118204
Let \(x_1\) and \(x_2\) be the roots of the equations \(x^2-\) \(p x-3=0\). If \(x_1^2+x_2^2=10\), then the value of \(p\) is equal to
1 -4 or 4
2 -3 or 3
3 -2 or 2
4 -1 or 1
5 0
Explanation:
C Given, \(\mathrm{x}_1\) and \(\mathrm{x}_2\) be the roots of the equation \(\mathrm{x}^2-\mathrm{px}-3=0\) \(\therefore \quad \mathrm{x}_1+\mathrm{x}_2=\mathrm{p}\) \(\mathrm{x}_1 \mathrm{x}_2=-3\) We know that, \(\left(\mathrm{x}_1+\mathrm{x}_2\right)^2=\mathrm{x}_1^2+\mathrm{x}_2^2+2 \mathrm{x}_1 \mathrm{x}_2\) \(\Rightarrow \mathrm{p}^2=10+2 \times(-3) \quad\left(\because \mathrm{x}_1^2+\mathrm{x}_2^2=10\right)\) \(=10-6=4\) \(\mathrm{p}=-2 \text { or } 2\)\(\therefore \quad \mathrm{p}=-2\) or 2
Kerala CEE-2017
Complex Numbers and Quadratic Equation
118205
If \(x_1\) and \(x_2\) are the roots of \(3 x^2-2 x-6=0\), then \(x_1^2+x_2^2=10\) is equal to
1 \(\frac{50}{9}\)
2 \(\frac{40}{9}\)
3 \(\frac{30}{9}\)
4 \(\frac{20}{9}\)
5 \(\frac{10}{9}\)
Explanation:
B Given, \(x_1\) and \(x_2\) are the roots of - \(3 x^2-2 x-6=0\) \(\therefore \quad \mathrm{x}_1+\mathrm{x}_2=\frac{2}{3}\) \(\mathrm{x}_1 \cdot \mathrm{x}_2=\frac{-6}{3}=-2\) \(\therefore \quad \mathrm{x}_1^2+\mathrm{x}_2^2=\left(\mathrm{x}_1+\mathrm{x}_2\right)^2-2 \mathrm{x}_1 \mathrm{x}_2\) \(=\frac{4}{9}-2(-2)\) \(=\frac{4}{9}+4=\frac{40}{9}\)
Kerala CEE-2017
Complex Numbers and Quadratic Equation
118208
If the sum of the roots of the equation \((a+1) x^2\) \(+(2 a+3) x+(3 a+4)=0\), where \(a \neq-1\), is -1 , then the product of the roots is
1 4
2 2
3 1
4 -2
5 -4
Explanation:
B Given the quadratic equation \((a+1) x^2+(2 a+3) x+(3 a+4)=0\), let its roots are \(\alpha\) and \(\beta\). \(\therefore \quad \alpha+\beta=\frac{-(2 a+3)}{a+1}=-1\) \(\Rightarrow 2 a+3=a+1 \text { or } a=-2\) \(\therefore \alpha \beta=\frac{(3 a+4)}{(a+1)}=\frac{3(-2)+4}{-2+1}\) \(=\frac{-6+4}{-1}=\frac{-2}{-1}=2\) \(\therefore\) The product of the roots is 2 . 766.The number of distinct real roots of the equation \(x^5\left(x^3-x^2-x+1\right)+x\left(3 x^3-4 x^2-2 x+4\right)-1=0\) is \(\qquad\) Ans: (3) Exp: (3) : Given equation, \(x^5\left(x^3-x^2-x+1\right)+\mathrm{x}\left(3 \mathrm{x}^3-4 \mathrm{x}^2-2 \mathrm{x}+4\right)-1=0\) \(\mathrm{x}^5\left(\mathrm{x}^3-\mathrm{x}^2-\mathrm{x}+1\right)+\mathrm{x}\left(3 \mathrm{x}^3-4 \mathrm{x}^2-2 \mathrm{x}+4\right)-1=0\) \(\mathrm{x}^7(\mathrm{x}-1)-\mathrm{x}^5(\mathrm{x}-1)+3 \mathrm{x}^3(\mathrm{x}-1)-\mathrm{x}\left(\mathrm{x}^2-1\right)+2 \mathrm{x}(1-\mathrm{x})\) \(\quad(\mathrm{x}-1)=0\) \((\mathrm{x}-1)\left(\mathrm{x}^7-\mathrm{x}^5+3 \mathrm{x}^3-\mathrm{x}(\mathrm{x}+1)-2 \mathrm{x}+1\right)=0\) \((\mathrm{x}-1)\left(\mathrm{x}^7-\mathrm{x}^5+3 \mathrm{x}^3-\mathrm{x}^2-3 \mathrm{x}+1\right)=0\) \((\mathrm{x}-1)\left(\mathrm{x}^5\left(\mathrm{x}^2-1\right)+3 \mathrm{x}\left(\mathrm{x}^2-1\right)-1\left(\mathrm{x}^2-1\right)\right)=0\) \((\mathrm{x}-1)\left(\mathrm{x}^2-1\right)\left(\mathrm{x}^5+3 \mathrm{x}-1\right)=0\) \(\therefore \mathrm{x}= \pm 1\) are roots of above equation and \(\mathrm{x}^5+3 \mathrm{x}-1\) is\) \( a monotonic terms hence vanishes at exactly one\) value of \(\mathrm{x}\) then 1 or -1 \(\therefore 3\) real roots.
JEE Main-26.07.2022
Complex Numbers and Quadratic Equation
118209
The equation \(e^{4 x}+8 e^{3 x}+13 e^{2 x}-8 e^x+1=0, x \in\) \(R\) has :
1 four solutions two of which are negative
2 two solutions and both are negative
3 no solution
4 two solutions and only one of them is negative
Explanation:
B Given eqution : - \(\mathrm{e}^{4 \mathrm{x}}+8 \mathrm{e}^{3 \mathrm{x}}+13 \mathrm{e}^{2 \mathrm{x}}-8 \mathrm{e}^{\mathrm{x}}+1=0,\) \(\mathrm{x} \in \mathrm{R}\) Let, consider \(e^x=y, \text { then }\) \(y^4+8 y^3+13 y^2-8 y+1=0\) \(y^2+8 y+13-\frac{8}{y}+\frac{1}{y^2}=0\) \(\left(y^2+\frac{1}{y^2}\right)+8\left(y-\frac{1}{y}\right)+13=0\) \(\left(y-\frac{1}{y}\right)^2+2+8\left(y-\frac{1}{y}\right)+13=0\) \(\left(y-\frac{1}{y}\right)^2+8\left(y+\frac{1}{y}\right)+15=0\) Again, Let, \(y-\frac{1}{y}=z\) we, get \(z^2+8 z+15=0\) \(z^2+3 z+5 z+15=0\) \(z(z+3)+5(z+3)=0\) \((\mathrm{z}+5)(\mathrm{z}+3)=0, \mathrm{z}=-5,-3\) \(\therefore \quad y-\frac{1}{y}=-3 \text { or } y-\frac{1}{y}=-5\) \(\mathrm{y}^2+3 \mathrm{y}-1=0 \text { or } \mathrm{y}^2+5 \mathrm{y}-1=0\) \(\mathrm{y}=\frac{-3 \pm \sqrt{3}}{2}, \frac{-3-\sqrt{13}}{2} \text { or }\) \(\frac{-5+\sqrt{29}}{2}, \frac{-5-\sqrt{29}}{2}\) So, \(\mathrm{e}^{\mathrm{x}}=\frac{-3+\sqrt{13}}{2}, \frac{-5+\sqrt{29}}{2} \quad\left[\because \mathrm{e}^{\mathrm{x}}>0\right]\) \(x=\log \left(\frac{-3+\sqrt{13}}{2}\right), \log \left(\frac{-5+\sqrt{29}}{2}\right)\) \(\therefore\) Equation has two solution and both are negative as \(\frac{-3+\sqrt{13}}{2} \text { and } \frac{-5+\sqrt{29}}{2} \text { both belong to }(0,1)\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Complex Numbers and Quadratic Equation
