For the first question,

 

Take f (x) = ( 1/x )^x  Therefore, ln f (x) = - x ln x

 

So, f  ' (x) = - f (x) [ 1 + ln x ] = - ( 1/x )^x [ ln (ex) ]

 

f ' (x) = 0 when ex = 1 i.e x = 1/e

 

Now, for f ' ' (1/e) < 0 i.e f (x) will be max at x = 1 / e

 

So, the maximum value of f (x) i.e ( 1/x )^x   is e ^{1 / e}

 

For the second question,

 

Take f (x) = (a² - 3a + 2 ) ( cos² x /4 - sin²x/ 4 ) (a- 1) + sin 1

              = ( a - 1 )( a - 2 ) ( cos x/2 ) ( a - 1 ) + sin 1

              = ( a - 1 )² ( a - 2 ) ( cos x / 2 ) + sin 1

 

f ' (x) = - 1 / 2  (a - 1)²(a - 2) sin x/2

 

Now, for a function, f (x) the points for which f ' (x) = 0 or f ' (x) does not exist are called critical points. Suppose for x₀ ,   f ' (x₀) = 0 or f ' (x₀) does not exist, then x₀ is called the critical point.

 

Now, f ' (x) will exist for all values of a & x. But f ' (x) will be zero for all x when

a = 1 or a = 2 .

 

But for any other values of a i.e for a  R - {1,2} where R is the set of real numbers, then also f ' (x) attains zero when sin x /2 = 0 i.e when x/2 = n

i.e for x = 2n

 

So,  x = 2n are the critical points of the given function independent of the values of a .

 

So, the question should be for what set of values of  ' a '  the function f (x) does not possess any critical points for all values of x . Then the answer will be:

 

a  R - {1,2}

 

So, ur second question is not completely correct.

Anyways, I hope I ' ve explained it fully.

 

Cheers !!!!!