Functions
Parameter range for real solution of equation
Grade Class 12
Question:
The set of all possible values of parameter $a$ such that the equation $(1+a)\left(\dfrac{x^2}{1+x^2}\right)^2 - 3a\left(\dfrac{x^2}{1+x^2}\right) + 4a = 0$ has a real solution is
$\left(-\frac{1}{2}, 0\right]$
$(-1, 1)$
$\left[-\frac{1}{2}, \frac{1}{2}\right]$
None of these
Step-by-Step Solution
Key Concept: Substitute $z = \frac{x^2}{1+x^2} \in [0,1)$. The equation becomes a quadratic/linear in $z$. Find the range of $a$ for which $z\in[0,1)$ has a solution.
$z\in[0,1)$, $f(z)$ decreasing from $1$ to $\frac{1}{2}$. So $1+a\in(\frac{1}{2},1]$, giving $a\in(-\frac{1}{2},0]$.
Correct Answer: 1