goIIT: IIT JEE AIEEE CET WBJEE VITJEE BITSAT IIIT AICET Engineering Entrance Experts, Student Discussion Community, IIT JEE Preparation Online

ans 90 degree

thanx sagarvaze

Find an equation of a hyperbola satisfying the following conditions ,asymptotes: y = (4/3)x, y = -(4/3)x, one vertex: (0,4)

thanks sir , but how these pictures are pasted here?

look the trick involved in this quest is u need to solve the eq. provided in quest first and that will simplify it as xsin² a=ycos²a,now simply differenciate both the sides and there u r

is ans tan²a

if yes then pls let me know

@eistien (or rather missing "n") i agree with your reply but for a biginner the ans by shaan2007ster is easy to understand so i have given vote to him. but anyways thanks a lot for your concern

but in rd sharma objective first solved example of chapter differentiation the ans is 1/x even the option is given 1/|x|

the differentiation of log|x| is 1/|x| or 1/x pls explain

whta is the projection of vector 4i-3j+k on the line joining point (2,3,-1) and (-2,-4,3)

you guys are really very helpful, thanks to everyone.

The function f is continuous on [0,1] and we know that f(0) = f(1). Show that there exists an x in [0,9/10] so that f(x) = f(x+1/10).

Suppose that f is a function defined on the real numbers for which f(a+b) = f(a) + f(b) + ab
for all a,b. What can f be?

no problem, can it be like x/|x|

@annihilator pls check it again