118158
The number of integer value (s) of \(k\) for which the expression \(x^2-2(4 k-1) x+15 k^2-2 k-7>\) 0 for every real number \(x\) is/are
1 none
2 one
3 finitely many greater than 1
4 infinitely many
Explanation:
B \(\mathrm{x}^2-2(4 \mathrm{k}-1) \mathrm{x}+15 \mathrm{k}^2-2 \mathrm{k}-7>0\) If value of quadratic function is greater then zero then discriminant is less than zero. \(\therefore \quad \mathrm{b}^2-4 \mathrm{ac}\lt 0\) \({[2(4 \mathrm{k}-1)]^2-4 \times 1 \times\left(15 \mathrm{k}^2-2 \mathrm{k}-7\right)\lt 0}\) \(4\left(16 \mathrm{k}^2+1-8 \mathrm{k}\right)-4 \times\left(15 \mathrm{k}^2-2 \mathrm{k}-7\right)\lt 0\) \(4\left(16 \mathrm{k}^2+1-8 \mathrm{k}-15 \mathrm{k}^2+2 \mathrm{k}+7\right)\lt 0\) \(\mathrm{k}^2-6 \mathrm{k}+8\lt 0\) \(\mathrm{k}^2-4 \mathrm{k}-2 \mathrm{k}+8\lt 0\) \(\mathrm{k}(\mathrm{k}-4)-2(\mathrm{k}-4)\lt 0\) \((\mathrm{k}-2)(\mathrm{k}-4)\lt 0\) \(\therefore \quad 2\lt \mathrm{k}\lt 4\)So, integer value of \(\mathrm{k}\) is 3 .
J and K CET-2015
Complex Numbers and Quadratic Equation
118159
Let \(a, b, c\) be positive real number. if \(\frac{x^2-b x}{a x-c}=\frac{m-1}{m+1}\) has two roots which are numerically equal but opposite in sign. Then the value of ' \(\mathrm{m}\) ' is
1 \(\mathrm{c}\)
2 \(\frac{1}{c}\)
3 \(\frac{a+b}{a-b}\)
4 \(\frac{a-b}{a+b}\)
Explanation:
D \(We have, \) \(\frac{x^2-b x}{a x-c}=\frac{m-1}{m+1}\) \(\left(\mathrm{x}^2-\mathrm{bx}\right)=(\mathrm{ax}-\mathrm{c})\left(\frac{\mathrm{m}-1}{\mathrm{~m}+1}\right)\) \(\left(\mathrm{x}^2-\mathrm{bx}\right)=\mathrm{ax}\left(\frac{\mathrm{m}-1}{\mathrm{~m}+1}\right)-\mathrm{c}\left(\frac{\mathrm{m}-1}{\mathrm{~m}+1}\right)\) \(x^2-\left(b+a\left(\frac{m-1}{m+1}\right)\right) x+c\left(\frac{m-1}{m+1}\right)=0\) \(\therefore \quad b+\left(\frac{m-1}{m+1}\right) a=0\) \(\mathrm{~m}=\frac{\mathrm{a}-\mathrm{b}}{\mathrm{a}+\mathrm{b}}\)If roots are equal in magnitude but opposite in sign then sum of roots is zero.
AP EAMCET-23.08.2021
Complex Numbers and Quadratic Equation
118160
If \(\alpha, \beta, \gamma\) are the roots of the equation \(x^3+4 x-\) \(19=0\). Then the value of \(\frac{\alpha^3}{19-4 \alpha}+\frac{\beta^3}{19-4 \beta}+\frac{\gamma^3}{19-4 \gamma}=\)
1 0
2 3
3 -3
4 2
Explanation:
B \(We have the equation, \mathrm{x}^3+4 \mathrm{x}-19=0\) \(\text { If, } \alpha, \beta, \gamma \text { are roots of the above equation then, }\) \(\alpha+\beta+\gamma=0\) \(\alpha \beta+\beta \gamma+\gamma \alpha=4\) \(\alpha \beta \gamma=19\) \(\therefore \alpha(\beta+\gamma)+\beta \gamma=4\) \(\alpha(-\alpha)+\beta \gamma=4\) \(-\alpha^2+\frac{19}{\alpha}=4\) \(-\alpha^3+19=4 \alpha\) \(\alpha^3=19-4 \alpha\) \(\left(\frac{\alpha^3}{19-4 \alpha}\right)=1\) \(\text { Similarly, }\left(\frac{\beta^3}{19-4 \beta}\right)=1 \text { and }\left(\frac{\gamma^3}{19-4 \gamma}\right)=1\) \(\frac{\alpha^3}{19-4 \alpha}+\frac{\beta^3}{19-4 \beta}+\frac{\gamma^3}{19-4 \gamma}=1+1+1\) \(\frac{\alpha^3}{19-4 \alpha}+\frac{\beta^3}{19-4 \beta}+\frac{\gamma^3}{19-4 \gamma}=3\)
AP EAMCET-23.08.2021
Complex Numbers and Quadratic Equation
118161
The number of real roots of \(3^{2 \mathrm{x}^2-7 \mathrm{x}+7}=9\) is
1 0
2 2
3 1
4 4
Explanation:
B Given the equation \(3^{2 \mathrm{x}^2-7 \mathrm{x}+7}=9=(3)^2\) \(\therefore \quad 2 \mathrm{x}^2-7 \mathrm{x}+7=2\) \(\begin{array}{ll}\therefore 2 x^2-7 x+5=0\end{array}\) Or \(\quad 2 x^2-5 x-2 x+5=0\) Or \(\quad x(2 x-5)-1(2 x-5)=0\) Or \(\quad(2 x-5)(x-1)=0\) Or \(\quad(2 \mathrm{x}-5)(\mathrm{x}-1)=0\) \(\therefore \quad \mathrm{x}=1, \quad \mathrm{x}=5 / 2\) \(\therefore\) Number of real roots are two \(\therefore\) Option (b) is correct.
