IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

studyid

@ saurabh ......i got the answer........but by a very lengthy method ...

sourab_MCA

np but atleast u tried genuinely , the method given by sagarvage was directly copied from book.thats y i asked if he can do it in ne other way.

studyid

do u want me to post the solution ............. but sincerely it is lengthy ...

i dont mind posting it ......... ..

sourab_MCA

if its too lengthy , then there is no use , coz in objective exams like AIEEE its better 2 leave such questions(if they come by chance).
otherwise u may miss other questions

ne way thnx.if u find ne short method than plz post

RyuAmakusa

SORRY FOR THE DELAY I ACTUALLY POSTED A DIFFERENT METHOD AT FIRST BUT LATER I THAUGHT THIS WAS MORE SIMPLE.IT COMES FROM THE BASIC DEFINITION OF DIFFERENTIATION

x₁=cos²a,y₁=sin²a. and also x₂=(cos²a)/n,y₂=(sin²a)/n. both satisfy the eq.(n is any no.) yhe diff is.

lim y₂-y₁/x₂-x₁.

(x_1,y₁)--->(x_2,y₂)

=> lim ((sin²a)/n-sin²a)/((cos²a)/n-(cos²a))

n--->0

=>lim tan²a(1-n)/(1-n) = tan²a

n--->0

WHAT DO U THINK.

little_genius

cos^4x /x + sin^4x /y = 1/(x+y)

put x=cos^2x,y=sin^2x

studyid

@ little_genius ...........

please see the question properly .........

in the numerator on the L.H.S. we have cos^4 (a)/x + sin ^4 (a)/y

and not cos^4x /x + sin^4x /y

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