Question

The value of a for which the function $f\left(x\right)=(4a-3)(x+\log 5)+2(a-7)\cot \frac{x}{2}{\sin }^2\frac{x}{2}$ does not possess critical points is

• A. $(-\infty ,-\frac{4}{3}\left)\cup \right(2,\infty )$
• B. $(-\infty ,-1)$
• C. $[1,\infty )$
• D. $(-2,\infty )$

Answer 1

Answer: (A)

$(-\infty ,-\frac{4}{3}\left)\cup \right(2,\infty )$

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

$f^{\prime }\left(x\right)=(4a-3)(1+0)+2(a-7)[\cot \frac{x}{2}\times 2\sin \frac{x}{2}\cos \frac{x}{2}\times \frac{1}{2}+{\sin }^2\frac{x}{2}\times (-{\csc }^2\frac{x}{2})\times \frac{1}{2}]$

$f^{\prime }\left(x\right)=(4a-3)+2(a-7)[{\cos }^2\frac{x}{2}-\frac{1}{2}]$

$f^{\prime }\left(x\right)=(4a-3)+2(a-7)\times \frac{\left(\cos x\right)}{2}[\because 1+\cos x=2{\cos }^2\frac{x}{2}]$

$f^{\prime }\left(x\right)=(4a-3)+(a-7)\cos x$

Now, for no critical points

$f^{\prime }\left(x\right)\ne 0$

$(4a-3)\ne (7-a)\cos x$

$\cos x\ne \frac{4a-3}{7-a}$

$\therefore \frac{4a-3}{7-a}>1$$4a-3>7-a$

$5a>10$

$a>2$

and

$\frac{4a-3}{7-a}<-1$

$4a-3<a-7$

$3a<-4$

$a<-4/3$

$\therefore aϵ(-\infty ,-4/3)\bigcup (2,\infty )$ is correct.