NEET Test Series from KOTA - 10 Papers In MS WORD
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Complex Numbers and Quadratic Equation
118188
The number of pairs of consecutive positive even integers such that the sum of their squares is 290 is
1 0
2 1
3 2
4 3
Explanation:
A Let the consecutive positive even integers are a, \(a+2\) \(\therefore \quad a^2+(a+2)^2=290\) \(a^2+a^2+4+4 a=290\) \(2 a^2+4 a-286=0\) \(\mathrm{a}^2+2 \mathrm{a}^2-143=0\) \(\mathrm{a}=\frac{-2 \pm \sqrt{4+4 \times 143}}{2}\) \(=\frac{-2 \pm \sqrt{576}}{2}=\frac{-2 \pm 24}{2}\) Taking \(+\mathrm{ve}\) sign, \(\mathrm{a}=\frac{-2+24}{2}=11\) - ve sign \(\mathrm{a}=-13\) \(\therefore \quad 11\) is not even integers -13 is also not + ve even integers \(\therefore\) We do not get any pairs as required.
AP EAMCET-08.07.2022
Complex Numbers and Quadratic Equation
118189
If \(\sqrt{x}+\frac{1}{\sqrt{x}}=2 \cos \theta\), then \(x^6+x^{-6}=\)
118207
If \(\alpha\) and \(\alpha^2\) are the roots of the equation \(x^2-6 x\) \(+c=0\), then the positive value of \(c\)
1 2
2 3
3 4
4 9
5 8
Explanation:
E Given, \(\alpha\) and \(\alpha^2\) are the roots of \(\mathrm{x}^2-6 \mathrm{x}+\mathrm{c}=0\) \(\therefore \quad \alpha+\alpha^2=6\) a. \(\alpha^2=\mathrm{c}\) From (i), \(\alpha^2+\alpha-6=0\) Or \(\quad(\alpha-2)(\alpha+3)=0\) \(\therefore \quad \alpha=2,-3\) \(\therefore \quad \mathrm{c}=2.4=8\) and \(\mathrm{c}=(-3) .9\) \(=-27\)\(\therefore\) The \(+\mathrm{ve}\) value of \(\mathrm{c}\) is 8 .
Kerala CEE-2016
Complex Numbers and Quadratic Equation
118190
Assertion (A) \(3 x^2-16 x+4>-16\) is satisfied for some values of real \(x\) in \(\left(0, \frac{10}{3}\right)\) Reason (R) \(a x^2+b x+c\) and a will have the same sign for some values of \(x \in R\) when \(b^2-4 \mathbf{a c}>\mathbf{0}\).
1 (A), is true, (R) is true and (R) is the correct explanation for (A).
2 (A) is true, (R) is true but (R) is not the correct explanation for (A).
3 (A) is true, but (R) is false.
4 (A) is fale, but (R) is true.
Explanation:
D Given, Assertion A: \(3 \mathrm{x}^2-16 \mathrm{x}+4>-16\) \(3 \mathrm{x}^2-16 \mathrm{x}+20>0\) \(\Rightarrow \quad 3 \mathrm{x}^2-6 \mathrm{x}-10 \mathrm{x}+20>0\) \(\Rightarrow \quad 3 \mathrm{x}(\mathrm{x}-2)-10(\mathrm{x}-2)>0\) or \(\quad(x-2)(3 x-10)>0\) Using method of interval \(x \in(-\infty, 2) \cup\left(\frac{10}{3}, \infty\right)\) The above does not contain \(\left(0, \frac{10}{3}\right)\) So, this assertion is false. Now, Assertion R: \(a x^2+b x+c=f(x)\) \(\therefore \quad f(x) =a\left[x^2+\frac{b}{a} x+\frac{c}{a}\right]\) \(=a\left[x^2+\left(\frac{b}{2 a}\right)^2+\frac{b}{a} x-\frac{b^2}{4 a^2}+\frac{c}{a}\right]\) \(=a\left[\left(x+\frac{b}{2 a}\right)^2-\frac{b^2}{4 a^2}+\frac{4 a c}{4 a^2}\right]\) \(=\mathrm{a}\left[\left(\mathrm{x}+\left(\frac{\mathrm{b}}{2 \mathrm{a}}\right)\right)^2-\frac{1}{4 \mathrm{a}^2}\left(\mathrm{~b}^2-4 \mathrm{ac}\right)\right]\) When \(\mathrm{b}^2-4 \mathrm{ac}\lt 0\), coefficient of a is always \(+v e\). So, \(\mathrm{f}(\mathrm{x})\) will have the same sign when \(b^2-4 \mathrm{ac}\lt 0\)\(\therefore\) This assertion is also false.
