NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Complex Numbers and Quadratic Equation
118134
If \(\alpha, \beta, \gamma\) are the roots of \(x_2^3-2 x_2^2+3 x-4=0\), then the value of \(\alpha^2 \beta^2+\beta^2 \gamma^2+\gamma^2 \alpha^2\) is
1 -7
2 -5
3 -3
4 0
Explanation:
A Given \( \alpha, \beta \text { and } \gamma \text { are the roots of the equation }\) \(\mathrm{x}^3-2 \mathrm{x}^2+3 \mathrm{x}-4=0\) \(\therefore \alpha+\beta+\gamma=2\) \(\alpha \beta+\beta \gamma+\alpha \gamma=3\) \(\alpha \beta \gamma=4\) \(\text { We know that }\) \((\alpha \beta+\beta \gamma+\alpha \gamma)^2=\alpha^2 \beta^2+\beta^2 \gamma^2+\alpha^2 \gamma^2+2 \alpha \beta \gamma(\alpha+\beta+\gamma)\) \(\therefore \alpha^2 \beta^2+\beta^2 \gamma^2+\alpha^2 \gamma^2=(\alpha \beta+\beta \gamma+\alpha \gamma)^2-2(\alpha \beta \gamma)(\alpha+\beta\) \(+\gamma)\) \(\text { Now, put the above values }\) \((\alpha \beta+\beta \gamma+\alpha \gamma)^2=3^2-2 \times 4 \times 2\) \((\alpha \beta+\beta \gamma+\alpha \gamma)^2=9-16=-7\)Now, put the above values
AP EAMCET-2007
Complex Numbers and Quadratic Equation
118135
The set of solutions satisfying both \(x^2+5 x+6 \geq 0 \text { and } x^2+3 x-4\lt 0 \text { is }\)
1 \((-4,1)\)
2 \((-4,-3] \cup[-2,1)\)
3 \((-4,-3) \cup(-2,1)\)
4 \([-4,-3] \cup[-2,1]\)
Explanation:
B We have the inequalities \(x^2+5 x+6 \geq 0\) and \(x^2+3 x-4\lt 0\) \(\mathrm{x}^2+3 \mathrm{x}+2 \mathrm{x}+6 \geq 0\) \((\mathrm{x}+3)(\mathrm{x}+2) \geq 0\) \(\Rightarrow \mathrm{x} \geq-2\) and \(\mathrm{x} \leq-3 \mathrm{x} \in(-\infty,-3] \cup[-2, \infty)\) Also, \(\quad x^2+3 x-4\lt 0\) \(x^2+4 x-x-4\lt 0\) \(x(x+4)-1(x+4)\lt 0\) \((x+4)(x-1)\lt 0\) \(-4\lt x\lt 1 \Rightarrow x \in(-4,1)\) Combination of (i) and (ii) is \(x \in(-4,-3] \cup[-2,1)\)\(\therefore\) Option (b) is correct.
