24Q1
i thought it is cauchy's eqn
f(x)=ax
f(5)=2 => 2=a5 -----(1)
f'(x)=ax.loga
f'(0)=3
=> 3=1.loga
putting a from (1)
=> 3=1.log21/5 ----(2)
f'(5)=a5.loga=2.log21/5 ----(3)
confused in both (2) and (3)..plz point out mistake
Q1. Let f(x+y) = f(x).f(y) for all x,y belongs to R and f(5) = 2 and f'(0) = 3. Then f'(5) = ?
Q2. Let f(x) be defined in R such that f(1) = 2 and f(2) = 8 and f(u+v) = f(u) + kuv - 2v2 for all u,v belongs to R and k is a fixed constant. Then f'(x) =
Check whether the questions are correct
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17 Answers
2424manish sir wrote what i aksed u in post#14...
ques is wrong anyways[3]
106Check this one
Q3. If a polynomial g(x) satisfies x.g(x+1) = (x-3).g(x) for all x and g(3) = 6, then the value of g(25) is _____
I think ans should be zero but it is given 13800
1357Yes in question 2
we get f(v)=f(0)-2v2
which does not satisfy f(1) and f(2) simultaneously
106i am saying that question itself is wrong.. both the conditions given do not satisfy simultaneously
106anyone checking if Q2 is correct?
there is one more which ill post after this one's verified
106thats y the thread title "check the correctness of the questions"
106exactly .. the question is ill framed ... either one of the conditions can hold at a time
eure see if both of ur (2) and (1) hold at the same time
24Q1 is my dbt also....did the same ques yesterday..didnt get rite answer
62well lets put it this way..
The question is not well defined and the answer is indeterminate :D :P
1so wat ans we will write is supossing this q comes in exam and we hav both 6 and 1
and btw r der any other values of f'(5) possible....
62LOL.. NOTHING IS WRONG...
i guess the problem is that this is one of those 1000;s of functional equation questions which are ill framed...
The mistake is probably that they have given 2 conditions.. which may not hold together!
21then wats wrong with this[2]
f(x+y)=f(x)f(y)
f'(x)= lim h→0 f(x+h)-f(x)h
=limh→0 f(x)f(h)-f(x) h
=f(x)(limh→0 f(h)-1h)
f'(x)=f(x)f'(0)
f'(x)=f(x).3
f'(5)=f(5).3
f'(5)=2.3=6
62Let f(x+y) = f(x).f(y) for all x,y belongs to R and f(5) = 2 and f'(0) = 3. Then f'(5) = ?
f(x+h)-f(h) = f(h)[f(x)-1]
f(x+h)-f(x)f(h) = [f(x)-1]
take limit h going to zero ...
f'(x) = f(x)-1
Thus,
f'(5) = 1
21wat ans u wer getting eureka for 1st 1....i m getting 6....its simple i guess