118188
The number of pairs of consecutive positive even integers such that the sum of their squares is 290 is
1 0
2 1
3 2
4 3
Explanation:
A Let the consecutive positive even integers are a, \(a+2\) \(\therefore \quad a^2+(a+2)^2=290\) \(a^2+a^2+4+4 a=290\) \(2 a^2+4 a-286=0\) \(\mathrm{a}^2+2 \mathrm{a}^2-143=0\) \(\mathrm{a}=\frac{-2 \pm \sqrt{4+4 \times 143}}{2}\) \(=\frac{-2 \pm \sqrt{576}}{2}=\frac{-2 \pm 24}{2}\) Taking \(+\mathrm{ve}\) sign, \(\mathrm{a}=\frac{-2+24}{2}=11\) - ve sign \(\mathrm{a}=-13\) \(\therefore \quad 11\) is not even integers -13 is also not + ve even integers \(\therefore\) We do not get any pairs as required.
AP EAMCET-08.07.2022
Complex Numbers and Quadratic Equation
118189
If \(\sqrt{x}+\frac{1}{\sqrt{x}}=2 \cos \theta\), then \(x^6+x^{-6}=\)
118207
If \(\alpha\) and \(\alpha^2\) are the roots of the equation \(x^2-6 x\) \(+c=0\), then the positive value of \(c\)
1 2
2 3
3 4
4 9
5 8
Explanation:
E Given, \(\alpha\) and \(\alpha^2\) are the roots of \(\mathrm{x}^2-6 \mathrm{x}+\mathrm{c}=0\) \(\therefore \quad \alpha+\alpha^2=6\) a. \(\alpha^2=\mathrm{c}\) From (i), \(\alpha^2+\alpha-6=0\) Or \(\quad(\alpha-2)(\alpha+3)=0\) \(\therefore \quad \alpha=2,-3\) \(\therefore \quad \mathrm{c}=2.4=8\) and \(\mathrm{c}=(-3) .9\) \(=-27\)\(\therefore\) The \(+\mathrm{ve}\) value of \(\mathrm{c}\) is 8 .
Kerala CEE-2016
Complex Numbers and Quadratic Equation
118190
Assertion (A) \(3 x^2-16 x+4>-16\) is satisfied for some values of real \(x\) in \(\left(0, \frac{10}{3}\right)\) Reason (R) \(a x^2+b x+c\) and a will have the same sign for some values of \(x \in R\) when \(b^2-4 \mathbf{a c}>\mathbf{0}\).
1 (A), is true, (R) is true and (R) is the correct explanation for (A).
2 (A) is true, (R) is true but (R) is not the correct explanation for (A).
3 (A) is true, but (R) is false.
4 (A) is fale, but (R) is true.
Explanation:
D Given, Assertion A: \(3 \mathrm{x}^2-16 \mathrm{x}+4>-16\) \(3 \mathrm{x}^2-16 \mathrm{x}+20>0\) \(\Rightarrow \quad 3 \mathrm{x}^2-6 \mathrm{x}-10 \mathrm{x}+20>0\) \(\Rightarrow \quad 3 \mathrm{x}(\mathrm{x}-2)-10(\mathrm{x}-2)>0\) or \(\quad(x-2)(3 x-10)>0\) Using method of interval \(x \in(-\infty, 2) \cup\left(\frac{10}{3}, \infty\right)\) The above does not contain \(\left(0, \frac{10}{3}\right)\) So, this assertion is false. Now, Assertion R: \(a x^2+b x+c=f(x)\) \(\therefore \quad f(x) =a\left[x^2+\frac{b}{a} x+\frac{c}{a}\right]\) \(=a\left[x^2+\left(\frac{b}{2 a}\right)^2+\frac{b}{a} x-\frac{b^2}{4 a^2}+\frac{c}{a}\right]\) \(=a\left[\left(x+\frac{b}{2 a}\right)^2-\frac{b^2}{4 a^2}+\frac{4 a c}{4 a^2}\right]\) \(=\mathrm{a}\left[\left(\mathrm{x}+\left(\frac{\mathrm{b}}{2 \mathrm{a}}\right)\right)^2-\frac{1}{4 \mathrm{a}^2}\left(\mathrm{~b}^2-4 \mathrm{ac}\right)\right]\) When \(\mathrm{b}^2-4 \mathrm{ac}\lt 0\), coefficient of a is always \(+v e\). So, \(\mathrm{f}(\mathrm{x})\) will have the same sign when \(b^2-4 \mathrm{ac}\lt 0\)\(\therefore\) This assertion is also false.
