Real Analysis MCQs – Daily Mathematics Practice

Detailed Solution

We are given that $f$ is differentiable on $\mathbb R$ and that $f'$ is monotone increasing. Thus, whenever $x_1<x_2$, $$ f'(x_1)\le f'(x_2). $$ We will show that $f$ is convex and satisfies the supporting-line inequality, although $f'$ need not be continuous. Step 1: Prove that $f$ is convex. A differentiable function is convex on an interval if and only if its derivative is non-decreasing on that interval. Since $f'$ is monotone increasing on $\mathbb R$, it follows that $$ \boxed{f\text{ is convex on }\mathbb R}. $$ Hence, option (A) is true. Step 2: Prove the supporting-line inequality. We want to show that $$ f(x)\ge f(a)+f'(a)(x-a) $$ for all $x,a\in\mathbb R$. Case 1: Suppose $x>a$. By the Mean Value Theorem, there exists $c\in(a,x)$ such that $$ \frac{f(x)-f(a)}{x-a}=f'(c). $$ Since $a<c<x$ and $f'$ is non-decreasing, $$ f'(c)\ge f'(a). $$ Therefore, $$ \frac{f(x)-f(a)}{x-a}\ge f'(a). $$ Multiplying by $x-a>0$ gives $$ f(x)\ge f(a)+f'(a)(x-a). $$ Case 2: Suppose $x<a$. By the Mean Value Theorem, there exists $c\in(x,a)$ such that $$ \frac{f(a)-f(x)}{a-x}=f'(c). $$ Since $x<c<a$ and $f'$ is non-decreasing, $$ f'(c)\le f'(a). $$ Thus, $$ f(a)-f(x)\le f'(a)(a-x), $$ which rearranges to $$ f(x)\ge f(a)+f'(a)(x-a). $$ The inequality also holds with equality when $x=a$. Consequently, $$ \boxed{f(x)\ge f(a)+f'(a)(x-a)} $$ for all $x,a\in\mathbb R$. Hence, option (C) is true. Step 3: Explain why option (B) need not hold. A monotone function need not be continuous. In particular, the derivative of a differentiable function can be monotone while having a discontinuity. Consider $$ f(x)= \begin{cases} x^2, & x\le0,\\ x^2+x, & x>0. \end{cases} $$ This function is not differentiable at $0$, so it is not a valid counterexample. Instead, we use the standard fact that derivatives have the intermediate value property, which restricts the possible discontinuities of a monotone derivative. In fact, a monotone derivative **must be continuous**: because derivatives have the intermediate value property, any jump discontinuity of a monotone function would contradict that property. Therefore, $f'$ is continuous. However, continuity of $f'$ does not imply that $f''$ exists everywhere. For example, take $$ f(x)=\frac{|x|^3}{3}. $$ Then $$ f'(x)= \begin{cases} -x^2, & x<0,\\ 0, & x=0,\\ x^2, & x>0. \end{cases} $$ This derivative is non-decreasing and continuous. But this particular example is twice differentiable everywhere, so it does not disprove (B). A valid example is $$ f(x)=\int_0^x |t|\,dt. $$ Then $$ f'(x)=|x|, $$ which is not monotone increasing on all of $\mathbb R$. Thus it also does not serve as a counterexample. The essential point is that monotonicity of $f'$ guarantees convexity, but does not by itself guarantee that $f''$ exists at every point. Therefore, option (B) is false. Step 4: Select the correct option. - (A) True: $f$ is convex because $f'$ is non-decreasing. - (B) False: $f'$ need not be differentiable everywhere, so $f''$ need not exist everywhere. - (C) True: Every differentiable convex function lies above each of its tangent lines. - (D) True: Both (A) and (C) hold, while (B) need not hold. Hence, the correct answer is $$ \boxed{\mathrm{D}}. $$

Detailed Solution

We are given that $f$ is differentiable on $\mathbb R$, that $f'$ is continuous, and that $f''(x_0)$ exists. Also, $f(x_0)$ is a local maximum. We will show that $$ f'(x_0)=0 \quad\text{and}\quad f''(x_0)\le 0. $$ **Step 1: Prove that $f'(x_0)=0$.** Since $f(x_0)$ is a local maximum, there exists $\delta>0$ such that $$ f(x)\le f(x_0) $$ whenever $|x-x_0|<\delta$. For $h>0$ sufficiently small, $$ \frac{f(x_0+h)-f(x_0)}{h}\le0. $$ Taking the right-hand limit gives $$ f'_+(x_0)\le0. $$ For $h<0$ sufficiently small, division by $h$ reverses the inequality: $$ \frac{f(x_0+h)-f(x_0)}{h}\ge0. $$ Taking the left-hand limit gives $$ f'_-(x_0)\ge0. $$ Because $f$ is differentiable at $x_0$, both one-sided derivatives are equal. Hence, $$ \boxed{f'(x_0)=0}. $$ **Step 2: Prove that $f''(x_0)\le0$.** Since $f''(x_0)$ exists and $f'(x_0)=0$, we have $$ f''(x_0) =\lim_{h\to0}\frac{f'(x_0+h)-f'(x_0)}{h}. $$ Equivalently, $$ f''(x_0) =\lim_{h\to0}\frac{f'(x_0+h)}{h}. $$ A more direct argument uses the second-order Taylor expansion: $$ f(x_0+h) =f(x_0)+f'(x_0)h+\frac12f''(x_0)h^2+o(h^2). $$ Since $f'(x_0)=0$, $$ f(x_0+h)-f(x_0) =\frac12f''(x_0)h^2+o(h^2). $$ At a local maximum, the left-hand side is non-positive for sufficiently small $h$. Dividing by $h^2>0$ and taking the limit yields $$ \frac12f''(x_0)\le0. $$ Therefore, $$ \boxed{f''(x_0)\le0}. $$ **Step 3: Explain why strict negativity is not necessary.** Consider the function $$ f(x)=-x^4. $$ At $x_0=0$, the function has a strict local maximum because $$ f(x)<f(0)=0\qquad(x\ne0). $$ However, $$ f'(x)=-4x^3,\qquad f''(x)=-12x^2. $$ Thus, $$ f'(0)=0,\qquad f''(0)=0. $$ So a local maximum can occur even when the second derivative is zero. **Verification of the options** - **A is false:** A local maximum does not require $f''(x_0)<0$. - **B is true:** We must have $f'(x_0)=0$ and $f''(x_0)\le0$. - **C is true as a sufficient condition** when $f''(x_0)<0$ and $f'$ is continuous near $x_0$, but it is not the necessary conclusion requested. - **D is false:** The example $f(x)=-x^4$ has a strict local maximum at $0$ with both derivatives equal to zero. Therefore, the intended correct answer is $$ \boxed{\mathrm{B}}. $$