AP EAMCET-2013
Complex Numbers and Quadratic Equation
118136
If the roots of \(x^3-42 x^2+336 x-512=0\), are in increasing geometric progression, then its common ratio is
1 \(2: 1\)
2 \(3: 1\)
3 \(4: 1\)
4 \(6: 1\)
Explanation:
C Given the equation. \(x^3-42 x^2+336 x-512=0\) If \(\alpha, \beta, \gamma\) are the roots of the given equation Then \(\alpha+\beta+\gamma=42\) \(\alpha \beta+\beta \gamma+\gamma \alpha=336\) \(\alpha \beta . \gamma=512\) Given \(\frac{\beta}{\alpha}=\frac{\gamma}{\beta}=\mathrm{r}\) \(\Rightarrow \beta=\mathrm{r} \cdot \alpha\) \(\therefore \gamma=\mathrm{r}^2 \alpha \quad(\because \gamma=\beta \times \mathrm{r})\) \(\therefore \alpha+\beta+\gamma=\alpha+\mathrm{r} \alpha+\mathrm{r}^2 \alpha=42\) \(\therefore \alpha\left(1+r+r^2\right)=42\) or \(\left(1+r+r^2\right)=\frac{42}{\alpha}\) and also, \(\alpha \cdot \mathrm{r} \alpha \cdot \mathrm{r}^2 \alpha=512\) \(\therefore(\text { r. } \alpha)^3=512\) \(\text { r. } \alpha=8\) \(\therefore \alpha=8 / r\) Solving equation (i) and equation (ii) we get \(\left(1+r+r^2\right)=\frac{42}{8} \times r=\frac{21}{4} r\) \(4+4 r+4 r^2=21 r\) \(4 r^2-17 r+4=0\) \(r=4,1 / 4 \text { But } r>1\) \(\therefore r=4 / 1\)
AP EAMCET-2013
Complex Numbers and Quadratic Equation
118137
If \(\alpha\) and \(\beta\) are the roots of the equation \(x^2-2 x\) \(+4=0\), then \(\alpha^9+\beta^9\) is equal to
118134
If \(\alpha, \beta, \gamma\) are the roots of \(x_2^3-2 x_2^2+3 x-4=0\), then the value of \(\alpha^2 \beta^2+\beta^2 \gamma^2+\gamma^2 \alpha^2\) is
1 -7
2 -5
3 -3
4 0
Explanation:
A Given \( \alpha, \beta \text { and } \gamma \text { are the roots of the equation }\) \(\mathrm{x}^3-2 \mathrm{x}^2+3 \mathrm{x}-4=0\) \(\therefore \alpha+\beta+\gamma=2\) \(\alpha \beta+\beta \gamma+\alpha \gamma=3\) \(\alpha \beta \gamma=4\) \(\text { We know that }\) \((\alpha \beta+\beta \gamma+\alpha \gamma)^2=\alpha^2 \beta^2+\beta^2 \gamma^2+\alpha^2 \gamma^2+2 \alpha \beta \gamma(\alpha+\beta+\gamma)\) \(\therefore \alpha^2 \beta^2+\beta^2 \gamma^2+\alpha^2 \gamma^2=(\alpha \beta+\beta \gamma+\alpha \gamma)^2-2(\alpha \beta \gamma)(\alpha+\beta\) \(+\gamma)\) \(\text { Now, put the above values }\) \((\alpha \beta+\beta \gamma+\alpha \gamma)^2=3^2-2 \times 4 \times 2\) \((\alpha \beta+\beta \gamma+\alpha \gamma)^2=9-16=-7\)Now, put the above values
AP EAMCET-2007
Complex Numbers and Quadratic Equation
118135
The set of solutions satisfying both \(x^2+5 x+6 \geq 0 \text { and } x^2+3 x-4\lt 0 \text { is }\)
1 \((-4,1)\)
2 \((-4,-3] \cup[-2,1)\)
3 \((-4,-3) \cup(-2,1)\)
4 \([-4,-3] \cup[-2,1]\)
Explanation:
B We have the inequalities \(x^2+5 x+6 \geq 0\) and \(x^2+3 x-4\lt 0\) \(\mathrm{x}^2+3 \mathrm{x}+2 \mathrm{x}+6 \geq 0\) \((\mathrm{x}+3)(\mathrm{x}+2) \geq 0\) \(\Rightarrow \mathrm{x} \geq-2\) and \(\mathrm{x} \leq-3 \mathrm{x} \in(-\infty,-3] \cup[-2, \infty)\) Also, \(\quad x^2+3 x-4\lt 0\) \(x^2+4 x-x-4\lt 0\) \(x(x+4)-1(x+4)\lt 0\) \((x+4)(x-1)\lt 0\) \(-4\lt x\lt 1 \Rightarrow x \in(-4,1)\) Combination of (i) and (ii) is \(x \in(-4,-3] \cup[-2,1)\)\(\therefore\) Option (b) is correct.
