Question

f:[0,8]→[0,8] be twice differentiable function with continuous and positive first derivative such that f(0) = 0; f (4) = 1; f(8) = 1, then

(P) There exist some $c \in (0,8)$ where $f'(c_1) = \frac{1}{4}$ (Q) There exist some $c \in (0,8)$ where $f'(c) = \frac{1}{12}$ (R) There exist $c_1,c_2 \in [0,8]$ where $8f'(c_1)f(c_2) = 1$ (S) There exist some $c \in [0,8]$ such that $f(f(8))-f(f(0)) = 8(f'(c))^2$

Then which of the following is correct

• A. P, Q are correct
• B. P, Q, R are correct
• C. P, R are correct
• D. P, Q, R, S are correct

Answer 1

Answer: (D)

P, Q, R, S are correct

Answer 2

The problem asks us to analyze a function $f:[0,8]\rightarrow[0,8]$ which is twice differentiable with a continuous and positive first derivative. We are given $f(0)=0$, $f(4)=1$, and $f(8)=1$.

First, let's address the condition "positive first derivative". If $f'(x) > 0$ for all $x \in [0,8]$, then $f(x)$ must be strictly increasing on $[0,8]$. However, we are given $f(4)=1$ and $f(8)=1$. Since $4 < 8$, if $f(x)$ were strictly increasing, we would have $f(4) < f(8)$, which means $1 < 1$, a contradiction. This implies that the condition "positive first derivative" must be interpreted as "non-negative first derivative", i.e., $f'(x) \ge 0$. This is a common interpretation in such problems where a strict inequality would lead to a contradiction with other given conditions, especially when differentiability and continuity are stated. If $f'(x) \ge 0$, then $f(x)$ is non-decreasing. The condition $f(4)=f(8)=1$ is then consistent, as it implies $f'(x)=0$ for some $x \in (4,8)$ by Rolle's Theorem.

Let's evaluate each statement under the assumption that $f'(x) \ge 0$.

Statement (P): There exist some $c_1 \in (0,8)$ where $f'(c_1) = \frac{1}{4}$ Applying the Mean Value Theorem (MVT) to $f(x)$ on the interval $[0,4]$: Since $f(x)$ is differentiable on $[0,4]$ and continuous on $[0,4]$, there exists some $c_1 \in (0,4)$ such that: $f'(c_1) = \frac{f(4) - f(0)}{4 - 0} = \frac{1 - 0}{4} = \frac{1}{4}$ Since $c_1 \in (0,4)$, it is also in $(0,8)$. Therefore, statement (P) is correct.

Statement (Q): There exist some $c \in (0,8)$ where $f'(c) = \frac{1}{12}$ From statement (P), we know there exists $c_1 \in (0,4)$ such that $f'(c_1) = \frac{1}{4}$. Applying Rolle's Theorem (a special case of MVT) to $f(x)$ on the interval $[4,8]$: Since $f(x)$ is differentiable on $[4,8]$ and $f(4)=f(8)=1$, there exists some $c_2 \in (4,8)$ such that: $f'(c_2) = \frac{f(8) - f(4)}{8 - 4} = \frac{1 - 1}{4} = 0$ Now, we have $f'(c_1) = \frac{1}{4}$ and $f'(c_2) = 0$. Since $f'(x)$ is continuous on $[c_1, c_2]$ (which is a subinterval of $[0,8]$), by the Intermediate Value Theorem (IVT) for continuous functions, $f'(x)$ must take on every value between $0$ and $\frac{1}{4}$. Since $0 < \frac{1}{12} < \frac{1}{4}$, there must exist some $c \in (c_1, c_2) \subset (0,8)$ such that $f'(c) = \frac{1}{12}$. Therefore, statement (Q) is correct.

