118113
If \(1-i\) is a root of the equation \(x^2+a x+b=0\), then \(b\) is equal to
1 1
2 -1
3 -2
4 2
Explanation:
D Given, (1-i) is a root of the equation \(x^2+a x+b=0\) then \((1+i)\) will be other root. \(\therefore \quad\) Product of the roots \(=\mathrm{b}\) \(\therefore \quad(1-\mathrm{i})(1+\mathrm{i})=\mathrm{b}\) or \(\quad b=2\)
AP EAMCET-2002
Complex Numbers and Quadratic Equation
118124
The roots of the equation \(\left[x^2-x-6\right]=x+2\) are
1 \(-2,1,4\)
2 \(0,2,4\)
3 \(0,1,4\)
4 \(-2,2,4\)
Explanation:
D Given, \(\left|x^2-x-6\right|=x+2\) \(\Rightarrow \mathrm{x}^2-\mathrm{x}-6=\mathrm{x}+2\) \(\Rightarrow \mathrm{x}^2-2 \mathrm{x}-8=0\) \(\Rightarrow(\mathrm{x}=-2,4) \text { or }(\mathrm{x}= \pm 2) \Rightarrow \mathrm{x}=(-2,2,4)\)
AP EAMCET-17.09.2020
Complex Numbers and Quadratic Equation
118128
If \(\tan A\) and \(\tan B\) are the roots of the quadratic equation \(x^2-p x+q=0\), then \(\sin ^2(A\) \(+\mathbf{B})\) is equal to
D We have } \tan \mathrm{A} \text { and } \tan \mathrm{B} \text { as the roots of the } \\ \text { equation } \\ \mathrm{x}^2-\mathrm{px}+\mathrm{q}=0 \\ \therefore \tan \mathrm{A}+\tan \mathrm{B}=\mathrm{p} \\ \tan \mathrm{A} \cdot \tan \mathrm{B}=\mathrm{q} \\ \therefore \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \cdot \tan B} \\ =\frac{p}{1-q} \\ \end{aligned}\) \(\therefore \sin (A+B)=\frac{p}{\sqrt{P^2+(1-q)^2}}\) \(\therefore \sin ^2(A+B)=\frac{p^2}{p^2+(1+q)^2}\)\(\therefore\) Option (d) is the correct answer.
AP EAMCET-2011
Complex Numbers and Quadratic Equation
118206
If one of the roots of the quadratic equation \(a x^2\) \(-b x+a=0\) is 6 , then value of \(\frac{b}{a}\) is equal to
1 \(\frac{1}{6}\)
2 \(\frac{11}{6}\)
3 \(\frac{37}{6}\)
4 \(\frac{6}{11}\)
5 \(\frac{6}{37}\)
Explanation:
C Given, one root of quadrant equation \(a x^2-b x+a=0\) is 6 . \(\therefore \quad \mathrm{a}(6)^2-\mathrm{b}(6)+\mathrm{a}=0\) \(\Rightarrow 36 \mathrm{a}+\mathrm{a}=6 \mathrm{~b}\) Or \(\quad 37 a=6 b \Rightarrow \frac{b}{a}=\frac{37}{6}\)
118113
If \(1-i\) is a root of the equation \(x^2+a x+b=0\), then \(b\) is equal to
1 1
2 -1
3 -2
4 2
Explanation:
D Given, (1-i) is a root of the equation \(x^2+a x+b=0\) then \((1+i)\) will be other root. \(\therefore \quad\) Product of the roots \(=\mathrm{b}\) \(\therefore \quad(1-\mathrm{i})(1+\mathrm{i})=\mathrm{b}\) or \(\quad b=2\)
AP EAMCET-2002
Complex Numbers and Quadratic Equation
118124
The roots of the equation \(\left[x^2-x-6\right]=x+2\) are
1 \(-2,1,4\)
2 \(0,2,4\)
3 \(0,1,4\)
4 \(-2,2,4\)
Explanation:
D Given, \(\left|x^2-x-6\right|=x+2\) \(\Rightarrow \mathrm{x}^2-\mathrm{x}-6=\mathrm{x}+2\) \(\Rightarrow \mathrm{x}^2-2 \mathrm{x}-8=0\) \(\Rightarrow(\mathrm{x}=-2,4) \text { or }(\mathrm{x}= \pm 2) \Rightarrow \mathrm{x}=(-2,2,4)\)
AP EAMCET-17.09.2020
Complex Numbers and Quadratic Equation
118128
If \(\tan A\) and \(\tan B\) are the roots of the quadratic equation \(x^2-p x+q=0\), then \(\sin ^2(A\) \(+\mathbf{B})\) is equal to
D We have } \tan \mathrm{A} \text { and } \tan \mathrm{B} \text { as the roots of the } \\ \text { equation } \\ \mathrm{x}^2-\mathrm{px}+\mathrm{q}=0 \\ \therefore \tan \mathrm{A}+\tan \mathrm{B}=\mathrm{p} \\ \tan \mathrm{A} \cdot \tan \mathrm{B}=\mathrm{q} \\ \therefore \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \cdot \tan B} \\ =\frac{p}{1-q} \\ \end{aligned}\) \(\therefore \sin (A+B)=\frac{p}{\sqrt{P^2+(1-q)^2}}\) \(\therefore \sin ^2(A+B)=\frac{p^2}{p^2+(1+q)^2}\)\(\therefore\) Option (d) is the correct answer.
