118252
The number of positive real roots of the equation \(3^{\mathrm{x+1}}+3^{-\mathrm{x}+\mathrm{f}}=\mathbf{1 0}\) is
1 3
2 2
3 1
4 Infinitely many
Explanation:
C Given, \(3^{x+1}+3^{-x+1}=10\) \(3^x \cdot 3+\frac{3}{3^x}=10\) Let, \(\quad 3^{\mathrm{x}}=\mathrm{t}\), then we have \(3 t+\frac{3}{t}=10\) \(3 t^2+3=10 t\) \(\text { or } 3 t^2-10 t+3=0\) \(\text { or } 3 t^2-9 t-t+3=0\) \(\text { or } 3 t(t-3)-1(t-3)=0\) \(\text { or } (3 t-1)(t-3)=0\) \(\text { or } t=\frac{1}{3}, 3\) \(\Rightarrow 3^x=\frac{1}{3}=3^{-1} \Rightarrow x=-1\) and \(\quad 3^{\mathrm{x}}=3 \Rightarrow \mathrm{x}=1\) \(\therefore\) Number of positive real root is 1
AP EAMCET-04.07.2021
Complex Numbers and Quadratic Equation
118253
If \(\alpha, \beta\) be the roots of \(x^2-a(x-1)+b=0\), then the value of \(\frac{1}{\alpha^2-a \alpha}+\frac{1}{\beta^2-a \beta}+\frac{2}{a+b}\)
118254
If the ratio of the roots of the equation \(p x^2+q x\) \(+r=0\) is \(a: b\), then \(\frac{a b}{(a+b)^2}=\)
1 \(\frac{\mathrm{p}^2}{\mathrm{qr}}\)
2 \(\frac{\mathrm{pr}}{\mathrm{q}^2}\)
3 \(\frac{\mathrm{q}^2}{\mathrm{pr}}\)
4 \(\frac{\mathrm{pq}}{\mathrm{r}^2}\)
Explanation:
B Given, Let the roots be am and bm. \(\text { Let the roots be am and bm. }\) \(\text { sum of roots }(a m+b m)=-\frac{q}{p}\) \((a+b)^2 m^2=\left(\frac{q}{p}right)^2\right.\) And product of roots \((\mathrm{am} \cdot \mathrm{bm})=\frac{\mathrm{r}}{\mathrm{p}}\) On dividing equation (ii) by (i), we get \(\frac{a \cdot b m^2}{(a+b)^2 m^2}=\frac{r \cdot p^2}{p \cdot q^2}\) \(\frac{a \cdot b}{(a+b)^2}=\frac{p \cdot r}{q^2}\)
WB JEE-2011
Complex Numbers and Quadratic Equation
118258
If \((\alpha+\sqrt{\beta})\) and \((\alpha-\sqrt{\beta})\) are the roots of the equation \(x^2+p x+q=0\), where \(\alpha, \beta\), p and \(q\) are real, then the roots of the equation \(\left(p^2-4 q\right)\left(p^2 x^2+4 p x\right)-16 q=0\) are
1 \(\left(\frac{1}{\alpha}+\frac{1}{\sqrt{\beta}}\right)\) and \(\left(\frac{1}{\alpha}-\frac{1}{\sqrt{\beta}}\right)\)
2 \(\left(\frac{1}{\sqrt{\alpha}}+\frac{1}{\beta}\right)\) and \(\left(\frac{1}{\sqrt{\alpha}}-\frac{1}{\beta}\right)\)
3 \(\left(\frac{1}{\sqrt{\alpha}}+\frac{1}{\sqrt{\beta}}\right)\) and \(\left(\frac{1}{\sqrt{\alpha}}-\frac{1}{\sqrt{\beta}}\right)\)
A Given, \((\alpha+\sqrt{\beta})\) and \((\alpha-\sqrt{\beta})\) are the roots of the equation \(\mathrm{x}^2+\mathrm{px}+\mathrm{q}=0\) sum of roots, \((\alpha+\sqrt{\beta})+(\alpha-\sqrt{\beta})=-p\) \(2 \alpha=-p\) \(\alpha=\frac{-p}{2}\) And, Product of roots, \((\alpha+\sqrt{\beta}) \cdot(\alpha-\sqrt{\beta})=\mathrm{q}\) \(\alpha^2-\beta=q\) \(\beta=\alpha^2-q=\left(-\frac{p}{2}\right)^2-q=\frac{p^2}{4}-q\) \(p^2-4 q=4 \beta\) \(\left(p^2-4 q\right)\left(p^2 x^2+4 p x\right)-16 q=0\) \(4 \beta\left(p^2 x^2+4 p x\right)-16\left(\alpha^2-\beta\right)=0\) \(\beta\left(p^2 x^2+4 p x\right)-4\left(\alpha^2-\beta\right)=0\) \(\beta\left(4 \alpha^2 