Messages posted by aditi garg

is its period 1/3.
im not very sure but i think so......
coz that sin part will always be 0
do lemme know the correct answer plzz

Re:khatarnaak questions (functions)

i think the answer should be p=q......

c whn we apply the formula tan @=q sin@/p+qcos@...........1

whn p ir reversed thn thn angle between p and q will be 180-@

and tan(90-@)=qsin(180-@)/ p+q(180-@).....2

==multiply 1 and 2

u get 1=q^2 sin^2@/p^2-q^ cos ^@

which further gives p^2=q^2

=p=q

was that ans correct??? in my opinion the ans should be 2/81... i think to calculate the number of cases such taht no box remains empty we should use the multinomial theorem... and frm that the cases come out to be onli 6 and total is 243 so probabilty come s to be 2/81.... tell me if im correct...

u hv solved the question and got the correct answer too but this could have been solved in just 4 lines.... use the formula... its simple... whn the angles are in AP and u hv to find the sum thn in case of cos series... the formula is... cos(A+[n-1]d/2) * sin(n*d/2)/sin(d/2).........in this case A=2pi/7 and n=3... so its cos(2pi/7+ [3-1]*pi/7) * sin(3pi/7)/sin(pi/7)....................... =coz(4pi/7)*sin(3pi/7)/sin(pi/7)= multiply and divide by 2 and apply C and D formula.....= sin(pi)-sin(pi/7)/2*sin(pi/7)= -1/2... i hope u hv got the method.... remember to rate me...

Re:cos2pi/7 + cos 6pi/7 + cos4pi/7 =??........

see its very simple....just multiply and divide the given expression by e^x and thn put e^x as t.... the expression becomes... cot ^ -1 t/t^2 dt

thn apply that by parts formula of integration and u can easily get the answer... option 3 is correct....and 1 piece of advice at the first attempt nvr try to solve any integration prob by differentiating the options... its just a waste of time... u can use it in case ur sitting for an examination and not getting any answer

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