118204
Let \(x_1\) and \(x_2\) be the roots of the equations \(x^2-\) \(p x-3=0\). If \(x_1^2+x_2^2=10\), then the value of \(p\) is equal to
1 -4 or 4
2 -3 or 3
3 -2 or 2
4 -1 or 1
5 0
Explanation:
C Given, \(\mathrm{x}_1\) and \(\mathrm{x}_2\) be the roots of the equation \(\mathrm{x}^2-\mathrm{px}-3=0\) \(\therefore \quad \mathrm{x}_1+\mathrm{x}_2=\mathrm{p}\) \(\mathrm{x}_1 \mathrm{x}_2=-3\) We know that, \(\left(\mathrm{x}_1+\mathrm{x}_2\right)^2=\mathrm{x}_1^2+\mathrm{x}_2^2+2 \mathrm{x}_1 \mathrm{x}_2\) \(\Rightarrow \mathrm{p}^2=10+2 \times(-3) \quad\left(\because \mathrm{x}_1^2+\mathrm{x}_2^2=10\right)\) \(=10-6=4\) \(\mathrm{p}=-2 \text { or } 2\)\(\therefore \quad \mathrm{p}=-2\) or 2
Kerala CEE-2017
Complex Numbers and Quadratic Equation
118205
If \(x_1\) and \(x_2\) are the roots of \(3 x^2-2 x-6=0\), then \(x_1^2+x_2^2=10\) is equal to
1 \(\frac{50}{9}\)
2 \(\frac{40}{9}\)
3 \(\frac{30}{9}\)
4 \(\frac{20}{9}\)
5 \(\frac{10}{9}\)
Explanation:
B Given, \(x_1\) and \(x_2\) are the roots of - \(3 x^2-2 x-6=0\) \(\therefore \quad \mathrm{x}_1+\mathrm{x}_2=\frac{2}{3}\) \(\mathrm{x}_1 \cdot \mathrm{x}_2=\frac{-6}{3}=-2\) \(\therefore \quad \mathrm{x}_1^2+\mathrm{x}_2^2=\left(\mathrm{x}_1+\mathrm{x}_2\right)^2-2 \mathrm{x}_1 \mathrm{x}_2\) \(=\frac{4}{9}-2(-2)\) \(=\frac{4}{9}+4=\frac{40}{9}\)
Kerala CEE-2017
Complex Numbers and Quadratic Equation
118208
If the sum of the roots of the equation \((a+1) x^2\) \(+(2 a+3) x+(3 a+4)=0\), where \(a \neq-1\), is -1 , then the product of the roots is
1 4
2 2
3 1
4 -2
5 -4
Explanation:
B Given the quadratic equation \((a+1) x^2+(2 a+3) x+(3 a+4)=0\), let its roots are \(\alpha\) and \(\beta\). \(\therefore \quad \alpha+\beta=\frac{-(2 a+3)}{a+1}=-1\) \(\Rightarrow 2 a+3=a+1 \text { or } a=-2\) \(\therefore \alpha \beta=\frac{(3 a+4)}{(a+1)}=\frac{3(-2)+4}{-2+1}\) \(=\frac{-6+4}{-1}=\frac{-2}{-1}=2\) \(\therefore\) The product of the roots is 2 . 766.The number of distinct real roots of the equation \(x^5\left(x^3-x^2-x+1\right)+x\left(3 x^3-4 x^2-2 x+4\right)-1=0\) is \(\qquad\) Ans: (3) Exp: (3) : Given equation, \(x^5\left(x^3-x^2-x+1\right)+\mathrm{x}\left(3 \mathrm{x}^3-4 \mathrm{x}^2-2 \mathrm{x}+4\right)-1=0\) \(\mathrm{x}^5\left(\mathrm{x}^3-\mathrm{x}^2-\mathrm{x}+1\right)+\mathrm{x}\left(3 \mathrm{x}^3-4 \mathrm{x}^2-2 \mathrm{x}+4\right)-1=0\) \(\mathrm{x}^7(\mathrm{x}-1)-\mathrm{x}^5(\mathrm{x}-1)+3 \mathrm{x}^3(\mathrm{x}-1)-\mathrm{x}\left(\mathrm{x}^2-1\right)+2 \mathrm{x}(1-\mathrm{x})\) \(\quad(\mathrm{x}-1)=0\) \((\mathrm{x}-1)\left(\mathrm{x}^7-\mathrm{x}^5+3 \mathrm{x}^3-\mathrm{x}(\mathrm{x}+1)-2 \mathrm{x}+1\right)=0\) \((\mathrm{x}-1)\left(\mathrm{x}^7-\mathrm{x}^5+3 \mathrm{x}^3-\mathrm{x}^2-3 \mathrm{x}+1\right)=0\) \((\mathrm{x}-1)\left(\mathrm{x}^5\left(\mathrm{x}^2-1\right)+3 \mathrm{x}\left(\mathrm{x}^2-1\right)-1\left(\mathrm{x}^2-1\right)\right)=0\) \((\mathrm{x}-1)\left(\mathrm{x}^2-1\right)\left(\mathrm{x}^5+3 \mathrm{x}-1\right)=0\) \(\therefore \mathrm{x}= \pm 1\) are roots of above equation and \(\mathrm{x}^5+3 \mathrm{x}-1\) is\) \( a monotonic terms hence vanishes at exactly one\) value of \(\mathrm{x}\) then 1 or -1 \(\therefore 3\) real roots.
JEE Main-26.07.2022
Complex Numbers and Quadratic Equation
118209
The equation \(e^{4 x}+8 e^{3 x}+13 e^{2 x}-8 e^x+1=0, x \in\) \(R\) has :
1 four solutions two of which are negative
2 two solutions and both are negative
3 no solution
4 two solutions and only one of them is negative
Explanation:
B Given eqution : - \(\mathrm{e}^{4 \mathrm{x}}+8 \mathrm{e}^{3 \mathrm{x}}+13 \mathrm{e}^{2 \mathrm{x}}-8 \mathrm{e}^{\mathrm{x}}+1=0,\) \(\mathrm{x} \in \mathrm{R}\) Let, consider \(e^x=y, \text { then }\) \(y^4+8 y^3+13 y^2-8 y+1=0\) \(y^2+8 y+13-\frac{8}{y}+\frac{1}{y^2}=0\) \(\left(y^2+\frac{1}{y^2}\right)+8\left(y-\frac{1}{y}\right)+13=0\) \(\left(y-\frac{1}{y}\right)^2+2+8\left(y-\frac{1}{y}\right)+13=0\) \(\left(y-\frac{1}{y}\right)^2+8\left(y+\frac{1}{y}\right)+15=0\) Again, Let, \(y-\frac{1}{y}=z\) we, get \(z^2+8 z+15=0\) \(z^2+3 z+5 z+15=0\) \(z(z+3)+5(z+3)=0\) \((\mathrm{z}+5)(\mathrm{z}+3)=0, \mathrm{z}=-5,-3\) \(\therefore \quad y-\frac{1}{y}=-3 \text { or } y-\frac{1}{y}=-5\) \(\mathrm{y}^2+3 \mathrm{y}-1=0 \text { or } \mathrm{y}^2+5 \mathrm{y}-1=0\) \(\mathrm{y}=\frac{-3 \pm \sqrt{3}}{2}, \frac{-3-\sqrt{13}}{2} \text { or }\) \(\frac{-5+\sqrt{29}}{2}, \frac{-5-\sqrt{29}}{2}\) So, \(\mathrm{e}^{\mathrm{x}}=\frac{-3+\sqrt{13}}{2}, \frac{-5+\sqrt{29}}{2} \quad\left[\because \mathrm{e}^{\mathrm{x}}>0\right]\) \(x=\log \left(\frac{-3+\sqrt{13}}{2}\right), \log \left(\frac{-5+\sqrt{29}}{2}\right)\) \(\therefore\) Equation has two solution and both are negative as \(\frac{-3+\sqrt{13}}{2} \text { and } \frac{-5+\sqrt{29}}{2} \text { both belong to }(0,1)\)
