Question

$f: [0,1] \rightarrow R$ is a function such that $f'(x)$ is continuous and $\int_0^1 f(x)dx = 0$, if maximum value of $|f'(x)|$ when $x \in [0, 1]$ is 24, then possible value of $|\int_0^x f(t)dt|$, $x \in [0, 1]$, is/are:

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (A), (B), (C)

1, 2, 3

Answer 2

Let $F(x) = \int_0^x f(t) dt$. We are given that $f: [0,1] \rightarrow R$ is a function such that $f'(x)$ is continuous on $[0,1]$. This implies $f(x)$ is continuously differentiable, and $F(x)$ is twice continuously differentiable. By the Fundamental Theorem of Calculus, $F'(x) = f(x)$ and $F''(x) = f'(x)$. We are given $\int_0^1 f(x) dx = 0$. This means $F(1) = \int_0^1 f(t) dt = 0$. Also, $F(0) = \int_0^0 f(t) dt = 0$. We are given that the maximum value of $|f'(x)|$ for $x \in [0,1]$ is 24. So, $|F''(x)| \le 24$ for all $x \in [0,1]$.

We have a function $F(x)$ defined on $[0,1]$ such that $F(0)=0$, $F(1)=0$, and $|F''(x)| \le 24$. We want to find the possible values of $|F(x)|$ for $x \in [0,1]$.

Consider the function $g(x) = F(x) - 12x(1-x)$. $g'(x) = F'(x) - 12(1-2x)$. $g''(x) = F''(x) - 12(-2) = F''(x) + 24$. Since $|F''(x)| \le 24$, we have $F''(x) \ge -24$. So $g''(x) = F''(x) + 24 \ge -24 + 24 = 0$. Since $g''(x) \ge 0$, the function $g(x)$ is convex on $[0,1]$. Also, $g(0) = F(0) - 12(0)(1-0) = 0 - 0 = 0$. And $g(1) = F(1) - 12(1)(1-1) = 0 - 0 = 0$. For a convex function on $[0,1]$ with $g(0)=g(1)=0$, we must have $g(x) \le 0$ for $x \in [0,1]$. Thus, $F(x) - 12x(1-x) \le 0$, which implies $F(x) \le 12x(1-x)$.

Consider the function $h(x) = F(x) + 12x(1-x)$. $h'(x) = F'(x) + 12(1-2x)$. $h''(x) = F''(x) + 12(-2) = F''(x) - 24$. Since $|F''(x)| \le 24$, we have $F''(x) \le 24$. So $h''(x) = F''(x) - 24 \le 24 - 24 = 0$. Since $h''(x) \le 0$, the function $h(x)$ is concave on $[0,1]$. Also, $h(0) = F(0) + 12(0)(1-0) = 0 + 0 = 0$. And $h(1) = F(1) + 12(1)(1-1) = 0 + 0 = 0$. For a concave function on $[0,1]$ with $h(0)=h(1)=0$, we must have $h(x) \ge 0$ for $x \in [0,1]$. Thus, $F(x) + 12x(1-x) \ge 0$, which implies $F(x) \ge -12x(1-x)$.

Combining the two inequalities, we have $-12x(1-x) \le F(x) \le 12x(1-x)$ for all $x \in [0,1]$. This means $|F(x)| \le 12x(1-x)$ for all $x \in [0,1]$.

The function $p(x) = 12x(1-x)$ is a parabola opening downwards with roots at $x=0$ and $x=1$. Its maximum value on $[0,1]$ occurs at the vertex $x = -12/(2 \times -12) = 1/2$. The maximum value is $12(1/2)(1-1/2) = 12(1/2)(1/2) = 12/4 = 3$. So, for any function $f$ satisfying the given conditions, we have $|F(x)| = |\int_0^x f(t) dt| \le 3$ for all $x \in [0,1]$. This means that any possible value of $|\int_0^x f(t) dt|$ must be less than or equal to 3.

To check if 3 is a possible value, we need to see if there exists a function $f$ satisfying the conditions such that $|F(x)|=3$ for some $x \in [0,1]$. Consider the function $F(x) = 12x(1-x)$. $F(0)=0$, $F(1)=0$. $F'(x) = 12(1-2x)$. Let $f(x) = F'(x) = 12(1-2x)$. $\int_0^1 f(x) dx = \int_0^1 12(1-2x) dx = 12[x-x^2]_0^1 = 12[(1-1)-(0-0)] = 0$. This condition is satisfied. $f'(x) = -24$. $|f'(x)| = 24$. The maximum value of $|f'(x)|$ on $[0,1]$ is indeed 24. So, the function $f(x) = 12(1-2x)$ satisfies all the given conditions. For this function, $F(x) = \int_0^x 12(1-2t) dt = 12[t-t^2]_0^x = 12(x-x^2) = 12x(1-x)$. $|F(x)| = |12x(1-x)| = 12x(1-x)$ for $x \in [0,1]$. The possible values of $|F(x)|$ for this specific function are the values in the range of $12x(1-x)$ for $x \in [0,1]$. The range of $12x(1-x)$ for $x \in [0,1]$ is $[0, 3]$. So, for this specific function, any value in $[0, 3]$ is a possible value of $|F(x)|$.

The question asks for "possible value of $|\int_0^x f(t)dt|$, $x \in [0, 1]$". This means a value that $|F(x)|$ can take for some $x \in [0,1]$ and for some function $f$ satisfying the conditions. Since we found a function $f$ for which the values of $|F(x)|$ cover the interval $[0, 3]$, any value in $[0, 3]$ is a possible value.

We check the given options: (1) 1: 1 is in $[0, 3]$. Possible. (2) 2: 2 is in $[0, 3]$. Possible. (3) 3: 3 is in $[0, 3]$. Possible. (4) 4: 4 is not in $[0, 3]$. Not possible.

The question asks for "possible value is/are", suggesting there might be multiple correct options. The possible values of $|\int_0^x f(t) dt|$ for $x \in [0,1]$ are the values in the interval $[0, 3]$. Options (1), (2), and (3) are all within this interval.