Statement (R): There exist $c_1,c_2 \in [0,8]$ where $8f'(c_1)f(c_2) = 1$ From statement (P), we know there exists $c_1 \in (0,4)$ such that $f'(c_1) = \frac{1}{4}$. Substitute this into the expression: $8 \times \left(\frac{1}{4}\right) \times f(c_2) = 1$, which simplifies to $2f(c_2) = 1$, or $f(c_2) = \frac{1}{2}$. Now we need to check if there exists $c_2 \in [0,8]$ such that $f(c_2) = \frac{1}{2}$. We know $f(0)=0$ and $f(4)=1$. Since $f(x)$ is continuous on $[0,4]$, by the IVT, for any value between $f(0)$ and $f(4)$, there must exist $c_2 \in (0,4)$ such that $f(c_2)$ equals that value. Since $0 < \frac{1}{2} < 1$, there exists some $c_2 \in (0,4)$ such that $f(c_2) = \frac{1}{2}$. Therefore, statement (R) is correct.

Statement (S): There exist some $c \in [0,8]$ such that $f(f(8))-f(f(0)) = 8(f'(c))^2$ First, evaluate the left side: $f(f(8)) - f(f(0)) = f(1) - f(0)$ (since $f(8)=1$ and $f(0)=0$) $f(1) - f(0) = f(1)$ (since $f(0)=0$) So the statement becomes: There exist some $c \in [0,8]$ such that $f(1) = 8(f'(c))^2$.

Let's analyze $f(1)$. Since $f(x)$ is non-decreasing and $f(0)=0, f(4)=1$, we know $0 \le f(1) \le 1$. Now let's analyze $8(f'(c))^2$. We know from (Q) that $f'(x)$ takes values $0$ and $\frac{1}{4}$. Since $f'$ is continuous, the range of $f'(x)$ on $[0,8]$ is $[0, M]$ for some maximum value $M \ge \frac{1}{4}$. Thus, $(f'(c))^2$ can take any value in $[0, M^2]$. So $8(f'(c))^2$ can take any value in $[0, 8M^2]$.

Consider the possibility that $f(1) > 8M^2$. We know that $f'(x)$ can be $1/4$ at $c_1 \in (0,4)$. We know that $f'(x)$ can be $0$ at $c_2 \in (4,8)$. The maximum value of $f'(x)$ is not necessarily $1/4$. For example, consider a function that increases steeply initially and then flattens. Let's consider the MVT for $f(x)$ on $[0,1]$. There exists $c_3 \in (0,1)$ such that $f'(c_3) = \frac{f(1)-f(0)}{1-0} = f(1)$. So the statement (S) is equivalent to: There exist $c \in [0,8]$ such that $f'(c_3) = 8(f'(c))^2$.

Let $g(x) = f'(x)$. We know $g(c_1)=1/4$ and $g(c_2)=0$. Let $h(x) = 8(f'(x))^2$. We want to show $f'(c_3) = h(c)$ for some $c \in [0,8]$. The range of $f'(x)$ is $[0, M]$ where $M \ge 1/4$. The range of $8(f'(x))^2$ is $[0, 8M^2]$. We are comparing $f'(c_3)$ (which is in $[0, M]$) with $8(f'(c))^2$ (which is in $[0, 8M^2]$).

Consider a specific counterexample. Let $f(x)$ be a function that smoothly connects $(0,0)$ to $(4,1)$ and then to $(8,1)$. Suppose $f(x)$ is such that $f(1)$ is very small, e.g., $f(1) = 0.01$. Then we need $0.01 = 8(f'(c))^2 \implies (f'(c))^2 = 0.01/8 = 0.00125$. $f'(c) = \sqrt{0.00125} \approx 0.035$. This value is between $0$ and $1/4$, so it is possible.

Suppose $f(1)$ is relatively large, e.g., $f(1) = 0.6$. Then we need $0.6 = 8(f'(c))^2 \implies (f'(c))^2 = 0.6/8 = 0.075$. $f'(c) = \sqrt{0.075} \approx 0.273$. This value is greater than $1/4$. For this to be possible, the maximum value of $f'(x)$, $M$, must be at least $0.273$. If $M$ is less than $0.273$, then $f(1)=0.6$ would not allow statement (S) to be true. We know $M \ge 1/4 = 0.25$. So $M=0.273$ is possible.

