Author Topic: [CIE] All Pure Mathematics (P1, P2 and P3) doubts here !  (Read 79253 times)

Offline Deadly_king

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #15 on: November 06, 2010, 05:53:35 am »
Chapter 10
Exercise 10 D
Q-1 (a) (i) and (ii)

try and tell me!!

Getting the answers, but not being confirmed with which are at the back.. :-\

Sorry for the late reply, I was busy in the science section.

Anyway it's obvious the answers from the book are not good. ;)
Let the length to be found be x.
Using pythagoras theorem :
x2 = 8.82 + (5,/3)2
x2 = 77.44 + 75
x2 = 3811/25 ----> x = 1/5(,/3811)

 sin theta = 8.8/x, cos theta = (5,/3)/x and tan theta = (5,/3)/8.8

NOTE : ,/ stands for square root. ;D

« Last Edit: November 06, 2010, 05:57:14 am by Deadly_king »

Offline Arthur Bon Zavi

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #16 on: November 06, 2010, 05:58:54 am »
Sorry for the late reply, I was busy in the science section.

Anyway it's obvious the answers from the book are not good. ;)
Let the length to be found be x.
Using pythagoras theorem :
x2 = 8.82 + (5,/3)2
x2 = 77.44 + 75
x2 = 4311/25 ----> x = 1/5(,/3811)

 sin theta = 8.8/x, cos theta = (5,/3)/x and tan theta = (5,/3)/8.8

NOTE : ,/ stands for square root. ;D



Yes, there is a printing mistake. Then I got same answers.
No logic has to be applied in this question, CIE is dumb. :D

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Offline Deadly_king

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #17 on: November 06, 2010, 06:01:40 am »
Yes, there is a printing mistake. Then I got same answers.
No logic has to be applied in this question, CIE is dumb. :D

Actually the book took x to be 4,/3 but although x is more or less equal to 4,/3, it is not exactly equal to 4,/3. ;D

Offline Arthur Bon Zavi

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #18 on: November 06, 2010, 06:09:06 am »
Actually the book took x to be 4,/3 but although x is more or less equal to 4,/3, it is not exactly equal to 4,/3. ;D

x is equal to ,/152.44

What they took is not near to x's real value also.

X is 12.34

Then how can they take it as X = 6.92 ?

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Offline Deadly_king

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #19 on: November 06, 2010, 06:17:36 am »
x is equal to ,/152.44

What they took is not near to x's real value also.

X is 12.34

Then how can they take it as X = 6.92 ?

I don't know. :-[

But according to its answers for sin, cos and tan, it seems they took 4,/3

Nevermind. Let's just forget about it. ;)

Offline Arthur Bon Zavi

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #20 on: November 06, 2010, 06:43:19 am »
I don't know. :-[

But according to its answers for sin, cos and tan, it seems they took 4,/3

Nevermind. Let's just forget about it. ;)

But answers given by the book are surely wrong, right ?

Continuous efforts matter more than the outcome.
- NU

Offline Deadly_king

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #21 on: November 06, 2010, 09:43:38 am »
But answers given by the book are surely wrong, right ?

Yeah they are wrong. ;)

Offline Arthur Bon Zavi

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #22 on: November 06, 2010, 01:59:27 pm »
Yeah they are wrong. ;)

Others are not wrong, they are getting on well with my answers. ;)

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Offline Arthur Bon Zavi

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #23 on: November 07, 2010, 03:25:04 pm »
Establish The Following :

\frac{1}{Cos x} + {Tan x}\equiv\frac{Cos x}{1 - Sin x}

What is this now ? Again some printing mistake ?

Question - 3 (c) Page Number - 152
« Last Edit: November 07, 2010, 03:32:58 pm by Ancestor »

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elemis

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #24 on: November 07, 2010, 03:48:36 pm »
Let me do it.

elemis

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #25 on: November 07, 2010, 03:51:44 pm »
No the book is correct.

Here's how (next post)

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #26 on: November 07, 2010, 03:54:53 pm »
\frac{1}{cosx} - \frac{sinx}{cosx}

\frac{cosx -cosxsinx}{cos^2x}

\frac{cosx(1-sinx)}{cos^2x}

\frac{1-sinx}{cosx}

Actually the textbook is wrong :P

Offline Arthur Bon Zavi

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #27 on: November 07, 2010, 03:58:48 pm »
\frac{1}{cosx} - \frac{sinx}{cosx}

\frac{cosx -cosxsinx}{cos^2x}

\frac{cosx(1-sinx)}{cos^2x}

\frac{1-sinx}{cosx}

Actually the textbook is wrong :P

Howzz That ? :P

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- NU

Offline SpongeBob

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #28 on: November 07, 2010, 04:36:56 pm »
1+sinx/ cosx should be.
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Offline Deadly_king

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #29 on: November 08, 2010, 03:54:36 am »
Establish The Following :

\frac{1}{Cos x} + {Tan x}\equiv\frac{Cos x}{1 - Sin x}

What is this now ? Again some printing mistake ?

Question - 3 (c) Page Number - 152

This identity CAN be solved. The book is NOT wrong.

Ari's working is good until the third step, except for the minus sign which is a plus sign. ;)

Then you should replace cos2x by 1 - sin2x which is then substituted with (1 + sin x)(1 - sin x)

The you can eliminate (1 + sin x) from both the numerator and denominator to obtain the required solution. :D