NEET Test Series from KOTA - 10 Papers In MS WORD
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Complex Numbers and Quadratic Equation
118219
If the sum of the roots of equation \(a x^2+b x+c\) \(=0\) is equal to the sum of the squares of their reciprocals, then \(\frac{\mathrm{a}}{\mathrm{c}}, \frac{\mathrm{b}}{\mathrm{a}}, \frac{\mathrm{c}}{\mathrm{b}}\) are in
1 A.P.
2 G.P
3 H.P.
4 None of these
Explanation:
C Let \(\alpha, \beta\) be the roots of \(a x^2+b x+c=0\). \(\alpha+\beta=-\frac{b}{a}, \alpha \cdot \beta=\frac{c}{a}\) by given condition, \(-\frac{b}{a}=\left(\frac{1}{\alpha^2}+\frac{1}{\beta^2}\right)=\frac{\alpha^2+\beta^2}{(\alpha \beta)^2}\) \(-\frac{b}{a}=\frac{(\alpha+\beta)^2-2 \alpha \beta}{(\alpha \beta)^2}\) \(-\frac{b}{a}=\frac{\frac{b^2}{a^2}-2\left(\frac{c}{a}\right)}{\left(\frac{c}{a}\right)^2}=\left(\frac{b^2-2 a c}{a^2 \cdot \frac{c^2}{a^2}}\right)\) \(-\frac{b}{a}=\frac{b^2-2 a c}{c^2}\) \(-b c^2=a^2-2 a^2 c \Rightarrow 2 a^2 c=b c^2+b^2 a\) divide the above equation by \(a b c\) \(\frac{2 \mathrm{a}}{\mathrm{b}}=\frac{\mathrm{c}}{\mathrm{a}}+\frac{\mathrm{b}}{\mathrm{c}}\)Hence, the reciprocals are in H.P.
COMEDK-2019
Complex Numbers and Quadratic Equation
118220
The roots of \((x-a)(x-a-1)+(x-a-1)(x-a-2)\) \(+(x-a)(x-a-2)=0\) \(a \in R\) are always
1 equal
2 imaginary
3 real and distinct
4 rational and equal
Explanation:
C Given \((\mathrm{x}-\mathrm{a})(\mathrm{x}-\mathrm{a}-1)+(\mathrm{x}-\mathrm{a}-1)(\mathrm{x}-\mathrm{a}-2)\) \(\quad+(\mathrm{x}-\mathrm{a})(\mathrm{x}-\mathrm{a}-2)=0\) \(\text { Let } \mathrm{x}-\mathrm{a}=\mathrm{t} \text {, then }\) \(\mathrm{t}(\mathrm{t}-1)+(\mathrm{t}-1)(\mathrm{t}-2)+\mathrm{t}(\mathrm{t}-2)=0\) \(\mathrm{t}^2-\mathrm{t}+\mathrm{t}^2-3 \mathrm{t}+2+\mathrm{t}^2-2 \mathrm{t}=0\) \(3 \mathrm{t}^2-6 \mathrm{t}+2=0\) \(\mathrm{t}=\frac{6 \pm \sqrt{36-24}}{2(3)}=\frac{6 \pm 2 \sqrt{3}}{2(3)}\) \(\mathrm{x}-\mathrm{a}=\frac{3 \pm \sqrt{3}}{3}\) \(\mathrm{x}=\mathrm{a}+\frac{3 \pm \sqrt{3}}{3}\) \(\text { Hence, } \mathrm{x} \text { is real and distinct. }\) Ans: c Exp: (C) : Given, Let \(\mathrm{x}-\mathrm{a}=\mathrm{t}\), then \(\mathrm{t}(\mathrm{t}-1)+(\mathrm{t}-1)(\mathrm{t}-2)+\mathrm{t}(\mathrm{t}-2)=0\) \(\mathrm{t}^2-\mathrm{t}+\mathrm{t}^2-3 \mathrm{t}+2+\mathrm{t}^2-2 \mathrm{t}=0\) \(3 \mathrm{t}^2-6 \mathrm{t}+2=0\) \(\mathrm{t}=\frac{6 \pm \sqrt{36-24}}{2(3)}=\frac{6 \pm 2 \sqrt{3}}{2(3)}\) \(\mathrm{x}-\mathrm{a}=\frac{3 \pm \sqrt{3}}{3}\) \(\mathrm{x}=\mathrm{a}+\frac{3 \pm \sqrt{3}}{3}\)Hence, \(\mathrm{x}\) is real and distinct.
