1hey.........have you created the 1st quest of ur own or is it from a book.....plzz refer the book and check the question........
q1.) f(x).f(1x) = f(x) + f1x)
f(3)=28
find f(4).
q2.) f(x)= x4+ ax3+bx2+cx+d
f(1)=10 , f(2)=20 , f(3)=30
find f(12) + f(-8)
plz tell me d general way to solve such functional equations.....i hv found such questions in many books but didn't find in any book any theory regarding them....is there any book which has theory for solving such equations....i can solve simple functional equations by exchanging d arguments of the function and by eliminating unknowns....
but what is d general method for solving such equations????
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9 Answers
11In the 2nd question.....
f(1) = 10
Put 1 in place of x.... and get an equation in terms of a,b,c...
f(2) = 20
Similarly...put 2 in place of x to form another equation....
And finally..... substitute 3 in place of x.....
It can be clearly seen that f(0) = d
Solve the above three equations to get the value of a,b,c and d....
And then solve f(12)+f(-8)....
There is no particular method to solve such questions......
it is just the basics.....
23http://www.targetiit.com/iit-jee-forum/posts/test-276-qno-1-7007.html
11st ques. is given in TMH....
i already have tried solving by d method u told...but its too lengthy....nd u cannot use this method in exam....
11For the 1st qsn, u must know that , given f(x) is a polynomial function, f(x) is of the form xn±1......always.
Further n can be evaluated from the given data.
1@Soumik
how can we tell that f(x) is polynomial of form x^n±1 ??? i hv seen in buks using the eq. u wrote...but without giving any explanation as to how it came......plz provide an explanation.....
1Q1> as rightly mentioned
it is always true for xn+-1
since f of 3=28
by observation f(x)=x3+1
now put x=4....hence answer is 65
1@gaurav , try to appreciate others effort , qwerty's solution is perfect