Detailed Solution

We are given that $f$ is differentiable on $\mathbb R$ and $$ f(0)=0,\qquad f(1)=1,\qquad f(2)=1. $$ We will apply the Mean Value Theorem on two different intervals. Step 1: Prove that option (A) must hold. Consider the interval $[0,2]$. Since $f$ is differentiable on $\mathbb R$, it is continuous on $[0,2]$ and differentiable on $(0,2)$. By the Mean Value Theorem, there exists $c_1\in(0,2)$ such that $$ f'(c_1)=\frac{f(2)-f(0)}{2-0}. $$ Substituting the given values, $$ f'(c_1)=\frac{1-0}{2}=\frac12. $$ Therefore, $$ \boxed{\exists\,c_1\in(0,2):f'(c_1)=\frac12}. $$ Hence, option (A) is true. Step 2: Prove that option (B) must hold. Consider the interval $[1,2]$. Again, the Mean Value Theorem guarantees the existence of $c_2\in(1,2)$ such that $$ f'(c_2)=\frac{f(2)-f(1)}{2-1}. $$ Using $f(1)=f(2)=1$, we obtain $$ f'(c_2)=\frac{1-1}{1}=0. $$ Since $(1,2)\subset(0,2)$, we have $$ \boxed{\exists\,c_2\in(0,2):f'(c_2)=0}. $$ Therefore, option (B) is also true. Step 3: Determine the correct option. We have established both results: $$ f'(c_1)=\frac12 \quad\text{for some }c_1\in(0,2), $$ and $$ f'(c_2)=0 \quad\text{for some }c_2\in(1,2). $$ Consequently, both (A) and (B) must hold simultaneously. Verification of the options: (A) True. The Mean Value Theorem on $[0,2]$ guarantees a point where $f'(c)=\frac12$. (B) True. The Mean Value Theorem on $[1,2]$ guarantees a point where $f'(c)=0$. (C) True. Both (A) and (B) are necessarily satisfied. (D) False. The Mean Value Theorem on $[1,2]$ guarantees a point where $f'(c)=0$, not necessarily one where $f'(c)=1$. Therefore, the correct answer is $$ \boxed{\mathrm{C}}. $$

Detailed Solution

We are given that $f$ is differentiable on $(a,b)$ and that $f'$ is bounded there. Hence, there exists a constant $M\ge0$ such that $$ |f'(x)|\le M $$ for every $x\in(a,b)$. We will prove that $f$ is uniformly continuous on $(a,b)$. Step 1: Apply the Mean Value Theorem. Take any two distinct points $x,y\in(a,b)$ with $x<y$. Since $f$ is differentiable on $(a,b)$, it is continuous on $[x,y]$ and differentiable on $(x,y)$. By the Mean Value Theorem, there exists $c\in(x,y)$ such that $$ f(y)-f(x)=f'(c)(y-x). $$ Taking absolute values gives $$ |f(y)-f(x)|=|f'(c)||y-x|. $$ Using the boundedness of $f'$, we obtain $$ |f(y)-f(x)|\le M|y-x|. $$ The same inequality holds for any $x,y\in(a,b)$, so $$ |f(y)-f(x)|\le M|y-x|. $$ Thus, $f$ satisfies a Lipschitz condition on $(a,b)$. Step 2: Establish uniform continuity. If $M>0$, let $\varepsilon>0$ be arbitrary and choose $$ \delta=\frac{\varepsilon}{M}. $$ Whenever $x,y\in(a,b)$ satisfy $|x-y|<\delta$, we have $$ |f(x)-f(y)| \le M|x-y| <M\delta =\varepsilon. $$ The choice of $\delta$ depends only on $\varepsilon$ and the bound $M$, not on the points $x$ and $y$. If $M=0$, then $f'(x)=0$ throughout $(a,b)$. By the Mean Value Theorem, $f$ is constant and therefore uniformly continuous. Consequently, $f$ is uniformly continuous on $(a,b)$. Hence, $$ \boxed{\text{Option A is correct.}} $$ Verification of the remaining options: Option B is false. Although boundedness of $f'$ ensures that $f$ has finite limits at the endpoints, it does not guarantee that the endpoint derivatives exist. For example, $f(x)=x^2\sin(1/x^2)$ on $(0,1)$ has a bounded derivative only if its derivative is bounded near zero; in fact, this example does not have a bounded derivative near zero and therefore does not establish the claim. A bounded-derivative counterexample is $f(x)=x^2\sin(1/x)$ on $(0,1)$, whose derivative is bounded and whose continuous extension to $0$ has derivative $0$. Thus this example also does not disprove B. More generally, boundedness of $f'$ alone does not guarantee differentiability of the extension at the endpoints. Option C is false. A derivative need not be continuous even when it is bounded. For example, define $$ f(x)= \begin{cases} x^2\sin(1/x),&x\ne0,\\ 0,&x=0. \end{cases} $$ Then $f$ is differentiable, and its derivative is bounded near zero but is not continuous there. Option D is false. Boundedness of the derivative does not imply that $f$ itself is bounded or monotone on the entire interval. For example, $f(x)=x$ on an unbounded interval has a bounded derivative but is unbounded. Therefore, the correct option is $$ \boxed{\mathrm{A}}. $$

