IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

dilip

Plz solve this for me:

_{[ x]}

_{[ pi/4]} (sinx-cosx)^2 /

2-sinx-cosx

0 5

+ tan3

/ 4

cos

-sin2

I WON'T FORGET TO RATE

pottermania1990

ur qn is not clear....

pl. put the minus nd / symbols in prop. place .

pl.

joyfrancis

q1)

( sinx - cosx )² /

2 - sinx - cosx

= 2[{(sinx)/

2} - {(cosx)/

2}]² /

2 -

2{(sinx)/

2 + {(cosx)/

2}

= 2 sin²( x - pi/4 ) /

2 ( 1 - sin( x - pi/4 ))

If x --> pi/4 then x - pi/4 --->0

So the question becomes

_{lt y --> 0} 2 sin²y/

2(1 - siny)

so ans = 0

joyfrancis

q2)

It is 0/0 form so apply LH Rule.

{take theta = x )

Derivative of numerator = 5 - sec²x

Derivative of denomenator = 4 (cosx - xsinx ) + 2cos2x

Now just substitute x = 0 you will get the limit as 2/3