AIEEE-2002
Complex Numbers and Quadratic Equation
118162
The value of ' \(a\) ' for which one root of the quadratic equation \(\left(a^2-5 a+3\right) x^2+(3 a-1) x+\) \(\mathbf{2}=\mathbf{0}\) is twice as large as the other, is
1 \(2 / 3\)
2 \(-2 / 3\)
3 \(1 / 3\)
4 \(-1 / 3\)
Explanation:
A Given the quadratic equation \(\left(a^2-5 a+3\right) x^2+(3 a-1) x+2=0\) Let \(\alpha\) and \(\beta\) are the roots to the equation and \(\alpha=2 \beta\). \(\therefore \quad \alpha+\beta=\frac{-(3 a-1)}{a^2-5 a+3}\) \(\alpha \cdot \beta=\frac{2}{a^2-5 a+3}\) \(\because \quad \alpha=2 \beta\) \(\therefore \quad 2 \beta+\beta=\frac{-(3 a-1)}{a^2-5 a+3}=3 \beta\) \(\text { And } 2 \beta^2=\frac{2}{a^2-5 a+3}\) \(\Rightarrow \beta^2=\frac{1}{a^2-5 a+3}\) \(\Rightarrow 9 \beta^2=\frac{9}{a^2-5 a+3}=\frac{(3 a-1)^2}{\left(a^2-5 a+3\right)^2}\) \(\Rightarrow 9\left(a^2-5 a+3\right)=9 a^2+1-6 a\) \(\Rightarrow 9 a^2-45 a+27=9 a^2+1-6 a\) \(\Rightarrow 39 a=26\) \(a=\frac{26}{39}=\frac{2}{3}\) \(\therefore\) Option (a) is correct.
118158
The number of integer value (s) of \(k\) for which the expression \(x^2-2(4 k-1) x+15 k^2-2 k-7>\) 0 for every real number \(x\) is/are
1 none
2 one
3 finitely many greater than 1
4 infinitely many
Explanation:
B \(\mathrm{x}^2-2(4 \mathrm{k}-1) \mathrm{x}+15 \mathrm{k}^2-2 \mathrm{k}-7>0\) If value of quadratic function is greater then zero then discriminant is less than zero. \(\therefore \quad \mathrm{b}^2-4 \mathrm{ac}\lt 0\) \({[2(4 \mathrm{k}-1)]^2-4 \times 1 \times\left(15 \mathrm{k}^2-2 \mathrm{k}-7\right)\lt 0}\) \(4\left(16 \mathrm{k}^2+1-8 \mathrm{k}\right)-4 \times\left(15 \mathrm{k}^2-2 \mathrm{k}-7\right)\lt 0\) \(4\left(16 \mathrm{k}^2+1-8 \mathrm{k}-15 \mathrm{k}^2+2 \mathrm{k}+7\right)\lt 0\) \(\mathrm{k}^2-6 \mathrm{k}+8\lt 0\) \(\mathrm{k}^2-4 \mathrm{k}-2 \mathrm{k}+8\lt 0\) \(\mathrm{k}(\mathrm{k}-4)-2(\mathrm{k}-4)\lt 0\) \((\mathrm{k}-2)(\mathrm{k}-4)\lt 0\) \(\therefore \quad 2\lt \mathrm{k}\lt 4\)So, integer value of \(\mathrm{k}\) is 3 .