x^2-8 \alpha x\right)-4\left(\alpha^2-\beta\right)=0\) \(\alpha^2 \beta x^2-2 \alpha \beta x+\beta=\alpha^2\) \((\alpha x \sqrt{\beta}-\sqrt{\beta})^2=\alpha^2\) \(\alpha x \sqrt{\beta}-\sqrt{\beta}= \pm \alpha\) \(\therefore x=\frac{1}{\alpha} \pm \frac{1}{\sqrt{\beta}}\) \(\left(\frac{1}{\alpha}+\frac{1}{\sqrt{\beta}}\right)\) and \(\left(\frac{1}{\alpha}-\frac{1}{\sqrt{\beta}}\right)\) are the roots of the given equation \(\left(p^2-4 q\right)\left(p^2 x^2+4 p x\right)-16 q=0\)
WB JEE-2012
Complex Numbers and Quadratic Equation
118259
The greatest real root of the equation \(6 x^4-35 x^3+62 x^2-35 x+6=0\) is
1 2
2 \(\frac{5}{2}\)
3 3
4 \(\frac{7}{2}\)
Explanation:
C Given, equation is \(6 x^4-35 x^3+62 x^2-35 x+6=0\) dividing equation \(\qquad\) (I) by \(x^2\) \(\left\{\left\{6\left(\mathrm{x}^2+\frac{1}{\mathrm{x}^2}\right)-35\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)+62\right\}=0\right.\) Let \(\mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{y}\), therefore, \(x\) \(6\left(y^2-2\right)-35 y+62=0\) \(6 y^2-12-35 y+62=0\) \(6 y^2-35 y+50=0\) \((2 y-5)(3 y-10)=0\) Now, Put back \(\mathrm{y}=\mathrm{x}+\frac{1}{\mathrm{x}}\) and multiply each factors by \(\mathrm{x}\), we get \(\left(2 \mathrm{x}^2-5 \mathrm{x}+2\right)\left(3 \mathrm{x}^2-10 \mathrm{x}+3\right)=0\) \((\mathrm{x}-2)(2 \mathrm{x}-1)(\mathrm{x}-3)(3 \mathrm{x}-1)=0\) \(\mathrm{x}=2, \frac{1}{2}, 3, \frac{1}{3}\)Hence, greatest value is 3 .
118252
The number of positive real roots of the equation \(3^{\mathrm{x+1}}+3^{-\mathrm{x}+\mathrm{f}}=\mathbf{1 0}\) is
1 3
2 2
3 1
4 Infinitely many
Explanation:
C Given, \(3^{x+1}+3^{-x+1}=10\) \(3^x \cdot 3+\frac{3}{3^x}=10\) Let, \(\quad 3^{\mathrm{x}}=\mathrm{t}\), then we have \(3 t+\frac{3}{t}=10\) \(3 t^2+3=10 t\) \(\text { or } 3 t^2-10 t+3=0\) \(\text { or } 3 t^2-9 t-t+3=0\) \(\text { or } 3 t(t-3)-1(t-3)=0\) \(\text { or } (3 t-1)(t-3)=0\) \(\text { or } t=\frac{1}{3}, 3\) \(\Rightarrow 3^x=\frac{1}{3}=3^{-1} \Rightarrow x=-1\) and \(\quad 3^{\mathrm{x}}=3 \Rightarrow \mathrm{x}=1\) \(\therefore\) Number of positive real root is 1
AP EAMCET-04.07.2021
Complex Numbers and Quadratic Equation
118253
If \(\alpha, \beta\) be the roots of \(x^2-a(x-1)+b=0\), then the value of \(\frac{1}{\alpha^2-a \alpha}+\frac{1}{\beta^2-a \beta}+\frac{2}{a+b}\)
118254
If the ratio of the roots of the equation \(p x^2+q x\) \(+r=0\) is \(a: b\), then \(\frac{a b}{(a+b)^2}=\)
1 \(\frac{\mathrm{p}^2}{\mathrm{qr}}\)
2 \(\frac{\mathrm{pr}}{\mathrm{q}^2}\)
3 \(\frac{\mathrm{q}^2}{\mathrm{pr}}\)
4 \(\frac{\mathrm{pq}}{\mathrm{r}^2}\)
Explanation:
B Given, Let the roots be am and bm. \(\text { Let the roots be am and bm. }\) \(\text { sum of roots }(a m+b m)=-\frac{q}{p}\) \((a+b)^2 m^2=\left(\frac{q}{p}right)^2\right.\) And product of roots \((\mathrm{am} \cdot \mathrm{bm})=\frac{\mathrm{r}}{\mathrm{p}}\) On dividing equation (ii) by (i), we get \(\frac{a \cdot b m^2}{(a+b)^2 m^2}=\frac{r \cdot p^2}{p \cdot q^2}\) \(\frac{a \cdot b}{(a+b)^2}=\frac{p \cdot r}{q^2}\)
