118116
Let \(a, b>0\) satisfy \(a^3+b^3=a-b\), then
1 \(a^2+b^2>1\)
2 \(\mathrm{a}^2+\mathrm{b}^2\lt 0\)
3 \(\mathrm{a}^2+\mathrm{b}^2=1\)
4 \(a^2+a b+b^2\lt 1\)
Explanation:
D Given that, \(\mathrm{a}^3+\mathrm{b}^3=\mathrm{a}-\mathrm{b}\) Let's us consider, \(\mathrm{a}=\frac{2}{3}, \mathrm{~b}=\frac{1}{3}\) So, \(\mathbf{a}^3+\mathrm{b}^3=\mathrm{a}-\mathrm{b}\) \(\left(\frac{2}{3}\right)^3+\left(\frac{1}{3}\right)^3=\frac{2}{3}-\frac{1}{3}\) \(\frac{8}{27}+\frac{1}{27}=\frac{1}{3}\) \(\frac{9}{27}=\frac{1}{3}\) \(\frac{1}{3}=\frac{1}{3}\) \(\therefore \mathrm{a}^2+\mathrm{ab}+\mathrm{b}^2=\frac{4}{9}+\frac{2}{3} \times \frac{1}{3}+\frac{1}{9}\) \(=\frac{4}{9}+\frac{2}{9}+\frac{1}{9}\) \(=\frac{7}{9}\lt 1\)So, \(a^2+a b+b^2\lt 1\)
AMU-2011
Complex Numbers and Quadratic Equation
118117
The number of solutions of equation \(\sin ^4 \theta-2 \sin ^2 \theta-1=0\) which lie between 0 and \(2 \pi\) is
1 0
2 1
3 2
4 4
Explanation:
A From the given equation we have, \(\sin ^2 \theta=\frac{2 \pm \sqrt{4+4}}{2}=1 \pm \sqrt{2}\) Now, \(\quad \sin ^2 \theta \neq 1+\sqrt{2}\) Also, \(\quad \sin ^2 \theta=1-\sqrt{2}\) \(\because 1-\sqrt{2}\lt 0\) and \(\sin ^2 \theta\) cannot be negative. \(\therefore\) No value of \(\theta\) satisfy the equation hence no root lies in 0 and \(2 \pi\).
AMU-2010
Complex Numbers and Quadratic Equation
118118
The number of solutions of the equation \(3 \sin ^2 x-7 \sin x+2=0\), in the interval \([0,5 \pi]\) is
1 0
2 5
3 6
4 10
Explanation:
C Given the equation, Let \(3 \sin ^2 \mathrm{x}-7 \sin \mathrm{x}+2=0\) \(\sin \mathrm{x}=\mathrm{t} \Rightarrow 3 \mathrm{t}^2-7 \mathrm{t}+2=0\) \(\Rightarrow\) \(3 \mathrm{t}^2-6 \mathrm{t}-\mathrm{t}+2=0\) Or \(\quad 3 t(t-2)-1(t-2)=0\) \(\Rightarrow\) \((\mathrm{t}-2)(3 \mathrm{t}-1)=0\) \(\Rightarrow\) But \(\mathrm{t}=\frac{1}{3}, 2\) \(\sin \mathrm{x}=\frac{1}{3}\) So, we get 6 solutions.
