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Differential Calculus

Blazing goIITian

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20 Jan 2007 23:25:05 IST

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1980

let's see who solves this..

Engineering Entrance , JEE Main , JEE Advanced , Mathematics , Differential Calculus

_{lim (cosa)^x-(sina)^x+sin2a/x-4}

_x->4

_{the whole expression is divided to x-4..}

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malay

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21 Jan 2007 10:04:04 IST

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I think numerator is finite at x=4 and denominatr is zero at x=4

gorakavipraveen

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21 Jan 2007 12:17:28 IST

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NISHANT , I THINK APPLYING L HOSPITALS RULE SOLVES IT .

ruhi agrawal

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21 Jan 2007 12:24:56 IST

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by applying l hospital rule i m getting............................
(cosa)^4 logcosa - (sina)^4 logsina + sina/2
is the answer correct.............plz tell me

CyBorG

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21 Jan 2007 12:25:26 IST

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But it is not of 0/0 form.So how can we apply L'Hospitals rule.

nrki99

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21 Jan 2007 15:58:19 IST

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hi nishant, the question given by u is incomplete.

if u simply put x=4 u get sin2a+cos2a in numerator & 0 in the denominator

if the numerator is positive quantity and limit is approaching from RHS then ans is +infinity and if the numerator is negative quantity and lmit is approaching from LHS then ans is -infinity as the interval of a is not defined by u so it does not give any proper solution.

nishant dash

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21 Jan 2007 17:44:27 IST

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well,sorry for not telling the interval to which a belongs.it is(0,

/2)

gorakavipraveen

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21 Jan 2007 20:29:43 IST

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nrki99, you failed in checking continuity of the limit before explaining the answere, i think.

nrki99

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22 Jan 2007 03:59:40 IST

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hi gkp, i think u can solve this problem better than me so

u can solve this problem.

message to nishant : is u made this problem ?

if yes, don't try make problems because there is already many problems

if no, then tell the source from which u got this problem.

nishant dash

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22 Jan 2007 17:31:08 IST

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i haven't made this problem.i am no genius to make problems.its frm a magazine called mathematics today..plz try to get the answer..

CyBorG

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22 Jan 2007 17:56:32 IST

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Try to take (cosa)^x-4-(sina)^x-4 outside so that we get

(((cosa)^x-4-1)-((sina)^x-4-1))/(x-4) and then simplify. I am not sure whether it will work or not.

rajiv_nag

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23 Jan 2007 00:48:42 IST

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numerator will be cos2a+sin2a.
denominator will be 0.
so no l'hospital's rule.
the ans is infinite.
infinite is also a limit,
for eg., limit tanx as x->90- is infinte.
not a gud problem

nrki99

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23 Jan 2007 03:52:31 IST

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hi rajiv, don't be oversmart, u didn't know whether the numerator is positive or

negative so before solving problem lets first read what argument is given by others

nishant dash

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23 Jan 2007 17:28:40 IST

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rajiv ,my friend.its cosa^x and sina^x.u can't write that as cos2a

nishant dash

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23 Jan 2007 23:57:01 IST

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KAB,i tried ur method but couldn't solve the question that way

hope someone else comes up with a soln. quickly.

i wish some expert could help..

AAKRITI

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24 Jan 2007 01:00:34 IST

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Let f (x) =[ (cosa)^x - (sina)^x + sin2a ] / (x-4)

=[ (cosa)^x - (sina)^x - cos2a ] / (x-4) + [sin2a + cos2a]/(x-4)

= g(x) + h(x) { let }

Then,

lim ( x--> 4 ) g(x) is of 0/0 form. ok?

use l'hospital's rule and find the value of g(x) .It is a finite value.

Now, f (x) = something finite + h(x)

And I think it is clear that h(x) is of form something finite/infinity

Therefore , lim ( x--> 4 ) f (x) doesn't exist

Though I believe there was no need to do this all, putting 4 in the expression of f(x) explains it as told by Malay.

nishant dash

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25 Jan 2007 10:33:39 IST

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thank u all for providing answer to my question..

nrki99

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26 Jan 2007 05:27:42 IST

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hey aakriti

the solution given by u is incomplete

because u consider the some finite value/ 0 =infinity

because some +ve finite value/0 = +infinity & some -ve finite value/ 0= -infinty

here may be the cases of 0+ and 0-. i am also very confused regarding that

problem.so u must use the interval here for solving the problem

i think the problem is wrong!

so nishant i advice u to ask this problem from any expert

jalanshruti

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26 Jan 2007 08:27:46 IST

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cant we directly substitute x=4 in the expression and since cos and sin has value btwn 0,1 so (cosa)^x-(sina)^x+sin2a/x cant be equal to 4 therefore cant be nethier indeterminate i think ans is infinite

nishant dash

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26 Jan 2007 13:06:19 IST

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will some forum expert kindly reply to my question........

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