Help...Quick

5 .. elements

nope...

yup i missed -1 n -ω .. total 7

yup...but watz the method??....only trial??

at first try i missed -ω,-ω²...:(

simplifying the bracket u will get 3 options
....(1)^m
....(1 + ω)^m
... (0)^m

so u get 2 ans 0 n 1

now (1 + ω)^m = (-ω²)^m = -ω² , ω² , -ω , ω , -1 , 1 for diff. values of m

easy one....

\lim_{x\rightarrow \frac{\pi }{2}}sinx^{tanx}

tan x ln sinx ... write it as ln sinx / cot x ... L hospital ... cot x / - cosec₂ x .. .this quantity is 0 as sin x cos x = 0 ....... so limit = ε⁰ = 1

thanks!

this one....

a,b,c real and distinct

no. of real soln of

(x-a)³+(x-b)³+(x-c)³=0

let F(x)=(x-a)3+(x-b)3+(x-c)3
lim x->+∞ F(x)-> +∞
lim x->-∞ F(x)-> -∞

F'(x)=3(x-a)²+3(x-b)²+3(x-c)²
we see that F'(x)>0 for all real x in the domain which means that
F(x) is increasing in the domain (-∞,+∞)
and F(x) is differentiable and continuous in its domain
=> F(x) has one and only one real root in the domain..
cheers!!!

:) ... yup u r correct gordo

a,b,c real and positive

prove that

\[ \frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)}\geq\frac{27}{2(a+b+c)^{2}}. \]

plz help!