118204
Let \(x_1\) and \(x_2\) be the roots of the equations \(x^2-\) \(p x-3=0\). If \(x_1^2+x_2^2=10\), then the value of \(p\) is equal to
1 -4 or 4
2 -3 or 3
3 -2 or 2
4 -1 or 1
5 0
Explanation:
C Given, \(\mathrm{x}_1\) and \(\mathrm{x}_2\) be the roots of the equation \(\mathrm{x}^2-\mathrm{px}-3=0\) \(\therefore \quad \mathrm{x}_1+\mathrm{x}_2=\mathrm{p}\) \(\mathrm{x}_1 \mathrm{x}_2=-3\) We know that, \(\left(\mathrm{x}_1+\mathrm{x}_2\right)^2=\mathrm{x}_1^2+\mathrm{x}_2^2+2 \mathrm{x}_1 \mathrm{x}_2\) \(\Rightarrow \mathrm{p}^2=10+2 \times(-3) \quad\left(\because \mathrm{x}_1^2+\mathrm{x}_2^2=10\right)\) \(=10-6=4\) \(\mathrm{p}=-2 \text { or } 2\)\(\therefore \quad \mathrm{p}=-2\) or 2
Kerala CEE-2017
Complex Numbers and Quadratic Equation
118205
If \(x_1\) and \(x_2\) are the roots of \(3 x^2-2 x-6=0\), then \(x_1^2+x_2^2=10\) is equal to
1 \(\frac{50}{9}\)
2 \(\frac{40}{9}\)
3 \(\frac{30}{9}\)
4 \(\frac{20}{9}\)
5 \(\frac{10}{9}\)
Explanation:
B Given, \(x_1\) and \(x_2\) are the roots of - \(3 x^2-2 x-6=0\) \(\therefore \quad \mathrm{x}_1+\mathrm{x}_2=\frac{2}{3}\) \(\mathrm{x}_1 \cdot \mathrm{x}_2=\frac{-6}{3}=-2\) \(\therefore \quad \mathrm{x}_1^2+\mathrm{x}_2^2=\left(\mathrm{x}_1+\mathrm{x}_2\right)^2-2 \mathrm{x}_1 \mathrm{x}_2\) \(=\frac{4}{9}-2(-2)\) \(=\frac{4}{9}+4=\frac{40}{9}\)
Kerala CEE-2017
Complex Numbers and Quadratic Equation
118208
If the sum of the roots of the equation \((a+1) x^2\) \(+(2 a+3) x+(3 a+4)=0\), where \(a \neq-1\), is -1 , then the product of the roots is
1 4
2 2
3 1
4 -2
5 -4
Explanation:
B Given the quadratic equation \((a+1) x^2+(2 a+3) x+(3 a+4)=0\), let its roots are \(\alpha\) and \(\beta\). \(\therefore \quad \alpha+\beta=\frac{-(2 a+3)}{a+1}=-1\) \(\Rightarrow 2 a+3=a+1 \text { or } a=-2\) \(\therefore \alpha \beta=\frac{(3 a+4)}{(a+1)}=\frac{3(-2)+4}{-2+1}\) \(=\frac{-6+4}{-1}=\frac{-2}{-1}=2\) \(\therefore\) The product of the roots is 2 . 766.The number of distinct real roots of the equation \(x^5\left(x^3-x^2-x+1\right)+x\left(3 x^3-4 x^2-2 x+4\right)-1=0\) is \(\qquad\) Ans: (3) Exp: (3) : Given equation, \(x^5\left(x^3-x^2-x+1\right)+\mathrm{x}\left(3 \mathrm{x}^3-4 \mathrm{x}^2-2 \mathrm{x}+4\right)-1=0\) \(\mathrm{x}^5\left(\mathrm{x}^3-\mathrm{x}^2-\mathrm{x}+1\right)+\mathrm{x}\left(3 \mathrm{x}^3-4 \mathrm{x}^2-2 \mathrm{x}+4\right)-1=0\) \(\mathrm{x}^7(\mathrm{x}-1)-\mathrm{x}^5(\mathrm{x}-1)+3 \mathrm{x}^3(\mathrm{x}-1)-\mathrm{x}\left(\mathrm{x}^2-1\right)+2 \mathrm{x}(1-\mathrm{x})\) \(\quad(\mathrm{x}-1)=0\) \((\mathrm{x}-1)\left(\mathrm{x}^7-\mathrm{x}^5+3 \mathrm{x}^3-\mathrm{x}(\mathrm{x}+1)-2 \mathrm{x}+1\right)=0\) \((\mathrm{x}-1)\left(\mathrm{x}^7-\mathrm{x}^5+3 \mathrm{x}^3-\mathrm{x}^2-3 \mathrm{x}+1\right)=0\) \((\mathrm{x}-1)\left(\mathrm{x}^5\left(\mathrm{x}^2-1\right)+3 \mathrm{x}\left(\mathrm{x}^2-1\right)-1\left(\mathrm{x}^2-1\right)\right)=0\) \((\mathrm{x}-1)\left(\mathrm{x}^2-1\right)\left(\mathrm{x}^5+3 \mathrm{x}-1\right)=0\) \(\therefore \mathrm{x}= \pm 1\) are roots of above equation and \(\mathrm{x}^5+3 \mathrm{x}-1\) is\) \( a monotonic terms hence vanishes at exactly one\) value of \(\mathrm{x}\) then 1 or -1 \(\therefore 3\) real roots.
JEE Main-26.07.2022
Complex Numbers and Quadratic Equation
118209
The equation \(e^{4 x}+8 e^{3 x}+13 e^{2 x}-8 e^x+1=0, x \in\) \(R\) has :
1 four solutions two of which are negative
2 two solutions and both are negative
3 no solution
4 two solutions and only one of them is negative
Explanation:
B Given eqution : - \(\mathrm{e}^{4 \mathrm{x}}+8 \mathrm{e}^{3 \mathrm{x}}+13 \mathrm{e}^{2 \mathrm{x}}-8 \mathrm{e}^{\mathrm{x}}+1=0,\) \(\mathrm{x} \in \mathrm{R}\) Let, consider \(e^x=y, \text { then }\) \(y^4+8 y^3+13 y^2-8 y+1=0\) \(y^2+8 y+13-\frac{8}{y}+\frac{1}{y^2}=0\) \(\left(y^2+\frac{1}{y^2}\right)+8\left(y-\frac{1}{y}\right)+13=0\) \(\left(y-\frac{1}{y}\right)^2+2+8\left(y-\frac{1}{y}\right)+13=0\) \(\left(y-\frac{1}{y}\right)^2+8\left(y+\frac{1}{y}\right)+15=0\) Again, Let, \(y-\frac{1}{y}=z\) we, get \(z^2+8 z+15=0\) \(z^2+3 z+5 z+15=0\) \(z(z+3)+5(z+3)=0\) \((\mathrm{z}+5)(\mathrm{z}+3)=0, \mathrm{z}=-5,-3\) \(\therefore \quad y-\frac{1}{y}=-3 \text { or } y-\frac{1}{y}=-5\) \(\mathrm{y}^2+3 \mathrm{y}-1=0 \text { or } \mathrm{y}^2+5 \mathrm{y}-1=0\) \(\mathrm{y}=\frac{-3 \pm \sqrt{3}}{2}, \frac{-3-\sqrt{13}}{2} \text { or }\) \(\frac{-5+\sqrt{29}}{2}, \frac{-5-\sqrt{29}}{2}\) So, \(\mathrm{e}^{\mathrm{x}}=\frac{-3+\sqrt{13}}{2}, \frac{-5+\sqrt{29}}{2} \quad\left[\because \mathrm{e}^{\mathrm{x}}>0\right]\) \(x=\log \left(\frac{-3+\sqrt{13}}{2}\right), \log \left(\frac{-5+\sqrt{29}}{2}\right)\) \(\therefore\) Equation has two solution and both are negative as \(\frac{-3+\sqrt{13}}{2} \text { and } \frac{-5+\sqrt{29}}{2} \text { both belong to }(0,1)\)