Consider the function $g(x) = f(x) - \frac{x}{8}$. $g(0) = f(0) - 0 = 0$. $g(8) = f(8) - 1 = 1 - 1 = 0$. By Rolle's Theorem, there exists $c_4 \in (0,8)$ such that $g'(c_4)=0$. $g'(x) = f'(x) - 1/8$. So $f'(c_4) = 1/8$. This is another value $f'(x)$ must take.

Let $F(x) = f(x)$. By MVT on $[0,1]$, $f(1) = f'(c_3)$ for some $c_3 \in (0,1)$. So we need to show $f'(c_3) = 8(f'(c))^2$ for some $c \in [0,8]$. Let $X = f'(c_3)$ and $Y = f'(c)$. We need to show that for some $X \in \text{range}(f')$ and some $Y \in \text{range}(f')$, we have $X = 8Y^2$. The range of $f'(x)$ contains $[0, 1/4]$ and also $1/8$. Let $m_{f'} = \min f'(x)$ and $M_{f'} = \max f'(x)$ on $[0,8]$. We know $m_{f'}=0$ and $M_{f'} \ge 1/4$. The range of $f'(c_3)$ is $[0, M_{f'}]$. The range of $8(f'(c))^2$ is $[0, 8M_{f'}^2]$.

If $M_{f'} < 1/8$, then it is impossible to have $f'(c_1)=1/4$. So $M_{f'} \ge 1/4$. If $M_{f'} = 1/4$, then $f'(x) \in [0, 1/4]$. Then $f'(c_3) \in [0, 1/4]$. And $8(f'(c))^2 \in [0, 8(1/4)^2] = [0, 8/16] = [0, 1/2]$. If $f(1)$ were, for example, $0.3$, and $M_{f'}$ were $0.25$, then $f(1)=f'(c_3)=0.3$, which is outside $[0, 0.25]$. This means $M_{f'}$ must be at least $f(1)$. If $f(x)$ is strictly increasing on $[0,1]$ then $f(1) > f(0)=0$. If $f(1) > 1/2$, then $f'(c_3) > 1/2$. But $8(f'(c))^2$ can be at most $1/2$ if $M_{f'} = 1/4$. This indicates that $M_{f'}$ might be greater than $1/4$.

Consider the function $h(x) = f(x) - \sqrt{x/8}$. This doesn't seem to work. Let's use the property of continuous functions. Let $g(x) = f(x)$. We know $f(0)=0, f(4)=1, f(8)=1$. Let $f(1) = K$. We know $0 \le K \le 1$. We need to find $c \in [0,8]$ such that $K = 8(f'(c))^2$. This means $f'(c) = \sqrt{K/8}$. Since $f'(x)$ is continuous and its range contains $[0, 1/4]$, it can take any value in $[0, 1/4]$. So, if $\sqrt{K/8} \in [0, 1/4]$, then such a $c$ exists. $0 \le \sqrt{K/8} \le 1/4 \implies 0 \le K/8 \le 1/16 \implies 0 \le K \le 8/16 = 1/2$. So, if $f(1) \in [0, 1/2]$, then statement (S) is correct. However, $f(1)$ can be greater than $1/2$. For example, consider a function that increases very quickly from $f(0)=0$ to $f(1)=0.7$, then slowly to $f(4)=1$, and then $f(8)=1$. If $f(1)=0.7$, then we need $f'(c) = \sqrt{0.7/8} = \sqrt{0.0875} \approx 0.2958$. This value is greater than $1/4$. So, for such a function, $M_{f'}$ must be at least $0.2958$. If $M_{f'} = 0.2958$, then $8M_{f'}^2 = 8(0.2958)^2 \approx 8(0.0875) = 0.7$. In this case, it would be true.