VITEEE-2009
Complex Numbers and Quadratic Equation
118221
If \(\alpha\) and \(\beta\) are the roots of the equation \(\mathrm{ax}^2+\mathrm{bx}\) \(+c=0\), then the value of \(\alpha^3+\beta^3\) is
1 \(\frac{3 a b c+b^3}{a^3}\)
2 \(\frac{a^3+b^3}{3 a b c}\)
3 \(\frac{3 a b c-b^3}{a^3}\)
4 \(\frac{-\left(3 a b c+b^3\right)}{a^3}\)
Explanation:
C Given : \(\alpha \& \beta\) are roots of the equation \(\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0\) \(\therefore \alpha+\beta=-\frac{\mathrm{b}}{\mathrm{a}} \& \alpha \beta=\frac{\mathrm{c}}{\mathrm{a}}\) Now, \(\alpha^3+\beta^3=(\alpha+\beta)^3-3 \alpha \beta(\alpha+\beta)\) \(\alpha^3+\beta^3=\left(-\frac{b}{a}\right)^3-3 \frac{c}{a}\left(-\frac{b}{a}\right)\) \(\alpha^3+\beta^3=-\frac{b^3}{a^3}+\frac{3 b c}{a^2}\) \(\alpha^3+\beta^3=\frac{-b^3+3 a b c}{a^3}\) \(\alpha^3+\beta^3=\frac{3 a b c-b^3}{a^3}\)
VITEEE-2006
Complex Numbers and Quadratic Equation
118222
It roots of equation \(x^2+x+1=0\) are \(a, b\) and roots of \(x^2+p x+q=0\) are \(\frac{a}{b}, \frac{b}{a}\) then the value of \(\mathbf{p}+\mathbf{q}\) is
1 2
2 \(\frac{\sqrt{2}+1}{2}\)
3 -1
4 1
Explanation:
A Given, \(\mathrm{x}^2+\mathrm{x}+1=0\) and roots are \(\mathrm{a}, \mathrm{b}\) then, \(a+b=-1\) \(\mathrm{a} \times \mathrm{b}=1\) other given equation is - \(\mathrm{x}^2+\mathrm{px}+\mathrm{q}=0\) and roots are \(\frac{\mathrm{a}}{\mathrm{b}}, \frac{\mathrm{b}}{\mathrm{a}}\) Then, Sum of roots \(\frac{\mathrm{a}}{\mathrm{b}}+\frac{\mathrm{b}}{\mathrm{a}}=-\mathrm{p}\) \(\frac{\mathrm{a}^2+\mathrm{b}^2}{\mathrm{ab}}=-\mathrm{p}\) \(\frac{(\mathrm{a}+\mathrm{b})^2-2 \mathrm{ab}}{\mathrm{ab}}=-\mathrm{p}\) \(\frac{(-1)^2-2}{1}=-\mathrm{p}\) \(\mathrm{p}=1\) Product of roots \(\frac{\mathrm{a}}{\mathrm{b}} \times \frac{\mathrm{b}}{\mathrm{a}}=\mathrm{q}\) \(\mathrm{q}=1\)So value of \(\mathrm{p}+\mathrm{q}=1+1=2\)
118219
If the sum of the roots of equation \(a x^2+b x+c\) \(=0\) is equal to the sum of the squares of their reciprocals, then \(\frac{\mathrm{a}}{\mathrm{c}}, \frac{\mathrm{b}}{\mathrm{a}}, \frac{\mathrm{c}}{\mathrm{b}}\) are in