118188
The number of pairs of consecutive positive even integers such that the sum of their squares is 290 is
1 0
2 1
3 2
4 3
Explanation:
A Let the consecutive positive even integers are a, \(a+2\) \(\therefore \quad a^2+(a+2)^2=290\) \(a^2+a^2+4+4 a=290\) \(2 a^2+4 a-286=0\) \(\mathrm{a}^2+2 \mathrm{a}^2-143=0\) \(\mathrm{a}=\frac{-2 \pm \sqrt{4+4 \times 143}}{2}\) \(=\frac{-2 \pm \sqrt{576}}{2}=\frac{-2 \pm 24}{2}\) Taking \(+\mathrm{ve}\) sign, \(\mathrm{a}=\frac{-2+24}{2}=11\) - ve sign \(\mathrm{a}=-13\) \(\therefore \quad 11\) is not even integers -13 is also not + ve even integers \(\therefore\) We do not get any pairs as required.
AP EAMCET-08.07.2022
Complex Numbers and Quadratic Equation
118189
If \(\sqrt{x}+\frac{1}{\sqrt{x}}=2 \cos \theta\), then \(x^6+x^{-6}=\)
118207
If \(\alpha\) and \(\alpha^2\) are the roots of the equation \(x^2-6 x\) \(+c=0\), then the positive value of \(c\)
1 2
2 3
3 4
4 9
5 8
Explanation:
E Given, \(\alpha\) and \(\alpha^2\) are the roots of \(\mathrm{x}^2-6 \mathrm{x}+\mathrm{c}=0\) \(\therefore \quad \alpha+\alpha^2=6\) a. \(\alpha^2=\mathrm{c}\) From (i), \(\alpha^2+\alpha-6=0\) Or \(\quad(\alpha-2)(\alpha+3)=0\) \(\therefore \quad \alpha=2,-3\) \(\therefore \quad \mathrm{c}=2.4=8\) and \(\mathrm{c}=(-3) .9\) \(=-27\)\(\therefore\) The \(+\mathrm{ve}\) value of \(\mathrm{c}\) is 8 .
Kerala CEE-2016
Complex Numbers and Quadratic Equation
118190
Assertion (A) \(3 x^2-16 x+4>-16\) is satisfied for some values of real \(x\) in \(\left(0, \frac{10}{3}\right)\) Reason (R) \(a x^2+b x+c\) and a will have the same sign for some values of \(x \in R\) when \(b^2-4 \mathbf{a c}>\mathbf{0}\).
1 (A), is true, (R) is true and (R) is the correct explanation for (A).
2 (A) is true, (R) is true but (R) is not the correct explanation for (A).
3 (A) is true, but (R) is false.
4 (A) is fale, but (R) is true.
Explanation:
D Given, Assertion A: \(3 \mathrm{x}^2-16 \mathrm{x}+4>-16\) \(3 \mathrm{x}^2-16 \mathrm{x}+20>0\) \(\Rightarrow \quad 3 \mathrm{x}^2-6 \mathrm{x}-10 \mathrm{x}+20>0\) \(\Rightarrow \quad 3 \mathrm{x}(\mathrm{x}-2)-10(\mathrm{x}-2)>0\) or \(\quad(x-2)(3 x-10)>0\) Using method of interval \(x \in(-\infty, 2) \cup\left(\frac{10}{3}, \infty\right)\) The above does not contain \(\left(0, \frac{10}{3}\right)\) So, this assertion is false. Now, Assertion R: \(a x^2+b x+c=f(x)\) \(\therefore \quad f(x) =a\left[x^2+\frac{b}{a} x+\frac{c}{a}\right]\) \(=a\left[x^2+\left(\frac{b}{2 a}\right)^2+\frac{b}{a} x-\frac{b^2}{4 a^2}+\frac{c}{a}\right]\) \(=a\left[\left(x+\frac{b}{2 a}\right)^2-\frac{b^2}{4 a^2}+\frac{4 a c}{4 a^2}\right]\) \(=\mathrm{a}\left[\left(\mathrm{x}+\left(\frac{\mathrm{b}}{2 \mathrm{a}}\right)\right)^2-\frac{1}{4 \mathrm{a}^2}\left(\mathrm{~b}^2-4 \mathrm{ac}\right)\right]\) When \(\mathrm{b}^2-4 \mathrm{ac}\lt 0\), coefficient of a is always \(+v e\). So, \(\mathrm{f}(\mathrm{x})\) will have the same sign when \(b^2-4 \mathrm{ac}\lt 0\)\(\therefore\) This assertion is also false.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Complex Numbers and Quadratic Equation
118188
The number of pairs of consecutive positive even integers such that the sum of their squares is 290 is
1 0
2 1
3 2
4 3
Explanation:
A Let the consecutive positive even integers are a, \(a+2\) \(\therefore \quad a^2+(a+2)^2=290\) \(a^2+a^2+4+4 a=290\) \(2 a^2+4 a-286=0\) \(\mathrm{a}^2+2 \mathrm{a}^2-143=0\) \(\mathrm{a}=\frac{-2 \pm \sqrt{4+4 \times 143}}{2}\) \(=\frac{-2 \pm \sqrt{576}}{2}=\frac{-2 \pm 24}{2}\) Taking \(+\mathrm{ve}\) sign, \(\mathrm{a}=\frac{-2+24}{2}=11\) - ve sign \(\mathrm{a}=-13\) \(\therefore \quad 11\) is not even integers -13 is also not + ve even integers \(\therefore\) We do not get any pairs as required.
