If $$\alpha \ne \beta $$ but $${\alpha ^2} = 5\alpha - 3$$ and $${\beta ^2} = 5\beta - 3$$ then the equation having $$\alpha /\beta $$ and $$\beta /\alpha \,\,$$ as its roots is
JEE · Math · previous-year question
- A.$$3{x^2} - 19x + 3 = 0$$correct
- B.$$3{x^2} + 19x - 3 = 0$$
- C.$$3{x^2} - 19x - 3 = 0$$
- D.$${x^2} - 5x + 3 = 0$$
Answer
A. $$3{x^2} - 19x + 3 = 0$$
Explanation
We have $${\alpha ^2} = 5\alpha - 3$$ and $${\beta ^2} = 5\beta - 3;$$ $$ \Rightarrow \alpha \,\,\& \,\,\beta $$ are roots of equation, $${x^2} = 5x - 3$$ or $${x^2} - 5x + 3 = 0$$ $$\therefore$$ $$\alpha + \beta = 5$$ and $$\alpha \beta = 3$$ Thus, the equation having $${\alpha \over \beta }\,\,\& \,\,{\beta \over \alpha }$$ as its roots is $${x^2} - x\left( {{\alpha \over \beta } + {\beta \over \alpha }} \right) + {{\alpha \beta } \over {\alpha \beta }} = 0$$ $$ \Rightarrow {x^2} - x\left( {{{{\alpha ^2} + {\beta ^2}} \over {\alpha \beta }}} \right) + 1 = 0$$ or $$3{x^2} - 19x + 3 = 0$$
Practice more JEE questions
Answer thousands more real JEE previous-year questions free, get graded instantly, and climb the global ranked leaderboard.
Practice JEE free →More JEE Math questions
- What is the total number of distinct x \in \mathbb{R} for which \left|\begin{array}{ccc}x …
- Let m be the smallest positive integer such that the coefficient of x^{2} in the expansion…
- What is the total number of distinct x \in[0,1] for which \int_{0}^{x} \frac{t^{2}}{1+t^{4…
- Let \alpha, \beta \in \mathbb{R} be such that \lim _{x \rightarrow 0} \frac{x^{2} \sin (\b…
- Let z=\frac{-1+\sqrt{3} i}{2}, where i=\sqrt{-1}, and r, s \in\{1,2,3\}. Let P=\left[\begi…
- For how many values of p, the circle x^{2}+y^{2}+2 x+4 y-p=0 and the coordinate axes have …
- Let f: \mathbb{R} \rightarrow \mathbb{R} be a differentiable function such that f(0)=0, f\…
- For a real number \alpha, if the system \[ \left[\begin{array}{ccc} 1 & \alpha & \alpha^{2…