The value of `a` for which the function `f(x)=(4a-3)(x+log5)+2(a-7)cot(x/2)sin^2(x/2)` does not possess critical points is (a)`(-oo,-4/3)` (b) `(-oo,-1)` (c)`[1,oo)` (d) `(2,oo)`

Options

A. `(-,oo,-4//3)`
B. `(-oo,-1)`
C. `[1,oo)`
D. `(2,oo)`

Answer

1,4

Explanation

we have `f(x)=(4a-3)(x+log5)+2(a-7)cot (x)/(2)sin^(2)(x)/(2)`
`=(4a-3)(x+log5)+(a-7)sinx)`
or `f(X)=(4a-3)+(a-7)cosx`
If f(X) does ot have critical ponts then f(x) =0 does not have any solution in R Now
`f(x)=0 cos x =(4a-3)/(7-a)`
or `|(4a-3)/(7-a)|le1`
or `-1le(4a-3)/(7-a)le1`
or `a-7le4a-3le7-a`
Thus f(x) =0m has solution in R if `-4//3leale2`
`ain(-oo,-4//3)cup(2,oo)`