Can you find this function?

Calculus Level 4

Let $f : \mathbb R \rightarrow \mathbb R$ be a continuous function such that $\displaystyle \int_0^1 f(xt) \, dt = 0$ for all $x\in \mathbb R$ .

Which of the following statements is/are true?

(A): There exists 3 integral values of $P$ such that the graph of $y=f(x)$ and $y=x^3-3x^2 + P$ intersects at 3 distinct points.

(B): $y= f(x)$ is a periodic function.

(C): If $A$ denotes the minimum area bounded by the curves $y=f(x)$ , $y =x^4-4x-a$ and the ordinates $x=2,x=4$ , then $a=69$ is true.

(D): $f(x)$ is neither even function nor odd function.

Notation: $\mathbb R$ denotes the set of real numbers .

A and C only A, C and D only A, B and C only A and D only

1 solution

Avi Solanki
Feb 16, 2017

The function ofcourse is equal to 0 . and the problem is finished

"Of course"... The condition tells us that $\tfrac{1}{x}\int_0^x f(t)\,dt \; = \; 0 \hspace{2cm} x \neq 0$ and so we deduce that $F(x) \; = \; \int_0^x f(t)\,dt \; = \; 0 \hspace{2cm} x \in \mathbb{R}$ and so, by the "mirror image" of the Fundamental Theorem of the Calculus, $f(x) \; = \; F'(x)\; = \; 0 \hspace{2cm} x \in \mathbb{R}$ Since $f \equiv 0$ , $f$ is certainly periodic, and so $B$ is true. That makes $A,B,C$ the only possible option (and indeed the correct answer). It is a pity that we don't need to check the other three conditions to find the right answer!

Mark Hennings - 4 years, 3 months ago

haha yes true that :)

avi solanki - 4 years, 3 months ago

Yeah thats called clever observation to avoid full work :P

Prakhar Bindal - 4 years, 3 months ago

Can you find this function?

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