Trigonometry & Inverse Trigonometry
Trigonometry
star_batch_jee_advanced_2025
Grade None
Question:
If $(\sin a)x^2 - 2x + b \geq 2$ for all real values of $x \leq 1$ and $a \in \left(0, \frac{\pi}{2}\right) \cup \left(\frac{7\pi}{2}, \pi\right)$, then $b$ can be equal to :
Step-by-Step Solution
Key Concept: For a quadratic with vertex beyond the domain boundary, the minimum on a restricted domain occurs at the boundary point.
The x-coordinate of the vertex is given by $x = \frac{1}{\sin a} = \csc a > 1$. For the function $f(x) = (\sin a)x^2 - 2x + (b-2)$ with $x \leq 1$, the minimum must be positive and occurs at $x = 1$. This gives $\sin a - 2 + b - 2 \geq 0$, so $b \geq 4 - \sin a$. Since $a \in (0, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi)$, we conclude $b > 3$.
Correct Answer: 3,4