functions

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\(\large \color{black}{\begin{align} & \normalsize \text{Let }\ f(x) \text{be function such that } \hspace{.33em}\\~\\ & f(x)f(x+1)=-f(x-1)f(x-2)f(x-3)f(x-4) ,\ \ x\geq 0 \hspace{.33em}\\~\\ & f(83)=81 \hspace{.33em}\\~\\ & f(77)=9 \hspace{.33em}\\~\\ & f(102)=? \hspace{.33em}\\~\\ & a.)\ 27 \hspace{.33em}\\~\\ & b.)\ 54 \hspace{.33em}\\~\\ & c.)\ 729 \hspace{.33em}\\~\\ & d.)\ \normalsize \text{Data insufficient} \hspace{.33em}\\~\\ \end{align}}\)
it should have a fast way , the time given to solve such problem is average 2 min
that negative sign should really be there right, on the right side of the eqn

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yes its there
is it C?
orr Data insufficient
@mathmath333 I always like your questions. Can I ask where do you get them from?
here is something we do know |dw:1435263889685:dw|
http://www.flipkart.com/quantitative-aptitude-quantum-cat-common-admission-test-into-iims-english/p/itmdygg8z2ggtuk2?pid=9789351416401&ppid=9789351416401
and every multiplication of 2 of them can be rewritten recursively
i have just checked into solution set , it has given solution for \(f(81)\) and not \(f(102)\) as described in the question
f(77)=9 f(83)=9^2 f(102)=? and 9^3 is 729 if this turns up
it could be lol, i feel like the separation is too much cant write anythign recursively
lol i think the question is this book has a typo \(\large \color{black}{\begin{align} & \normalsize \text{Let }\ f(x) \text{be function such that } \hspace{.33em}\\~\\ & f(x)f(x+1)=-f(x-1)f(x-2)f(x-3)f(x-4) ,\ \ x\geq 0 \hspace{.33em}\\~\\ & f(83)=81 \hspace{.33em}\\~\\ & f(77)=9 \hspace{.33em}\\~\\ &\color{red}{ f(81)}=? \hspace{.33em}\\~\\ & a.)\ 27 \hspace{.33em}\\~\\ & b.)\ 54 \hspace{.33em}\\~\\ & c.)\ 729 \hspace{.33em}\\~\\ & d.)\ \normalsize \text{Data insufficient} \hspace{.33em}\\~\\ \end{align}}\)
27 is given as the answer.
okay lets see so 3^2, 3^3 and 3^4 for f(77), f(77+4), f(77+6)
actually I think we get \[(f(81))^2=729\]
let me post my work
can we do another question which we know doesnt have a typo xD
\(\large \color{black}{\begin{align} & f(82)f(83)=-f(81)f(80)f(79)f(78) \ -\color{red}{(1)}\hspace{.33em}\\~\\ & f(81)f(82)=-f(80)f(79)f(78)f(77) \ -\color{red}{(2)}\hspace{.33em}\\~\\ \end{align}}\) dividing 1 and 2 works
oh beauty
\[f(82)f(83)=-f(81)f(80)f(79)f(78) \\ f(82) \cdot 81 =-f(81)f(80)f(79)f(78) \\ 81=\frac{-f(81)f(80)f(79)f(78)}{f(82)} \\ \text{ now we plug \in } 81 \\ f(81)f(82)=-f(80)f(79)f(78)f(77) \\ f(81)=\frac{-f(80)f(79)f(78)}{f(82)} f(77) \\ \text{ multiply both sides by} f(81) \\ f(81) \cdot f(81)=\frac{-f(81)f(80)f(79)f(78)}{f(82)} f(77)\]
thats really cool, give me another one!
we already said that one thingy was 81 so you have \[(f(81))^2=81 f(77)\]
and I agree with @dan815 more please!
I better go to sleep! very clever @freckles :-)
well i only post the one's i stuck at, let's see if i have some tough ones
\(\large \color{black}{\begin{align} & \normalsize \text{let}\ f(x)=121-x^2,\ g(x)=|x-8|+|x+8| \hspace{.33em}\\~\\ & \normalsize \text{and}\ h(x)=\text{min}(f(x),g(x)). \hspace{.33em}\\~\\ & \normalsize \text{What is the number of integer values of x for which } \hspace{.33em}\\~\\ & h(x)\ \normalsize \text{is equal to a positive integral value? } \hspace{.33em}\\~\\ & a.)\ 17 \hspace{.33em}\\~\\ & b.)\ 19 \hspace{.33em}\\~\\ & c.)\ 21 \hspace{.33em}\\~\\ & d.)\ 23 \hspace{.33em}\\~\\ \end{align}}\)
ok I'm here and looking now
yep good luck
