(i) we have
`f(x) {:{(1=x , , 0 le x le 2), (3-x , , 2 lt x le 3):}`
`f(f(x))= {:{(1+f(x), 0 le f(x) le 2),(3-f(x),2lt f(x)le3):}`
`{:{(1+(1=x), 0le 1 + x le2 and0 lex le2),(3-(1+x), 2 lt 1 =x le 3 and 0 le x le 2), (1 +(3-x), 0 le 3 - x le 2 and 2 le x le 3 ) , (3-(3-x), 2le 3 -x le 3 and 2 lt x le 3):}`
Thus,
`Rightarrow f(f(x))= {:{(2+x, 0 lex le1),(2-x , 1 lt x le 2),(4-x, 2 lt x lt 3):}`
Let g (x) = f(f(x))
Now , we find that
` underset(x to1)lim g(x) = underset( h to 0)lim g(1-h) = underset(h to 0)lim 2=1 -h =3`
`and underset(g to 1)lim g(x) = underset(h to 0)limg(1+h) = underset(h to 0)lim 2-(1+h)=1`
g(1)=3
Here we observe that
`Rightarrow g(x) ` is discontinuous at x =1
Again , `lim_(x to 2) g(x) = lim_( h to0)=g(2-h) = lim_(h to0) 2-(2-h)=0`
`lim_(x to2), g(x) = lim_(h to 0)(2+h) = lim(h to 0) 4 - (2+h) =2`
g(2)=0
Here we find that
` Rightarrow g(x) ` is discontinuous at x =2
At all other point , g(x) is continuous as it is a linear polynomial in x .
(ii) The given function is
f(x) ` [x]^(2) - [x^(2)]`
Which may have discontinuities only at points where x is an integer or `x^(2)` is an integer .
Case - I
Let x = n , n `in Z`
We find that
`lim_(h to 0) f( n -h) = lim_( h to 0) [ n-h]^(2) - [(n-h)^(2)]`
` lim_(h to 0) [ (n -1)^(2) - (n^(2) -2nh +h^(2))]`
` = (n-1)^(2) - (n^(2)-1) [ as h^(2) - 2nh lt 0` for small h ]
2-2h
Also, ` lim_( h to 0) f( n + h) = lim_( h to 0) [ n +h]^(2) - [ (n + h)^(2)]`
` = lim_(h to 0) (n^(2)- [ n^(2) + 2h + h^(2))]`
` lim_(h to 0) (n^(2)-h^(2)) = lim_(h to 0) (0)=0`
` f(n) = [n]^(2) -[n]^(2) = n^(2) -n^(2)=0`
Here we observe that
`lim_(h to 0) f( n -h) ne lim(h to 0) f(n +h) = f(n)`
`Rightarrow f(x) ` is discontinuous at n
We can see that` lim_( h to 0) f (1-h) = lim_( h to 0) f (1+h) =f (1)`
` Rightarrow f(x)` is continuous at x =1
thus the given function is discontinuous at all integral point except 1.
Case II
Let ` x = +- sqrtn , n in N `
` lim_(h to 0) f( sqrtn -h) = lim_( h to 0) [ sqrtn - h]^(2) - [ sqrtn -h)^(2)]`
` = lim_(h to 0) [ sqrtn]^(2) - [ n +h^(2) -2h sqrtn ]`
` = [sqrtn]^(2) - (n -1) " " ( because h^(2) -2h sqrtn lt 0 ` for small h)
` [ sqrtn]^(2) - n +1 `
`lim_(h to 0) f(sqrtn + h) = lim_(h to 0) [ sqrtn + h]^(2) - [( sqrtn + h)^(2)]`
` = lim_(h to 0) [sqrtn)^(2) - [ n + h^(2) + 2h sqrtn]`
` = [ sqrtn]^(2) -n`
` f (sqrtn) = [ sqrtn]^(2) - [n ] = [ sqrtn]^(2) -n`
Here we find that ,
` lim_( h to 0) f (sqrtn -h) ne lim_( h to 0) f ( sqrtn +h) = f( sqrtn)`
` Rightarrow f(x)` is discontinuous at x `= sqrtn , n in N`
Combining the above two case we can conculude that teh given function is discontinuous at all point ` x in Z - {1} cup { +- sqrtn , n in N }`
(iii) we have,
` f( pi/2) = a+ 4`
`lim_(h to 0) f(pi/2 -h) = lim_(h to 0) (8/5)^((tan8(pi/2-h))/(tan5(pi/2-h)))= (8/5) ^(lim_(h to0)(tan(4 pi 8h))/(tan((5pi)/2 - 5h)))`
` (8/5)^(lim_(h to 0)(-tan 8h)/(cot 5 h) = (8/5) ^(-lim_(h to 0) tan 8h tan 5h) = (8/5)^(0) =1`
`lim_(h to 0) f(pi/2 +h)= lim_(h to 0) (1+|cot(pi/2+h)|)^((b|tan(pi/2+h)|)/a`
` = lim_(h to 0) (1+tan h)^(( b cot h)/a ) = e^(lim_( h to a) (b cot h)/a ( 1 + tan h -1))`
` = e^(b/a)`
For the function to be continuous at = pi/2 , we must have ` lim_(h to 0) f(pi/2 +h)=f(pi/2)`
` Rightarrow 1=a +4=e^(b/a)`
` Rightarrow a = -3 , b=0`