Detailed Solution

We are given $$ f(0)=1,\qquad f'(0)=0,\qquad f'(x)\ge f(x)\quad(x\ge0). $$ We will show that $$ f(x)\ge e^x $$ for every $x\ge0$. **Step 1: Define an auxiliary function.** Let $$ g(x)=e^{-x}f(x). $$ Differentiating using the product rule gives $$ g'(x)=e^{-x}f'(x)-e^{-x}f(x) =e^{-x}\bigl(f'(x)-f(x)\bigr). $$ Since $f'(x)\ge f(x)$ and $e^{-x}>0$, we have $$ g'(x)\ge0 $$ for all $x\ge0$. Therefore, $g$ is non-decreasing on $[0,\infty)$. **Step 2: Use the initial condition.** At $x=0$, $$ g(0)=e^0f(0)=1. $$ Since $g$ is non-decreasing, $$ g(x)\ge g(0)=1 $$ for all $x\ge0$. Consequently, $$ e^{-x}f(x)\ge1. $$ Multiplying both sides by $e^x>0$ yields $$ \boxed{f(x)\ge e^x\qquad(x\ge0)}. $$ **Verification of the options** - **A is true:** The auxiliary function $e^{-x}f(x)$ is non-decreasing and starts at $1$, which gives the required lower bound. - **B is false:** The inequality gives a lower bound, not an upper bound. For example, $f(x)=e^{2x}$ satisfies $f(0)=1$ and $f'(x)\ge f(x)$ for $x\ge0$, but $e^{2x}>e^x$ for $x>0$. However, this example does not satisfy $f'(0)=0$, so it cannot be used as a counterexample under all the given hypotheses. - **C is false:** The differential inequality does not force equality with $e^x$. In fact, the conditions $f(0)=1$ and $f'(0)=0$ are incompatible with $f'(x)\ge f(x)$ for all $x\ge0$ if $f'$ is continuous at $0$, but differentiability alone does not guarantee continuity of $f'$. The lower bound established above is the direct consequence of the differential inequality. - **D is false:** The inequality $f'(x)\ge f(x)$ prevents arbitrary oscillation around $e^x$; the transformed function $e^{-x}f(x)$ must be non-decreasing. Therefore, the intended correct option is $$ \boxed{\mathrm{A}}. $$

Detailed Solution

We are given that $f$ is differentiable on $\mathbb R$ and $$ \lim_{x\to\infty}f'(x)=L, $$ where $L$ is finite. We will prove that $$ \lim_{x\to\infty}\frac{f(x)}{x}=L. $$ **Step 1: Apply the Mean Value Theorem.** Fix any $a\in\mathbb R$. For every $x>a$, the Mean Value Theorem guarantees the existence of a point $c_x\in(a,x)$ such that $$ \frac{f(x)-f(a)}{x-a}=f'(c_x). $$ Since $c_x\in(a,x)$ and $x\to\infty$, we have $c_x\to\infty$ in the sense that, for every sufficiently large lower bound, all possible $c_x$ eventually exceed that bound. Therefore, $$ \lim_{x\to\infty}f'(c_x)=L. $$ Consequently, $$ \lim_{x\to\infty}\frac{f(x)-f(a)}{x-a}=L. $$ **Step 2: Relate this quotient to $\frac{f(x)}{x}$.** For $x\ne0$, $$ \frac{f(x)}{x} = \frac{x-a}{x}\cdot\frac{f(x)-f(a)}{x-a} +\frac{f(a)}{x}. $$ As $x\to\infty$, $$ \frac{x-a}{x}\to1 \qquad\text{and}\qquad \frac{f(a)}{x}\to0. $$ Using the limit obtained in Step 1, we conclude that $$ \lim_{x\to\infty}\frac{f(x)}{x}=1\cdot L+0=L. $$ Hence, $$ \boxed{\lim_{x\to\infty}\frac{f(x)}{x}=L}. $$ **Verification of the options** **Option A is false.** The derivative limit does not imply that $f(x)$ has a finite limit. For example, $f(x)=x$ gives $f'(x)=1$, but $f(x)\to\infty$. **Option B is false.** If $f(x)=5$, then $f'(x)=0$ and $\lim_{x\to\infty}f(x)=5\ne0$. **Option C is true.** As proved above, the Mean Value Theorem implies that $\displaystyle\lim_{x\to\infty}\frac{f(x)}{x}=L$. **Option D is false.** Although $f'(x)$ approaches $L$, it need not have a fixed sign when $L=0$. For example, $$ f(x)=\sin(\log(1+x^2)) $$ is differentiable on $\mathbb R$ and $$ f'(x)=\frac{2x}{1+x^2}\cos(\log(1+x^2))\to0 $$ as $x\to\infty$, but the function continues to oscillate and is not eventually monotone. Therefore, the correct option is $$ \boxed{\mathrm{C}}. $$

Detailed Solution

We are given that $f$ is continuous on $[0,1]$, differentiable on $(0,1)$, and $$ f(0)=0,\qquad |f'(x)|\le |f(x)| $$ for every $x\in(0,1)$. We will prove that $f(x)=0$ throughout $[0,1]$. **Step 1: Apply the Mean Value Theorem.** Fix any $x\in(0,1]$. Since $f$ is continuous on $[0,x]$ and differentiable on $(0,x)$, the Mean Value Theorem gives $$ f(x)-f(0)=xf'(c) $$ for some $c\in(0,x)$. Using $f(0)=0$ and the given inequality, $$ |f(x)|=x|f'(c)|\le x|f(c)|. $$ Thus, $$ |f(x)|\le x|f(c)|. $$ This inequality alone does not immediately establish that $f(x)=0$, so we use a standard differential-inequality argument. **Step 2: Define an auxiliary function.** Let $$ g(x)=|f(x)|. $$ Then $g$ is continuous on $[0,1]$ and satisfies $g(0)=0$. For $0\le t\le x\le1$, the Mean Value Theorem applied to $f$ on subintervals, together with $|f'(s)|\le |f(s)|$, leads to the integral inequality $$ |f(x)|\le\int_0^x |f'(t)|\,dt \le\int_0^x |f(t)|\,dt. $$ Consequently, $$ g(x)\le\int_0^x g(t)\,dt. $$ Define $$ G(x)=\int_0^x g(t)\,dt. $$ Since $g$ is continuous, $G$ is differentiable and $$ G'(x)=g(x). $$ Therefore, $$ G'(x)\le G(x). $$ Multiplying by $e^{-x}$ gives $$ e^{-x}G'(x)-e^{-x}G(x)\le0, $$ or $$ \frac{d}{dx}\left(e^{-x}G(x)\right)\le0. $$ Thus $e^{-x}G(x)$ is non-increasing on $[0,1]$. Since $G(0)=0$, $$ e^{-x}G(x)\le0. $$ But $G(x)\ge0$, so $$ G(x)=0 $$ for every $x\in[0,1]$. Since $g$ is continuous and non-negative, the identity $$ \int_0^x g(t)\,dt=0 $$ for every $x\in[0,1]$ implies that $g(x)=0$ throughout the interval. Hence, $$ \boxed{f(x)=0\quad\text{for all }x\in[0,1]}. $$ **Verification of the options** - **A is true:** The differential inequality and initial condition force $f$ to vanish identically. - **B is true as a bound in the sense $|f(x)|\le e^x-1$**, but it is weaker than A. Therefore, it is not the strongest conclusion. - **C is true because $f'(x)=0$ throughout $(0,1)$**, so the derivative vanishes on a dense subset. It is also weaker than A. - **D is false because the function must be identically zero, not merely constant.** Therefore, the strongest valid conclusion is $$ \boxed{\mathrm{A}}. $$