J and K CET-2015
Complex Numbers and Quadratic Equation
118159
Let \(a, b, c\) be positive real number. if \(\frac{x^2-b x}{a x-c}=\frac{m-1}{m+1}\) has two roots which are numerically equal but opposite in sign. Then the value of ' \(\mathrm{m}\) ' is
1 \(\mathrm{c}\)
2 \(\frac{1}{c}\)
3 \(\frac{a+b}{a-b}\)
4 \(\frac{a-b}{a+b}\)
Explanation:
D \(We have, \) \(\frac{x^2-b x}{a x-c}=\frac{m-1}{m+1}\) \(\left(\mathrm{x}^2-\mathrm{bx}\right)=(\mathrm{ax}-\mathrm{c})\left(\frac{\mathrm{m}-1}{\mathrm{~m}+1}\right)\) \(\left(\mathrm{x}^2-\mathrm{bx}\right)=\mathrm{ax}\left(\frac{\mathrm{m}-1}{\mathrm{~m}+1}\right)-\mathrm{c}\left(\frac{\mathrm{m}-1}{\mathrm{~m}+1}\right)\) \(x^2-\left(b+a\left(\frac{m-1}{m+1}\right)\right) x+c\left(\frac{m-1}{m+1}\right)=0\) \(\therefore \quad b+\left(\frac{m-1}{m+1}\right) a=0\) \(\mathrm{~m}=\frac{\mathrm{a}-\mathrm{b}}{\mathrm{a}+\mathrm{b}}\)If roots are equal in magnitude but opposite in sign then sum of roots is zero.
AP EAMCET-23.08.2021
Complex Numbers and Quadratic Equation
118160
If \(\alpha, \beta, \gamma\) are the roots of the equation \(x^3+4 x-\) \(19=0\). Then the value of \(\frac{\alpha^3}{19-4 \alpha}+\frac{\beta^3}{19-4 \beta}+\frac{\gamma^3}{19-4 \gamma}=\)
1 0
2 3
3 -3
4 2
Explanation:
B \(We have the equation, \mathrm{x}^3+4 \mathrm{x}-19=0\) \(\text { If, } \alpha, \beta, \gamma \text { are roots of the above equation then, }\) \(\alpha+\beta+\gamma=0\) \(\alpha \beta+\beta \gamma+\gamma \alpha=4\) \(\alpha \beta \gamma=19\) \(\therefore \alpha(\beta+\gamma)+\beta \gamma=4\) \(\alpha(-\alpha)+\beta \gamma=4\) \(-\alpha^2+\frac{19}{\alpha}=4\) \(-\alpha^3+19=4 \alpha\) \(\alpha^3=19-4 \alpha\) \(\left(\frac{\alpha^3}{19-4 \alpha}\right)=1\) \(\text { Similarly, }\left(\frac{\beta^3}{19-4 \beta}\right)=1 \text { and }\left(\frac{\gamma^3}{19-4 \gamma}\right)=1\) \(\frac{\alpha^3}{19-4 \alpha}+\frac{\beta^3}{19-4 \beta}+\frac{\gamma^3}{19-4 \gamma}=1+1+1\) \(\frac{\alpha^3}{19-4 \alpha}+\frac{\beta^3}{19-4 \beta}+\frac{\gamma^3}{19-4 \gamma}=3\)
AP EAMCET-23.08.2021
Complex Numbers and Quadratic Equation
118161
The number of real roots of \(3^{2 \mathrm{x}^2-7 \mathrm{x}+7}=9\) is
1 0
2 2
3 1
4 4
Explanation:
B Given the equation \(3^{2 \mathrm{x}^2-7 \mathrm{x}+7}=9=(3)^2\) \(\therefore \quad 2 \mathrm{x}^2-7 \mathrm{x}+7=2\) \(\begin{array}{ll}\therefore 2 x^2-7 x+5=0\end{array}\) Or \(\quad 2 x^2-5 x-2 x+5=0\) Or \(\quad x(2 x-5)-1(2 x-5)=0\) Or \(\quad(2 x-5)(x-1)=0\) Or \(\quad(2 \mathrm{x}-5)(\mathrm{x}-1)=0\) \(\therefore \quad \mathrm{x}=1, \quad \mathrm{x}=5 / 2\) \(\therefore\) Number of real roots are two \(\therefore\) Option (b) is correct.