AP EAMCET-08.07.2022
Complex Numbers and Quadratic Equation
118189
If \(\sqrt{x}+\frac{1}{\sqrt{x}}=2 \cos \theta\), then \(x^6+x^{-6}=\)
118207
If \(\alpha\) and \(\alpha^2\) are the roots of the equation \(x^2-6 x\) \(+c=0\), then the positive value of \(c\)
1 2
2 3
3 4
4 9
5 8
Explanation:
E Given, \(\alpha\) and \(\alpha^2\) are the roots of \(\mathrm{x}^2-6 \mathrm{x}+\mathrm{c}=0\) \(\therefore \quad \alpha+\alpha^2=6\) a. \(\alpha^2=\mathrm{c}\) From (i), \(\alpha^2+\alpha-6=0\) Or \(\quad(\alpha-2)(\alpha+3)=0\) \(\therefore \quad \alpha=2,-3\) \(\therefore \quad \mathrm{c}=2.4=8\) and \(\mathrm{c}=(-3) .9\) \(=-27\)\(\therefore\) The \(+\mathrm{ve}\) value of \(\mathrm{c}\) is 8 .
Kerala CEE-2016
Complex Numbers and Quadratic Equation
118190
Assertion (A) \(3 x^2-16 x+4>-16\) is satisfied for some values of real \(x\) in \(\left(0, \frac{10}{3}\right)\) Reason (R) \(a x^2+b x+c\) and a will have the same sign for some values of \(x \in R\) when \(b^2-4 \mathbf{a c}>\mathbf{0}\).
1 (A), is true, (R) is true and (R) is the correct explanation for (A).
2 (A) is true, (R) is true but (R) is not the correct explanation for (A).
3 (A) is true, but (R) is false.
4 (A) is fale, but (R) is true.
Explanation:
D Given, Assertion A: \(3 \mathrm{x}^2-16 \mathrm{x}+4>-16\) \(3 \mathrm{x}^2-16 \mathrm{x}+20>0\) \(\Rightarrow \quad 3 \mathrm{x}^2-6 \mathrm{x}-10 \mathrm{x}+20>0\) \(\Rightarrow \quad 3 \mathrm{x}(\mathrm{x}-2)-10(\mathrm{x}-2)>0\) or \(\quad(x-2)(3 x-10)>0\) Using method of interval \(x \in(-\infty, 2) \cup\left(\frac{10}{3}, \infty\right)\) The above does not contain \(\left(0, \frac{10}{3}\right)\) So, this assertion is false. Now, Assertion R: \(a x^2+b x+c=f(x)\) \(\therefore \quad f(x) =a\left[x^2+\frac{b}{a} x+\frac{c}{a}\right]\) \(=a\left[x^2+\left(\frac{b}{2 a}\right)^2+\frac{b}{a} x-\frac{b^2}{4 a^2}+\frac{c}{a}\right]\) \(=a\left[\left(x+\frac{b}{2 a}\right)^2-\frac{b^2}{4 a^2}+\frac{4 a c}{4 a^2}\right]\) \(=\mathrm{a}\left[\left(\mathrm{x}+\left(\frac{\mathrm{b}}{2 \mathrm{a}}\right)\right)^2-\frac{1}{4 \mathrm{a}^2}\left(\mathrm{~b}^2-4 \mathrm{ac}\right)\right]\) When \(\mathrm{b}^2-4 \mathrm{ac}\lt 0\), coefficient of a is always \(+v e\). So, \(\mathrm{f}(\mathrm{x})\) will have the same sign when \(b^2-4 \mathrm{ac}\lt 0\)\(\therefore\) This assertion is also false.
