118051
The number of real solution of the equation \(\left|\mathbf{x}^2\right|-3|x|+2=0\) is
1 3
2 2
3 1
4 4
Explanation:
D We have :- \(\left|x^2\right|-3|x|+2=0\) Let \(\mathrm{y}=|\mathrm{x}|\) \(\mathrm{x}= \pm \mathrm{y}\) So, eq \({ }^n\) (i) becomes :- \(\mathrm{y}^2-3 \mathrm{y}+2=0\) \((\mathrm{y}-2)(\mathrm{y}-1)=0\) \(\mathrm{y}=2,1\) From (ii), we have \(x= \pm 2, \pm 1\)\(\therefore 4\) solutions exist.
SRM JEEE-2013
Complex Numbers and Quadratic Equation
118052
The solution of the equation \(9^x+78=3^{2 x+3}\) is
1 2
2 3
3 \(1 / 3\)
4 \(1 / 2\)
Explanation:
D Given :- \(9^x+78=3^{2 x+3}\) \(3^{2 x}+78=3^{2 x} \cdot 27\) Put \(3^{2 x}=y\) \(\therefore\) \(y+78=27 y\) \(y=3\) \(3^{2 x}=3\) \(So, 2 x=1\) \(x=\frac{1}{2}\)
SRM JEEE-2012
Complex Numbers and Quadratic Equation
118055
If the roots of the equation \(x^3+a x^2+b x+c=0\) are in A.P., then \(2 a^3-9 a b=\)
1 \(9 \mathrm{c}\)
2 \(18 \mathrm{c}\)
3 \(27 \mathrm{c}\)
4 \(-27 \mathrm{c}\)
Explanation:
D Given, equation is : \(\mathrm{x}^3+\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0\) Then, sum of roots, \(\alpha+\beta+\gamma=-a\) sum of product of roots taking two at a time, \((\alpha \beta+\beta \gamma+\gamma \alpha)=\mathrm{b}\) product of roots, \(\alpha \beta \gamma=-\mathrm{c}\) According to the question, roots are in A.P. \(\therefore \quad \beta-\alpha=\gamma-\beta\) \(2 \beta=\alpha+\gamma\) \(\because \quad \alpha+\beta+\gamma=-\mathrm{a}\) \(\therefore \quad(\alpha+\gamma)+\beta=-a\) \(2 \beta+\beta=-\mathrm{a}\) \(3 \beta=-a \Rightarrow \beta=\frac{-a}{3}\) Since, \(\beta\) is root of \(\mathrm{eq}^{\mathrm{n}} \mathrm{x}^3+\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0\) \(\therefore \beta^3+\mathrm{a} \beta^2+\mathrm{b} \beta+\mathrm{c}=0\) put \(\beta=\frac{-\mathrm{a}}{3}\) in above eq \({ }^{\mathrm{n}}\) \(\left(\frac{-a}{3}\right)^3+a\left(\frac{-a}{3}\right)^2+b\left(\frac{-a}{3}\right)+c=0\) \(\frac{-a^3}{27}+\frac{a^3}{9}-\frac{a b}{3}+c=0\) \(\frac{-a^3+3 a^3-9 a b+27 c}{27}=0\) \(2 a^3-9 a b=-27 c\)
Karnataka CET-2013
Complex Numbers and Quadratic Equation
118057
If \(\alpha, \beta\) and \(\gamma\) are roots of \(x^3-2 x+1=0\), then the value of \(\Sigma\left(\frac{1}{\alpha+\beta-\gamma}\right)\) is
118051
The number of real solution of the equation \(\left|\mathbf{x}^2\right|-3|x|+2=0\) is
1 3
2 2
3 1
4 4
Explanation:
D We have :- \(\left|x^2\right|-3|x|+2=0\) Let \(\mathrm{y}=|\mathrm{x}|\) \(\mathrm{x}= \pm \mathrm{y}\) So, eq \({ }^n\) (i) becomes :- \(\mathrm{y}^2-3 \mathrm{y}+2=0\) \((\mathrm{y}-2)(\mathrm{y}-1)=0\) \(\mathrm{y}=2,1\) From (ii), we have \(x= \pm 2, \pm 1\)\(\therefore 4\) solutions exist.