WB JEE-2011
Complex Numbers and Quadratic Equation
118258
If \((\alpha+\sqrt{\beta})\) and \((\alpha-\sqrt{\beta})\) are the roots of the equation \(x^2+p x+q=0\), where \(\alpha, \beta\), p and \(q\) are real, then the roots of the equation \(\left(p^2-4 q\right)\left(p^2 x^2+4 p x\right)-16 q=0\) are
1 \(\left(\frac{1}{\alpha}+\frac{1}{\sqrt{\beta}}\right)\) and \(\left(\frac{1}{\alpha}-\frac{1}{\sqrt{\beta}}\right)\)
2 \(\left(\frac{1}{\sqrt{\alpha}}+\frac{1}{\beta}\right)\) and \(\left(\frac{1}{\sqrt{\alpha}}-\frac{1}{\beta}\right)\)
3 \(\left(\frac{1}{\sqrt{\alpha}}+\frac{1}{\sqrt{\beta}}\right)\) and \(\left(\frac{1}{\sqrt{\alpha}}-\frac{1}{\sqrt{\beta}}\right)\)
A Given, \((\alpha+\sqrt{\beta})\) and \((\alpha-\sqrt{\beta})\) are the roots of the equation \(\mathrm{x}^2+\mathrm{px}+\mathrm{q}=0\) sum of roots, \((\alpha+\sqrt{\beta})+(\alpha-\sqrt{\beta})=-p\) \(2 \alpha=-p\) \(\alpha=\frac{-p}{2}\) And, Product of roots, \((\alpha+\sqrt{\beta}) \cdot(\alpha-\sqrt{\beta})=\mathrm{q}\) \(\alpha^2-\beta=q\) \(\beta=\alpha^2-q=\left(-\frac{p}{2}\right)^2-q=\frac{p^2}{4}-q\) \(p^2-4 q=4 \beta\) \(\left(p^2-4 q\right)\left(p^2 x^2+4 p x\right)-16 q=0\) \(4 \beta\left(p^2 x^2+4 p x\right)-16\left(\alpha^2-\beta\right)=0\) \(\beta\left(p^2 x^2+4 p x\right)-4\left(\alpha^2-\beta\right)=0\) \(\beta\left(4 \alpha^2 x^2-8 \alpha x\right)-4\left(\alpha^2-\beta\right)=0\) \(\alpha^2 \beta x^2-2 \alpha \beta x+\beta=\alpha^2\) \((\alpha x \sqrt{\beta}-\sqrt{\beta})^2=\alpha^2\) \(\alpha x \sqrt{\beta}-\sqrt{\beta}= \pm \alpha\) \(\therefore x=\frac{1}{\alpha} \pm \frac{1}{\sqrt{\beta}}\) \(\left(\frac{1}{\alpha}+\frac{1}{\sqrt{\beta}}\right)\) and \(\left(\frac{1}{\alpha}-\frac{1}{\sqrt{\beta}}\right)\) are the roots of the given equation \(\left(p^2-4 q\right)\left(p^2 x^2+4 p x\right)-16 q=0\)
WB JEE-2012
Complex Numbers and Quadratic Equation
118259
The greatest real root of the equation \(6 x^4-35 x^3+62 x^2-35 x+6=0\) is
1 2
2 \(\frac{5}{2}\)
3 3
4 \(\frac{7}{2}\)
Explanation:
C Given, equation is \(6 x^4-35 x^3+62 x^2-35 x+6=0\) dividing equation \(\qquad\) (I) by \(x^2\) \(\left\{\left\{6\left(\mathrm{x}^2+\frac{1}{\mathrm{x}^2}\right)-35\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)+62\right\}=0\right.\) Let \(\mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{y}\), therefore, \(x\) \(6\left(y^2-2\right)-35 y+62=0\) \(6 y^2-12-35 y+62=0\) \(6 y^2-35 y+50=0\) \((2 y-5)(3 y-10)=0\) Now, Put back \(\mathrm{y}=\mathrm{x}+\frac{1}{\mathrm{x}}\) and multiply each factors by \(\mathrm{x}\), we get \(\left(2 \mathrm{x}^2-5 \mathrm{x}+2\right)\left(3 \mathrm{x}^2-10 \mathrm{x}+3\right)=0\) \((\mathrm{x}-2)(2 \mathrm{x}-1)(\mathrm{x}-3)(3 \mathrm{x}-1)=0\) \(\mathrm{x}=2, \frac{1}{2}, 3, \frac{1}{3}\)Hence, greatest value is 3 .
