A function `f` is continuous for all `x` (and not everywhere zero) such that `f^2(x)=int_0^xf(t)(cost)/(2+sint)dtdot` Then `f(x)` is (a) `1/2 1n((x+cosx)/2); x!=0` (b) `1/2 1n(3/(x+cosx)); x!=0` (c) `1/2 1n((2+sinx)/2); x!=npi,n in I` (d) `(cosx+sinx)/(2+sinx); x!=npi+(3pi)/4,n in I`

Options

A. `1/2 In((x+cosx)/2)`
B. `1/2 In(3/(2+cosx))`
C. `1/2In((2+sinx)/2)`
D. `(cosx+sinx)/(2+sinx)`

Answer

C

Explanation

`f^(2)(x)=int_(0)^(x)f(t) (cost)/(2+sint)dt`
or `2f(x)f'(x)=f(x)(cosx)/(2+sinx)` (Differentiating w.r.t `x` using Leibniz rule)
or `2f'(x)=(cosx)/(2+sinx)` [As `f(x)` is nor zero everywhere]
or `2intf'(x)dx+int(cosx)/(2+sinx)dx` ltbgt or `2f(x)=log_(e)(2+sinx)+logC`
Put `x=0`. Then `2f(0)=log2+logC` or `logC=-log2`
`:.f(x)=1/2 In((2+sinx)/2)`