AP EAMCET-2013
Complex Numbers and Quadratic Equation
118136
If the roots of \(x^3-42 x^2+336 x-512=0\), are in increasing geometric progression, then its common ratio is
1 \(2: 1\)
2 \(3: 1\)
3 \(4: 1\)
4 \(6: 1\)
Explanation:
C Given the equation. \(x^3-42 x^2+336 x-512=0\) If \(\alpha, \beta, \gamma\) are the roots of the given equation Then \(\alpha+\beta+\gamma=42\) \(\alpha \beta+\beta \gamma+\gamma \alpha=336\) \(\alpha \beta . \gamma=512\) Given \(\frac{\beta}{\alpha}=\frac{\gamma}{\beta}=\mathrm{r}\) \(\Rightarrow \beta=\mathrm{r} \cdot \alpha\) \(\therefore \gamma=\mathrm{r}^2 \alpha \quad(\because \gamma=\beta \times \mathrm{r})\) \(\therefore \alpha+\beta+\gamma=\alpha+\mathrm{r} \alpha+\mathrm{r}^2 \alpha=42\) \(\therefore \alpha\left(1+r+r^2\right)=42\) or \(\left(1+r+r^2\right)=\frac{42}{\alpha}\) and also, \(\alpha \cdot \mathrm{r} \alpha \cdot \mathrm{r}^2 \alpha=512\) \(\therefore(\text { r. } \alpha)^3=512\) \(\text { r. } \alpha=8\) \(\therefore \alpha=8 / r\) Solving equation (i) and equation (ii) we get \(\left(1+r+r^2\right)=\frac{42}{8} \times r=\frac{21}{4} r\) \(4+4 r+4 r^2=21 r\) \(4 r^2-17 r+4=0\) \(r=4,1 / 4 \text { But } r>1\) \(\therefore r=4 / 1\)
AP EAMCET-2013
Complex Numbers and Quadratic Equation
118137
If \(\alpha\) and \(\beta\) are the roots of the equation \(x^2-2 x\) \(+4=0\), then \(\alpha^9+\beta^9\) is equal to
118134
If \(\alpha, \beta, \gamma\) are the roots of \(x_2^3-2 x_2^2+3 x-4=0\), then the value of \(\alpha^2 \beta^2+\beta^2 \gamma^2+\gamma^2 \alpha^2\) is
1 -7
2 -5
3 -3
4 0
Explanation:
A Given \( \alpha, \beta \text { and } \gamma \text { are the roots of the equation }\) \(\mathrm{x}^3-2 \mathrm{x}^2+3 \mathrm{x}-4=0\) \(\therefore \alpha+\beta+\gamma=2\) \(\alpha \beta+\beta \gamma+\alpha \gamma=3\) \(\alpha \beta \gamma=4\) \(\text { We know that }\) \((\alpha \beta+\beta \gamma+\alpha \gamma)^2=\alpha^2 \beta^2+\beta^2 \gamma^2+\alpha^2 \gamma^2+2 \alpha \beta \gamma(\alpha+\beta+\gamma)\) \(\therefore \alpha^2 \beta^2+\beta^2 \gamma^2+\alpha^2 \gamma^2=(\alpha \beta+\beta \gamma+\alpha \gamma)^2-2(\alpha \beta \gamma)(\alpha+\beta\) \(+\gamma)\) \(\text { Now, put the above values }\) \((\alpha \beta+\beta \gamma+\alpha \gamma)^2=3^2-2 \times 4 \times 2\) \((\alpha \beta+\beta \gamma+\alpha \gamma)^2=9-16=-7\)Now, put the above values
AP EAMCET-2007
Complex Numbers and Quadratic Equation
118135
The set of solutions satisfying both \(x^2+5 x+6 \geq 0 \text { and } x^2+3 x-4\lt 0 \text { is }\)
1 \((-4,1)\)
2 \((-4,-3] \cup[-2,1)\)
3 \((-4,-3) \cup(-2,1)\)
4 \([-4,-3] \cup[-2,1]\)
Explanation:
B We have the inequalities \(x^2+5 x+6 \geq 0\) and \(x^2+3 x-4\lt 0\) \(\mathrm{x}^2+3 \mathrm{x}+2 \mathrm{x}+6 \geq 0\) \((\mathrm{x}+3)(\mathrm{x}+2) \geq 0\) \(\Rightarrow \mathrm{x} \geq-2\) and \(\mathrm{x} \leq-3 \mathrm{x} \in(-\infty,-3] \cup[-2, \infty)\) Also, \(\quad x^2+3 x-4\lt 0\) \(x^2+4 x-x-4\lt 0\) \(x(x+4)-1(x+4)\lt 0\) \((x+4)(x-1)\lt 0\) \(-4\lt x\lt 1 \Rightarrow x \in(-4,1)\) Combination of (i) and (ii) is \(x \in(-4,-3] \cup[-2,1)\)\(\therefore\) Option (b) is correct.