1 A.P.
2 G.P
3 H.P.
4 None of these
Explanation:
C Let \(\alpha, \beta\) be the roots of \(a x^2+b x+c=0\). \(\alpha+\beta=-\frac{b}{a}, \alpha \cdot \beta=\frac{c}{a}\) by given condition, \(-\frac{b}{a}=\left(\frac{1}{\alpha^2}+\frac{1}{\beta^2}\right)=\frac{\alpha^2+\beta^2}{(\alpha \beta)^2}\) \(-\frac{b}{a}=\frac{(\alpha+\beta)^2-2 \alpha \beta}{(\alpha \beta)^2}\) \(-\frac{b}{a}=\frac{\frac{b^2}{a^2}-2\left(\frac{c}{a}\right)}{\left(\frac{c}{a}\right)^2}=\left(\frac{b^2-2 a c}{a^2 \cdot \frac{c^2}{a^2}}\right)\) \(-\frac{b}{a}=\frac{b^2-2 a c}{c^2}\) \(-b c^2=a^2-2 a^2 c \Rightarrow 2 a^2 c=b c^2+b^2 a\) divide the above equation by \(a b c\) \(\frac{2 \mathrm{a}}{\mathrm{b}}=\frac{\mathrm{c}}{\mathrm{a}}+\frac{\mathrm{b}}{\mathrm{c}}\)Hence, the reciprocals are in H.P.
COMEDK-2019
Complex Numbers and Quadratic Equation
118220
The roots of \((x-a)(x-a-1)+(x-a-1)(x-a-2)\) \(+(x-a)(x-a-2)=0\) \(a \in R\) are always
1 equal
2 imaginary
3 real and distinct
4 rational and equal
Explanation:
C Given \((\mathrm{x}-\mathrm{a})(\mathrm{x}-\mathrm{a}-1)+(\mathrm{x}-\mathrm{a}-1)(\mathrm{x}-\mathrm{a}-2)\) \(\quad+(\mathrm{x}-\mathrm{a})(\mathrm{x}-\mathrm{a}-2)=0\) \(\text { Let } \mathrm{x}-\mathrm{a}=\mathrm{t} \text {, then }\) \(\mathrm{t}(\mathrm{t}-1)+(\mathrm{t}-1)(\mathrm{t}-2)+\mathrm{t}(\mathrm{t}-2)=0\) \(\mathrm{t}^2-\mathrm{t}+\mathrm{t}^2-3 \mathrm{t}+2+\mathrm{t}^2-2 \mathrm{t}=0\) \(3 \mathrm{t}^2-6 \mathrm{t}+2=0\) \(\mathrm{t}=\frac{6 \pm \sqrt{36-24}}{2(3)}=\frac{6 \pm 2 \sqrt{3}}{2(3)}\) \(\mathrm{x}-\mathrm{a}=\frac{3 \pm \sqrt{3}}{3}\) \(\mathrm{x}=\mathrm{a}+\frac{3 \pm \sqrt{3}}{3}\) \(\text { Hence, } \mathrm{x} \text { is real and distinct. }\) Ans: c Exp: (C) : Given, Let \(\mathrm{x}-\mathrm{a}=\mathrm{t}\), then \(\mathrm{t}(\mathrm{t}-1)+(\mathrm{t}-1)(\mathrm{t}-2)+\mathrm{t}(\mathrm{t}-2)=0\) \(\mathrm{t}^2-\mathrm{t}+\mathrm{t}^2-3 \mathrm{t}+2+\mathrm{t}^2-2 \mathrm{t}=0\) \(3 \mathrm{t}^2-6 \mathrm{t}+2=0\) \(\mathrm{t}=\frac{6 \pm \sqrt{36-24}}{2(3)}=\frac{6 \pm 2 \sqrt{3}}{2(3)}\) \(\mathrm{x}-\mathrm{a}=\frac{3 \pm \sqrt{3}}{3}\) \(\mathrm{x}=\mathrm{a}+\frac{3 \pm \sqrt{3}}{3}\)Hence, \(\mathrm{x}\) is real and distinct.