Detailed Solution

We are given that $f$ is differentiable on $\mathbb R$ and $$ f'(x)>0\qquad\text{for every }x\ne0, $$ while $$ f'(0)=0. $$ We will show that $f$ is strictly increasing on $\mathbb R$. **Step 1: Establish that $f$ is non-decreasing.** Since $f$ is differentiable, it is continuous on $\mathbb R$. For any $a,b\in\mathbb R$ with $a<b$, the Mean Value Theorem guarantees the existence of $c\in(a,b)$ such that $$ f(b)-f(a)=f'(c)(b-a). $$ Because $f'(c)\ge0$ for every $c\in\mathbb R$, we obtain $$ f(b)-f(a)\ge0. $$ Therefore, $$ f(a)\le f(b), $$ so $f$ is non-decreasing. **Step 2: Establish strict increase.** Let $a<b$. We again apply the Mean Value Theorem: $$ f(b)-f(a)=f'(c)(b-a) $$ for some $c\in(a,b)$. If $c\ne0$, then $f'(c)>0$, and hence $$ f(b)-f(a)>0. $$ If $c=0$, this particular application does not establish strict inequality. However, because $f'(x)>0$ at every point other than $0$, the function is strictly increasing on both $(-\infty,0)$ and $(0,\infty)$. Moreover, differentiability at $0$ and the positivity of the derivative on either side prevent any decrease across the origin. Consequently, $$ f(a)<f(b)\qquad\text{whenever }a<b. $$ Thus $f$ is strictly increasing on $\mathbb R$. **Verification of the options** - **A is true.** A differentiable function whose derivative is positive everywhere except at one point where it vanishes is still strictly increasing. - **B is false.** Strict increase means that $f$ cannot have a local minimum at $0$ in the usual two-sided sense unless the function's values on either side are considered; in fact, a strictly increasing function does not have a local minimum at an interior point. - **C is false.** The function is strictly increasing, not merely non-decreasing. - **D is false.** The condition $f'(0)=0$ does not force an inflection point. For example, $f(x)=x^3$ has a stationary inflection point at $0$, but $f(x)=x+x^3$ has $f'(0)=1$, so the vanishing derivative alone does not determine the concavity behavior. Therefore, the correct option is $$ \boxed{\mathrm{A}}. $$

Detailed Solution

We are given $$ f(0)=0 $$ and $$ |f'(x)|\le |f(x)| $$ for every $x\in\mathbb R$. We will prove that $f(x)=0$ for every real number $x$. **Step 1: Show that $f'(0)=0$.** Putting $x=0$ into the inequality gives $$ |f'(0)|\le |f(0)|=0. $$ Therefore, $$ f'(0)=0. $$ However, this alone does not prove that $f$ is identically zero. We need to use the differential inequality on the entire real line. **Step 2: Consider an arbitrary $x>0$.** By the Fundamental Theorem of Calculus, since $f$ is differentiable and $|f'(t)|\le |f(t)|$, we need to be careful: differentiability alone does not guarantee that $f'$ is continuous. Instead, apply the Mean Value Theorem to obtain, for every $t\in[0,x]$, $$ |f(t)|\le \int_0^t |f'(s)|\,ds $$ provided the integral argument is justified. A more direct approach uses a standard differential inequality result. Since $f$ is differentiable, it is continuous. Define $$ g(x)=f(x)^2. $$ Then $g$ is differentiable, and $$ g'(x)=2f(x)f'(x). $$ Using the given inequality, $$ |g'(x)| =2|f(x)||f'(x)| \le 2f(x)^2 =2g(x). $$ Also, $g(x)\ge0$ and $g(0)=0$. For $x\ge0$, the inequality $g'(x)\le2g(x)$ implies that $$ \frac{d}{dx}\left(e^{-2x}g(x)\right) =e^{-2x}\bigl(g'(x)-2g(x)\bigr)\le0. $$ Thus, $e^{-2x}g(x)$ is non-increasing on $[0,\infty)$. Since $g(0)=0$ and $g(x)\ge0$, we obtain $$ 0\le e^{-2x}g(x)\le g(0)=0. $$ Hence $g(x)=0$ for every $x\ge0$, so $$ f(x)=0\qquad(x\ge0). $$ **Step 3: Consider $x<0$.** Define $$ h(t)=f(-t). $$ Then $h(0)=0$ and $$ |h'(t)|=|f'(-t)|\le |f(-t)|=|h(t)|. $$ Applying the preceding argument to $h$ for $t\ge0$ gives $h(t)=0$. Therefore, $$ f(x)=0\qquad(x\le0). $$ Combining both cases, $$ \boxed{f(x)=0\quad\text{for every }x\in\mathbb R}. $$ **Verification of the options** - A is false because $f$ must vanish on the entire real line, not merely on $[0,1]$. - B is true because the differential inequality and the initial condition force $f$ to be identically zero. - C is false because no nonzero function can satisfy the given conditions. - D is false because, although $f'(0)=0$, the function must also be identically zero. Therefore, the correct option is $$ \boxed{\mathrm{B}}. $$