SRM JEEE-2013
Complex Numbers and Quadratic Equation
118052
The solution of the equation \(9^x+78=3^{2 x+3}\) is
1 2
2 3
3 \(1 / 3\)
4 \(1 / 2\)
Explanation:
D Given :- \(9^x+78=3^{2 x+3}\) \(3^{2 x}+78=3^{2 x} \cdot 27\) Put \(3^{2 x}=y\) \(\therefore\) \(y+78=27 y\) \(y=3\) \(3^{2 x}=3\) \(So, 2 x=1\) \(x=\frac{1}{2}\)
SRM JEEE-2012
Complex Numbers and Quadratic Equation
118055
If the roots of the equation \(x^3+a x^2+b x+c=0\) are in A.P., then \(2 a^3-9 a b=\)
1 \(9 \mathrm{c}\)
2 \(18 \mathrm{c}\)
3 \(27 \mathrm{c}\)
4 \(-27 \mathrm{c}\)
Explanation:
D Given, equation is : \(\mathrm{x}^3+\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0\) Then, sum of roots, \(\alpha+\beta+\gamma=-a\) sum of product of roots taking two at a time, \((\alpha \beta+\beta \gamma+\gamma \alpha)=\mathrm{b}\) product of roots, \(\alpha \beta \gamma=-\mathrm{c}\) According to the question, roots are in A.P. \(\therefore \quad \beta-\alpha=\gamma-\beta\) \(2 \beta=\alpha+\gamma\) \(\because \quad \alpha+\beta+\gamma=-\mathrm{a}\) \(\therefore \quad(\alpha+\gamma)+\beta=-a\) \(2 \beta+\beta=-\mathrm{a}\) \(3 \beta=-a \Rightarrow \beta=\frac{-a}{3}\) Since, \(\beta\) is root of \(\mathrm{eq}^{\mathrm{n}} \mathrm{x}^3+\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0\) \(\therefore \beta^3+\mathrm{a} \beta^2+\mathrm{b} \beta+\mathrm{c}=0\) put \(\beta=\frac{-\mathrm{a}}{3}\) in above eq \({ }^{\mathrm{n}}\) \(\left(\frac{-a}{3}\right)^3+a\left(\frac{-a}{3}\right)^2+b\left(\frac{-a}{3}\right)+c=0\) \(\frac{-a^3}{27}+\frac{a^3}{9}-\frac{a b}{3}+c=0\) \(\frac{-a^3+3 a^3-9 a b+27 c}{27}=0\) \(2 a^3-9 a b=-27 c\)
Karnataka CET-2013
Complex Numbers and Quadratic Equation
118057
If \(\alpha, \beta\) and \(\gamma\) are roots of \(x^3-2 x+1=0\), then the value of \(\Sigma\left(\frac{1}{\alpha+\beta-\gamma}\right)\) is
118051
The number of real solution of the equation \(\left|\mathbf{x}^2\right|-3|x|+2=0\) is
1 3
2 2
3 1
4 4
Explanation:
D We have :- \(\left|x^2\right|-3|x|+2=0\) Let \(\mathrm{y}=|\mathrm{x}|\) \(\mathrm{x}= \pm \mathrm{y}\) So, eq \({ }^n\) (i) becomes :- \(\mathrm{y}^2-3 \mathrm{y}+2=0\) \((\mathrm{y}-2)(\mathrm{y}-1)=0\) \(\mathrm{y}=2,1\) From (ii), we have \(x= \pm 2, \pm 1\)\(\therefore 4\) solutions exist.
SRM JEEE-2013
Complex Numbers and Quadratic Equation
118052
The solution of the equation \(9^x+78=3^{2 x+3}\) is
1 2
2 3
3 \(1 / 3\)
4 \(1 / 2\)
Explanation:
D Given :- \(9^x+78=3^{2 x+3}\) \(3^{2 x}+78=3^{2 x} \cdot 27\) Put \(3^{2 x}=y\) \(\therefore\) \(y+78=27 y\) \(y=3\) \(3^{2 x}=3\) \(So, 2 x=1\) \(x=\frac{1}{2}\)
SRM JEEE-2012
Complex Numbers and Quadratic Equation
118055
If the roots of the equation \(x^3+a x^2+b x+c=0\) are in A.P., then \(2 a^3-9 a b=\)
1 \(9 \mathrm{c}\)
2 \(18 \mathrm{c}\)
3 \(27 \mathrm{c}\)
4 \(-27 \mathrm{c}\)
Explanation:
