Complex Numbers
Complex Numbers
star_batch_jee_advanced_2025
Grade 11
Question:
The equation $\left(1+a\right)x^2+2a^2x+a^2+b^2-1=0$ has roots of opposite sign, if $a+b$ lies, $(a>-1)$ :
On straight line $x+y=1$
Inside a circle of centre $(0,0)$ and radius '1'
On a parabola of vertex $(0,0)$ and focal length '1'
None of the above
Step-by-Step Solution
Key Concept: Roots of opposite sign require negative product of roots, which translates to the constraint $a^2 + b^2 < 1$ after applying Vieta's formulas.
For a quadratic equation $(1+a)x^2 + 2a^2x + a^2 + b^2 - 1 = 0$ to have roots of opposite sign, the product of roots must be negative. Using Vieta's formulas, the product of roots is $\frac{a^2 + b^2 - 1}{1+a} < 0$. Since $a > -1$, we have $1 + a > 0$, so we need $a^2 + b^2 - 1 < 0$, which gives $a^2 + b^2 < 1$. This is the interior of a circle centered at the origin with radius 1.
Correct Answer: 2