by induction, I got this: \[\Large {\left\{ {f\left( n \right)} \right\}^2} = f\left( {n + 2} \right) \times f\left( {n - 4} \right)\]
setting n=81, we get: \[\Large {\left\{ {f\left( {81} \right)} \right\}^2} = f\left( {83} \right)f\left( {77} \right)\] namely the result of @freckles
I have to come back and look this one crab time sorry
i dont quite understand what does it mean min(f(x),g(x))
u need to graph that f(x) and g(x) and find the intersection , and as it asks minimum u have to choose the lowest part with respect to that
it can also be done without graphing
oh ok i see now
example min(f(x),g(x)) for |dw:1435266966304:dw| f(x)=-x,g(x) =-x^2+5
|dw:1435267032841:dw|
i think to start off look at g1(x) = |x-8| and g2(x) = |x+8|
this will give us a constant when 0<|x| < 8
it will be 16 in that domain
from -8 to 8
it looks line this|dw:1435267447526:dw|
yes exactly
|dw:1435267527733:dw|
|dw:1435267567368:dw|
u want the number of integers of x in that intersection?
|dw:1435267618964:dw|
between the intersection?
|dw:1435267649589:dw|
or just the min on the abs value?
well i m still confused on interpreeting the quetion let me think
ya me too the min(f(x),g(x)) still dunno what that means exacttllyy
there are 17 points from that flat line, for hte integers then we have y=2x and y=-2x lines from the abs value
anser given is 21
yeah there will be 2 more from the intersection on each side
we see 11^2-x^2=2x x^2+2x-11^2=0 solve for the roots and see how many integers we can fit
you get sqrt 122 -1 so its greater than 10 barely that means another 2 from the right side and another 2 from the left side
17+4=21
it is not giving rational roots hmm
we just want to see the bound
yea we need to count the number line of x of the intersection part.
tbh i think the question is worded wrong it should just say.. how many integer solutions are there that are less than the quadratic
these are framed to confuse students majority are confuzing
lol thats annoying -.- they should confuse us with hard questions, not easy ones worded badly
my head spins now going to sleep
okay cya! thanks for the questions
thnks
looks like you already found this but ill leave this here, just in case on how we got the equations 2x and -2x for the lines
|dw:1435268655727:dw|
|dw:1435268725528:dw|
there are 21 integers between -11 and 11 (not including the endpoints) because at the endpoints h=0 which isn't a positive number
like i didn't want to include the endpoints because I want h positive
and anything outside the interval I mentioned h would be negative
|dw:1435269002394:dw|
for example h(-10)=g(-10) (since f(-10)>g(-10)) ... h(-5)=g(-5) (since f((-5)>g(-5)) ... h(0)=g(0) (since f(0)>g(0)) ..so on... h(x)=g(x) for integer solutions between -10 and 10 (inclusive) h(x)=f(x) for integers solutions on (-inf,-11] union [11,inf) but h(x)<=0 there so what we want to look at is: h(x)=g(x) for integer solutions between -10 and 10 (inclusive) since h(x)>0 here \[|x-8|+|x+8|=121-x^2 \\ \text{ from our graph one solution occurs on } (-\inf,-8) \\ -(x-8)-(x+8)=121-x^2 \\ (x-1)^2=122 \\ x= 1 \pm \sqrt{122} \\ x=1-\sqrt{122} \text{ is only valid on that interval } \\ x \approx -10.045 \\ |x-8|+|x+8|=121-x^2 \\ \text{ the other solution \to this occurs on } (8,\infty) \\ x-8+x+8=121-x^2 \\ (x+1)^2=122 \ x=-1 \pm \sqrt{121} \\ x=-1 +\sqrt{122} \text{ is only valid on that interval } \\ x=10.045\] |dw:1435269711175:dw|
anyways just saying all of this just in case you guys were still confused on the h=min(f,g) thing

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