17 > Let g ( x ) = f ( x + 12 ) - f ( x ) , where " x " belongs from " 0 " to " 12 ".
Needless to say , " g " is also continuos .
g ( 0 ) = f ( 12 ) - f ( 0 )
g ( 12 ) = f ( 1 ) - f ( 12 ) = f ( 0 ) - f ( 12 ) ..............by given condition
Since , g ( 0 ) and g ( 12 ) , have opposite signs , then by IVT , we must have g ( k ) = 0 for some " k " lying between " 0 " to " 12 " .
So , g ( k ) = 0 implies ,
f ( k ) = f ( k + 12 ) for some " k " lying between " 0 " to " 12 " .
16 >
f ( x + y ) = e y f ( x ) - e x f ( y )
If I just set " x = y " , then ,
f ( 2 x ) = e x f ( x ) - e x f ( x ) = 0
So " f ( x ) " is identically " 0 " for all " x " , which does not co - inside with the condition given !
I guess you made some typo !
23in 6 i fogot to mention that y≠x is given
btw tnx for the 7th one ( nice soln ricky [1] )
yes nishant sir most of them are doubts
btw even hints are enough
14 >
If " x ≤ y " implies that " f ( x ) ≤ f ( y ) " , then it means that the function " f " is increasing .
In other words , f ' ( x ) ≥ 0 for all x for this function " f " .
Let the minimum value of " f ' ( x ) " be " c2 " , where " c " is obviously ≥ 0 .
Lets apply LMVT to f ( x ) , in the interval [ a + x , a - x ] .
We get , f ' ( k ) = f ( a + x ) - f ( a - x )2 x ≥ c2 ...................as it is the minimum value of f ' ( x )
So , f ( a + x ) - f ( a - x ) ≥ c x , where c ≥ 0 .
66Q8)
Define
g(x) = a0 + a1 (x-a) + a2(x-a)2 + . . . + an (x-a)n
with
a0 = 1f(a)
a1 = &ndash f'(a) g(a)f(a)
a2 = &ndash f''(a) g(a) + f'(a) g'(a)f(a)
It can be seen that g(x) will then satisfy the criteria specified in the problem.
23hey i hav confirmed dat questions 1,2,3,6 have some discrepancy abt them
so thread closed
tnx all for ur help
