Calculus - BITSAT 2

1

sushrut17 ·2009-05-02 22:22:31

c

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1

AARTHI ·2009-05-02 22:23:30

yup c -1

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1

kamalendu ghosh ·2009-05-02 22:30:09

yup it is c but..my doubt is u get something like -sinx//sinx/...so how can u say it is -1 for sure?

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1

AARTHI ·2009-05-02 22:35:48

okie if +1 is also der den der mite b an ambiguity

so here we hav -1

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1

AARTHI ·2009-05-02 22:35:55

okie if +1 is also der den der mite b an ambiguity

so here we hav -1

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1

kamalendu ghosh ·2009-05-02 22:36:47

why not none of these then....

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1

AARTHI ·2009-05-02 23:07:45

haan...im stck

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1

AARTHI ·2009-05-02 23:07:48

haan...im stck

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1

Bhaskar Mukherjee ·2009-05-04 07:12:14

yesss!!!
it shld be none of dese bcoz u r gettin smthin lik [-sinx/mod(sinx)]
so when we sinx 2 b +ve we hv answer 2 b (+1) && if sinx is -ve den d answer is (-1)

do u guys agree???

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1

kamalendu ghosh ·2009-05-04 07:15:03

i agree (post #4)

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1

skygirl ·2009-05-04 18:34:56

but i got -1 only.... [12]

y = sin^-1cos(x)
= sin^-1sin(Î /2 - x) [hope x is between 0 to pi ]

dy/dx = -1.

i think they didn take any limits into account.

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1

sindhu br ·2009-05-19 01:32:26

even i got -1

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11

Mani Pal Singh ·2009-05-19 01:38:07

here's a other look at this perspective
see it like

sin^-1(sin√1-x²)
=√1-x²
dy/dx= none of these
but #12 is perfect and so as mine

so the question is ambigious[1]

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1

neil.dhruva ·2009-05-19 02:03:46

how is cos x = sin√1-x² ??

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1

kamalendu ghosh ·2009-05-19 05:19:13

I am commenting on post #12,

according to sky sin^-1(sin(Î /2-x))=Ï€/2 - x....
but that is true for a certain limit of interval.. (as sky said "hope x is between 0 to pi") . so the q is really ambiguous

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skygirl ·2009-05-20 23:29:53

yes i the q IS ambiguous.. they should have specified the limit..

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