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Options

A. `f''(x)` exists for al `x epsilon (0,oo)`
B. `f'(x)` exists for all `x epsilon(0,oo)` and `f'` is continuous on `(0,oo)` but not differentiable on `(0,oo)`.
C. There exists `alpha gt1` suchthat `|f'(x)|lt|f(x)+` for all `x epsilon(alpha, oo)`
D. There exists `beta gt 0` such that `|f(x)|+|f'(x)+lebeta` for all `x epsilon (0,oo)`

Answer

B, C

Explanation

`f(x)=In x+int_(0)^(x)sqrt(1+sint)dt`
`f'(x)=1/x+sqrt(1+sinx)`
`f''(x)=-1/(x^(2))+(cosx)/(2sqrt(1+sinx))`
`f''` isnot defined for `x=-(pi)/2+2npi, n epsilonZ` so 1 is wrong
2. `f'(x)` always exists for `x gt0` ltbgt 3. `|f'(x)+lt|f(x)|`
Since `f'gt0` and `fgt0`
we have `f'(x)ltf(x)`
`implies 1/x+sqrt(1+sinx)lt In x+int_(0)^(x)sqrt(1+six) dx`
Now LHS isbounded by R.H.S is incresing with value approaching to infinity.
So there exist some `alpha` beyond which RHS is greater than LHS.`|f(x)|+|f'(x)|le beta` is wrong is `f` is increasing andits range isnot bound while `beta` is finite.