A) \[(-\infty ,\,\,2)\]
B) \[(-\infty ,\,\,-1)\]
C) \[[1,\,\,\infty )\]
D) \[\left( -\infty ,\,\,\frac{-4}{3} \right)\cup (2,\,\,\infty )\]
Correct Answer: B
Solution :
We have,\[f(x)=(4a-3)(x+\log 5)+2(a-7)\] \[\cot \frac{x}{2}\cdot {{\sin }^{2}}\frac{x}{2}\] \[=(4a-3)(x+\log 5)+(a-7)\sin x\] \[\Rightarrow \] \[f'(x)=(4a-3)(a-7)\cos x\] If \[f(x)\] does not have critical points, then\[f'(x)=0\] does not have any solution in\[R\]. Now,\[f'(x)=0\Rightarrow \cos x=\frac{4a-3}{7-a}\] \[\Rightarrow \] \[\left| \frac{4a-3}{7-a} \right|\le 1\] \[(\because \,\,|\cos x|\,\,\le 1)\] \[\Rightarrow \] \[-\le \frac{4a-3}{7-a}\le 1\] \[\Rightarrow \] \[a-7\le 4a-3\le 7-a\] \[\Rightarrow \] \[a\ge -4/3\]and\[a\le 2\]. Thus, \[f'(x)=0\] has solution in\[R\], if\[-4/3\le a\le 2\]. So,\[f'(x)=0\]is not solvable in\[R\], if\[a<-4/3\]or\[a>2\], \[i.e.,\]\[a\in (-\infty ,\,\,-4/3)\cup (2,\,\,\infty )\].
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