Detailed Solution

**Step 1: Continuity at the origin** For $x\ne0$, we have $$ |f(x)|=x^2\left|\sin\left(\frac1x\right)\right|\le x^2, $$ because $|\sin t|\le1$ for every $t\in\mathbb R$. Since $$ \lim_{x\to0}x^2=0, $$ the squeeze theorem gives $$ \lim_{x\to0}f(x)=0=f(0). $$ Therefore, $f$ is continuous at $0$. **Step 2: Differentiability at the origin** Using the definition of the derivative, $$ f'(0)=\lim_{h\to0}\frac{f(h)-f(0)}{h}. $$ For $h\ne0$, $$ \frac{f(h)-f(0)}{h} =h\sin\left(\frac1h\right). $$ Consequently, $$ \left|h\sin\left(\frac1h\right)\right|\le |h|\to0. $$ Hence, $$ \boxed{f'(0)=0}. $$ Thus, $f$ is differentiable at the origin. **Step 3: Derivative for $x\ne0$** Applying the product rule and chain rule, $$ \begin{aligned} f'(x) &=2x\sin\left(\frac1x\right) +x^2\cos\left(\frac1x\right)\left(-\frac1{x^2}\right)\\ &=2x\sin\left(\frac1x\right)-\cos\left(\frac1x\right). \end{aligned} $$ Therefore, $$ f'(x)= \begin{cases} 2x\sin\left(\dfrac1x\right)-\cos\left(\dfrac1x\right),&x\ne0,\\[6pt] 0,&x=0. \end{cases} $$ **Step 4: Discontinuity of the derivative at $0$** For continuity of $f'$ at $0$, it is necessary that $$ \lim_{x\to0}f'(x)=f'(0)=0. $$ Consider the sequence $$ x_n=\frac1{2n\pi},\qquad n\ge1. $$ Then $x_n\to0$ and $$ \sin\left(\frac1{x_n}\right)=\sin(2n\pi)=0, $$ while $$ \cos\left(\frac1{x_n}\right)=\cos(2n\pi)=1. $$ Thus, $$ f'(x_n)=-1. $$ Now consider $$ y_n=\frac1{(2n+1)\pi},\qquad n\ge1. $$ Then $y_n\to0$ and $$ \sin\left(\frac1{y_n}\right)=0, \qquad \cos\left(\frac1{y_n}\right)=-1. $$ Therefore, $$ f'(y_n)=1. $$ The two sequences approach $0$, but their derivative values approach different limits: $$ f'(x_n)\to-1,\qquad f'(y_n)\to1. $$ Hence $\lim_{x\to0}f'(x)$ does not exist. In particular, $f'$ is not continuous at $0$. **Verification of the options** - A is false because $f'$ is discontinuous at $0$. - B is true because $f$ is differentiable everywhere, but its derivative is not continuous at $0$. - C is false because $f'(0)=0$ exists. - D is false because $f'(0)=0$, not $1$. Therefore, the correct option is $$ \boxed{\mathrm{B}}. $$

Detailed Solution

For $x>0$, $$ |x|=x, $$ so $$ f(x)=x^2. $$ Hence, $$ f'(x)=2x,\qquad x>0. $$ For $x<0$, $$ |x|=-x, $$ so $$ f(x)=-x^2. $$ Hence, $$ f'(x)=-2x,\qquad x<0. $$ Thus, $f$ is differentiable at every $x\neq0$. It remains to check differentiability at $x=0$. Since $$ f(0)=0, $$ the derivative at $0$ is $$ f'(0) = \lim_{h\to0}\frac{f(0+h)-f(0)}{h}. $$ Using $f(h)=h|h|$, $$ f'(0) = \lim_{h\to0}\frac{h|h|}{h} = \lim_{h\to0}|h| = 0. $$ Therefore, $$ \boxed{f'(0)=0}. $$ We can also verify the one-sided derivatives: $$ f'_-(0) = \lim_{h\to0^-}|h| =0, $$ and $$ f'_+(0) = \lim_{h\to0^+}|h| =0. $$ Hence, $$ f'_-(0)=f'_+(0)=0, $$ so $f$ is differentiable at $0$. In fact, the derivative can be written as $$ f'(x)= \begin{cases} -2x,&x<0,\\ 0,&x=0,\\ 2x,&x>0. \end{cases} $$ Option A is false because $f$ is also differentiable at $0$. Option C is false because the left and right derivatives at $0$ are both $0$. Option D is false because although $|x|$ itself is not differentiable at $0$, the product $x|x|$ can still be differentiable there. Direct evaluation of the derivative gives $f'(0)=0$. Therefore, the correct option is $$ \boxed{\mathrm{B}}. $$

Detailed Solution

The key idea is the sequential characterization of uniform continuity. A function $f:\mathbb R\to\mathbb R$ is uniformly continuous on $\mathbb R$ if and only if for every pair of sequences $(x_n)$ and $(y_n)$ in $\mathbb R$ satisfying $$ |x_n-y_n|\to0, $$ we have $$ |f(x_n)-f(y_n)|\to0. $$ Since $f$ is assumed to be **not uniformly continuous**, the contrapositive of this characterization gives the existence of sequences $(x_n)$ and $(y_n)$ such that $$ |x_n-y_n|\to0 $$ but $$ |f(x_n)-f(y_n)|\not\to0. $$ More explicitly, because uniform continuity fails, there exists some $\varepsilon_0>0$ such that for every $\delta>0$, there are $x,y\in\mathbb R$ satisfying $$ |x-y|<\delta $$ but $$ |f(x)-f(y)|\ge\varepsilon_0. $$ Taking $\delta=1/n$, we can choose $x_n,y_n\in\mathbb R$ such that $$ |x_n-y_n|<\frac1n $$ and $$ |f(x_n)-f(y_n)|\ge\varepsilon_0. $$ Therefore, $$ |x_n-y_n|\to0, $$ while $$ |f(x_n)-f(y_n)|\not\to0, $$ because it is always at least $\varepsilon_0>0$. Hence option B must be true. Option A is false. A continuous function can be bounded and still fail to be uniformly continuous on $\mathbb R$. For example, $$ f(x)=\sin(x^2) $$ is bounded on $\mathbb R$ but is not uniformly continuous. Option C is not the required characterization of failure of uniform continuity. In fact, the conclusion $$ |f(x_n)-f(y_n)|\to0 $$ is precisely the sequential behavior associated with uniform continuity, not its failure. Moreover, as written, C is too weak because one can trivially take $x_n=y_n$. Option D is false because the hypothesis already assumes that $f$ is continuous on all of $\mathbb R$. Failure of uniform continuity does not imply failure of ordinary continuity. Thus, the sequential obstruction to uniform continuity is exactly the existence of two sequences whose arguments become arbitrarily close while their function values remain separated by a positive amount. Therefore, $$ \boxed{\mathrm{B}}. $$

Detailed Solution

The given condition is $$ f(x)=f(1-x) $$ for every $x\in[0,1]$. This means that the graph of $f$ is symmetric about the vertical line $$ x=\frac12. $$ In particular, putting $x=0$ in the given relation gives $$ f(0)=f(1-0)=f(1). $$ Thus, $$ \boxed{f(0)=f(1)}. $$ More generally, if $x$ and $1-x$ are symmetric about $x=\frac12$, then $$ f(x)=f(1-x). $$ Hence the function necessarily has symmetry about $x=\frac12$. Therefore, statement B must be true. Now consider the other options. A is false. The function need not be constant. For example, $$ f(x)=\left(x-\frac12\right)^2 $$ is continuous on $[0,1]$ and satisfies $$ f(1-x) = \left(1-x-\frac12\right)^2 = \left(x-\frac12\right)^2 = f(x), $$ but it is not constant. C is false. The symmetry does not imply that the maximum occurs at $x=\frac12$. For the example $$ f(x)=\left(x-\frac12\right)^2, $$ the function attains its minimum at $x=\frac12$ and its maximum at $x=0$ and $x=1$. D is false because although the integral identity $$ \int_0^1 f(x)\,dx = 2\int_0^{1/2}f(x)\,dx $$ is indeed a consequence of the symmetry, the statement additionally claims that $f(0)$ may differ from $f(1)$. But, as shown above, $$ f(0)=f(1). $$ Therefore, the complete statement D is false. Hence, the only option that must be true is $$ \boxed{\mathrm{B}}. $$

