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

A, D

Explanation

We have
`f(x)=(4a-3)(x+log5)+2(a-7) cot""x/2 sin^2""x/2`
`rArrf(x)=(4a-3)(x+log5)+(a-7)sinx`
`thereforef'(x)=(4a-3)+(a-7)cosx`
If f(x) does not have critical points , then f'(x) =0 does not have many solution in R
Now
f'(x)=0
`rArr cosx=(4a-3)/(7-a)`
`rArr|(4a-3)/(7-a)|le1 " " [therefore|cosx|le1]`
`rArr -1le(4a-3)/(7-a)le1`
`rArr a-7le4a-3le7-arArrage-4//3 " or " a le2`
Thus, f'(x)=0 has solutions in R if `age-4//3 or ale2`
So , f'(x) =0 is not soluble in R if `agt-4//3 or agt2 " i.e. "a in(-oo,-4//3)cup(2,oo)`