mathmath333
  • mathmath333
functions
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} & \normalsize \text{Let }\ f(x) \text{be function such that } \hspace{.33em}\\~\\ & f(x)f(x+1)=-f(x-1)f(x-2)f(x-3)f(x-4) ,\ \ x\geq 0 \hspace{.33em}\\~\\ & f(83)=81 \hspace{.33em}\\~\\ & f(77)=9 \hspace{.33em}\\~\\ & f(102)=? \hspace{.33em}\\~\\ & a.)\ 27 \hspace{.33em}\\~\\ & b.)\ 54 \hspace{.33em}\\~\\ & c.)\ 729 \hspace{.33em}\\~\\ & d.)\ \normalsize \text{Data insufficient} \hspace{.33em}\\~\\ \end{align}}\)
mathmath333
  • mathmath333
it should have a fast way , the time given to solve such problem is average 2 min
dan815
  • dan815
that negative sign should really be there right, on the right side of the eqn

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