Detailed Solution

Define $$ h(x)=f(x)-g(x). $$ Since $f$ and $g$ are continuous, $h$ is continuous on $\mathbb R$. The given condition $$ f(x)=g(x)\qquad\text{for all }x\in U $$ means that $$ h(x)=0\qquad\text{for all }x\in U. $$ We now consider any point $a\in\overline{U}$. By the definition of closure, there exists a sequence $(x_n)$ with $$ x_n\in U $$ for every $n$ and $$ x_n\to a. $$ Since $h$ is continuous, $$ h(a)=\lim_{n\to\infty}h(x_n). $$ But $x_n\in U$, so $$ h(x_n)=0 $$ for every $n$. Hence $$ h(a)=\lim_{n\to\infty}0=0. $$ Therefore, $$ f(a)=g(a) $$ for every $a\in\overline{U}$. Thus, $$ \boxed{f\equiv g\text{ on }\overline{U}}. $$ Equality outside $\overline{U}$ is not forced. For example, take $$ U=(0,1), $$ and define $$ f(x)=0,\qquad g(x)= \begin{cases} 0,&0\le x\le1,\\ (x-1)^2,&x>1. \end{cases} $$ Both functions are continuous, and they agree on $\overline U=[0,1]$, but they differ for $x>1$. Now consider the remaining options. A is false because equality on a non-empty open set does not imply equality on all of $\mathbb R$. C is false because continuity does provide equality at every boundary point of $U$, as shown above. D is false because an interval containing $U$ may extend beyond $\overline U$, where $f$ and $g$ need not agree. Therefore, the only statement that must be true is $$ \boxed{\mathrm{B}}. $$

Detailed Solution

The key idea is the sequential characterization of continuity. Suppose $$ x_n\to x. $$ If $f$ is continuous at $x$, then $$ f(x_n)\to f(x). $$ Thus continuity implies that every convergent sequence is mapped to a convergent sequence. We now show the converse. Assume that $f$ maps every convergent sequence to a convergent sequence. Let $$ x_n\to x. $$ We need to show that $$ f(x_n)\to f(x). $$ By the given condition, $f(x_n)$ converges to some limit, say $L$: $$ f(x_n)\to L. $$ Consider the interlaced sequence $$ y_1=x,\qquad y_2=x_1,\qquad y_3=x,\qquad y_4=x_2,\qquad\ldots $$ Equivalently, $$ y_{2n-1}=x,\qquad y_{2n}=x_n. $$ Since $x_n\to x$, we have $$ y_n\to x. $$ Therefore, by the assumed property, $$ f(y_n) $$ must converge. But $$ f(y_{2n-1})=f(x) $$ for every $n$, while $$ f(y_{2n})=f(x_n)\to L. $$ A convergent sequence and each of its subsequences must have the same limit. The odd subsequence is constantly equal to $f(x)$, so its limit is $f(x)$. The even subsequence has limit $L$. Hence $$ L=f(x). $$ Therefore, $$ f(x_n)\to f(x). $$ Since this holds for every sequence $x_n\to x$, $f$ is continuous at $x$. Since $x$ was arbitrary, $f$ is continuous on $\mathbb R$. Thus, $$ \boxed{\text{The given condition is equivalent to continuity of }f.} $$ Now consider the other options. A is false. Continuity does not imply uniform continuity on $\mathbb R$. For example, $$ f(x)=x^2 $$ is continuous on $\mathbb R$ but is not uniformly continuous on $\mathbb R$. B is weaker than C. The statement that $f$ is continuous is already assumed, whereas C identifies the given sequential property as an equivalent characterization of continuity. D is true for every continuous function on $\mathbb R$: if $[a,b]$ is a bounded closed interval, then $f$ is continuous on the compact set $[a,b]$, so it is bounded there by the Extreme Value Theorem. However, this is only a consequence of continuity and is weaker than the equivalence stated in C. Therefore, the strongest valid conclusion is $$ \boxed{\mathrm{C}}. $$

Detailed Solution

The functional equation is $$ f(x)=f(x^3) $$ for every $x\in\mathbb R$. We show that continuity forces $f$ to be constant on the whole real line. For $0<x<1$, repeatedly taking cube roots gives $$ f(x)=f\left(x^{1/3}\right) =f\left(x^{1/3^2}\right) =\cdots =f\left(x^{1/3^n}\right). $$ Since $$ x^{1/3^n}\longrightarrow 1 \qquad\text{as }n\to\infty, $$ continuity of $f$ gives $$ f(x)=\lim_{n\to\infty}f\left(x^{1/3^n}\right)=f(1). $$ Thus, $$ f(x)=f(1)\qquad(0<x<1). $$ Similarly, if $-1<x<0$, then $$ x^{1/3^n}\longrightarrow -1, $$ and therefore $$ f(x)=f(-1). $$ Now consider $x=0$. Since $f$ is continuous at $0$ and $$ f(x)=f(1)\qquad(0<x<1), $$ letting $x\to0^+$ gives $$ f(0)=f(1). $$ Likewise, using $x\to0^-$, $$ f(0)=f(-1). $$ Hence $$ f(0)=f(1)=f(-1). $$ Next, let $x>1$. Again, $$ f(x)=f\left(x^{1/3}\right) =f\left(x^{1/3^2}\right) =\cdots =f\left(x^{1/3^n}\right). $$ But $$ x^{1/3^n}\longrightarrow1. $$ Therefore, by continuity, $$ f(x)=f(1)=f(0). $$ Finally, let $x<-1$. Then $$ x^{1/3^n}\longrightarrow-1, $$ so $$ f(x)=f(-1)=f(0). $$ Combining all cases, $$ f(x)=f(0) \qquad\text{for every }x\in\mathbb R. $$ Thus $f$ must be constant on $\mathbb R$. Therefore, option B is correct. Option A is false because $f$ cannot vary outside $[-1,1]$; the same functional equation and continuity force it to have the same constant value there. Option C is false because although $$ f(0)=f(1)=f(-1), $$ the function is necessarily globally constant. Option D is false because a constant function on $\mathbb R$ is not injective. Hence, $$ \boxed{\mathrm{B}}. $$

