Doubt from function

  • let f(x)=x^2-x+1 \ x \in \mathcal{R}\\g(x) is also quadratic polynomial such that [f(x)]=[g(x)]
    \ x \in \mathcal{R} then find value of
    1- f(2)-f(1)
    2-f(3)-f(2)
    [ ] denotes gif

Will the answer be '-3' and '-9' ?

not correct

Both the answers are between -1 and 0 right?

Answer will be a range right ?

nope exact constant number

no

1 and 4?

2 and 4. what is ur approach???

Yeah right 2 and 4

approach??

Can you show that f(x) = g(x)?

sir its means f(x) and g(x) are exactly same
how to prove that i m thinking steps
like i tried i m getting if two box r equal then diffrence of them must lie in -1 to 1 but not confirm it will be close interval on or not
nd thinking to find coefficent of a,b in g(x) but not getting proper mthd sir
i dont have solution of this i m little bit confused on accurate value what should be g(x)

@Hari_Shankar sir how to prove please give hint or solution of ur..

I don't think it is necessary to prove it for all x, but it can be proven for the values of x required.

yep that may be but for learning purpose i want to know the proper mthd nd if we will use hit nd trail their will create confusion is this only function or not...

. Prove the other case..

If [f(x_0)]=[g(x_0)]=n then both n \le f(x_0), g(x_0)<n+1 and hence |g(x_0)-f(x_0)|<1 and hence

|g(x)-f(x)|<1 \ \forall \ x \in \mathbb{R}

So if g(x)=ax^2-bx+c then |(a-1)x^2-(b-1)x+(c-1)|<1 which is not possible as for any polynomial P(x), we have |P(\pm \infty)|=\infty, and so a=b=1. Hence g(x)=x^2-x+c where 0 \le c<2. (Note that even if they had not specified that g(x) is a quadratic we would have arrived at this form for the quadratic)

Now since f(x) is continuous we have for some arbitrary integer n a real number x_1 such that f(x_1)= n- \epsilon where \epsilon < \dfrac{c-1}{2} so that [f(x_1)]=n-1
Then g(x_1) = x^2-x+c = (x^2-x+1) + (c-1) =n-\epsilon+c-1>n - \dfrac{c-1}{2}+c-1 = n+\dfrac{c-1}{2}

so that [g(x_1)] = n \ne [f(x_1)]

Hence we see that we need c=1

Thus g(x)=x^2-x+1 = f(x)

2 Likes

thanks sir its too impressive solution...