AP EAMCET-2013
Complex Numbers and Quadratic Equation
118136
If the roots of \(x^3-42 x^2+336 x-512=0\), are in increasing geometric progression, then its common ratio is
1 \(2: 1\)
2 \(3: 1\)
3 \(4: 1\)
4 \(6: 1\)
Explanation:
C Given the equation. \(x^3-42 x^2+336 x-512=0\) If \(\alpha, \beta, \gamma\) are the roots of the given equation Then \(\alpha+\beta+\gamma=42\) \(\alpha \beta+\beta \gamma+\gamma \alpha=336\) \(\alpha \beta . \gamma=512\) Given \(\frac{\beta}{\alpha}=\frac{\gamma}{\beta}=\mathrm{r}\) \(\Rightarrow \beta=\mathrm{r} \cdot \alpha\) \(\therefore \gamma=\mathrm{r}^2 \alpha \quad(\because \gamma=\beta \times \mathrm{r})\) \(\therefore \alpha+\beta+\gamma=\alpha+\mathrm{r} \alpha+\mathrm{r}^2 \alpha=42\) \(\therefore \alpha\left(1+r+r^2\right)=42\) or \(\left(1+r+r^2\right)=\frac{42}{\alpha}\) and also, \(\alpha \cdot \mathrm{r} \alpha \cdot \mathrm{r}^2 \alpha=512\) \(\therefore(\text { r. } \alpha)^3=512\) \(\text { r. } \alpha=8\) \(\therefore \alpha=8 / r\) Solving equation (i) and equation (ii) we get \(\left(1+r+r^2\right)=\frac{42}{8} \times r=\frac{21}{4} r\) \(4+4 r+4 r^2=21 r\) \(4 r^2-17 r+4=0\) \(r=4,1 / 4 \text { But } r>1\) \(\therefore r=4 / 1\)
AP EAMCET-2013
Complex Numbers and Quadratic Equation
118137
If \(\alpha\) and \(\beta\) are the roots of the equation \(x^2-2 x\) \(+4=0\), then \(\alpha^9+\beta^9\) is equal to
118134
If \(\alpha, \beta, \gamma\) are the roots of \(x_2^3-2 x_2^2+3 x-4=0\), then the value of \(\alpha^2 \beta^2+\beta^2 \gamma^2+\gamma^2 \alpha^2\) is
1 -7
2 -5
3 -3
4 0
Explanation:
A Given \( \alpha, \beta \text { and } \gamma \text { are the roots of the equation }\) \(\mathrm{x}^3-2 \mathrm{x}^2+3 \mathrm{x}-4=0\) \(\therefore \alpha+\beta+\gamma=2\) \(\alpha \beta+\beta \gamma+\alpha \gamma=3\) \(\alpha \beta \gamma=4\) \(\text { We know that }\) \((\alpha \beta+\beta \gamma+\alpha \gamma)^2=\alpha^2 \beta^2+\beta^2 \gamma^2+\alpha^2 \gamma^2+2 \alpha \beta \gamma(\alpha+\beta+\gamma)\) \(\therefore \alpha^2 \beta^2+\beta^2 \gamma^2+\alpha^2 \gamma^2=(\alpha \beta+\beta \gamma+\alpha \gamma)^2-2(\alpha \beta \gamma)(\alpha+\beta\) \(+\gamma)\) \(\text { Now, put the above values }\) \((\alpha \beta+\beta \gamma+\alpha \gamma)^2=3^2-2 \times 4 \times 2\) \((\alpha \beta+\beta \gamma+\alpha \gamma)^2=9-16=-7\)Now, put the above values
