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Differential Calculus
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delhitushar
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14 Jul 2007 11:28:26 IST
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answer is 36
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14 Jul 2007 12:05:34 IST
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A box function, say [ x ], where [x] denotes the greatest integer less than or equal to x, is not differentiable at the integral points or in other words, [ x ] is not differentiable when x is an integer.
Therefore, f (x) = [ n + p sinx ] will not be differentiable when n + p sinx is an integer.
0 < x <
, i.e 0 < sin x
1
Since, n is an integer, n + p sinx will be an integer only when p sinx is an integer.
Since, p is a prime = 19, p sinx will be an integer when,
sin x = 1/19, 2/19, 3/19,...........................18 / 19, 19/19
There are 18 values of x lying between 0 and
/ 2, for which
sin x = 1/19, 2/19, 3/19 ......................18/ 19
At
/ 2 , sin x = 19 / 19 = 1
Again there are 18 values of x lying between
/ 2 and
, for which
sin x = 1/19, 2/19, 3/19 ......................18/ 19
So, there are in total (18 + 1 + 18) = 37 values of x for given value of n and p = 19, for which f (x) is not differentiable.
Cheers !!!!!
Therefore, f (x) = [ n + p sinx ] will not be differentiable when n + p sinx is an integer.
0 < x <
, i.e 0 < sin x
1 Since, n is an integer, n + p sinx will be an integer only when p sinx is an integer.
Since, p is a prime = 19, p sinx will be an integer when,
sin x = 1/19, 2/19, 3/19,...........................18 / 19, 19/19
There are 18 values of x lying between 0 and
/ 2, for which sin x = 1/19, 2/19, 3/19 ......................18/ 19
At
/ 2 , sin x = 19 / 19 = 1Again there are 18 values of x lying between
/ 2 and
, for which sin x = 1/19, 2/19, 3/19 ......................18/ 19
So, there are in total (18 + 1 + 18) = 37 values of x for given value of n and p = 19, for which f (x) is not differentiable.
Cheers !!!!!
14 Jul 2007 16:19:50 IST
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Good solution Titun , but check out my solution here it is :
f(x) = [ n + pSinx ]
f(x) = n + [ pSinx ]
So , [ pSinx ] has to be integral .
f(x) will be discontinuous of " pSinx " will take integral values.
Therefore ,
Sinx = 1 , 1/p , 2/p , 3/p , 4/p , ................ ( p-1 ) / p { (p-1) ways }
Or
Sinx = -1 , -1/p , -2/p , -3/p , -4/p , .............- (p-1) / p { (p-1) ways }
So overall there can be
(2p - 2) ways and +1 way when pSinx = p/p at
/2.
/2.so total of (2p - 1) ways .
Therefore the number of values for which f(x) will be Discontinuous is
(2p-1) ,
Value of p = 19
Therefore , The number of values will be
=[ 2(19) - 1 ]
=37 values.
plZ rate me if you like it.
Cheers !!!!!!!!!!!!!!

14 Jul 2007 16:24:10 IST
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I think there is a slight mistake water demon. psinx can never be zero in my opinion, because according to the question, x lies under a strict inequality, ie, it is strictly greater than zero, and strictly less than pi. So, the extra one comes from the situation where sinx = p/p at pi/2.
Anyway, the rest of the solution is convincing and different.
Anyway, the rest of the solution is convincing and different.


I [ integers ]