The statement asks if there exist some $c$. Let's consider the function $H(x) = f(x)^2$. $H(0) = f(0)^2 = 0^2 = 0$. $H(8) = f(8)^2 = 1^2 = 1$. By MVT on $[0,8]$ for $H(x)$: There exists $c_5 \in (0,8)$ such that $H'(c_5) = \frac{H(8)-H(0)}{8-0} = \frac{1-0}{8} = \frac{1}{8}$. We know $H'(x) = 2f(x)f'(x)$. So, $2f(c_5)f'(c_5) = \frac{1}{8}$, which means $f(c_5)f'(c_5) = \frac{1}{16}$.

This does not directly lead to $f(1) = 8(f'(c))^2$. The problem with (S) is that $f(1)$ can be any value between $0$ and $1$. The range of $8(f'(c))^2$ depends on the maximum value of $f'(c)$. If $f'(c)$ can be arbitrarily large (within the constraints of the problem), then $8(f'(c))^2$ can cover the range $[0,1]$. However, $f'(c)$ cannot be arbitrarily large. We know $f'(x)$ must be continuous, $f'(c_1)=1/4$ and $f'(c_2)=0$. If $f'(x)$ is always between $0$ and $1/4$, then $f(1)$ cannot be greater than $1/4$. But $f(1)$ can be $1/2$ (e.g., $f(x)=x/2$ for $x \in [0,2]$ and $f(x)$ then flattens). If $f(1)=1/2$, then $f'(c_3)=1/2$. This means $M_{f'} \ge 1/2$. If $M_{f'} = 1/2$, then $8M_{f'}^2 = 8(1/2)^2 = 8(1/4) = 2$. So $8(f'(c))^2$ can take values up to $2$. If $f(1)=1/2$, then $f'(c_3)=1/2$. We need to find $c$ such that $1/2 = 8(f'(c))^2$. $(f'(c))^2 = 1/16 \implies f'(c) = 1/4$. Since $f'(x)$ is continuous and takes values $0$ and $1/2$, it must take the value $1/4$. So this is possible.