AMU-2021
Complex Numbers and Quadratic Equation
118119
If one of the roots of the equation \(\mathbf{x}^2+\mathbf{p x}+\mathbf{q}=\) 0 is equal to the square of the other then \(\qquad\)
118116
Let \(a, b>0\) satisfy \(a^3+b^3=a-b\), then
1 \(a^2+b^2>1\)
2 \(\mathrm{a}^2+\mathrm{b}^2\lt 0\)
3 \(\mathrm{a}^2+\mathrm{b}^2=1\)
4 \(a^2+a b+b^2\lt 1\)
Explanation:
D Given that, \(\mathrm{a}^3+\mathrm{b}^3=\mathrm{a}-\mathrm{b}\) Let's us consider, \(\mathrm{a}=\frac{2}{3}, \mathrm{~b}=\frac{1}{3}\) So, \(\mathbf{a}^3+\mathrm{b}^3=\mathrm{a}-\mathrm{b}\) \(\left(\frac{2}{3}\right)^3+\left(\frac{1}{3}\right)^3=\frac{2}{3}-\frac{1}{3}\) \(\frac{8}{27}+\frac{1}{27}=\frac{1}{3}\) \(\frac{9}{27}=\frac{1}{3}\) \(\frac{1}{3}=\frac{1}{3}\) \(\therefore \mathrm{a}^2+\mathrm{ab}+\mathrm{b}^2=\frac{4}{9}+\frac{2}{3} \times \frac{1}{3}+\frac{1}{9}\) \(=\frac{4}{9}+\frac{2}{9}+\frac{1}{9}\) \(=\frac{7}{9}\lt 1\)So, \(a^2+a b+b^2\lt 1\)
AMU-2011
Complex Numbers and Quadratic Equation
118117
The number of solutions of equation \(\sin ^4 \theta-2 \sin ^2 \theta-1=0\) which lie between 0 and \(2 \pi\) is
1 0
2 1
3 2
4 4
Explanation:
A From the given equation we have, \(\sin ^2 \theta=\frac{2 \pm \sqrt{4+4}}{2}=1 \pm \sqrt{2}\) Now, \(\quad \sin ^2 \theta \neq 1+\sqrt{2}\) Also, \(\quad \sin ^2 \theta=1-\sqrt{2}\) \(\because 1-\sqrt{2}\lt 0\) and \(\sin ^2 \theta\) cannot be negative. \(\therefore\) No value of \(\theta\) satisfy the equation hence no root lies in 0 and \(2 \pi\).
AMU-2010
Complex Numbers and Quadratic Equation
118118
The number of solutions of the equation \(3 \sin ^2 x-7 \sin x+2=0\), in the interval \([0,5 \pi]\) is
1 0
2 5
3 6
4 10
Explanation:
C Given the equation, Let \(3 \sin ^2 \mathrm{x}-7 \sin \mathrm{x}+2=0\) \(\sin \mathrm{x}=\mathrm{t} \Rightarrow 3 \mathrm{t}^2-7 \mathrm{t}+2=0\) \(\Rightarrow\) \(3 \mathrm{t}^2-6 \mathrm{t}-\mathrm{t}+2=0\) Or \(\quad 3 t(t-2)-1(t-2)=0\) \(\Rightarrow\) \((\mathrm{t}-2)(3 \mathrm{t}-1)=0\) \(\Rightarrow\) But \(\mathrm{t}=\frac{1}{3}, 2\) \(\sin \mathrm{x}=\frac{1}{3}\) So, we get 6 solutions.
AMU-2021
Complex Numbers and Quadratic Equation
118119
If one of the roots of the equation \(\mathbf{x}^2+\mathbf{p x}+\mathbf{q}=\) 0 is equal to the square of the other then \(\qquad\)
118116
Let \(a, b>0\) satisfy \(a^3+b^3=a-b\), then
1 \(a^2+b^2>1\)
2 \(\mathrm{a}^2+\mathrm{b}^2\lt 0\)
3 \(\mathrm{a}^2+\mathrm{b}^2=1\)
4 \(a^2+a b+b^2\lt 1\)
Explanation:
D Given that, \(\mathrm{a}^3+\mathrm{b}^3=\mathrm{a}-\mathrm{b}\) Let's us consider, \(\mathrm{a}=\frac{2}{3}, \mathrm{~b}=\frac{1}{3}\) So, \(\mathbf{a}^3+\mathrm{b}^3=\mathrm{a}-\mathrm{b}\) \(\left(\frac{2}{3}\right)^3+\left(\frac{1}{3}\right)^3=\frac{2}{3}-\frac{1}{3}\) \(\frac{8}{27}+\frac{1}{27}=\frac{1}{3}\) \(\frac{9}{27}=\frac{1}{3}\) \(\frac{1}{3}=\frac{1}{3}\) \(\therefore \mathrm{a}^2+\mathrm{ab}+\mathrm{b}^2=\frac{4}{9}+\frac{2}{3} \times \frac{1}{3}+\frac{1}{9}\) \(=\frac{4}{9}+\frac{2}{9}+\frac{1}{9}\) \(=\frac{7}{9}\lt 1\)So, \(a^2+a b+b^2\lt 1\)
AMU-2011
Complex Numbers and Quadratic Equation
118117
The number of solutions of equation \(\sin ^4 \theta-2 \sin ^2 \theta-1=0\) which lie between 0 and \(2 \pi\) is
1 0
2 1
3 2
4 4
Explanation:
A From the given equation we have, \(\sin ^2 \theta=\frac{2 \pm \sqrt{4+4}}{2}=1 \pm \sqrt{2}\) Now, \(\quad \sin ^2 \theta \neq 1+\sqrt{2}\) Also, \(\quad \sin ^2 \theta=1-\sqrt{2}\) \(\because 1-\sqrt{2}\lt 0\) and \(\sin ^2 \theta\) cannot be negative. \(\therefore\) No value of \(\theta\) satisfy the equation hence no root lies in 0 and \(2 \pi\).