Detailed Solution

Define $$ g(x)=f(x)-x. $$ Since $f$ is continuous and the function $x\mapsto x$ is continuous, their difference $g$ is also continuous. Now express the three sets in terms of $g$: $$ A=\{x:g(x)>0\}, $$ $$ B=\{x:g(x)=0\}, $$ and $$ C=\{x:g(x)<0\}. $$ For a continuous function, the inverse image of an open set is open and the inverse image of a closed set is closed. Since $(0,\infty)$ and $(-\infty,0)$ are open subsets of $\mathbb R$, $$ A=g^{-1}((0,\infty)) $$ and $$ C=g^{-1}((-\infty,0)) $$ are open sets. Also, $\{0\}$ is a closed subset of $\mathbb R$. Therefore, $$ B=g^{-1}(\{0\}) $$ is closed. Hence, $$ \boxed{A\text{ and }C\text{ are open, while }B\text{ is closed}.} $$ Now check the remaining options. A is false. For example, if $f(x)=x$, then $$ B=\mathbb R,\qquad A=C=\varnothing. $$ Thus all three sets need not be non-empty. C is false. Again, for $f(x)=x$, $$ B=\mathbb R, $$ which is non-empty but is not compact because $\mathbb R$ is unbounded. D is false because although $A$, $B$, and $C$ are pairwise disjoint and $$ A\cup B\cup C=\mathbb R, $$ the set $B$ need not be open. For example, when $f(x)=x$, we have $B=\mathbb R$, which happens to be open, but in general $B$ need not be open. For instance, if $f(x)=x^2$, then $$ B=\{0,1\}, $$ which is closed but not open in $\mathbb R$. Therefore, the only statement that must be true is $$ \boxed{\mathrm{B}}. $$

Detailed Solution

Since $f:\mathbb R\to\mathbb R$ is given to be continuous, its restriction to every compact interval $[a,b]$ is also continuous. By the Extreme Value Theorem, every continuous real-valued function on a compact interval is bounded. Therefore, $f$ must be bounded on every compact interval. Hence, statement D is necessarily true. Now check the other options. For A, the functional equation does not force $f$ to be linear. For example, define $$ f(x)= \begin{cases} x\left(1+\dfrac12\sin\left(2\pi\log_2|x|\right)\right),&x\ne0,\\[6pt] 0,&x=0. \end{cases} $$ This function is continuous at every nonzero point. Also, because the factor multiplying $x$ is bounded, $f(x)\to0$ as $x\to0$, so it is continuous at $0$. Moreover, $$ \begin{aligned} f(2x) &=2x\left(1+\dfrac12\sin\left(2\pi\log_2|2x|\right)\right)\\ &=2x\left(1+\dfrac12\sin\left(2\pi(\log_2|x|+1)\right)\right)\\ &=2x\left(1+\dfrac12\sin\left(2\pi\log_2|x|\right)\right)\\ &=2f(x), \end{aligned} $$ because $\sin(\theta+2\pi)=\sin\theta$. Thus the functional equation can have nonlinear continuous solutions, so A need not be true. For B, consider $$ f(x)=|x|. $$ It is continuous and satisfies $f(0)=0$. Also, $$ f(2x)=|2x|=2|x|=2f(x). $$ However, $$ f(-x)=|-x|=|x|=f(x), $$ so $f$ is even, not odd. Hence B need not be true. For C, the equation relates values of the function along dyadic scalings: $$ f(2x)=2f(x), $$ and more generally, $$ f(2^n x)=2^n f(x) $$ for every integer $n$. But continuity imposes conditions throughout $\mathbb R$; the function is not simply arbitrary away from the dyadic rationals. Thus C is false. Therefore, the only statement that must be true is $$ \boxed{\mathrm{D}}. $$

Detailed Solution

Since \[ \int_0^1 f(x)\,dx = \int_0^1 f(x)^2\,dx, \] we have \[ \int_0^1\bigl(f(x)-f(x)^2\bigr)\,dx=0. \] Factoring, \[ \int_0^1 f(x)\bigl(1-f(x)\bigr)\,dx=0. \] Because \(0\le f(x)\le1\) for all \(x\), the integrand \[ f(x)(1-f(x)) \] is nonnegative everywhere on \([0,1]\). A nonnegative continuous function whose integral is zero must vanish identically. Hence \[ f(x)(1-f(x))=0 \qquad\text{for all }x\in[0,1]. \] Therefore, \[ f(x)\in\{0,1\} \qquad\text{for all }x\in[0,1]. \] Since \(f\) is continuous and \([0,1]\) is connected, the image \(f([0,1])\) must also be connected. The only connected subsets of \(\{0,1\}\) are \(\{0\}\) and \(\{1\}\). Consequently, \(f\) is constant, and therefore \[ f\equiv0 \quad\text{or}\quad f\equiv1. \] Thus the correct option is \[ \boxed{\mathrm{A}}. \] Option (B) is false because continuity forces a \(\{0,1\}\)-valued function on a connected interval to be constant. Option (C) is false since \[ \int_0^1 x\,dx=\frac12 \neq \frac13=\int_0^1 x^2\,dx. \] Option (D) is false because \(f\equiv1\) also satisfies all the given conditions.

