The range of a in R for which the function f( | vedclass

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(B) Given $f(x)=(4 a-3)\left(x+\log _{e} 5\right)+2(a-7) \cot \left(\frac{x}{2}\right) \sin ^{2}\left(\frac{x}{2}\right)$.Since $\cot \left(\frac{x}{2

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$[-\frac{4}{3}, 2]$

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(B) Given $f(x)=(4 a-3)\left(x+\log _{e} 5ight)+2(a-7) \cot \left(\frac{x}{2}ight) \sin ^{2}\left(\frac{x}{2}ight)$.Since $\cot \left(\frac{x}{2}ight) \sin ^{2}\left(\frac{x}{2}ight) = \frac{\cos(x/2)}{\sin(x/2)} \cdot \sin^2(x/2) = \sin(x/2)\cos(x/2) = \frac{1}{2} \sin x$,the function simplifies to:$f(x)=(4 a-3)\left(x+\log _{e} 5ight)+(a-7) \sin x$.For critical points,we set $f'(x) = 0$:$f'(x)=(4 a-3)(1)+(a-7) \cos x = 0$.This implies $\cos x = \frac{3-4 a}{a-7}$.Since $-1 \leq \cos x \leq 1$ and $x 
eq 2n\pi$ (which means $\cos x 
eq 1$),we have:$-1 \leq \frac{3-4 a}{a-7} Case $1$: $\frac{3-4 a}{a-7} \geq -1 \Rightarrow \frac{3-4 a + a - 7}{a-7} \geq 0 \Rightarrow \frac{-3 a - 4}{a-7} \geq 0 \Rightarrow \frac{3 a + 4}{a-7} \leq 0$.This gives $a \in [-\frac{4}{3}, 7)$.Case $2$: $\frac{3-4 a}{a-7}  0$.This gives $a \in (-\infty, 2) \cup (7, \infty)$.Taking the intersection of Case $1$ and Case $2$,we get $a \in [-\frac{4}{3}, 2)$.However,checking the boundary $a=2$,$\cos x = \frac{3-8}{2-7} = \frac{-5}{-5} = 1$,which is excluded as $x 
eq 2n\pi$. Thus,the range is $[-\frac{4}{3}, 2)$.

The range of $a \in R$ for which the function $f(x)=(4 a-3)\left(x+\log _{e} 5\right)+2(a-7) \cot \left(\frac{x}{2}\right) \sin ^{2}\left(\frac{x}{2}\right)$ where $x \neq 2 n \pi, n \in N$,has critical points,is

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