AMU-2010
Complex Numbers and Quadratic Equation
118118
The number of solutions of the equation \(3 \sin ^2 x-7 \sin x+2=0\), in the interval \([0,5 \pi]\) is
1 0
2 5
3 6
4 10
Explanation:
C Given the equation, Let \(3 \sin ^2 \mathrm{x}-7 \sin \mathrm{x}+2=0\) \(\sin \mathrm{x}=\mathrm{t} \Rightarrow 3 \mathrm{t}^2-7 \mathrm{t}+2=0\) \(\Rightarrow\) \(3 \mathrm{t}^2-6 \mathrm{t}-\mathrm{t}+2=0\) Or \(\quad 3 t(t-2)-1(t-2)=0\) \(\Rightarrow\) \((\mathrm{t}-2)(3 \mathrm{t}-1)=0\) \(\Rightarrow\) But \(\mathrm{t}=\frac{1}{3}, 2\) \(\sin \mathrm{x}=\frac{1}{3}\) So, we get 6 solutions.
AMU-2021
Complex Numbers and Quadratic Equation
118119
If one of the roots of the equation \(\mathbf{x}^2+\mathbf{p x}+\mathbf{q}=\) 0 is equal to the square of the other then \(\qquad\)
118116
Let \(a, b>0\) satisfy \(a^3+b^3=a-b\), then
1 \(a^2+b^2>1\)
2 \(\mathrm{a}^2+\mathrm{b}^2\lt 0\)
3 \(\mathrm{a}^2+\mathrm{b}^2=1\)
4 \(a^2+a b+b^2\lt 1\)
Explanation:
D Given that, \(\mathrm{a}^3+\mathrm{b}^3=\mathrm{a}-\mathrm{b}\) Let's us consider, \(\mathrm{a}=\frac{2}{3}, \mathrm{~b}=\frac{1}{3}\) So, \(\mathbf{a}^3+\mathrm{b}^3=\mathrm{a}-\mathrm{b}\) \(\left(\frac{2}{3}\right)^3+\left(\frac{1}{3}\right)^3=\frac{2}{3}-\frac{1}{3}\) \(\frac{8}{27}+\frac{1}{27}=\frac{1}{3}\) \(\frac{9}{27}=\frac{1}{3}\) \(\frac{1}{3}=\frac{1}{3}\) \(\therefore \mathrm{a}^2+\mathrm{ab}+\mathrm{b}^2=\frac{4}{9}+\frac{2}{3} \times \frac{1}{3}+\frac{1}{9}\) \(=\frac{4}{9}+\frac{2}{9}+\frac{1}{9}\) \(=\frac{7}{9}\lt 1\)So, \(a^2+a b+b^2\lt 1\)
AMU-2011
Complex Numbers and Quadratic Equation
118117
The number of solutions of equation \(\sin ^4 \theta-2 \sin ^2 \theta-1=0\) which lie between 0 and \(2 \pi\) is
1 0
2 1
3 2
4 4
Explanation:
A From the given equation we have, \(\sin ^2 \theta=\frac{2 \pm \sqrt{4+4}}{2}=1 \pm \sqrt{2}\) Now, \(\quad \sin ^2 \theta \neq 1+\sqrt{2}\) Also, \(\quad \sin ^2 \theta=1-\sqrt{2}\) \(\because 1-\sqrt{2}\lt 0\) and \(\sin ^2 \theta\) cannot be negative. \(\therefore\) No value of \(\theta\) satisfy the equation hence no root lies in 0 and \(2 \pi\).
