JEE MAIN - Mathematics (2022 - 28th July Evening Shift - No. 2)
Let $$\alpha$$, $$\beta$$ be the roots of the equation $$x^{2}-\sqrt{2} x+\sqrt{6}=0$$ and $$\frac{1}{\alpha^{2}}+1, \frac{1}{\beta^{2}}+1$$ be the roots of the equation $$x^{2}+a x+b=0$$. Then the roots of the equation $$x^{2}-(a+b-2) x+(a+b+2)=0$$ are :
non-real complex numbers
real and both negative
real and both positive
real and exactly one of them is positive
Explanation
$$\alpha + \beta = \sqrt 2 $$, $$\alpha \beta = \sqrt 6 $$
$${1 \over {{\alpha ^2}}} + 1 + {1 \over {{\beta ^2}}} + 1 = 2 + {{{\alpha ^2} + {\beta ^2}} \over 6}$$
$$ = 2 + {{2 - 2\sqrt 6 } \over 6} = - a$$
$$\left( {{1 \over {{\alpha ^2}}} + 1} \right)\left( {{1 \over {{\beta ^2}}} + 1} \right) = 1 + {1 \over {{\alpha ^2}}} + {1 \over {{\beta ^2}}} + {1 \over {{\alpha ^2}{\beta ^2}}}$$
$$ = {7 \over 6} + {{2 - 2\sqrt 6 } \over 6} = b$$
$$ \Rightarrow a + b = {{ - 5} \over 6}$$
So, equation is $${x^2} + {{17x} \over 6} + {7 \over 6} = 0$$
OR $$6{x^2} + 17x + 7 = 0$$
Both roots of equation are $$-$$ve and distinct
Comments (0)
