1For the question shows that f(x) is continuous about not only x=1,2, but about x=n, which is nowhere in the question.
chatle chalte ye sawaal bhj dekhe:
q1)the graph of y=f(x) is symmetrical abt the lines x=1 and x=2,then
a)f(x+1)=f(x)
b)f(x+2)=f(x)
c)f(x+3)=f(x)
d)none
q2)lim x->oo ( (x+5)tan-1(x+5) + (x+1)tan-1(x+1)
a)Ï€/2
b)Î
c)2Î
d)none
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38 Answers
1@Nishant Bhaiya "abhishek.. you wanted to know why the periodicity is not one.."
Yes bhaiya plz respond and tell whether my graph is correct or not and if correct then why the ans is not option a)and c)
1Is the third answer correct??[7]
So the solution to the third one is...........
13actually f(x) is a constant function here ....periodic yet has no fundamental period
1
11q2)lim x->oo ( (x+5)tan-1(x+5) + (x+1)tan-1(x+1)
a)Ï€/2
b)Î
c)2Î
d)none
put x+3=t
so
(t+2)(tan-1(t+2))+(t-2)(tan-1(t-2)
limit t(tan-1(t+2) + tan-1(t-2)) +2(tan-1(t+2)-tan-1(t-2))
t-->∞
ON SOLVING WE GET t=-2
SO THE ANSWER IS
(D)
11Mr Grandmaster wat is the answer of 2
please tell!!!!!!!!!!!!!!!!!!!!
11THEN I NEED UR SOLUTION TO IT
AND PLEASE FIND THE FLAW IN MINE TOO..............
1actually kalyan the ans given in the book is "none of these" so i don't actually know
sorry yaar!!!but i couldn't help
11grandmaster
kabhie tu kehta chalte -2 ques aur tu yeh ques book se uthata hai
joootha huh!!!!!!
62just because for one situation the periodicity becomes 1 does not mean that it is true across the board.
I knew that you will post that graph [6] so i posted mine even before that..
btw if it was a multiple correct answer i wud have marked b and c
because periodicity 2 means that f(x+4)=f(x) as well..
11sir i said #26
answer of Q2 given by me
i think u checked another thing[2][2][2]
11but sir how one can get the answer pi
it looks perfect by intuition
BUT FAILS MATHEMATICALLY AS SHOWN!!!!!!!!!!!!!
1q3)if f| |(x)>0,for all real x,and f|(3)=0 and g(x)= f(tan2x - 2tan x+ 4)then for 0<x<Î /2.
g(x) increases on:
(subjective type)
621 is a very good questin
amit thnk in terms of graph
if i am not wrong then the answer should be all b
1q4)minimum value of [ (x1-x2)2 + (12 + (1-4x2)1/2 - 2√x2 )2 ]1/2
is
a)4√5+1
a)4√5-1
a)√5+1
a)√5-1
1ya nishant bhaiya!!! i also think so, the curve is some what repetative abt x= (integer) ,some what like y =|sin (Î x)|
Whats your graph bhaiya