118252
The number of positive real roots of the equation \(3^{\mathrm{x+1}}+3^{-\mathrm{x}+\mathrm{f}}=\mathbf{1 0}\) is
1 3
2 2
3 1
4 Infinitely many
Explanation:
C Given, \(3^{x+1}+3^{-x+1}=10\) \(3^x \cdot 3+\frac{3}{3^x}=10\) Let, \(\quad 3^{\mathrm{x}}=\mathrm{t}\), then we have \(3 t+\frac{3}{t}=10\) \(3 t^2+3=10 t\) \(\text { or } 3 t^2-10 t+3=0\) \(\text { or } 3 t^2-9 t-t+3=0\) \(\text { or } 3 t(t-3)-1(t-3)=0\) \(\text { or } (3 t-1)(t-3)=0\) \(\text { or } t=\frac{1}{3}, 3\) \(\Rightarrow 3^x=\frac{1}{3}=3^{-1} \Rightarrow x=-1\) and \(\quad 3^{\mathrm{x}}=3 \Rightarrow \mathrm{x}=1\) \(\therefore\) Number of positive real root is 1
AP EAMCET-04.07.2021
Complex Numbers and Quadratic Equation
118253
If \(\alpha, \beta\) be the roots of \(x^2-a(x-1)+b=0\), then the value of \(\frac{1}{\alpha^2-a \alpha}+\frac{1}{\beta^2-a \beta}+\frac{2}{a+b}\)
118254
If the ratio of the roots of the equation \(p x^2+q x\) \(+r=0\) is \(a: b\), then \(\frac{a b}{(a+b)^2}=\)
1 \(\frac{\mathrm{p}^2}{\mathrm{qr}}\)
2 \(\frac{\mathrm{pr}}{\mathrm{q}^2}\)
3 \(\frac{\mathrm{q}^2}{\mathrm{pr}}\)
4 \(\frac{\mathrm{pq}}{\mathrm{r}^2}\)
Explanation:
B Given, Let the roots be am and bm. \(\text { Let the roots be am and bm. }\) \(\text { sum of roots }(a m+b m)=-\frac{q}{p}\) \((a+b)^2 m^2=\left(\frac{q}{p}right)^2\right.\) And product of roots \((\mathrm{am} \cdot \mathrm{bm})=\frac{\mathrm{r}}{\mathrm{p}}\) On dividing equation (ii) by (i), we get \(\frac{a \cdot b m^2}{(a+b)^2 m^2}=\frac{r \cdot p^2}{p \cdot q^2}\) \(\frac{a \cdot b}{(a+b)^2}=\frac{p \cdot r}{q^2}\)
WB JEE-2011
Complex Numbers and Quadratic Equation
118258
If \((\alpha+\sqrt{\beta})\) and \((\alpha-\sqrt{\beta})\) are the roots of the equation \(x^2+p x+q=0\), where \(\alpha, \beta\), p and \(q\) are real, then the roots of the equation \(\left(p^2-4 q\right)\left(p^2 x^2+4 p x\right)-16 q=0\) are
1 \(\left(\frac{1}{\alpha}+\frac{1}{\sqrt{\beta}}\right)\) and \(\left(\frac{1}{\alpha}-\frac{1}{\sqrt{\beta}}\right)\)
2 \(\left(\frac{1}{\sqrt{\alpha}}+\frac{1}{\beta}\right)\) and \(\left(\frac{1}{\sqrt{\alpha}}-\frac{1}{\beta}\right)\)