AIEEE-2002
Complex Numbers and Quadratic Equation
118162
The value of ' \(a\) ' for which one root of the quadratic equation \(\left(a^2-5 a+3\right) x^2+(3 a-1) x+\) \(\mathbf{2}=\mathbf{0}\) is twice as large as the other, is
1 \(2 / 3\)
2 \(-2 / 3\)
3 \(1 / 3\)
4 \(-1 / 3\)
Explanation:
A Given the quadratic equation \(\left(a^2-5 a+3\right) x^2+(3 a-1) x+2=0\) Let \(\alpha\) and \(\beta\) are the roots to the equation and \(\alpha=2 \beta\). \(\therefore \quad \alpha+\beta=\frac{-(3 a-1)}{a^2-5 a+3}\) \(\alpha \cdot \beta=\frac{2}{a^2-5 a+3}\) \(\because \quad \alpha=2 \beta\) \(\therefore \quad 2 \beta+\beta=\frac{-(3 a-1)}{a^2-5 a+3}=3 \beta\) \(\text { And } 2 \beta^2=\frac{2}{a^2-5 a+3}\) \(\Rightarrow \beta^2=\frac{1}{a^2-5 a+3}\) \(\Rightarrow 9 \beta^2=\frac{9}{a^2-5 a+3}=\frac{(3 a-1)^2}{\left(a^2-5 a+3\right)^2}\) \(\Rightarrow 9\left(a^2-5 a+3\right)=9 a^2+1-6 a\) \(\Rightarrow 9 a^2-45 a+27=9 a^2+1-6 a\) \(\Rightarrow 39 a=26\) \(a=\frac{26}{39}=\frac{2}{3}\) \(\therefore\) Option (a) is correct.
NEET Test Series from KOTA - 10 Papers In MS WORD
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Complex Numbers and Quadratic Equation
118158
The number of integer value (s) of \(k\) for which the expression \(x^2-2(4 k-1) x+15 k^2-2 k-7>\) 0 for every real number \(x\) is/are
1 none
2 one
3 finitely many greater than 1
4 infinitely many
Explanation:
B \(\mathrm{x}^2-2(4 \mathrm{k}-1) \mathrm{x}+15 \mathrm{k}^2-2 \mathrm{k}-7>0\) If value of quadratic function is greater then zero then discriminant is less than zero. \(\therefore \quad \mathrm{b}^2-4 \mathrm{ac}\lt 0\) \({[2(4 \mathrm{k}-1)]^2-4 \times 1 \times\left(15 \mathrm{k}^2-2 \mathrm{k}-7\right)\lt 0}\) \(4\left(16 \mathrm{k}^2+1-8 \mathrm{k}\right)-4 \times\left(15 \mathrm{k}^2-2 \mathrm{k}-7\right)\lt 0\) \(4\left(16 \mathrm{k}^2+1-8 \mathrm{k}-15 \mathrm{k}^2+2 \mathrm{k}+7\right)\lt 0\) \(\mathrm{k}^2-6 \mathrm{k}+8\lt 0\) \(\mathrm{k}^2-4 \mathrm{k}-2 \mathrm{k}+8\lt 0\) \(\mathrm{k}(\mathrm{k}-4)-2(\mathrm{k}-4)\lt 0\) \((\mathrm{k}-2)(\mathrm{k}-4)\lt 0\) \(\therefore \quad 2\lt \mathrm{k}\lt 4\)So, integer value of \(\mathrm{k}\) is 3 .
J and K CET-2015
Complex Numbers and Quadratic Equation
118159
Let \(a, b, c\) be positive real number. if \(\frac{x^2-b x}{a x-c}=\frac{m-1}{m+1}\) has two roots which are numerically equal but opposite in sign. Then the value of ' \(\mathrm{m}\) ' is
1 \(\mathrm{c}\)
2 \(\frac{1}{c}\)
3 \(\frac{a+b}{a-b}\)
4 \(\frac{a-b}{a+b}\)
Explanation:
D \(We have, \) \(\frac{x^2-b x}{a x-c}=\frac{m-1}{m+1}\) \(\left(\mathrm{x}^2-\mathrm{bx}\right)=(\mathrm{ax}-\mathrm{c})\left(\frac{\mathrm{m}-1}{\mathrm{~m}+1}\right)\) \(\left(\mathrm{x}^2-\mathrm{bx}\right)=\mathrm{ax}\left(\frac{\mathrm{m}-1}{\mathrm{~m}+1}\right)-\mathrm{c}\left(\frac{\mathrm{m}-1}{\mathrm{~m}+1}\right)\) \(x^2-\left(b+a\left(\frac{m-1}{m+1}\right)\right) x+c\left(\frac{m-1}{m+1}\right)=0\) \(\therefore \quad b+\left(\frac{m-1}{m+1}\right) a=0\) \(\mathrm{~m}=\frac{\mathrm{a}-\mathrm{b}}{\mathrm{a}+\mathrm{b}}\)If roots are equal in magnitude but opposite in sign then sum of roots is zero.