D Given, equation is : \(\mathrm{x}^3+\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0\) Then, sum of roots, \(\alpha+\beta+\gamma=-a\) sum of product of roots taking two at a time, \((\alpha \beta+\beta \gamma+\gamma \alpha)=\mathrm{b}\) product of roots, \(\alpha \beta \gamma=-\mathrm{c}\) According to the question, roots are in A.P. \(\therefore \quad \beta-\alpha=\gamma-\beta\) \(2 \beta=\alpha+\gamma\) \(\because \quad \alpha+\beta+\gamma=-\mathrm{a}\) \(\therefore \quad(\alpha+\gamma)+\beta=-a\) \(2 \beta+\beta=-\mathrm{a}\) \(3 \beta=-a \Rightarrow \beta=\frac{-a}{3}\) Since, \(\beta\) is root of \(\mathrm{eq}^{\mathrm{n}} \mathrm{x}^3+\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0\) \(\therefore \beta^3+\mathrm{a} \beta^2+\mathrm{b} \beta+\mathrm{c}=0\) put \(\beta=\frac{-\mathrm{a}}{3}\) in above eq \({ }^{\mathrm{n}}\) \(\left(\frac{-a}{3}\right)^3+a\left(\frac{-a}{3}\right)^2+b\left(\frac{-a}{3}\right)+c=0\) \(\frac{-a^3}{27}+\frac{a^3}{9}-\frac{a b}{3}+c=0\) \(\frac{-a^3+3 a^3-9 a b+27 c}{27}=0\) \(2 a^3-9 a b=-27 c\)
Karnataka CET-2013
Complex Numbers and Quadratic Equation
118057
If \(\alpha, \beta\) and \(\gamma\) are roots of \(x^3-2 x+1=0\), then the value of \(\Sigma\left(\frac{1}{\alpha+\beta-\gamma}\right)\) is
118051
The number of real solution of the equation \(\left|\mathbf{x}^2\right|-3|x|+2=0\) is
1 3
2 2
3 1
4 4
Explanation:
D We have :- \(\left|x^2\right|-3|x|+2=0\) Let \(\mathrm{y}=|\mathrm{x}|\) \(\mathrm{x}= \pm \mathrm{y}\) So, eq \({ }^n\) (i) becomes :- \(\mathrm{y}^2-3 \mathrm{y}+2=0\) \((\mathrm{y}-2)(\mathrm{y}-1)=0\) \(\mathrm{y}=2,1\) From (ii), we have \(x= \pm 2, \pm 1\)\(\therefore 4\) solutions exist.
SRM JEEE-2013
Complex Numbers and Quadratic Equation
118052
The solution of the equation \(9^x+78=3^{2 x+3}\) is
1 2
2 3
3 \(1 / 3\)
4 \(1 / 2\)
Explanation:
D Given :- \(9^x+78=3^{2 x+3}\) \(3^{2 x}+78=3^{2 x} \cdot 27\) Put \(3^{2 x}=y\) \(\therefore\) \(y+78=27 y\) \(y=3\) \(3^{2 x}=3\) \(So, 2 x=1\) \(x=\frac{1}{2}\)
SRM JEEE-2012
Complex Numbers and Quadratic Equation
118055
If the roots of the equation \(x^3+a x^2+b x+c=0\) are in A.P., then \(2 a^3-9 a b=\)
1 \(9 \mathrm{c}\)
2 \(18 \mathrm{c}\)
3 \(27 \mathrm{c}\)
4 \(-27 \mathrm{c}\)
Explanation:
D Given, equation is : \(\mathrm{x}^3+\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0\) Then, sum of roots, \(\alpha+\beta+\gamma=-a\) sum of product of roots taking two at a time, \((\alpha \beta+\beta \gamma+\gamma \alpha)=\mathrm{b}\) product of roots, \(\alpha \beta \gamma=-\mathrm{c}\) According to the question, roots are in A.P. \(\therefore \quad \beta-\alpha=\gamma-\beta\) \(2 \beta=\alpha+\gamma\) \(\because \quad \alpha+\beta+\gamma=-\mathrm{a}\) \(\therefore \quad(\alpha+\gamma)+\beta=-a\) \(2 \beta+\beta=-\mathrm{a}\) \(3 \beta=-a \Rightarrow \beta=\frac{-a}{3}\) Since, \(\beta\) is root of \(\mathrm{eq}^{\mathrm{n}} \mathrm{x}^3+\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0\) \(\therefore \beta^3+\mathrm{a} \beta^2+\mathrm{b} \beta+\mathrm{c}=0\) put \(\beta=\frac{-\mathrm{a}}{3}\) in above eq \({ }^{\mathrm{n}}\) \(\left(\frac{-a}{3}\right)^3+a\left(\frac{-a}{3}\right)^2+b\left(\frac{-a}{3}\right)+c=0\) \(\frac{-a^3}{27}+\frac{a^3}{9}-\frac{a b}{3}+c=0\) \(\frac{-a^3+3 a^3-9 a b+27 c}{27}=0\) \(2 a^3-9 a b=-27 c\)
Karnataka CET-2013
Complex Numbers and Quadratic Equation
118057
If \(\alpha, \beta\) and \(\gamma\) are roots of \(x^3-2 x+1=0\), then the value of \(\Sigma\left(\frac{1}{\alpha+\beta-\gamma}\right)\) is