3 \(\left(\frac{1}{\sqrt{\alpha}}+\frac{1}{\sqrt{\beta}}\right)\) and \(\left(\frac{1}{\sqrt{\alpha}}-\frac{1}{\sqrt{\beta}}\right)\)
A Given, \((\alpha+\sqrt{\beta})\) and \((\alpha-\sqrt{\beta})\) are the roots of the equation \(\mathrm{x}^2+\mathrm{px}+\mathrm{q}=0\) sum of roots, \((\alpha+\sqrt{\beta})+(\alpha-\sqrt{\beta})=-p\) \(2 \alpha=-p\) \(\alpha=\frac{-p}{2}\) And, Product of roots, \((\alpha+\sqrt{\beta}) \cdot(\alpha-\sqrt{\beta})=\mathrm{q}\) \(\alpha^2-\beta=q\) \(\beta=\alpha^2-q=\left(-\frac{p}{2}\right)^2-q=\frac{p^2}{4}-q\) \(p^2-4 q=4 \beta\) \(\left(p^2-4 q\right)\left(p^2 x^2+4 p x\right)-16 q=0\) \(4 \beta\left(p^2 x^2+4 p x\right)-16\left(\alpha^2-\beta\right)=0\) \(\beta\left(p^2 x^2+4 p x\right)-4\left(\alpha^2-\beta\right)=0\) \(\beta\left(4 \alpha^2 x^2-8 \alpha x\right)-4\left(\alpha^2-\beta\right)=0\) \(\alpha^2 \beta x^2-2 \alpha \beta x+\beta=\alpha^2\) \((\alpha x \sqrt{\beta}-\sqrt{\beta})^2=\alpha^2\) \(\alpha x \sqrt{\beta}-\sqrt{\beta}= \pm \alpha\) \(\therefore x=\frac{1}{\alpha} \pm \frac{1}{\sqrt{\beta}}\) \(\left(\frac{1}{\alpha}+\frac{1}{\sqrt{\beta}}\right)\) and \(\left(\frac{1}{\alpha}-\frac{1}{\sqrt{\beta}}\right)\) are the roots of the given equation \(\left(p^2-4 q\right)\left(p^2 x^2+4 p x\right)-16 q=0\)
WB JEE-2012
Complex Numbers and Quadratic Equation
118259
The greatest real root of the equation \(6 x^4-35 x^3+62 x^2-35 x+6=0\) is
1 2
2 \(\frac{5}{2}\)
3 3
4 \(\frac{7}{2}\)
Explanation:
C Given, equation is \(6 x^4-35 x^3+62 x^2-35 x+6=0\) dividing equation \(\qquad\) (I) by \(x^2\) \(\left\{\left\{6\left(\mathrm{x}^2+\frac{1}{\mathrm{x}^2}\right)-35\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)+62\right\}=0\right.\) Let \(\mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{y}\), therefore, \(x\) \(6\left(y^2-2\right)-35 y+62=0\) \(6 y^2-12-35 y+62=0\) \(6 y^2-35 y+50=0\) \((2 y-5)(3 y-10)=0\) Now, Put back \(\mathrm{y}=\mathrm{x}+\frac{1}{\mathrm{x}}\) and multiply each factors by \(\mathrm{x}\), we get \(\left(2 \mathrm{x}^2-5 \mathrm{x}+2\right)\left(3 \mathrm{x}^2-10 \mathrm{x}+3\right)=0\) \((\mathrm{x}-2)(2 \mathrm{x}-1)(\mathrm{x}-3)(3 \mathrm{x}-1)=0\) \(\mathrm{x}=2, \frac{1}{2}, 3, \frac{1}{3}\)Hence, greatest value is 3 .
