A) \[k>6\]
B) \[2\le k\le 6\]
C) \[k<2\]
D) \[-6\le k\le -2\]
E) \[k\le -6\]
Correct Answer: B
Solution :
\[k\text{ }sin\text{ }x+cos\text{ }2x=2k-7\] \[k\text{ }sin\text{ }x+1-2si{{n}^{2}}x=2k-7\] \[\Rightarrow \] \[2{{\sin }^{2}}x-k\sin x+2k-8=0\] \[\Rightarrow \] \[\sin x=\frac{k\pm \sqrt{{{k}^{2}}-8(2k-8)}}{4}\] \[\Rightarrow \] \[\sin x=\frac{k\pm \sqrt{{{k}^{2}}-16k+64}}{4}\] \[\Rightarrow \] \[=\frac{k\pm \sqrt{{{(k-8)}^{2}}}}{4}\] \[\Rightarrow \] \[\sin x=\frac{k\pm (k-8)}{4}\] \[\Rightarrow \] \[\sin x=\frac{k-4}{2},2\] \[(\because \sin x\ne 2)\] \[\because \] \[-1\le \sin x\le 1\] \[\Rightarrow \] \[-1\le \frac{k-4}{2}\le 1\] \[\Rightarrow \] \[-2\le k-4\le 2\] \[\Rightarrow \] \[2\le k\le 6\]or\[k\in [2,6]\]
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