I found these choices to be true. Please let me know if mistakes are present.
Updated with Sir's answer key.
14. Question not correct
Are you implying this because of the equality sign present?
Since it is mentioned that function is twice differentiable I have doubt whether this means higher derivatives don't exist or are equal to 0. In second case second derivative must be nonzero where first derivative is 0 and it will be necessary and sufficient condition.However if third or higher derivatives exist then it will only be a necessary condition as the point can be inflection point. My doubt lies on meaning of twice differentiable.
x^3 is both twice and thrice differentiable, we're only given it's a twice differentiable fuction nothing about possibilities of higher order derivatives, it may or may not exist. But again I am not sure.
14.Since cumulative distribution function attains value 1 at x=infinity so it is only possible if a=0 and b=1
F(x)= 1 for x>=0
=x^2-x +1 for x<0
Since x is defined for [0,1] so second part becomes redundant.
so the distribution is discrete having p(x)=1 at x=0 and p(x)=0 elsewhere
problem is question mentions that 'a' belongs to (0,1) so a=0 is not a viable solution.I am stuck in here.
If ' a ' is considered to be 0 then F becomes continuous in (0,1).
If we sum up the two value over two intervals and equate it to one ...and then in that eq replacing X by a ...we will get a=1 but 1 Doesn't belong to the given interval...so it might be the case that it's not continuous at a