118252
The number of positive real roots of the equation \(3^{\mathrm{x+1}}+3^{-\mathrm{x}+\mathrm{f}}=\mathbf{1 0}\) is
1 3
2 2
3 1
4 Infinitely many
Explanation:
C Given, \(3^{x+1}+3^{-x+1}=10\) \(3^x \cdot 3+\frac{3}{3^x}=10\) Let, \(\quad 3^{\mathrm{x}}=\mathrm{t}\), then we have \(3 t+\frac{3}{t}=10\) \(3 t^2+3=10 t\) \(\text { or } 3 t^2-10 t+3=0\) \(\text { or } 3 t^2-9 t-t+3=0\) \(\text { or } 3 t(t-3)-1(t-3)=0\) \(\text { or } (3 t-1)(t-3)=0\) \(\text { or } t=\frac{1}{3}, 3\) \(\Rightarrow 3^x=\frac{1}{3}=3^{-1} \Rightarrow x=-1\) and \(\quad 3^{\mathrm{x}}=3 \Rightarrow \mathrm{x}=1\) \(\therefore\) Number of positive real root is 1
AP EAMCET-04.07.2021
Complex Numbers and Quadratic Equation
118253
If \(\alpha, \beta\) be the roots of \(x^2-a(x-1)+b=0\), then the value of \(\frac{1}{\alpha^2-a \alpha}+\frac{1}{\beta^2-a \beta}+\frac{2}{a+b}\)
118254
If the ratio of the roots of the equation \(p x^2+q x\) \(+r=0\) is \(a: b\), then \(\frac{a b}{(a+b)^2}=\)
1 \(\frac{\mathrm{p}^2}{\mathrm{qr}}\)
2 \(\frac{\mathrm{pr}}{\mathrm{q}^2}\)
3 \(\frac{\mathrm{q}^2}{\mathrm{pr}}\)
4 \(\frac{\mathrm{pq}}{\mathrm{r}^2}\)
Explanation:
B Given, Let the roots be am and bm. \(\text { Let the roots be am and bm. }\) \(\text { sum of roots }(a m+b m)=-\frac{q}{p}\) \((a+b)^2 m^2=\left(\frac{q}{p}right)^2\right.\) And product of roots \((\mathrm{am} \cdot \mathrm{bm})=\frac{\mathrm{r}}{\mathrm{p}}\) On dividing equation (ii) by (i), we get \(\frac{a \cdot b m^2}{(a+b)^2 m^2}=\frac{r \cdot p^2}{p \cdot q^2}\) \(\frac{a \cdot b}{(a+b)^2}=\frac{p \cdot r}{q^2}\)
WB JEE-2011
Complex Numbers and Quadratic Equation
118258
If \((\alpha+\sqrt{\beta})\) and \((\alpha-\sqrt{\beta})\) are the roots of the equation \(x^2+p x+q=0\), where \(\alpha, \beta\), p and \(q\) are real, then the roots of the equation \(\left(p^2-4 q\right)\left(p^2 x^2+4 p x\right)-16 q=0\) are
1 \(\left(\frac{1}{\alpha}+\frac{1}{\sqrt{\beta}}\right)\) and \(\left(\frac{1}{\alpha}-\frac{1}{\sqrt{\beta}}\right)\)
2 \(\left(\frac{1}{\sqrt{\alpha}}+\frac{1}{\beta}\right)\) and \(\left(\frac{1}{\sqrt{\alpha}}-\frac{1}{\beta}\right)\)
3 \(\left(\frac{1}{\sqrt{\alpha}}+\frac{1}{\sqrt{\beta}}\right)\) and \(\left(\frac{1}{\sqrt{\alpha}}-\frac{1}{\sqrt{\beta}}\right)\)
A Given, \((\alpha+\sqrt{\beta})\) and \((\alpha-\sqrt{\beta})\) are the roots of the equation \(\mathrm{x}^2+\mathrm{px}+\mathrm{q}=0\) sum of roots, \((\alpha+\sqrt{\beta})+(\alpha-\sqrt{\beta})=-p\) \(2 \alpha=-p\) \(\alpha=\frac{-p}{2}\) And, Product of roots, \((\alpha+\sqrt{\beta}) \cdot(\alpha-\sqrt{\beta})=\mathrm{q}\) \(\alpha^2-\beta=q\) \(\beta=\alpha^2-q=\left(-\frac{p}{2}\right)^2-q=\frac{p^2}{4}-q\) \(p^2-4 q=4 \beta\) \(\left(p^2-4 q\right)\left(p^2 x^2+4 p x\right)-16 q=0\) \(4 \beta\left(p^2 x^2+4 p x\right)-16\left(\alpha^2-\beta\right)=0\) \(\beta\left(p^2 x^2+4 p x\right)-4\left(\alpha^2-\beta\right)=0\) \(\beta\left(4 \alpha^2 