AP EAMCET-23.08.2021
Complex Numbers and Quadratic Equation
118160
If \(\alpha, \beta, \gamma\) are the roots of the equation \(x^3+4 x-\) \(19=0\). Then the value of \(\frac{\alpha^3}{19-4 \alpha}+\frac{\beta^3}{19-4 \beta}+\frac{\gamma^3}{19-4 \gamma}=\)
1 0
2 3
3 -3
4 2
Explanation:
B \(We have the equation, \mathrm{x}^3+4 \mathrm{x}-19=0\) \(\text { If, } \alpha, \beta, \gamma \text { are roots of the above equation then, }\) \(\alpha+\beta+\gamma=0\) \(\alpha \beta+\beta \gamma+\gamma \alpha=4\) \(\alpha \beta \gamma=19\) \(\therefore \alpha(\beta+\gamma)+\beta \gamma=4\) \(\alpha(-\alpha)+\beta \gamma=4\) \(-\alpha^2+\frac{19}{\alpha}=4\) \(-\alpha^3+19=4 \alpha\) \(\alpha^3=19-4 \alpha\) \(\left(\frac{\alpha^3}{19-4 \alpha}\right)=1\) \(\text { Similarly, }\left(\frac{\beta^3}{19-4 \beta}\right)=1 \text { and }\left(\frac{\gamma^3}{19-4 \gamma}\right)=1\) \(\frac{\alpha^3}{19-4 \alpha}+\frac{\beta^3}{19-4 \beta}+\frac{\gamma^3}{19-4 \gamma}=1+1+1\) \(\frac{\alpha^3}{19-4 \alpha}+\frac{\beta^3}{19-4 \beta}+\frac{\gamma^3}{19-4 \gamma}=3\)
AP EAMCET-23.08.2021
Complex Numbers and Quadratic Equation
118161
The number of real roots of \(3^{2 \mathrm{x}^2-7 \mathrm{x}+7}=9\) is
1 0
2 2
3 1
4 4
Explanation:
B Given the equation \(3^{2 \mathrm{x}^2-7 \mathrm{x}+7}=9=(3)^2\) \(\therefore \quad 2 \mathrm{x}^2-7 \mathrm{x}+7=2\) \(\begin{array}{ll}\therefore 2 x^2-7 x+5=0\end{array}\) Or \(\quad 2 x^2-5 x-2 x+5=0\) Or \(\quad x(2 x-5)-1(2 x-5)=0\) Or \(\quad(2 x-5)(x-1)=0\) Or \(\quad(2 \mathrm{x}-5)(\mathrm{x}-1)=0\) \(\therefore \quad \mathrm{x}=1, \quad \mathrm{x}=5 / 2\) \(\therefore\) Number of real roots are two \(\therefore\) Option (b) is correct.
AIEEE-2002
Complex Numbers and Quadratic Equation
118162
The value of ' \(a\) ' for which one root of the quadratic equation \(\left(a^2-5 a+3\right) x^2+(3 a-1) x+\) \(\mathbf{2}=\mathbf{0}\) is twice as large as the other, is
1 \(2 / 3\)
2 \(-2 / 3\)
3 \(1 / 3\)
4 \(-1 / 3\)
Explanation:
A Given the quadratic equation \(\left(a^2-5 a+3\right) x^2+(3 a-1) x+2=0\) Let \(\alpha\) and \(\beta\) are the roots to the equation and \(\alpha=2 \beta\). \(\therefore \quad \alpha+\beta=\frac{-(3 a-1)}{a^2-5 a+3}\) \(\alpha \cdot \beta=\frac{2}{a^2-5 a+3}\) \(\because \quad \alpha=2 \beta\) \(\therefore \quad 2 \beta+\beta=\frac{-(3 a-1)}{a^2-5 a+3}=3 \beta\) \(\text { And } 2 \beta^2=\frac{2}{a^2-5 a+3}\) \(\Rightarrow \beta^2=\frac{1}{a^2-5 a+3}\) \(\Rightarrow 9 \beta^2=\frac{9}{a^2-5 a+3}=\frac{(3 a-1)^2}{\left(a^2-5 a+3\right)^2}\) \(\Rightarrow 9\left(a^2-5 a+3\right)=9 a^2+1-6 a\) \(\Rightarrow 9 a^2-45 a+27=9 a^2+1-6 a\) \(\Rightarrow 39 a=26\) \(a=\frac{26}{39}=\frac{2}{3}\) \(\therefore\) Option (a) is correct.
118158
The number of integer value (s) of \(k\) for which the expression \(x^2-2(4 k-1) x+15 k^2-2 k-7>\) 0 for every real number \(x\) is/are
1 none
2 one
3 finitely many greater than 1
4 infinitely many
Explanation:
B \(\mathrm{x}^2-2(4 \mathrm{k}-1) \mathrm{x}+15 \mathrm{k}^2-2 \mathrm{k}-7>0\) If value of quadratic function is greater then zero then discriminant is less than zero. \(\therefore \quad \mathrm{b}^2-4 \mathrm{ac}\lt 0\) \({[2(4 \mathrm{k}-1)]^2-4 \times 1 \times\left(15 \mathrm{k}^2-2 \mathrm{k}-7\right)\lt 0}\) \(4\left(16 \mathrm{k}^2+1-8 \mathrm{k}\right)-4 \times\left(15 \mathrm{k}^2-2 \mathrm{k}-7\right)\lt 0\) \(4\left(16 \mathrm{k}^2+1-8 \mathrm{k}-15 \mathrm{k}^2+2 \mathrm{k}+7\right)\lt 0\) \(\mathrm{k}^2-6 \mathrm{k}+8\lt 0\) \(\mathrm{k}^2-4 \mathrm{k}-2 \mathrm{k}+8\lt 0\) \(\mathrm{k}(\mathrm{k}-4)-2(\mathrm{k}-4)\lt 0\) \((\mathrm{k}-2)(\mathrm{k}-4)\lt 0\) \(\therefore \quad 2\lt \mathrm{k}\lt 4\)So, integer value of \(\mathrm{k}\) is 3 .
