1057@sambit.....wrong?there obviusly is a unique value each for a,b,c,d,e,f such that P(1)=1;P(2)=2;P(3)=3;P(4)=4;P(5)=5 and P(6)=6....how can it be wrong????
1)f(x) and g(x) are linear functions for all x such that f(g(x)) and g(f(x)) are Identity functions.If f(0)=4 and g(5)=17 find f(2006)?
2)let f(x)=(x+1)(x+2)(x+3)(x+4)+5 where x belongs to [-6,6].If range of f(x) is [a,b] where a,b belongs to naturals the find(a+b)?
3)Let P(x)=x6+ax5+bx4+cx3+dx2+ex+f such that P(1)=1;P(2)=2;P(3)=3;P(4)=4;P(5)=5 and P(6)=6,find the value of P(7)
(plz give a method where we dont have to solve 6 equations....)
4)Suppose f(x) is a polynomial with integere coefficients.The remainder when f(x) is divided by x-1 is 1 and the remainder when f(x) is divided by x-4 is 10.If r(x) is the remainder when f(x) is divided by (x-1)(x-4),find r(2006) ?
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13 Answers
1057
341Hint: look at the polynomial F(x) = P(x)-x. What are the roots of F(x)?
13) using the hint given by bhat sair,
F(x) = P(x)-x = (x-1)(x-2)(x-3)(x-4)(x-5)(x-6)
since leading coeff. has to be 1 as P(x) = x6 +....
=> p(7)-7 = 6!
=> P(7) = 727
14) is simple remainder theorum,
f(x) = (x-1)(x-4) + ax+b
=> solving f(1) and f(4) we get a = 3 ; b=-2
=> r(x) = 3x-2
=> r(2006) = 6016
712. It can be simplified to f(z) = z2+10z+29 where z = x2+5x
391
f(x)=ax+b
f(0=4 so b=4
f(g(5))=f(17)=1=17a+4
a=-3/17
now u can get f(2006)
39i took the defn of identity function wrong and even did multiplication wrong >.<
anyways gng byt he same logic as above
f(g(5))=g(5) =17 since its an identity function
from here proceed on as above
71Solution 1:
We have, as pointed out by Dr. House the definition of Identity Functions as " a function that always returns the same value that was used as its argument. In terms of equations, the function is given by f(x) = x. "
So we have here f(g(x)) = x and g(f(x)) = x
Now Let us assume two functions as f(x) = ax+4 and g(x) = cx + d
So, f(g(5))= 5 => f(17) = 5 => 17a+4 = 5 => a = 1/17
Hence, f(x) = x/17 + 4
f(2006) = 122
Now for g(x) :
g(f(0)) = 0 => g(4) = 0 => 4c+d = 0 ...... (i)
g(5) = 17 => 17 = 5c + d ....... (ii)
Solving both g(x) = 17x-68