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Differential Calculus

Blazing goIITian

Joined:	9 Mar 2008
Post:	555

6 May 2009 23:45:05 IST

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349

bansal notes question

None

Q1 limit -----pi/2 (sin(x))/cos-1(1/4(3 sinx -sin3x))

Q2 f(x)=x/lnx

and g(x)=lnx/x

then (1) 1/g(x) and f(x) are identical function

(2) 1/f(x) and g(x) are identical function

(3) f(x) .g(x)=1

(4) 1/f(x).g(x) =1 for all x>0

hese all statements looks very same but answer given is 1 option
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Comments (4)

XYZ

Hot goIITian

Joined: 18 Jan 2009

Posts: 193

7 May 2009 01:18:58 IST

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We have to consider for x>0 only because the functions are defined for x>0 only.

I think the option (iv) that you have given is . Since there was some confusion in that option, so I gave how I have understood it.

Option (iv) is incorrect because it is not true for x=1.

Option (iii) is incorrect because f(x) is not defined for x=1. So, option (iii) is not always equal to 1.

Option (ii) is incorrect because of the following reason:-

At x=1, f(x) tends to infinity. So, 1/f(x) tends to zero. Whereas, at x=1, g(x) is equal to one (it is not approximate one, but it's an exact one)

But option (i) is correct.

Hope it helps!!!

If you have any doubts, then you may ask!!!

And regarding your first question, please retype it using Insert Equation. The question is not clear. And please mention if there's a use of greatest integer function (if any).

ayush

Blazing goIITian

Joined: 9 Mar 2008

Posts: 555

8 May 2009 07:10:32 IST

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thanks .second question is edited pl check

ayush

Blazing goIITian

Joined: 9 Mar 2008

Posts: 555

8 May 2009 07:10:33 IST

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thanks .second question is edited pl check

Mirka

Blazing goIITian

Joined: 13 Aug 2008

Posts: 1313

15 May 2009 21:01:04 IST

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ur first Question .....

now,

as x--> pi/2, numerator --> 1

and denominator --> cos^-11 = 0

So, the answer im guessing shud be infinity .... what say ???

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