J and K CET-2015
Complex Numbers and Quadratic Equation
118159
Let \(a, b, c\) be positive real number. if \(\frac{x^2-b x}{a x-c}=\frac{m-1}{m+1}\) has two roots which are numerically equal but opposite in sign. Then the value of ' \(\mathrm{m}\) ' is
1 \(\mathrm{c}\)
2 \(\frac{1}{c}\)
3 \(\frac{a+b}{a-b}\)
4 \(\frac{a-b}{a+b}\)
Explanation:
D \(We have, \) \(\frac{x^2-b x}{a x-c}=\frac{m-1}{m+1}\) \(\left(\mathrm{x}^2-\mathrm{bx}\right)=(\mathrm{ax}-\mathrm{c})\left(\frac{\mathrm{m}-1}{\mathrm{~m}+1}\right)\) \(\left(\mathrm{x}^2-\mathrm{bx}\right)=\mathrm{ax}\left(\frac{\mathrm{m}-1}{\mathrm{~m}+1}\right)-\mathrm{c}\left(\frac{\mathrm{m}-1}{\mathrm{~m}+1}\right)\) \(x^2-\left(b+a\left(\frac{m-1}{m+1}\right)\right) x+c\left(\frac{m-1}{m+1}\right)=0\) \(\therefore \quad b+\left(\frac{m-1}{m+1}\right) a=0\) \(\mathrm{~m}=\frac{\mathrm{a}-\mathrm{b}}{\mathrm{a}+\mathrm{b}}\)If roots are equal in magnitude but opposite in sign then sum of roots is zero.
AP EAMCET-23.08.2021
Complex Numbers and Quadratic Equation
118160
If \(\alpha, \beta, \gamma\) are the roots of the equation \(x^3+4 x-\) \(19=0\). Then the value of \(\frac{\alpha^3}{19-4 \alpha}+\frac{\beta^3}{19-4 \beta}+\frac{\gamma^3}{19-4 \gamma}=\)
1 0
2 3
3 -3
4 2
Explanation:
B \(We have the equation, \mathrm{x}^3+4 \mathrm{x}-19=0\) \(\text { If, } \alpha, \beta, \gamma \text { are roots of the above equation then, }\) \(\alpha+\beta+\gamma=0\) \(\alpha \beta+\beta \gamma+\gamma \alpha=4\) \(\alpha \beta \gamma=19\) \(\therefore \alpha(\beta+\gamma)+\beta \gamma=4\) \(\alpha(-\alpha)+\beta \gamma=4\) \(-\alpha^2+\frac{19}{\alpha}=4\) \(-\alpha^3+19=4 \alpha\) \(\alpha^3=19-4 \alpha\) \(\left(\frac{\alpha^3}{19-4 \alpha}\right)=1\) \(\text { Similarly, }\left(\frac{\beta^3}{19-4 \beta}\right)=1 \text { and }\left(\frac{\gamma^3}{19-4 \gamma}\right)=1\) \(\frac{\alpha^3}{19-4 \alpha}+\frac{\beta^3}{19-4 \beta}+\frac{\gamma^3}{19-4 \gamma}=1+1+1\) \(\frac{\alpha^3}{19-4 \alpha}+\frac{\beta^3}{19-4 \beta}+\frac{\gamma^3}{19-4 \gamma}=3\)
AP EAMCET-23.08.2021
Complex Numbers and Quadratic Equation
118161
The number of real roots of \(3^{2 \mathrm{x}^2-7 \mathrm{x}+7}=9\) is
1 0
2 2
3 1
4 4
Explanation:
B Given the equation \(3^{2 \mathrm{x}^2-7 \mathrm{x}+7}=9=(3)^2\) \(\therefore \quad 2 \mathrm{x}^2-7 \mathrm{x}+7=2\) \(\begin{array}{ll}\therefore 2 x^2-7 x+5=0\end{array}\) Or \(\quad 2 x^2-5 x-2 x+5=0\) Or \(\quad x(2 x-5)-1(2 x-5)=0\) Or \(\quad(2 x-5)(x-1)=0\) Or \(\quad(2 \mathrm{x}-5)(\mathrm{x}-1)=0\) \(\therefore \quad \mathrm{x}=1, \quad \mathrm{x}=5 / 2\) \(\therefore\) Number of real roots are two \(\therefore\) Option (b) is correct.
