39ya sorry i was thinking 27x4=81. kaha na no brain working
then there is no answer in options
1. If tan x > cot y , where tan x , cot y \epsilon N such that tan2x + 3 cot y and cot2y + 3 tan x are both perfect squares then
A>x=\tan^{-1} (1)
B>x= \tan^{-1} (4)
C>x=\tan^{-1} (16)
D>x=\tan^{-1} (64)
2.If the perimeter of a traingle is equal to its in radius then the traingle is necessarily
A> acute angled
B>obtuse angled
C> right angled
D> no such traingle
3. Let f and g be continuous and diffrentiable functions . if f(0)=f(2)=f(4) ; f(1)+f(3)=0 ; g(0)=g(2)=g(4)=0 and if f(x)=0 and g'(x) = 0 do not have a common root , then the minimum no. of zeros of f'(x)g'(x) +f(x)g"(x) in [0,4] is ________
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27 Answers
39
62and latex was done after your suggestion :)
I was kind of against introducing it.. but after i actually put it, it has made life far easier :)
62which browser are you using!
i am surprised that you cant start a new thread..
When you click "New Topic"
you have to scroll down to the bottom of the page.
That should do the trick
because the new topic box opens right at the bottom.
I know it is a bit weird but I will try to fix that too :)
341But I am still unable to start a new thread, so I am unable to post questions. Any remedies, please?
341nishant sir, latex has made posting in this forum a delight. Many many thanks.
3412nd one:
From Heron's formula, the area of the triangle
\bigtriangleup = \sqrt{s(s-a)(s-b)(s-c)} \le \sqrt{s^4} = s^2
where s is the semi-perimeter
We also have \bigtriangleup = rs where r is the in-radius
Hence we obtain rs \le s^2 \Rightarrow r<s
Hence the inradius can never equal even the semi-perimeter let alone perimeter
62Let f and g be continuous and diffrentiable functions . if f(0)=f(2)=f(4) ; f(1)+f(3)=0 ; g(0)=g(2)=g(4)=0 and if f(x)=0 and g'(x) = 0 do not have a common root , then the minimum no. of zeros of f'(x)g'(x) +f(x)g"(x) in [0,4] is ________
minimum number of zeroes of f'(x)g'(x) +f(x)g"(x)
is same as
number of zeroes of d/dx{ f(x)g'(x) } is the number of maxima minimas of f(x)g'(x)
f(1) and f(3) are opposite in sign.
f(x) and g' dont have a common root. so they are never zero together.
now there are some more conclusions to make...
Should I leave this for you guys here? or you want the complete soln?
Try this it is a good question!
622nd part was the easiest..
The answer is none of these.
just draw the triangle once.
@bhargav: the mistake in the 2nd part was that If i prove that
x is not less than 5 does not mean that x > 5
it could also mean that no value satisfies.
3411st one: done at goiit: http://www.goiit.com/posts/list/algebra-challenging-question-1-72596.htm#358205 [Please see only my (hsbhatt's) third post there. after that some needless arguments follow)
1one dot over f(x) suggets dy/dx two dots d2y/dx2 .
and for the 2ns question answer is not OBTUSE
39IN THE THIRD ONE THE LAST EQUATION IS NOT CLEAR. PLEASE USE ATEX OR POST IT MORE CLEARLY
39as sky said cot y = (M- tan2x) /3 ..M=a perfect sqr.
substitute this in 2nd equation,
we get [M-tan2x]2 +27tanx=9N
from this we see that 27tanx should be a perfect square
39sky u r doing a mistake, substitute for coty in 2nd equation, see the answers , we need to eliminate y and not x.
39aur ha first waala substitute karlo na. nahiin toh formal proof bana padega, abhi utna dimaag kaam nahiin karra.
1q.1) see the constraints are:
cot y = (M- tan2x) /3 ..M=a perfect sqr.
(M- tan2x) >=0 and has to b a multiple of 3.
similarly , tanx = (N - cot2y) /3 .
this shud help... but afer this,,,, i need help ;P.
39method is as follows
r=[s-a]tanA/2
2s=[s-a]tanA/2
2cotA/2 = 1-a/s in this eqtn, rhs <1
therefore tanA/2>2
tanA= 2tanA/2/[1-tan2A/2] becomes negative.
hence triangle is obtuse
1I think for second it is D....no triangle
inradius>perimeter.... "impractical" according my thinking..