AP EAMCET-2011
Complex Numbers and Quadratic Equation
118206
If one of the roots of the quadratic equation \(a x^2\) \(-b x+a=0\) is 6 , then value of \(\frac{b}{a}\) is equal to
1 \(\frac{1}{6}\)
2 \(\frac{11}{6}\)
3 \(\frac{37}{6}\)
4 \(\frac{6}{11}\)
5 \(\frac{6}{37}\)
Explanation:
C Given, one root of quadrant equation \(a x^2-b x+a=0\) is 6 . \(\therefore \quad \mathrm{a}(6)^2-\mathrm{b}(6)+\mathrm{a}=0\) \(\Rightarrow 36 \mathrm{a}+\mathrm{a}=6 \mathrm{~b}\) Or \(\quad 37 a=6 b \Rightarrow \frac{b}{a}=\frac{37}{6}\)
118113
If \(1-i\) is a root of the equation \(x^2+a x+b=0\), then \(b\) is equal to
1 1
2 -1
3 -2
4 2
Explanation:
D Given, (1-i) is a root of the equation \(x^2+a x+b=0\) then \((1+i)\) will be other root. \(\therefore \quad\) Product of the roots \(=\mathrm{b}\) \(\therefore \quad(1-\mathrm{i})(1+\mathrm{i})=\mathrm{b}\) or \(\quad b=2\)
AP EAMCET-2002
Complex Numbers and Quadratic Equation
118124
The roots of the equation \(\left[x^2-x-6\right]=x+2\) are
1 \(-2,1,4\)
2 \(0,2,4\)
3 \(0,1,4\)
4 \(-2,2,4\)
Explanation:
D Given, \(\left|x^2-x-6\right|=x+2\) \(\Rightarrow \mathrm{x}^2-\mathrm{x}-6=\mathrm{x}+2\) \(\Rightarrow \mathrm{x}^2-2 \mathrm{x}-8=0\) \(\Rightarrow(\mathrm{x}=-2,4) \text { or }(\mathrm{x}= \pm 2) \Rightarrow \mathrm{x}=(-2,2,4)\)
AP EAMCET-17.09.2020
Complex Numbers and Quadratic Equation
118128
If \(\tan A\) and \(\tan B\) are the roots of the quadratic equation \(x^2-p x+q=0\), then \(\sin ^2(A\) \(+\mathbf{B})\) is equal to
D We have } \tan \mathrm{A} \text { and } \tan \mathrm{B} \text { as the roots of the } \\ \text { equation } \\ \mathrm{x}^2-\mathrm{px}+\mathrm{q}=0 \\ \therefore \tan \mathrm{A}+\tan \mathrm{B}=\mathrm{p} \\ \tan \mathrm{A} \cdot \tan \mathrm{B}=\mathrm{q} \\ \therefore \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \cdot \tan B} \\ =\frac{p}{1-q} \\ \end{aligned}\) \(\therefore \sin (A+B)=\frac{p}{\sqrt{P^2+(1-q)^2}}\) \(\therefore \sin ^2(A+B)=\frac{p^2}{p^2+(1+q)^2}\)\(\therefore\) Option (d) is the correct answer.