AIEEE-2002
Complex Numbers and Quadratic Equation
118162
The value of ' \(a\) ' for which one root of the quadratic equation \(\left(a^2-5 a+3\right) x^2+(3 a-1) x+\) \(\mathbf{2}=\mathbf{0}\) is twice as large as the other, is
1 \(2 / 3\)
2 \(-2 / 3\)
3 \(1 / 3\)
4 \(-1 / 3\)
Explanation:
A Given the quadratic equation \(\left(a^2-5 a+3\right) x^2+(3 a-1) x+2=0\) Let \(\alpha\) and \(\beta\) are the roots to the equation and \(\alpha=2 \beta\). \(\therefore \quad \alpha+\beta=\frac{-(3 a-1)}{a^2-5 a+3}\) \(\alpha \cdot \beta=\frac{2}{a^2-5 a+3}\) \(\because \quad \alpha=2 \beta\) \(\therefore \quad 2 \beta+\beta=\frac{-(3 a-1)}{a^2-5 a+3}=3 \beta\) \(\text { And } 2 \beta^2=\frac{2}{a^2-5 a+3}\) \(\Rightarrow \beta^2=\frac{1}{a^2-5 a+3}\) \(\Rightarrow 9 \beta^2=\frac{9}{a^2-5 a+3}=\frac{(3 a-1)^2}{\left(a^2-5 a+3\right)^2}\) \(\Rightarrow 9\left(a^2-5 a+3\right)=9 a^2+1-6 a\) \(\Rightarrow 9 a^2-45 a+27=9 a^2+1-6 a\) \(\Rightarrow 39 a=26\) \(a=\frac{26}{39}=\frac{2}{3}\) \(\therefore\) Option (a) is correct.
118158
The number of integer value (s) of \(k\) for which the expression \(x^2-2(4 k-1) x+15 k^2-2 k-7>\) 0 for every real number \(x\) is/are
1 none
2 one
3 finitely many greater than 1
4 infinitely many
Explanation:
B \(\mathrm{x}^2-2(4 \mathrm{k}-1) \mathrm{x}+15 \mathrm{k}^2-2 \mathrm{k}-7>0\) If value of quadratic function is greater then zero then discriminant is less than zero. \(\therefore \quad \mathrm{b}^2-4 \mathrm{ac}\lt 0\) \({[2(4 \mathrm{k}-1)]^2-4 \times 1 \times\left(15 \mathrm{k}^2-2 \mathrm{k}-7\right)\lt 0}\) \(4\left(16 \mathrm{k}^2+1-8 \mathrm{k}\right)-4 \times\left(15 \mathrm{k}^2-2 \mathrm{k}-7\right)\lt 0\) \(4\left(16 \mathrm{k}^2+1-8 \mathrm{k}-15 \mathrm{k}^2+2 \mathrm{k}+7\right)\lt 0\) \(\mathrm{k}^2-6 \mathrm{k}+8\lt 0\) \(\mathrm{k}^2-4 \mathrm{k}-2 \mathrm{k}+8\lt 0\) \(\mathrm{k}(\mathrm{k}-4)-2(\mathrm{k}-4)\lt 0\) \((\mathrm{k}-2)(\mathrm{k}-4)\lt 0\) \(\therefore \quad 2\lt \mathrm{k}\lt 4\)So, integer value of \(\mathrm{k}\) is 3 .
J and K CET-2015
Complex Numbers and Quadratic Equation
118159
Let \(a, b, c\) be positive real number. if \(\frac{x^2-b x}{a x-c}=\frac{m-1}{m+1}\) has two roots which are numerically equal but opposite in sign. Then the value of ' \(\mathrm{m}\) ' is
1 \(\mathrm{c}\)
2 \(\frac{1}{c}\)
3 \(\frac{a+b}{a-b}\)
4 \(\frac{a-b}{a+b}\)
Explanation:
D \(We have, \) \(\frac{x^2-b x}{a x-c}=\frac{m-1}{m+1}\) \(\left(\mathrm{x}^2-\mathrm{bx}\right)=(\mathrm{ax}-\mathrm{c})\left(\frac{\mathrm{m}-1}{\mathrm{~m}+1}\right)\) \(\left(\mathrm{x}^2-\mathrm{bx}\right)=\mathrm{ax}\left(\frac{\mathrm{m}-1}{\mathrm{~m}+1}\right)-\mathrm{c}\left(\frac{\mathrm{m}-1}{\mathrm{~m}+1}\right)\) \(x^2-\left(b+a\left(\frac{m-1}{m+1}\right)\right) x+c\left(\frac{m-1}{m+1}\right)=0\) \(\therefore \quad b+\left(\frac{m-1}{m+1}\right) a=0\) \(\mathrm{~m}=\frac{\mathrm{a}-\mathrm{b}}{\mathrm{a}+\mathrm{b}}\)If roots are equal in magnitude but opposite in sign then sum of roots is zero.