x^2-8 \alpha x\right)-4\left(\alpha^2-\beta\right)=0\) \(\alpha^2 \beta x^2-2 \alpha \beta x+\beta=\alpha^2\) \((\alpha x \sqrt{\beta}-\sqrt{\beta})^2=\alpha^2\) \(\alpha x \sqrt{\beta}-\sqrt{\beta}= \pm \alpha\) \(\therefore x=\frac{1}{\alpha} \pm \frac{1}{\sqrt{\beta}}\) \(\left(\frac{1}{\alpha}+\frac{1}{\sqrt{\beta}}\right)\) and \(\left(\frac{1}{\alpha}-\frac{1}{\sqrt{\beta}}\right)\) are the roots of the given equation \(\left(p^2-4 q\right)\left(p^2 x^2+4 p x\right)-16 q=0\)
WB JEE-2012
Complex Numbers and Quadratic Equation
118259
The greatest real root of the equation \(6 x^4-35 x^3+62 x^2-35 x+6=0\) is
1 2
2 \(\frac{5}{2}\)
3 3
4 \(\frac{7}{2}\)
Explanation:
C Given, equation is \(6 x^4-35 x^3+62 x^2-35 x+6=0\) dividing equation \(\qquad\) (I) by \(x^2\) \(\left\{\left\{6\left(\mathrm{x}^2+\frac{1}{\mathrm{x}^2}\right)-35\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)+62\right\}=0\right.\) Let \(\mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{y}\), therefore, \(x\) \(6\left(y^2-2\right)-35 y+62=0\) \(6 y^2-12-35 y+62=0\) \(6 y^2-35 y+50=0\) \((2 y-5)(3 y-10)=0\) Now, Put back \(\mathrm{y}=\mathrm{x}+\frac{1}{\mathrm{x}}\) and multiply each factors by \(\mathrm{x}\), we get \(\left(2 \mathrm{x}^2-5 \mathrm{x}+2\right)\left(3 \mathrm{x}^2-10 \mathrm{x}+3\right)=0\) \((\mathrm{x}-2)(2 \mathrm{x}-1)(\mathrm{x}-3)(3 \mathrm{x}-1)=0\) \(\mathrm{x}=2, \frac{1}{2}, 3, \frac{1}{3}\)Hence, greatest value is 3 .
118252
The number of positive real roots of the equation \(3^{\mathrm{x+1}}+3^{-\mathrm{x}+\mathrm{f}}=\mathbf{1 0}\) is
1 3
2 2
3 1
4 Infinitely many
Explanation:
C Given, \(3^{x+1}+3^{-x+1}=10\) \(3^x \cdot 3+\frac{3}{3^x}=10\) Let, \(\quad 3^{\mathrm{x}}=\mathrm{t}\), then we have \(3 t+\frac{3}{t}=10\) \(3 t^2+3=10 t\) \(\text { or } 3 t^2-10 t+3=0\) \(\text { or } 3 t^2-9 t-t+3=0\) \(\text { or } 3 t(t-3)-1(t-3)=0\) \(\text { or } (3 t-1)(t-3)=0\) \(\text { or } t=\frac{1}{3}, 3\) \(\Rightarrow 3^x=\frac{1}{3}=3^{-1} \Rightarrow x=-1\) and \(\quad 3^{\mathrm{x}}=3 \Rightarrow \mathrm{x}=1\) \(\therefore\) Number of positive real root is 1
AP EAMCET-04.07.2021
Complex Numbers and Quadratic Equation
118253
If \(\alpha, \beta\) be the roots of \(x^2-a(x-1)+b=0\), then the value of \(\frac{1}{\alpha^2-a \alpha}+\frac{1}{\beta^2-a \beta}+\frac{2}{a+b}\)
118254
If the ratio of the roots of the equation \(p x^2+q x\) \(+r=0\) is \(a: b\), then \(\frac{a b}{(a+b)^2}=\)
1 \(\frac{\mathrm{p}^2}{\mathrm{qr}}\)
2 \(\frac{\mathrm{pr}}{\mathrm{q}^2}\)
3 \(\frac{\mathrm{q}^2}{\mathrm{pr}}\)
4 \(\frac{\mathrm{pq}}{\mathrm{r}^2}\)
Explanation:
B Given, Let the roots be am and bm. \(\text { Let the roots be am and bm. }\) \(\text { sum of roots }(a m+b m)=-\frac{q}{p}\) \((a+b)^2 m^2=\left(\frac{q}{p}right)^2\right.\) And product of roots \((\mathrm{am} \cdot \mathrm{bm})=\frac{\mathrm{r}}{\mathrm{p}}\) On dividing equation (ii) by (i), we get \(\frac{a \cdot b m^2}{(a+b)^2 m^2}=\frac{r \cdot p^2}{p \cdot q^2}\) \(\frac{a \cdot b}{(a+b)^2}=\frac{p \cdot r}{q^2}\)
