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f(0)=96,f(1) =1, f(2)= 4, f(3)=9, f(4) =16 then find the value of f(5)=?
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- raviteja please explain this?how f(5)=25 when f(0)=96
- 10 years agoHelpfull: Yes(8) No(0)
- f(1)=1^2=1,f(2)=2^2=4 so squares of given number f(5)=25
- 10 years agoHelpfull: Yes(3) No(1)
- f(5)= 121, by theory of equations
- 10 years agoHelpfull: Yes(2) No(0)
- this function could be like
f(x)=96 if x=0
otherwise x2
so f(5)=25 is a correct solution
- 10 years agoHelpfull: Yes(2) No(0)
- raviteja is dere not any use of f(0)=96????
- 10 years agoHelpfull: Yes(1) No(0)
- 1^2=1
2^2=4
3^2=9
4^4=16
5^2=25 - 10 years agoHelpfull: Yes(1) No(1)
- hi SOMI
wat abt f(0)=96 - 10 years agoHelpfull: Yes(1) No(1)
- Tell me answer pls..........
wht about f(0)=96 - 10 years agoHelpfull: Yes(1) No(1)
- let f(x)=ax^n+bx^n-1+........+constant
as 5 values are known let n=4 as many values can found
so from f(0) constant=96
From f(1) a+b+c+d=-96
form f(2) 16a+8b+4c+2d=-95
form f(3) 81a+27b+9c+3d=-92
form f(4) 256a+64b+16c+4d=-80
solving 4 equations simultaneously
we get a,b,c,d
and hence f(5)=
- 10 years agoHelpfull: Yes(0) No(1)
- f(0)=96 is just meant to misguide people dear @priya
- 10 years agoHelpfull: Yes(0) No(1)
- f(5)=5^5=25 as f(1)=1^1=1
f(2)=2^2=4
f(3)=3^3=9
f(4)=4^4=16 - 10 years agoHelpfull: Yes(0) No(2)
- 25 start from 1 4 9 16 u will find difference is 3,5,7 and then 9 so 16+9=25.
- 10 years agoHelpfull: Yes(0) No(0)
- putting the values on this equation you will able to find any values of given function,
49x^2-144x+96
therefore,
f(5)=601 - 10 years agoHelpfull: Yes(0) No(1)
- 1^2=1
2^2=4
3^2=9
4^2=16
5^2=25 - 8 years agoHelpfull: Yes(0) No(0)
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