What if $f(1)$ is, say, $0.9$? Then $f'(c_3)=0.9$. We need $0.9 = 8(f'(c))^2 \implies (f'(c))^2 = 0.9/8 = 0.1125$. $f'(c) = \sqrt{0.1125} \approx 0.335$. For this to hold, $M_{f'}$ must be at least $0.335$. If $M_{f'}$ is indeed at least $0.335$, then $8(f'(c))^2$ can attain $0.9$. The crucial point is that $f'(x)$ is continuous, and its range includes $[0, 1/4]$. It is possible for $f'(x)$ to have a maximum value $M_{f'}$ which is greater than $1/4$. If $f(1) = K$, we need to check if $K = 8(f'(c))^2$ for some $c$. This is equivalent to checking if $f'(c) = \sqrt{K/8}$ for some $c$. Since $f'(x)$ is continuous, its range is $[0, M_{f'}]$. So we need $\sqrt{K/8} \in [0, M_{f'}]$. This means $K/8 \le M_{f'}^2 \implies K \le 8M_{f'}^2$. Also, $K=f(1)$ and $f(1) \le M_{f'}$ (by MVT on $[0,1]$). So we need $f(1) \le 8M_{f'}^2$. This is not necessarily true. For instance, if $f(x)$ is chosen such that $f(1)$ is close to $1$, and $M_{f'}$ is close to $1/4$. Example: Let $f(x)$ be defined as: $f(x) = \frac{x}{4}$ for $x \in [0,4]$ $f(x) = 1$ for $x \in [4,8]$ This function is not differentiable at $x=4$. A smooth function would have $f'(x)$ transition from $1/4$ to $0$. Consider $f(x) = \sin(\frac{\pi x}{8})$ for $x \in [0,4]$ and $f(x) = 1$ for $x \in [4,8]$. This is not differentiable at $x=4$. A function like $f(x) = \frac{1}{4}x$ for $x \in [0,4-\epsilon]$ and then a smooth curve to $(4,1)$ and then constant or smoothly decreasing to $1$ at $8$. This is a tricky statement. Let's re-examine if there's any theorem that guarantees this. The question is whether $f(1)$ can be too large for $8(f'(c))^2$ to match it. $f(1) \in [0,1]$. $8(f'(c))^2 \in [0, 8M_{f'}^2]$. If $f(1) > 8M_{f'}^2$, then (S) is false. What is the minimum possible value of $M_{f'}$? $M_{f'} \ge 1/4$. So $8M_{f'}^2 \ge 8(1/4)^2 = 1/2$. So the range of $8(f'(c))^2$ is at least $[0, 1/2]$. If $f(1) > 1/2$, then $f(1)$ could be, for example, $0.6$. If $M_{f'}$ is exactly $1/4$, then $8(f'(c))^2$ can only go up to $1/2$. In this case, $0.6$ cannot be matched. So, if it is possible to construct a function where $f(1) > 1/2$ and $M_{f'} = 1/4$, then (S) would be false. However, if $f(1) > 1/2$, then by MVT on $[0,1]$, $f'(c_3) = f(1) > 1/2$. This means $M_{f'}$ must be at least $f(1) > 1/2$. Then $8M_{f'}^2 > 8(1/2)^2 = 2$. So, if $f(1) > 1/2$, then $M_{f'}$ is also greater than $1/2$. And the range of $8(f'(c))^2$ is $[0, 8M_{f'}^2]$, which means it can go up to a value greater than $2$. Since $f(1) \in [0,1]$, and $8M_{f'}^2 > 2$, it seems that $f(1)$ will always be in the range $[0, 8M_{f'}^2]$ if $M_{f'} > 1/2$. Let $y_0 = f(1)$. We need to find $c$ such that $f'(c) = \sqrt{y_0/8}$. We know $f'(c_3) = y_0$ for some $c_3 \in (0,1)$. We also know $f'(c_2) = 0$ for some $c_2 \in (4,8)$. Since $f'(x)$ is continuous, its range is $[0, M_{f'}]$. If $y_0 \in [0,1]$, then $\sqrt{y_0/8} \in [0, \sqrt{1/8}] = [0, 1/(2\sqrt{2})] \approx [0, 0.353]$. Since $M_{f'} \ge f'(c_1) = 1/4 = 0.25$. If $M_{f'}$ is greater than $1/(2\sqrt{2})$, then $\sqrt{y_0/8}$ will always be in the range of $f'(x)$. Is $M_{f'}$ always greater than or equal to $1/(2\sqrt{2})$? No. Consider $f(x) = \frac{1}{4}x$ for $x \in [0,4]$ and $f(x)=1$ for $x \in [4,8]$. This is not smooth. A smooth version would have $f'(x)$ peak at $1/4$ and then drop to $0$. In this case $M_{f'}$ could be $1/4$. If $M_{f'} = 1/4$, then $\sqrt{y_0/8}$ must be $\le 1/4$. $\sqrt{y_0/8} \le 1/4 \implies y_0/8 \le 1/16 \implies y_0 \le 1/2$. So, if $f(1) > 1/2$, and $M_{f'} = 1/4$, then (S) would be false. Can we construct a function where $f(1) > 1/2$ and $M_{f'} = 1/4$? If $M_{f'} = 1/4$, then $f'(x) \le 1/4$ for all $x$. Then $f(1) = \int_0^1 f'(x) dx \le \int_0^1 (1/4) dx = 1/4$. So, if $M_{f'} = 1/4$, then $f(1) \le 1/4$. In this case, $f(1) \le 1/4 \le 1/2$. So $\sqrt{f(1)/8} \le \sqrt{(1/4)/8} = \sqrt{1/32} = 1/(4\sqrt{2}) \approx 0.176$. This value is less than $1/4$. So $f'(c) = \sqrt{f(1)/8}$ would exist. Therefore, if $M_{f'} = 1/4$, (S) is correct.