AP EAMCET-23.08.2021
Complex Numbers and Quadratic Equation
118160
If \(\alpha, \beta, \gamma\) are the roots of the equation \(x^3+4 x-\) \(19=0\). Then the value of \(\frac{\alpha^3}{19-4 \alpha}+\frac{\beta^3}{19-4 \beta}+\frac{\gamma^3}{19-4 \gamma}=\)
1 0
2 3
3 -3
4 2
Explanation:
B \(We have the equation, \mathrm{x}^3+4 \mathrm{x}-19=0\) \(\text { If, } \alpha, \beta, \gamma \text { are roots of the above equation then, }\) \(\alpha+\beta+\gamma=0\) \(\alpha \beta+\beta \gamma+\gamma \alpha=4\) \(\alpha \beta \gamma=19\) \(\therefore \alpha(\beta+\gamma)+\beta \gamma=4\) \(\alpha(-\alpha)+\beta \gamma=4\) \(-\alpha^2+\frac{19}{\alpha}=4\) \(-\alpha^3+19=4 \alpha\) \(\alpha^3=19-4 \alpha\) \(\left(\frac{\alpha^3}{19-4 \alpha}\right)=1\) \(\text { Similarly, }\left(\frac{\beta^3}{19-4 \beta}\right)=1 \text { and }\left(\frac{\gamma^3}{19-4 \gamma}\right)=1\) \(\frac{\alpha^3}{19-4 \alpha}+\frac{\beta^3}{19-4 \beta}+\frac{\gamma^3}{19-4 \gamma}=1+1+1\) \(\frac{\alpha^3}{19-4 \alpha}+\frac{\beta^3}{19-4 \beta}+\frac{\gamma^3}{19-4 \gamma}=3\)
AP EAMCET-23.08.2021
Complex Numbers and Quadratic Equation
118161
The number of real roots of \(3^{2 \mathrm{x}^2-7 \mathrm{x}+7}=9\) is
1 0
2 2
3 1
4 4
Explanation:
B Given the equation \(3^{2 \mathrm{x}^2-7 \mathrm{x}+7}=9=(3)^2\) \(\therefore \quad 2 \mathrm{x}^2-7 \mathrm{x}+7=2\) \(\begin{array}{ll}\therefore 2 x^2-7 x+5=0\end{array}\) Or \(\quad 2 x^2-5 x-2 x+5=0\) Or \(\quad x(2 x-5)-1(2 x-5)=0\) Or \(\quad(2 x-5)(x-1)=0\) Or \(\quad(2 \mathrm{x}-5)(\mathrm{x}-1)=0\) \(\therefore \quad \mathrm{x}=1, \quad \mathrm{x}=5 / 2\) \(\therefore\) Number of real roots are two \(\therefore\) Option (b) is correct.
AIEEE-2002
Complex Numbers and Quadratic Equation
118162
The value of ' \(a\) ' for which one root of the quadratic equation \(\left(a^2-5 a+3\right) x^2+(3 a-1) x+\) \(\mathbf{2}=\mathbf{0}\) is twice as large as the other, is
1 \(2 / 3\)
2 \(-2 / 3\)
3 \(1 / 3\)
4 \(-1 / 3\)
Explanation:
A Given the quadratic equation \(\left(a^2-5 a+3\right) x^2+(3 a-1) x+2=0\) Let \(\alpha\) and \(\beta\) are the roots to the equation and \(\alpha=2 \beta\). \(\therefore \quad \alpha+\beta=\frac{-(3 a-1)}{a^2-5 a+3}\) \(\alpha \cdot \beta=\frac{2}{a^2-5 a+3}\) \(\because \quad \alpha=2 \beta\) \(\therefore \quad 2 \beta+\beta=\frac{-(3 a-1)}{a^2-5 a+3}=3 \beta\) \(\text { And } 2 \beta^2=\frac{2}{a^2-5 a+3}\) \(\Rightarrow \beta^2=\frac{1}{a^2-5 a+3}\) \(\Rightarrow 9 \beta^2=\frac{9}{a^2-5 a+3}=\frac{(3 a-1)^2}{\left(a^2-5 a+3\right)^2}\) \(\Rightarrow 9\left(a^2-5 a+3\right)=9 a^2+1-6 a\) \(\Rightarrow 9 a^2-45 a+27=9 a^2+1-6 a\) \(\Rightarrow 39 a=26\) \(a=\frac{26}{39}=\frac{2}{3}\) \(\therefore\) Option (a) is correct.