VITEEE-2009
Complex Numbers and Quadratic Equation
118221
If \(\alpha\) and \(\beta\) are the roots of the equation \(\mathrm{ax}^2+\mathrm{bx}\) \(+c=0\), then the value of \(\alpha^3+\beta^3\) is
1 \(\frac{3 a b c+b^3}{a^3}\)
2 \(\frac{a^3+b^3}{3 a b c}\)
3 \(\frac{3 a b c-b^3}{a^3}\)
4 \(\frac{-\left(3 a b c+b^3\right)}{a^3}\)
Explanation:
C Given : \(\alpha \& \beta\) are roots of the equation \(\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0\) \(\therefore \alpha+\beta=-\frac{\mathrm{b}}{\mathrm{a}} \& \alpha \beta=\frac{\mathrm{c}}{\mathrm{a}}\) Now, \(\alpha^3+\beta^3=(\alpha+\beta)^3-3 \alpha \beta(\alpha+\beta)\) \(\alpha^3+\beta^3=\left(-\frac{b}{a}\right)^3-3 \frac{c}{a}\left(-\frac{b}{a}\right)\) \(\alpha^3+\beta^3=-\frac{b^3}{a^3}+\frac{3 b c}{a^2}\) \(\alpha^3+\beta^3=\frac{-b^3+3 a b c}{a^3}\) \(\alpha^3+\beta^3=\frac{3 a b c-b^3}{a^3}\)
VITEEE-2006
Complex Numbers and Quadratic Equation
118222
It roots of equation \(x^2+x+1=0\) are \(a, b\) and roots of \(x^2+p x+q=0\) are \(\frac{a}{b}, \frac{b}{a}\) then the value of \(\mathbf{p}+\mathbf{q}\) is
1 2
2 \(\frac{\sqrt{2}+1}{2}\)
3 -1
4 1
Explanation:
A Given, \(\mathrm{x}^2+\mathrm{x}+1=0\) and roots are \(\mathrm{a}, \mathrm{b}\) then, \(a+b=-1\) \(\mathrm{a} \times \mathrm{b}=1\) other given equation is - \(\mathrm{x}^2+\mathrm{px}+\mathrm{q}=0\) and roots are \(\frac{\mathrm{a}}{\mathrm{b}}, \frac{\mathrm{b}}{\mathrm{a}}\) Then, Sum of roots \(\frac{\mathrm{a}}{\mathrm{b}}+\frac{\mathrm{b}}{\mathrm{a}}=-\mathrm{p}\) \(\frac{\mathrm{a}^2+\mathrm{b}^2}{\mathrm{ab}}=-\mathrm{p}\) \(\frac{(\mathrm{a}+\mathrm{b})^2-2 \mathrm{ab}}{\mathrm{ab}}=-\mathrm{p}\) \(\frac{(-1)^2-2}{1}=-\mathrm{p}\) \(\mathrm{p}=1\) Product of roots \(\frac{\mathrm{a}}{\mathrm{b}} \times \frac{\mathrm{b}}{\mathrm{a}}=\mathrm{q}\) \(\mathrm{q}=1\)So value of \(\mathrm{p}+\mathrm{q}=1+1=2\)
118219
If the sum of the roots of equation \(a x^2+b x+c\) \(=0\) is equal to the sum of the squares of their reciprocals, then \(\frac{\mathrm{a}}{\mathrm{c}}, \frac{\mathrm{b}}{\mathrm{a}}, \frac{\mathrm{c}}{\mathrm{b}}\) are in
1 A.P.
2 G.P
3 H.P.
4 None of these
Explanation:
C Let \(\alpha, \beta\) be the roots of \(a x^2+b x+c=0\). \(\alpha+\beta=-\frac{b}{a}, \alpha \cdot \beta=\frac{c}{a}\) by given condition, \(-\frac{b}{a}=\left(\frac{1}{\alpha^2}+\frac{1}{\beta^2}\right)=\frac{\alpha^2+\beta^2}{(\alpha \beta)^2}\) \(-\frac{b}{a}=\frac{(\alpha+\beta)^2-2 \alpha \beta}{(\alpha \beta)^2}\) \(-\frac{b}{a}=\frac{\frac{b^2}{a^2}-2\left(\frac{c}{a}\right)}{\left(\frac{c}{a}\right)^2}=\left(\frac{b^2-2 a c}{a^2 \cdot \frac{c^2}{a^2}}\right)\) \(-\frac{b}{a}=\frac{b^2-2 a c}{c^2}\) \(-b c^2=a^2-2 a^2 c \Rightarrow 2 a^2 c=b c^2+b^2 a\) divide the above equation by \(a b c\) \(\frac{2 \mathrm{a}}{\mathrm{b}}=\frac{\mathrm{c}}{\mathrm{a}}+\frac{\mathrm{b}}{\mathrm{c}}\)Hence, the reciprocals are in H.P.