118204
Let \(x_1\) and \(x_2\) be the roots of the equations \(x^2-\) \(p x-3=0\). If \(x_1^2+x_2^2=10\), then the value of \(p\) is equal to
1 -4 or 4
2 -3 or 3
3 -2 or 2
4 -1 or 1
5 0
Explanation:
C Given, \(\mathrm{x}_1\) and \(\mathrm{x}_2\) be the roots of the equation \(\mathrm{x}^2-\mathrm{px}-3=0\) \(\therefore \quad \mathrm{x}_1+\mathrm{x}_2=\mathrm{p}\) \(\mathrm{x}_1 \mathrm{x}_2=-3\) We know that, \(\left(\mathrm{x}_1+\mathrm{x}_2\right)^2=\mathrm{x}_1^2+\mathrm{x}_2^2+2 \mathrm{x}_1 \mathrm{x}_2\) \(\Rightarrow \mathrm{p}^2=10+2 \times(-3) \quad\left(\because \mathrm{x}_1^2+\mathrm{x}_2^2=10\right)\) \(=10-6=4\) \(\mathrm{p}=-2 \text { or } 2\)\(\therefore \quad \mathrm{p}=-2\) or 2
Kerala CEE-2017
Complex Numbers and Quadratic Equation
118205
If \(x_1\) and \(x_2\) are the roots of \(3 x^2-2 x-6=0\), then \(x_1^2+x_2^2=10\) is equal to
1 \(\frac{50}{9}\)
2 \(\frac{40}{9}\)
3 \(\frac{30}{9}\)
4 \(\frac{20}{9}\)
5 \(\frac{10}{9}\)
Explanation:
B Given, \(x_1\) and \(x_2\) are the roots of - \(3 x^2-2 x-6=0\) \(\therefore \quad \mathrm{x}_1+\mathrm{x}_2=\frac{2}{3}\) \(\mathrm{x}_1 \cdot \mathrm{x}_2=\frac{-6}{3}=-2\) \(\therefore \quad \mathrm{x}_1^2+\mathrm{x}_2^2=\left(\mathrm{x}_1+\mathrm{x}_2\right)^2-2 \mathrm{x}_1 \mathrm{x}_2\) \(=\frac{4}{9}-2(-2)\) \(=\frac{4}{9}+4=\frac{40}{9}\)
Kerala CEE-2017
Complex Numbers and Quadratic Equation
118208
If the sum of the roots of the equation \((a+1) x^2\) \(+(2 a+3) x+(3 a+4)=0\), where \(a \neq-1\), is -1 , then the product of the roots is
1 4
2 2
3 1
4 -2
5 -4
Explanation:
B Given the quadratic equation \((a+1) x^2+(2 a+3) x+(3 a+4)=0\), let its roots are \(\alpha\) and \(\beta\). \(\therefore \quad \alpha+\beta=\frac{-(2 a+3)}{a+1}=-1\) \(\Rightarrow 2 a+3=a+1 \text { or } a=-2\) \(\therefore \alpha \beta=\frac{(3 a+4)}{(a+1)}=\frac{3(-2)+4}{-2+1}\) \(=\frac{-6+4}{-1}=\frac{-2}{-1}=2\) \(\therefore\) The product of the roots is 2 . 766.The number of distinct real roots of the equation \(x^5\left(x^3-x^2-x+1\right)+x\left(3 x^3-4 x^2-2 x+4\right)-1=0\) is \(\qquad\) Ans: (3) Exp: (3) : Given equation, \(x^5\left(x^3-x^2-x+1\right)+\mathrm{x}\left(3 \mathrm{x}^3-4 \mathrm{x}^2-2 \mathrm{x}+4\right)-1=0\) \(\mathrm{x}^5\left(\mathrm{x}^3-\mathrm{x}^2-\mathrm{x}+1\right)+\mathrm{x}\left(3 \mathrm{x}^3-4 \mathrm{x}^2-2 \mathrm{x}+4\right)-1=0\) \(\mathrm{x}^7(\mathrm{x}-1)-\mathrm{x}^5(\mathrm{x}-1)+3 \mathrm{x}^3(\mathrm{x}-1)-\mathrm{x}\left(\mathrm{x}^2-1\right)+2 \mathrm{x}(1-\mathrm{x})\) \(\quad(\mathrm{x}-1)=0\) \((\mathrm{x}-1)\left(\mathrm{x}^7-\mathrm{x}^5+3 \mathrm{x}^3-\mathrm{x}(\mathrm{x}+1)-2 \mathrm{x}+1\right)=0\) \((\mathrm{x}-1)\left(\mathrm{x}^7-\mathrm{x}^5+3 \mathrm{x}^3-\mathrm{x}^2-3 \mathrm{x}+1\right)=0\) \((\mathrm{x}-1)\left(\mathrm{x}^5\left(\mathrm{x}^2-1\right)+3 \mathrm{x}\left(\mathrm{x}^2-1\right)-1\left(\mathrm{x}^2-1\right)\right)=0\) \((\mathrm{x}-1)\left(\mathrm{x}^2-1\right)\left(\mathrm{x}^5+3 \mathrm{x}-1\right)=0\) \(\therefore \mathrm{x}= \pm 1\) are roots of above equation and \(\mathrm{x}^5+3 \mathrm{x}-1\) is\) \( a monotonic terms hence vanishes at exactly one\) value of \(\mathrm{x}\) then 1 or -1 \(\therefore 3\) real roots.
JEE Main-26.07.2022
Complex Numbers and Quadratic Equation
118209
The equation \(e^{4 x}+8 e^{3 x}+13 e^{2 x}-8 e^x+1=0, x \in\) \(R\) has :
1 four solutions two of which are negative
2 two solutions and both are negative
3 no solution
4 two solutions and only one of them is negative
Explanation:
B Given eqution : - \(\mathrm{e}^{4 \mathrm{x}}+8 \mathrm{e}^{3 \mathrm{x}}+13 \mathrm{e}^{2 \mathrm{x}}-8 \mathrm{e}^{\mathrm{x}}+1=0,\) \(\mathrm{x} \in \mathrm{R}\) Let, consider \(e^x=y, \text { then }\) \(y^4+8 y^3+13 y^2-8 y+1=0\) \(y^2+8 y+13-\frac{8}{y}+\frac{1}{y^2}=0\) \(\left(y^2+\frac{1}{y^2}\right)+8\left(y-\frac{1}{y}\right)+13=0\) \(\left(y-\frac{1}{y}\right)^2+2+8\left(y-\frac{1}{y}\right)+13=0\) \(\left(y-\frac{1}{y}\right)^2+8\left(y+\frac{1}{y}\right)+15=0\) Again, Let, \(y-\frac{1}{y}=z\) we, get \(z^2+8 z+15=0\) \(z^2+3 z+5 z+15=0\) \(z(z+3)+5(z+3)=0\) \((\mathrm{z}+5)(\mathrm{z}+3)=0, \mathrm{z}=-5,-3\) \(\therefore \quad y-\frac{1}{y}=-3 \text { or } y-\frac{1}{y}=-5\) \(\mathrm{y}^2+3 \mathrm{y}-1=0 \text { or } \mathrm{y}^2+5 \mathrm{y}-1=0\) \(\mathrm{y}=\frac{-3 \pm \sqrt{3}}{2}, \frac{-3-\sqrt{13}}{2} \text { or }\) \(\frac{-5+\sqrt{29}}{2}, \frac{-5-\sqrt{29}}{2}\) So, \(\mathrm{e}^{\mathrm{x}}=\frac{-3+\sqrt{13}}{2}, \frac{-5+\sqrt{29}}{2} \quad\left[\because \mathrm{e}^{\mathrm{x}}>0\right]\) \(x=\log \left(\frac{-3+\sqrt{13}}{2}\right), \log \left(\frac{-5+\sqrt{29}}{2}\right)\) \(\therefore\) Equation has two solution and both are negative as \(\frac{-3+\sqrt{13}}{2} \text { and } \frac{-5+\sqrt{29}}{2} \text { both belong to }(0,1)\)
