33Kitni bar post karna padega... [3]
This is a old trick question.. (not brillaint if u try to solve it first hand...)
what is the number of solutions of
x2+4x+1 = √x+3-2
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43 Answers
62chalo this became more than a trick question...
caught me on the wrong foot atleast.......
33No i am saying Akshay...
He said if f(x)=f-1(x)
then he is saying soln is same as x=f(x)
i.e. f(x) and f-1(x) intersects at y=x only .... this is wrong
62arrey he is not saying only!!!
Nor did i use the word only priyam...
The easier solution of this wud be found by putting that f(x)=x
The others wont exist! (that is said by the graph method!)
There is no contradiction...
33i was telling akshay kahi galti se to nahi sahi kiya ho isliye...
to finaly no of soln by graph and soln by inverse.. :)
62x2+4x+1
celestine.. either i am sleeping or u are.. there is a solution...
x2+3x+1=0
x= (-3±√5)/2
9dude sub it in org eq and chk if it is satisfies
-3 is a sol
9-3 is a sol but u arent gettin it here y
cos stat is rong
sol cant be found puttin fx = x
its f(x)= f-1(x)
i m repeatin these statements agn again ;)
62x2+4x+1 = √(x+3)-2
9-12+1 = -2
oops i am caught....
then i think it is because of the old question priyam gave....
There will be equations other than the ones on x=y
and so we will first have to find the biquadratic equation..
then use the two roots we found...
annd then find the other two roots using these two roots!!
Good that you pointed this out celestine...
62oh and in all this i forgot to tell one more thing.... which is very improtatn..
not all roots that you find will satisy...
bcos of the root that we have taken..
so the RHS is only half the curve.. and not the full parabola...
9yes ur second root -3-√5/2 isnt applicable ;)
thats y i din see ne point in usin this trick cos drawin graph uld do bettr
1hehe nice discussions only wud we have stuck to the basics the problem would have been long over ;D
62well celestine may be it din work in this question.. but there are definiely a lot of places where it works....
Just that right now i am not able to "construct" a good question..
tihs will come in some question fo the day soon :P
9oh sry i dint understand akshays point now understood
actually i think he needed to derive that
fx = f-1x
now let x=fx implyin f-x=x confirmin fx = f-1x
but u surely cant find all roots usin this method
62actually celestine the point is that you can for a very very large number of questions!!!
9thats bcos all sol may lie within
x = fx which u assume bfore findin roots
62:D
well how about spilling the beans already...??
Hint: Can you figure out why i have given the equations in the form that i have and not in the form that priyam wrote after that?
Is there some relation between the RHS and LHS?
62yeah there are 2 solutions only
even i started to wonder.. how :)
1please tell me i there is any trick involved coz i did it the simple way
1ya [1] most definitely [11]
why din i think of them b4[11]
[4]thanx
62well there is a slight different trick...
It is about something more than these two ..
what if i asked to find the roots too?