AP EAMCET-2011
Complex Numbers and Quadratic Equation
118206
If one of the roots of the quadratic equation \(a x^2\) \(-b x+a=0\) is 6 , then value of \(\frac{b}{a}\) is equal to
1 \(\frac{1}{6}\)
2 \(\frac{11}{6}\)
3 \(\frac{37}{6}\)
4 \(\frac{6}{11}\)
5 \(\frac{6}{37}\)
Explanation:
C Given, one root of quadrant equation \(a x^2-b x+a=0\) is 6 . \(\therefore \quad \mathrm{a}(6)^2-\mathrm{b}(6)+\mathrm{a}=0\) \(\Rightarrow 36 \mathrm{a}+\mathrm{a}=6 \mathrm{~b}\) Or \(\quad 37 a=6 b \Rightarrow \frac{b}{a}=\frac{37}{6}\)
118113
If \(1-i\) is a root of the equation \(x^2+a x+b=0\), then \(b\) is equal to
1 1
2 -1
3 -2
4 2
Explanation:
D Given, (1-i) is a root of the equation \(x^2+a x+b=0\) then \((1+i)\) will be other root. \(\therefore \quad\) Product of the roots \(=\mathrm{b}\) \(\therefore \quad(1-\mathrm{i})(1+\mathrm{i})=\mathrm{b}\) or \(\quad b=2\)
AP EAMCET-2002
Complex Numbers and Quadratic Equation
118124
The roots of the equation \(\left[x^2-x-6\right]=x+2\) are
1 \(-2,1,4\)
2 \(0,2,4\)
3 \(0,1,4\)
4 \(-2,2,4\)
Explanation:
D Given, \(\left|x^2-x-6\right|=x+2\) \(\Rightarrow \mathrm{x}^2-\mathrm{x}-6=\mathrm{x}+2\) \(\Rightarrow \mathrm{x}^2-2 \mathrm{x}-8=0\) \(\Rightarrow(\mathrm{x}=-2,4) \text { or }(\mathrm{x}= \pm 2) \Rightarrow \mathrm{x}=(-2,2,4)\)
AP EAMCET-17.09.2020
Complex Numbers and Quadratic Equation
118128
If \(\tan A\) and \(\tan B\) are the roots of the quadratic equation \(x^2-p x+q=0\), then \(\sin ^2(A\) \(+\mathbf{B})\) is equal to
D We have } \tan \mathrm{A} \text { and } \tan \mathrm{B} \text { as the roots of the } \\ \text { equation } \\ \mathrm{x}^2-\mathrm{px}+\mathrm{q}=0 \\ \therefore \tan \mathrm{A}+\tan \mathrm{B}=\mathrm{p} \\ \tan \mathrm{A} \cdot \tan \mathrm{B}=\mathrm{q} \\ \therefore \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \cdot \tan B} \\ =\frac{p}{1-q} \\ \end{aligned}\) \(\therefore \sin (A+B)=\frac{p}{\sqrt{P^2+(1-q)^2}}\) \(\therefore \sin ^2(A+B)=\frac{p^2}{p^2+(1+q)^2}\)\(\therefore\) Option (d) is the correct answer.
AP EAMCET-2011
Complex Numbers and Quadratic Equation
118206
If one of the roots of the quadratic equation \(a x^2\) \(-b x+a=0\) is 6 , then value of \(\frac{b}{a}\) is equal to
1 \(\frac{1}{6}\)
2 \(\frac{11}{6}\)
3 \(\frac{37}{6}\)
4 \(\frac{6}{11}\)
5 \(\frac{6}{37}\)
Explanation:
C Given, one root of quadrant equation \(a x^2-b x+a=0\) is 6 . \(\therefore \quad \mathrm{a}(6)^2-\mathrm{b}(6)+\mathrm{a}=0\) \(\Rightarrow 36 \mathrm{a}+\mathrm{a}=6 \mathrm{~b}\) Or \(\quad 37 a=6 b \Rightarrow \frac{b}{a}=\frac{37}{6}\)