WB JEE-2011
Complex Numbers and Quadratic Equation
118258
If \((\alpha+\sqrt{\beta})\) and \((\alpha-\sqrt{\beta})\) are the roots of the equation \(x^2+p x+q=0\), where \(\alpha, \beta\), p and \(q\) are real, then the roots of the equation \(\left(p^2-4 q\right)\left(p^2 x^2+4 p x\right)-16 q=0\) are
1 \(\left(\frac{1}{\alpha}+\frac{1}{\sqrt{\beta}}\right)\) and \(\left(\frac{1}{\alpha}-\frac{1}{\sqrt{\beta}}\right)\)
2 \(\left(\frac{1}{\sqrt{\alpha}}+\frac{1}{\beta}\right)\) and \(\left(\frac{1}{\sqrt{\alpha}}-\frac{1}{\beta}\right)\)
3 \(\left(\frac{1}{\sqrt{\alpha}}+\frac{1}{\sqrt{\beta}}\right)\) and \(\left(\frac{1}{\sqrt{\alpha}}-\frac{1}{\sqrt{\beta}}\right)\)
A Given, \((\alpha+\sqrt{\beta})\) and \((\alpha-\sqrt{\beta})\) are the roots of the equation \(\mathrm{x}^2+\mathrm{px}+\mathrm{q}=0\) sum of roots, \((\alpha+\sqrt{\beta})+(\alpha-\sqrt{\beta})=-p\) \(2 \alpha=-p\) \(\alpha=\frac{-p}{2}\) And, Product of roots, \((\alpha+\sqrt{\beta}) \cdot(\alpha-\sqrt{\beta})=\mathrm{q}\) \(\alpha^2-\beta=q\) \(\beta=\alpha^2-q=\left(-\frac{p}{2}\right)^2-q=\frac{p^2}{4}-q\) \(p^2-4 q=4 \beta\) \(\left(p^2-4 q\right)\left(p^2 x^2+4 p x\right)-16 q=0\) \(4 \beta\left(p^2 x^2+4 p x\right)-16\left(\alpha^2-\beta\right)=0\) \(\beta\left(p^2 x^2+4 p x\right)-4\left(\alpha^2-\beta\right)=0\) \(\beta\left(4 \alpha^2 x^2-8 \alpha x\right)-4\left(\alpha^2-\beta\right)=0\) \(\alpha^2 \beta x^2-2 \alpha \beta x+\beta=\alpha^2\) \((\alpha x \sqrt{\beta}-\sqrt{\beta})^2=\alpha^2\) \(\alpha x \sqrt{\beta}-\sqrt{\beta}= \pm \alpha\) \(\therefore x=\frac{1}{\alpha} \pm \frac{1}{\sqrt{\beta}}\) \(\left(\frac{1}{\alpha}+\frac{1}{\sqrt{\beta}}\right)\) and \(\left(\frac{1}{\alpha}-\frac{1}{\sqrt{\beta}}\right)\) are the roots of the given equation \(\left(p^2-4 q\right)\left(p^2 x^2+4 p x\right)-16 q=0\)
WB JEE-2012
Complex Numbers and Quadratic Equation
118259
The greatest real root of the equation \(6 x^4-35 x^3+62 x^2-35 x+6=0\) is
1 2
2 \(\frac{5}{2}\)
3 3
4 \(\frac{7}{2}\)
Explanation:
C Given, equation is \(6 x^4-35 x^3+62 x^2-35 x+6=0\) dividing equation \(\qquad\) (I) by \(x^2\) \(\left\{\left\{6\left(\mathrm{x}^2+\frac{1}{\mathrm{x}^2}\right)-35\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)+62\right\}=0\right.\) Let \(\mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{y}\), therefore, \(x\) \(6\left(y^2-2\right)-35 y+62=0\) \(6 y^2-12-35 y+62=0\) \(6 y^2-35 y+50=0\) \((2 y-5)(3 y-10)=0\) Now, Put back \(\mathrm{y}=\mathrm{x}+\frac{1}{\mathrm{x}}\) and multiply each factors by \(\mathrm{x}\), we get \(\left(2 \mathrm{x}^2-5 \mathrm{x}+2\right)\left(3 \mathrm{x}^2-10 \mathrm{x}+3\right)=0\) \((\mathrm{x}-2)(2 \mathrm{x}-1)(\mathrm{x}-3)(3 \mathrm{x}-1)=0\) \(\mathrm{x}=2, \frac{1}{2}, 3, \frac{1}{3}\)Hence, greatest value is 3 .