AP EAMCET-2007
Complex Numbers and Quadratic Equation
118135
The set of solutions satisfying both \(x^2+5 x+6 \geq 0 \text { and } x^2+3 x-4\lt 0 \text { is }\)
1 \((-4,1)\)
2 \((-4,-3] \cup[-2,1)\)
3 \((-4,-3) \cup(-2,1)\)
4 \([-4,-3] \cup[-2,1]\)
Explanation:
B We have the inequalities \(x^2+5 x+6 \geq 0\) and \(x^2+3 x-4\lt 0\) \(\mathrm{x}^2+3 \mathrm{x}+2 \mathrm{x}+6 \geq 0\) \((\mathrm{x}+3)(\mathrm{x}+2) \geq 0\) \(\Rightarrow \mathrm{x} \geq-2\) and \(\mathrm{x} \leq-3 \mathrm{x} \in(-\infty,-3] \cup[-2, \infty)\) Also, \(\quad x^2+3 x-4\lt 0\) \(x^2+4 x-x-4\lt 0\) \(x(x+4)-1(x+4)\lt 0\) \((x+4)(x-1)\lt 0\) \(-4\lt x\lt 1 \Rightarrow x \in(-4,1)\) Combination of (i) and (ii) is \(x \in(-4,-3] \cup[-2,1)\)\(\therefore\) Option (b) is correct.
AP EAMCET-2013
Complex Numbers and Quadratic Equation
118136
If the roots of \(x^3-42 x^2+336 x-512=0\), are in increasing geometric progression, then its common ratio is
1 \(2: 1\)
2 \(3: 1\)
3 \(4: 1\)
4 \(6: 1\)
Explanation:
C Given the equation. \(x^3-42 x^2+336 x-512=0\) If \(\alpha, \beta, \gamma\) are the roots of the given equation Then \(\alpha+\beta+\gamma=42\) \(\alpha \beta+\beta \gamma+\gamma \alpha=336\) \(\alpha \beta . \gamma=512\) Given \(\frac{\beta}{\alpha}=\frac{\gamma}{\beta}=\mathrm{r}\) \(\Rightarrow \beta=\mathrm{r} \cdot \alpha\) \(\therefore \gamma=\mathrm{r}^2 \alpha \quad(\because \gamma=\beta \times \mathrm{r})\) \(\therefore \alpha+\beta+\gamma=\alpha+\mathrm{r} \alpha+\mathrm{r}^2 \alpha=42\) \(\therefore \alpha\left(1+r+r^2\right)=42\) or \(\left(1+r+r^2\right)=\frac{42}{\alpha}\) and also, \(\alpha \cdot \mathrm{r} \alpha \cdot \mathrm{r}^2 \alpha=512\) \(\therefore(\text { r. } \alpha)^3=512\) \(\text { r. } \alpha=8\) \(\therefore \alpha=8 / r\) Solving equation (i) and equation (ii) we get \(\left(1+r+r^2\right)=\frac{42}{8} \times r=\frac{21}{4} r\) \(4+4 r+4 r^2=21 r\) \(4 r^2-17 r+4=0\) \(r=4,1 / 4 \text { But } r>1\) \(\therefore r=4 / 1\)
AP EAMCET-2013
Complex Numbers and Quadratic Equation
118137
If \(\alpha\) and \(\beta\) are the roots of the equation \(x^2-2 x\) \(+4=0\), then \(\alpha^9+\beta^9\) is equal to
