JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations
The number of solutions of \(\sin ^2 \mathrm{x}+\left(2+2 \mathrm{x}-\mathrm{x}^2\right) \sin \mathrm{x}-3(\mathrm{x}-1)^2=0\), where \(-\pi \leq \mathrm{x} \leq \pi\), is ..........
- A \(6\)
- B \(7\)
- C \(2\)
- D \(4\)
Answer & Solution
Correct Answer
(C) \(2\)
Step-by-step Solution
Detailed explanation
\( \sin ^2 x-\left(x^2-2 x-2\right) \sin x-3(x-1)^2=0 \) \( \left.\sin ^2 x-(x-1)^2\right) \sin x-3(x-1)^2=0\) \( \sin x=-3( reject) or (x-1)^2 \) \( \sin x=(x-1)^2\)
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