\) \(\text { or } y^2+16-8 y-4\left(y^2+1-2 y\right) \geq 0\) \(-3 y^2+12 \geq 0\) \(3 y^2 \leq 12\) \(\text { or }-3 y^2 \leq 12\) \(\text { or } 3 y^2-12 \leq 0\) \(3\left(y^2-4\right) \leq 0 \Rightarrow(y-2)(y+2) \leq 0\) \(\Rightarrow-2 \leq y \leq 2\) \(\therefore\) The minimum value is -2 and max value is 2 Sum of minimum and maximum, \(-2+2=0\)So, option (d) is correct.
118150
The number of solutions for the equation \(x^2-5\) \(|x|+6=0\) is
1 4
2 3
3 2
4 1
Explanation:
A Given the equation. \(\mathrm{x}^2-5|\mathrm{x}|+6=0\) case \(1:-\mathrm{x}>0 \Rightarrow|\mathrm{x}|=\mathrm{x}\) \(\therefore \mathrm{x}^2-5 \mathrm{x}+6=0\) \(\mathrm{x}^2-3 \mathrm{x}-2 \mathrm{x}+6=0\) \(\mathrm{x}(\mathrm{x}-3)-2(\mathrm{x}-3)=0\) \((\mathrm{x}-2)(\mathrm{x}-3)=0\) \(\mathrm{x}=2,3\) \(\therefore\) case 2 : \(\mathrm{x}\lt 0 \Rightarrow|\mathrm{x}|=-\mathrm{x}\), then the given equation becomes. \(\mathrm{x}^2-5(-\mathrm{x})+6=0\) \(\Rightarrow \mathrm{x}^2+5 \mathrm{x}+6=0\) \(\Rightarrow \mathrm{x}^2+3 \mathrm{x}+2 \mathrm{x}+6=0\) \(\Rightarrow \mathrm{x}(\mathrm{x}+3)+2(\mathrm{x}+3)=0\) \((\mathrm{x}+2)(\mathrm{x}+3)=0\) \(\Rightarrow \mathrm{x}=-2,-3\) So, the number of solutions is 4
AP EAMCET-22.09.2020
Complex Numbers and Quadratic Equation
118151
The roots of the equation \(x^3-6 x^2+3 x+10=0\) are
1 in geometric progression
2 in arithmetic progression
3 in harmonic progression
4 Positive
Explanation:
B Given the equation \(\mathrm{x}^3-6 \mathrm{x}^2+3 \mathrm{x}+10=0\) has roots \(\alpha, \beta, \gamma\) where \(\alpha+\beta+\gamma=6\) \(\alpha \beta+\beta \gamma+\alpha \gamma=3\) \(\alpha \beta \gamma=-10\) \(\alpha, \beta, \gamma\) are in A. P if \(\beta=\frac{\alpha+\gamma}{2}\) From equation (i) \(\beta+2 \beta=6 \Rightarrow \beta=2\) From equation (iii) \(\alpha \beta . \gamma=-10\) \(\Rightarrow \alpha \gamma=\frac{-10}{2}=-5\) Now from equation to, \((\alpha+\gamma) \beta+\alpha \gamma=3\) \(4 \times 2+(-5)=3\), which satisfy equation (ii) \(\therefore \alpha, \beta, \gamma\) are in A.P
AP EAMCET-05.10.2021
Complex Numbers and Quadratic Equation
118152
The number of distinct real solutions for the equation \(\left|x^2+2 x-8\right|+x-2=0\) is
1 1
2 2
3 3
4 4
Explanation:
C Given the equation\(\left|x^2+2 x-8\right|+x-2=0\) \(\text { case } 1: x^2+2 x-8>0 \Rightarrow\left|x^2+2 x-8\right|=x^2+2 x-8\) \(x^2+2 x-8+x-2=0\) \(x^2+3 x-10=0\) \(\Rightarrow x^2+5 x-2 x-10=0\) \(\Rightarrow \mathrm{x}(\mathrm{x}+5)-2(\mathrm{x}+5)=0right.\)Ans: c Exp: (C) : Given the equation \(\therefore(\mathrm{x}-2)(\mathrm{x}+5)=0 \Rightarrow \mathrm{x}=(2,-5)\) \(\text { Now, } \mathrm{x}^2+2 \mathrm{x}-8>0\) \(\Rightarrow \mathrm{x}^2+4 \mathrm{x}-2 \mathrm{x}-8>0\) \(\Rightarrow \mathrm{x}(\mathrm{x}+4)-2(\mathrm{x}+4)>0 \Rightarrow(\mathrm{x}-2)(\mathrm{x}+4)>0\) \(\Rightarrow \mathrm{x}>2 \text { and } \mathrm{x}\lt -4\) \(\therefore 2,-5 \text { are the solutions }\) \(\text { case } 2:-\mathrm{x}^2+2 \mathrm{x}-8\lt 0\) \(\Rightarrow-4\lt \mathrm{x}\lt 2 \text { and }\left|\mathrm{x}^2+2 \mathrm{x}-8\right|=-\left(\mathrm{x}^2+2 \mathrm{x}-8\right)\) \(\therefore \text { The given equation becomes, }\) \(-\left(\mathrm{x}^2+2 \mathrm{x}-8\right)+\mathrm{x}-2=0\) \(0 \mathrm{r}-\mathrm{x}^2-2 \mathrm{x}+8+\mathrm{x}-2=0\) \(\mathrm{x}^2+\mathrm{x}-6=0\) \(\Rightarrow \mathrm{x}^2+3 \mathrm{x}-2 \mathrm{x}-6=0\) \(\Rightarrow \mathrm{x}(\mathrm{x}+3)-2(\mathrm{x}+3)=0\) \(\Rightarrow(\mathrm{x}+3)(\mathrm{x}-2)=0\) \(\therefore \mathrm{x}=2,-3\) \(\therefore 2,-3 \text { are the solutions. }\) \(\therefore \text { The solution are }\) \(2,-3,-5\) \(\therefore \text { The total No. of solutions are } 3 .\)\(\therefore\) The total No. of solutions are 3 .
AP EAMCET-24.04.2018
Complex Numbers and Quadratic Equation
118153
If \(x\) is real, then the sum of the maximum and the minimum values of the expression \(\frac{\mathrm{x}^2+4 \mathrm{x}+1}{\mathrm{x}^2+\mathrm{x}+1}\) is
1 -2
2 2
3 1
4 0
Explanation:
D Let, \(y=\frac{x^2+4 x+1}{x^2+x+1}\) \(\Rightarrow y x^2+y x+y=x^2+4 x+1\) \(\Rightarrow(y-1) x^2+(y-4) x+y-1=0\) \(\because \mathrm{x}\) is real, therefore. \((y-4)^2-4(y-1)(y-1) \geq 0\) \(\text { or } y^2+16-8 y-4\left(y^2+1-2 y\right) \geq 0\) \(-3 y^2+12 \geq 0\) \(3 y^2 \leq 12\) \(\text { or }-3 y^2 \leq 12\) \(\text { or } 3 y^2-12 \leq 0\) \(3\left(y^2-4\right) \leq 0 \Rightarrow(y-2)(y+2) \leq 0\) \(\Rightarrow-2 \leq y \leq 2\) \(\therefore\) The minimum value is -2 and max value is 2 Sum of minimum and maximum, \(-2+2=0\)So, option (d) is correct.