COMEDK-2019
Complex Numbers and Quadratic Equation
118220
The roots of \((x-a)(x-a-1)+(x-a-1)(x-a-2)\) \(+(x-a)(x-a-2)=0\) \(a \in R\) are always
1 equal
2 imaginary
3 real and distinct
4 rational and equal
Explanation:
C Given \((\mathrm{x}-\mathrm{a})(\mathrm{x}-\mathrm{a}-1)+(\mathrm{x}-\mathrm{a}-1)(\mathrm{x}-\mathrm{a}-2)\) \(\quad+(\mathrm{x}-\mathrm{a})(\mathrm{x}-\mathrm{a}-2)=0\) \(\text { Let } \mathrm{x}-\mathrm{a}=\mathrm{t} \text {, then }\) \(\mathrm{t}(\mathrm{t}-1)+(\mathrm{t}-1)(\mathrm{t}-2)+\mathrm{t}(\mathrm{t}-2)=0\) \(\mathrm{t}^2-\mathrm{t}+\mathrm{t}^2-3 \mathrm{t}+2+\mathrm{t}^2-2 \mathrm{t}=0\) \(3 \mathrm{t}^2-6 \mathrm{t}+2=0\) \(\mathrm{t}=\frac{6 \pm \sqrt{36-24}}{2(3)}=\frac{6 \pm 2 \sqrt{3}}{2(3)}\) \(\mathrm{x}-\mathrm{a}=\frac{3 \pm \sqrt{3}}{3}\) \(\mathrm{x}=\mathrm{a}+\frac{3 \pm \sqrt{3}}{3}\) \(\text { Hence, } \mathrm{x} \text { is real and distinct. }\) Ans: c Exp: (C) : Given, Let \(\mathrm{x}-\mathrm{a}=\mathrm{t}\), then \(\mathrm{t}(\mathrm{t}-1)+(\mathrm{t}-1)(\mathrm{t}-2)+\mathrm{t}(\mathrm{t}-2)=0\) \(\mathrm{t}^2-\mathrm{t}+\mathrm{t}^2-3 \mathrm{t}+2+\mathrm{t}^2-2 \mathrm{t}=0\) \(3 \mathrm{t}^2-6 \mathrm{t}+2=0\) \(\mathrm{t}=\frac{6 \pm \sqrt{36-24}}{2(3)}=\frac{6 \pm 2 \sqrt{3}}{2(3)}\) \(\mathrm{x}-\mathrm{a}=\frac{3 \pm \sqrt{3}}{3}\) \(\mathrm{x}=\mathrm{a}+\frac{3 \pm \sqrt{3}}{3}\)Hence, \(\mathrm{x}\) is real and distinct.
VITEEE-2009
Complex Numbers and Quadratic Equation
118221
If \(\alpha\) and \(\beta\) are the roots of the equation \(\mathrm{ax}^2+\mathrm{bx}\) \(+c=0\), then the value of \(\alpha^3+\beta^3\) is
1 \(\frac{3 a b c+b^3}{a^3}\)
2 \(\frac{a^3+b^3}{3 a b c}\)
3 \(\frac{3 a b c-b^3}{a^3}\)
4 \(\frac{-\left(3 a b c+b^3\right)}{a^3}\)
Explanation:
C Given : \(\alpha \& \beta\) are roots of the equation \(\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0\) \(\therefore \alpha+\beta=-\frac{\mathrm{b}}{\mathrm{a}} \& \alpha \beta=\frac{\mathrm{c}}{\mathrm{a}}\) Now, \(\alpha^3+\beta^3=(\alpha+\beta)^3-3 \alpha \beta(\alpha+\beta)\) \(\alpha^3+\beta^3=\left(-\frac{b}{a}\right)^3-3 \frac{c}{a}\left(-\frac{b}{a}\right)\) \(\alpha^3+\beta^3=-\frac{b^3}{a^3}+\frac{3 b c}{a^2}\) \(\alpha^3+\beta^3=\frac{-b^3+3 a b c}{a^3}\) \(\alpha^3+\beta^3=\frac{3 a b c-b^3}{a^3}\)
VITEEE-2006
Complex Numbers and Quadratic Equation
118222
It roots of equation \(x^2+x+1=0\) are \(a, b\) and roots of \(x^2+p x+q=0\) are \(\frac{a}{b}, \frac{b}{a}\) then the value of \(\mathbf{p}+\mathbf{q}\) is
1 2
2 \(\frac{\sqrt{2}+1}{2}\)
3 -1
4 1
Explanation:
A Given, \(\mathrm{x}^2+\mathrm{x}+1=0\) and roots are \(\mathrm{a}, \mathrm{b}\) then, \(a+b=-1\) \(\mathrm{a} \times \mathrm{b}=1\) other given equation is - \(\mathrm{x}^2+\mathrm{px}+\mathrm{q}=0\) and roots are \(\frac{\mathrm{a}}{\mathrm{b}}, \frac{\mathrm{b}}{\mathrm{a}}\) Then, Sum of roots \(\frac{\mathrm{a}}{\mathrm{b}}+\frac{\mathrm{b}}{\mathrm{a}}=-\mathrm{p}\) \(\frac{\mathrm{a}^2+\mathrm{b}^2}{\mathrm{ab}}=-\mathrm{p}\) \(\frac{(\mathrm{a}+\mathrm{b})^2-2 \mathrm{ab}}{\mathrm{ab}}=-\mathrm{p}\) \(\frac{(-1)^2-2}{1}=-\mathrm{p}\) \(\mathrm{p}=1\) Product of roots \(\frac{\mathrm{a}}{\mathrm{b}} \times \frac{\mathrm{b}}{\mathrm{a}}=\mathrm{q}\) \(\mathrm{q}=1\)So value of \(\mathrm{p}+\mathrm{q}=1+1=2\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Complex Numbers and Quadratic Equation
118219
If the sum of the roots of equation \(a x^2+b x+c\) \(=0\) is equal to the sum of the squares of their reciprocals, then \(\frac{\mathrm{a}}{\mathrm{c}}, \frac{\mathrm{b}}{\mathrm{a}}, \frac{\mathrm{c}}{\mathrm{b}}\) are in
1 A.P.
2 G.P
3 H.P.
4 None of these
Explanation:
C Let \(\alpha, \beta\) be the roots of \(a x^2+b x+c=0\). \(\alpha+\beta=-\frac{b}{a}, \alpha \cdot \beta=\frac{c}{a}\) by given condition, \(-\frac{b}{a}=\left(\frac{1}{\alpha^2}+\frac{1}{\beta^2}\right)=\frac{\alpha^2+\beta^2}{(\alpha \beta)^2}\) \(-\frac{b}{a}=\frac{(\alpha+\beta)^2-2 \alpha \beta}{(\alpha \beta)^2}\) \(-\frac{b}{a}=\frac{\frac{b^2}{a^2}-2\left(\frac{c}{a}\right)}{\left(\frac{c}{a}\right)^2}=\left(\frac{b^2-2 a c}{a^2 \cdot \frac{c^2}{a^2}}\right)\) \(-\frac{b}{a}=\frac{b^2-2 a c}{c^2}\) \(-b c^2=a^2-2 a^2 c \Rightarrow 2 a^2 c=b c^2+b^2 a\) divide the above equation by \(a b c\) \(\frac{2 \mathrm{a}}{\mathrm{b}}=\frac{\mathrm{c}}{\mathrm{a}}+\frac{\mathrm{b}}{\mathrm{c}}\)Hence, the reciprocals are in H.P.
COMEDK-2019
Complex Numbers and Quadratic Equation
118220
The roots of \((x-a)(x-a-1)+(x-a-1)(x-a-2)\) \(+(x-a)(x-a-2)=0\) \(a \in R\) are always
1 equal
2 imaginary
3 real and distinct
4 rational and equal
Explanation:
C Given \((\mathrm{x}-\mathrm{a})(\mathrm{x}-\mathrm{a}-1)+(\mathrm{x}-\mathrm{a}-1)(\mathrm{x}-\mathrm{a}-2)\) \(\quad+(\mathrm{x}-\mathrm{a})(\mathrm{x}-\mathrm{a}-2)=0\) \(\text { Let } \mathrm{x}-\mathrm{a}=\mathrm{t} \text {, then }\) \(\mathrm{t}(\mathrm{t}-1)+(\mathrm{t}-1)(\mathrm{t}-2)+\mathrm{t}(\mathrm{t}-2)=0\) \(\mathrm{t}^2-\mathrm{t}+\mathrm{t}^2-3 \mathrm{t}+2+\mathrm{t}^2-2 \mathrm{t}=0\) \(3 \mathrm{t}^2-6 \mathrm{t}+2=0\) \(\mathrm{t}=\frac{6 \pm \sqrt{36-24}}{2(3)}=\frac{6 \pm 2 \sqrt{3}}{2(3)}\) \(\mathrm{x}-\mathrm{a}=\frac{3 \pm \sqrt{3}}{3}\) \(\mathrm{x}=\mathrm{a}+\frac{3 \pm \sqrt{3}}{3}\) \(\text { Hence, } \mathrm{x} \text { is real and distinct. }\) Ans: c Exp: (C) : Given, Let \(\mathrm{x}-\mathrm{a}=\mathrm{t}\), then \(\mathrm{t}(\mathrm{t}-1)+(\mathrm{t}-1)(\mathrm{t}-2)+\mathrm{t}(\mathrm{t}-2)=0\) \(\mathrm{t}^2-\mathrm{t}+\mathrm{t}^2-3 \mathrm{t}+2+\mathrm{t}^2-2 \mathrm{t}=0\) \(3 \mathrm{t}^2-6 \mathrm{t}+2=0\) \(\mathrm{t}=\frac{6 \pm \sqrt{36-24}}{2(3)}=\frac{6 \pm 2 \sqrt{3}}{2(3)}\) \(\mathrm{x}-\mathrm{a}=\frac{3 \pm \sqrt{3}}{3}\) \(\mathrm{x}=\mathrm{a}+\frac{3 \pm \sqrt{3}}{3}\)Hence, \(\mathrm{x}\) is real and distinct.