AMU-2010
Complex Numbers and Quadratic Equation
118118
The number of solutions of the equation \(3 \sin ^2 x-7 \sin x+2=0\), in the interval \([0,5 \pi]\) is
1 0
2 5
3 6
4 10
Explanation:
C Given the equation, Let \(3 \sin ^2 \mathrm{x}-7 \sin \mathrm{x}+2=0\) \(\sin \mathrm{x}=\mathrm{t} \Rightarrow 3 \mathrm{t}^2-7 \mathrm{t}+2=0\) \(\Rightarrow\) \(3 \mathrm{t}^2-6 \mathrm{t}-\mathrm{t}+2=0\) Or \(\quad 3 t(t-2)-1(t-2)=0\) \(\Rightarrow\) \((\mathrm{t}-2)(3 \mathrm{t}-1)=0\) \(\Rightarrow\) But \(\mathrm{t}=\frac{1}{3}, 2\) \(\sin \mathrm{x}=\frac{1}{3}\) So, we get 6 solutions.
AMU-2021
Complex Numbers and Quadratic Equation
118119
If one of the roots of the equation \(\mathbf{x}^2+\mathbf{p x}+\mathbf{q}=\) 0 is equal to the square of the other then \(\qquad\)
118116
Let \(a, b>0\) satisfy \(a^3+b^3=a-b\), then
1 \(a^2+b^2>1\)
2 \(\mathrm{a}^2+\mathrm{b}^2\lt 0\)
3 \(\mathrm{a}^2+\mathrm{b}^2=1\)
4 \(a^2+a b+b^2\lt 1\)
Explanation:
D Given that, \(\mathrm{a}^3+\mathrm{b}^3=\mathrm{a}-\mathrm{b}\) Let's us consider, \(\mathrm{a}=\frac{2}{3}, \mathrm{~b}=\frac{1}{3}\) So, \(\mathbf{a}^3+\mathrm{b}^3=\mathrm{a}-\mathrm{b}\) \(\left(\frac{2}{3}\right)^3+\left(\frac{1}{3}\right)^3=\frac{2}{3}-\frac{1}{3}\) \(\frac{8}{27}+\frac{1}{27}=\frac{1}{3}\) \(\frac{9}{27}=\frac{1}{3}\) \(\frac{1}{3}=\frac{1}{3}\) \(\therefore \mathrm{a}^2+\mathrm{ab}+\mathrm{b}^2=\frac{4}{9}+\frac{2}{3} \times \frac{1}{3}+\frac{1}{9}\) \(=\frac{4}{9}+\frac{2}{9}+\frac{1}{9}\) \(=\frac{7}{9}\lt 1\)So, \(a^2+a b+b^2\lt 1\)
AMU-2011
Complex Numbers and Quadratic Equation
118117
The number of solutions of equation \(\sin ^4 \theta-2 \sin ^2 \theta-1=0\) which lie between 0 and \(2 \pi\) is
1 0
2 1
3 2
4 4
Explanation:
A From the given equation we have, \(\sin ^2 \theta=\frac{2 \pm \sqrt{4+4}}{2}=1 \pm \sqrt{2}\) Now, \(\quad \sin ^2 \theta \neq 1+\sqrt{2}\) Also, \(\quad \sin ^2 \theta=1-\sqrt{2}\) \(\because 1-\sqrt{2}\lt 0\) and \(\sin ^2 \theta\) cannot be negative. \(\therefore\) No value of \(\theta\) satisfy the equation hence no root lies in 0 and \(2 \pi\).
AMU-2010
Complex Numbers and Quadratic Equation
118118
The number of solutions of the equation \(3 \sin ^2 x-7 \sin x+2=0\), in the interval \([0,5 \pi]\) is
1 0
2 5
3 6
4 10
Explanation:
C Given the equation, Let \(3 \sin ^2 \mathrm{x}-7 \sin \mathrm{x}+2=0\) \(\sin \mathrm{x}=\mathrm{t} \Rightarrow 3 \mathrm{t}^2-7 \mathrm{t}+2=0\) \(\Rightarrow\) \(3 \mathrm{t}^2-6 \mathrm{t}-\mathrm{t}+2=0\) Or \(\quad 3 t(t-2)-1(t-2)=0\) \(\Rightarrow\) \((\mathrm{t}-2)(3 \mathrm{t}-1)=0\) \(\Rightarrow\) But \(\mathrm{t}=\frac{1}{3}, 2\) \(\sin \mathrm{x}=\frac{1}{3}\) So, we get 6 solutions.
AMU-2021
Complex Numbers and Quadratic Equation
118119
If one of the roots of the equation \(\mathbf{x}^2+\mathbf{p x}+\mathbf{q}=\) 0 is equal to the square of the other then \(\qquad\)
