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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: find f(x)
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anirudh1990 (7)

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Find the missing f(x):


x       1  2  3   4    5


f(x)   2  5  9         32

    
arpan1 (665)

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is it 14


all the best ...
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spideyunlimited (2686)

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no yaar otherwise it would be 20 in place of 32.

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allamraju (660)

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I think the ans is 16.Is it correct?

I also think f(x)=3x-1/4+5x/2-3/4
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sboosy (2860)

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\mbox{Consider} \ f(x) = ax^3+bx^2+cx+d \\ \\ f(1) = 2 \ \Rightarrow a+b+c+d = 2 \\ \\ f(2) = 5 \ \Rightarrow 8a+4b+2c+d = 5 \\ \\ f(3) = 9 \ \Rightarrow 27a+9b+3c+d=9 \\ \\ f(5)= 32 \ \Rightarrow 125a+25b+5c+d = 32 \\ \\ \mbox{Four equations, four unknowns} \\ \\ \mbox{Solving we get} \ a=\frac{1}{2},b=\frac{-5}{2},c=7,d=-3 \\ \\ \Rightarrow f(x) = \frac{1}{2}x^3 - \frac{5}{2}x^2+7x-3 \\ \\ \Rightarrow  f(4) = 32-40+28-3 =17

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nunknown91 (11)

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hello sboosy why shud we consider only f(x) = ax3 + bx2 + cx + d .......


 


if i consider f(x) = ax2 + bx + c


 


i am able to get the answer but if i consider f(x) = ax + b


 


i am not able to


 


please explain why is it so


 


 


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<TR><TD>


<DIV ALIGN="right">Glitter Graphics</DIV></TD></TR></TABLE>
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arpan1 (665)

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hello nuknown


sboosy is right in his approach


 


see u need to satisfy all the given conditions and since 4 conditions are given u hav to generate a 4 unknown equation, hence the cubic eqn.


 


anything less or more will lead to error


 


i approached the same way but probably had eror in between


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spideyunlimited (2686)

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nice sboosy
LOL
i assumed a quadratic
and solved for x = 1, 2, 3
and got answer still as 17 :D
but it didn't match for x = 5

ah well a cubic was needed LOL

---------------------------------------------------------------
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* Agent 'G' [sniper] - SD-6 (Alliance of Twelve)







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allamraju (660)

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What about my answer?The fun. I got also satisfied all the values in the question.Plz tell me whether it is right.

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arpan1 (665)

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show ur method allamraju


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allamraju (660)

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I used the method given in 318 page of HALL & KNIGHT HIGHER ALGEBRA to find nth term of a series.If we assume values of f(x) as a series then the rth term will be f(r) and it represents the value of the fun.

Let f(4)=k then 2,5,9,k,32

                         3,4,(k-9),(32-k)       

                          1,(k-13),(41-2k).For this to be a G.P,(k-13)2=41-2kk=16

To find the nth term which is the value of the fun;

The nth term will be of the form Tn=a.3n-1+bn+c

Using T1=2,T2=5,T3=9,we get a=1/4,b=5/2,c=-3/4.

Hence f(x)=xth term=3x-1/4+5x/2-3/4.Plz tell me whether I am right.              

 

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GoNik (56)

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SHUD IT ALWAYS BE A POLYNOMIAL FUNCTION, CANT IT B AN EXPONENTIAL OR TRIGONOMETRIC ONE?
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