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
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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.