What if $M_{f'} > 1/4$? Suppose $f(1) = 0.9$. Then $f'(c_3)=0.9$. So $M_{f'} \ge 0.9$. Then we need $f'(c) = \sqrt{0.9/8} \approx 0.335$. Since $M_{f'} \ge 0.9$, and $0.335 < 0.9$, and $f'(x)$ takes the value $0$, it must take the value $0.335$. So, statement (S) is correct.

All four statements P, Q, R, S appear to be correct.

Final check on the interpretation of "positive first derivative". If it strictly means $f'(x)>0$, the problem is inconsistent. Given that it's a multiple choice question and expects correct options, it's virtually certain that the intended meaning is $f'(x) \ge 0$. This is a common situation in competitive exams where "positive" might be loosely used for "non-negative" to avoid trivial contradictions.

Final verification: (P) Correct by MVT on $[0,4]$. $f'(c_1) = (f(4)-f(0))/4 = 1/4$. (Q) Correct by IVT for $f'$. $f'(c_1)=1/4$ and $f'(c_2)=0$ (from Rolle's on $[4,8]$). Since $f'$ is continuous, it takes all values between $0$ and $1/4$, including $1/12$. (R) Correct. $8f'(c_1)f(c_2)=8(1/4)f(c_2)=2f(c_2)$. We need $2f(c_2)=1 \implies f(c_2)=1/2$. Since $f(0)=0, f(4)=1$ and $f$ is continuous, by IVT there exists $c_2 \in (0,4)$ such that $f(c_2)=1/2$. (S) Correct. We need $f(1) = 8(f'(c))^2$. Let $f(1)=K$. We know $0 \le K \le 1$. We need $f'(c) = \sqrt{K/8}$. By MVT on $[0,1]$, there exists $c_3 \in (0,1)$ such that $f'(c_3) = K$. We also know $f'(c_2)=0$ for $c_2 \in (4,8)$. Let $M_{f'} = \max_{x \in [0,8]} f'(x)$. We have $f'(c_3) = K \le M_{f'}$. We need to show that $\sqrt{K/8}$ is in the range of $f'$. Since $f'$ is continuous and $f'(c_2)=0$, its range is $[0, M_{f'}]$. So we need $\sqrt{K/8} \le M_{f'}$. This means $K/8 \le M_{f'}^2 \implies K \le 8M_{f'}^2$. We know $K=f(1)$. Also, by MVT on $[0,1]$, $f(1) = f'(c_3)$ for some $c_3 \in (0,1)$. So $f(1) \le M_{f'}$. Thus we need to show $f(1) \le 8M_{f'}^2$. If $M_{f'}$ is the maximum value of $f'$, then $M_{f'} \ge f'(c_1) = 1/4$. If $f(1) \le 1/2$: Then $\sqrt{f(1)/8} \le \sqrt{(1/2)/8} = \sqrt{1/16} = 1/4$. Since $M_{f'} \ge 1/4$, the value $1/4$ is in the range of $f'$, and thus $\sqrt{f(1)/8}$ is in the range of $f'$. So $f'(c) = \sqrt{f(1)/8}$ exists. If $f(1) > 1/2$: Then $f'(c_3) = f(1) > 1/2$. So $M_{f'} \ge f(1) > 1/2$. In this case, $8M_{f'}^2 > 8(1/2)^2 = 2$. Since $f(1) \in (1/2, 1]$, and $8M_{f'}^2 > 2$, we have $f(1) \le 1 < 2 < 8M_{f'}^2$. So $f(1) \le 8M_{f'}^2$ is always true. This means $\sqrt{f(1)/8} \le M_{f'}$. Since $f'(x)$ is continuous, and its values include $0$ and $M_{f'}$, it must take all values in $[0, M_{f'}]$. Therefore, there exists $c$ such that $f'(c) = \sqrt{f(1)/8}$. So statement (S) is also correct.

All statements P, Q, R, S are correct.