VITEEE-2009
Complex Numbers and Quadratic Equation
118221
If \(\alpha\) and \(\beta\) are the roots of the equation \(\mathrm{ax}^2+\mathrm{bx}\) \(+c=0\), then the value of \(\alpha^3+\beta^3\) is
1 \(\frac{3 a b c+b^3}{a^3}\)
2 \(\frac{a^3+b^3}{3 a b c}\)
3 \(\frac{3 a b c-b^3}{a^3}\)
4 \(\frac{-\left(3 a b c+b^3\right)}{a^3}\)
Explanation:
C Given : \(\alpha \& \beta\) are roots of the equation \(\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0\) \(\therefore \alpha+\beta=-\frac{\mathrm{b}}{\mathrm{a}} \& \alpha \beta=\frac{\mathrm{c}}{\mathrm{a}}\) Now, \(\alpha^3+\beta^3=(\alpha+\beta)^3-3 \alpha \beta(\alpha+\beta)\) \(\alpha^3+\beta^3=\left(-\frac{b}{a}\right)^3-3 \frac{c}{a}\left(-\frac{b}{a}\right)\) \(\alpha^3+\beta^3=-\frac{b^3}{a^3}+\frac{3 b c}{a^2}\) \(\alpha^3+\beta^3=\frac{-b^3+3 a b c}{a^3}\) \(\alpha^3+\beta^3=\frac{3 a b c-b^3}{a^3}\)
VITEEE-2006
Complex Numbers and Quadratic Equation
118222
It roots of equation \(x^2+x+1=0\) are \(a, b\) and roots of \(x^2+p x+q=0\) are \(\frac{a}{b}, \frac{b}{a}\) then the value of \(\mathbf{p}+\mathbf{q}\) is
1 2
2 \(\frac{\sqrt{2}+1}{2}\)
3 -1
4 1
Explanation:
A Given, \(\mathrm{x}^2+\mathrm{x}+1=0\) and roots are \(\mathrm{a}, \mathrm{b}\) then, \(a+b=-1\) \(\mathrm{a} \times \mathrm{b}=1\) other given equation is - \(\mathrm{x}^2+\mathrm{px}+\mathrm{q}=0\) and roots are \(\frac{\mathrm{a}}{\mathrm{b}}, \frac{\mathrm{b}}{\mathrm{a}}\) Then, Sum of roots \(\frac{\mathrm{a}}{\mathrm{b}}+\frac{\mathrm{b}}{\mathrm{a}}=-\mathrm{p}\) \(\frac{\mathrm{a}^2+\mathrm{b}^2}{\mathrm{ab}}=-\mathrm{p}\) \(\frac{(\mathrm{a}+\mathrm{b})^2-2 \mathrm{ab}}{\mathrm{ab}}=-\mathrm{p}\) \(\frac{(-1)^2-2}{1}=-\mathrm{p}\) \(\mathrm{p}=1\) Product of roots \(\frac{\mathrm{a}}{\mathrm{b}} \times \frac{\mathrm{b}}{\mathrm{a}}=\mathrm{q}\) \(\mathrm{q}=1\)So value of \(\mathrm{p}+\mathrm{q}=1+1=2\)