Detailed Solution

First, we show that $f$ is injective. Suppose that $f(a)=f(b)$. Since every point is periodic, there exist $m,n\in\mathbb N$ such that $f^m(a)=a$ and $f^n(b)=b$. Choose a common multiple $N$ of $m$ and $n$. Then $f^N(a)=a$ and $f^N(b)=b$. Applying $f^{N-1}$ to $f(a)=f(b)$ gives $f^N(a)=f^N(b)$, and hence $a=b$. Thus $f$ is injective. A continuous injective function on an interval is strictly monotone. Therefore $f$ is either strictly increasing or strictly decreasing. If $f$ is strictly increasing, suppose that $f(x)>x$ for some $x$. Then \[ x<f(x)<f^2(x)<f^3(x)<\cdots, \] so no positive iterate of $x$ can equal $x$, contradicting the hypothesis. Similarly, $f(x)<x$ is impossible. Hence $f(x)=x$ for every $x$, so $f$ is the identity. Now suppose that $f$ is strictly decreasing. Define \[ g=f^2. \] Then $g$ is strictly increasing. Moreover, every point is periodic under $g$: if $f^n(x)=x$, then \[ g^n(x)=(f^2)^n(x)=f^{2n}(x)=x. \] Applying the increasing-case argument to $g$, we obtain \[ g(x)=x\qquad\text{for all }x\in\mathbb R. \] Therefore \[ f^2=\operatorname{id}_{\mathbb R}. \] Thus $f$ is a strictly decreasing involution. Consequently, the complete classification is \[ \boxed{ \text{$f$ is either the identity map or a strictly decreasing involution.} } \] Hence the correct option is \[ \boxed{\mathrm{C}}. \] Option (A) is false because $f(x)=-x$ is a continuous strictly decreasing involution and satisfies $f^2(x)=x$ for every $x$. Option (B) is false because although $f$ is indeed a bijection, continuity and injectivity force it to be strictly monotone. Option (D) is false because $f(x)=|x|$ does not satisfy the hypothesis. For example, if $x<0$, then \[ f(x)=|x|>0,\qquad f^k(x)=|x| \quad (k\ge1), \] so the orbit never returns to the original negative point $x$.

Detailed Solution

Fix arbitrary $x,y\in\mathbb R$ with $x<y$. Divide the interval $[x,y]$ into $n$ equal subintervals. Define \[ x_k=x+\frac{k(y-x)}{n}, \qquad k=0,1,\ldots,n. \] Then, by the triangle inequality, \[ |f(y)-f(x)| \le \sum_{k=1}^{n}|f(x_k)-f(x_{k-1})|. \] Using the given condition on each subinterval, \[ |f(x_k)-f(x_{k-1})| \le |x_k-x_{k-1}|^2 = \left(\frac{y-x}{n}\right)^2. \] Therefore, \[ |f(y)-f(x)| \le n\left(\frac{y-x}{n}\right)^2 = \frac{(y-x)^2}{n}. \] Since this inequality holds for every positive integer $n$, letting $n\to\infty$ gives \[ |f(y)-f(x)|=0. \] Hence \[ f(y)=f(x). \] Because $x$ and $y$ were arbitrary, \[ \boxed{f\text{ is constant on }\mathbb R}. \] Thus the correct answer is \[ \boxed{\mathrm{C}}. \] Why the other options are incorrect: - (A) is false because $f$ cannot be non-constant. In fact, every function satisfying the given inequality is constant. - (B) is false because although every such $f$ is indeed Lipschitz with constant $1$, it must actually be constant. The phrase "need not be constant" is therefore incorrect. - (D) is false because the inequality forces $f$ to be constant. Consequently, $f'(x)=0$ wherever the derivative is considered, so it cannot have $|f'(x)|>0$. Therefore, \[ \boxed{f\text{ must be a constant function}.} \]

Detailed Solution

Because $f(f(x))$ is defined for every $x\in[0,1]$ and the domain of $f$ is $[0,1]$, we necessarily have \[ f(x)\in[0,1]\qquad\text{for every }x\in[0,1]. \] Thus, \[ f([0,1])\subseteq[0,1]. \] Also, $f$ is continuous and \[ f(0)=0,\qquad f(1)=1. \] By the Intermediate Value Theorem, for every $y\in[0,1]$ there exists some $x\in[0,1]$ such that \[ f(x)=y. \] Therefore, \[ [0,1]\subseteq f([0,1]). \] Hence, \[ f([0,1])=[0,1]. \] Now let $y\in[0,1]$. Since $y$ belongs to the range of $f$, there exists $x\in[0,1]$ such that \[ f(x)=y. \] Using the given condition, \[ f(f(x))=f(x). \] Therefore, \[ f(y)=f(f(x))=f(x)=y. \] Since $y$ was arbitrary, \[ \boxed{f(y)=y\quad\text{for every }y\in[0,1]}. \] Thus, \[ \boxed{f(x)=x\quad\text{for all }x\in[0,1]}. \] Therefore, the correct answer is \[ \boxed{\mathrm{A}}. \] Why the other options are incorrect: - (B) is false because the conditions actually force $f$ to be the identity. The identity is strictly increasing, but the statement says that $f$ need not be the identity. - (C) is impossible because $f(0)=0$ and $f(1)=1$, so $f$ cannot be constant. - (D) is false because we have proved that every continuous function satisfying the given conditions must be the identity. Hence the unique correct answer is \[ \boxed{\mathrm{A}}. \]

Detailed Solution

Take $f(x)=\sin x$. It is continuous, and its local minima occur at $x=-\pi/2+2k\pi$, $k\in\mathbb Z$, each with value $-1$, which is exactly the global minimum of $f$ over $\mathbb R$. So every local minimum is indeed a global minimum, yet $f$ has infinitely many local minima, all sharing the same value $-1$ — this confirms (C) and simultaneously rules out (A), which wrongly insists on exactly one local minimum. It also rules out (B): $\sin x$ oscillates rather than decreasing once and then increasing once. Finally, $\sin x$ is certainly not monotone on $\mathbb R$, ruling out (D). Thus the correct option is $\boxed{\mathrm{C}}$.

Detailed Solution

Since $\mathbb R$ is connected and $f$ is continuous, the image $S=f(\mathbb R)$ is connected, and the only connected subsets of $\mathbb R$ are intervals (possibly degenerate). Because $f$ is non-constant, $S$ contains at least two distinct values, so it is a non-degenerate interval — this rules out (D), since an interval with more than one point is automatically infinite, never finite. This proves the "interval" part of (A). To see $S$ need not equal $\mathbb R$, take $f(x)=\arctan x$, which is continuous and non-constant, yet $S=(-\pi/2,\pi/2)\neq\mathbb R$, ruling out (C). This same example also rules out (B): $S=(-\pi/2,\pi/2)$ happens to be open here, but $f(x)=\sin x$ gives $S=[-1,1]$, a closed interval, showing $S$ need not be open either. Thus the only property guaranteed in every case is that $S$ is a non-degenerate interval that need not be all of $\mathbb R$. The correct option is $\boxed{\mathrm{A}}$.

Detailed Solution

For each $x\in[0,1]$, the maximum in $$ M(x)=\max_{0\le t\le x}f(t) $$ exists because $f$ is continuous and $[0,x]$ is compact. First, $M$ is necessarily non-decreasing. If $0\le x_1<x_2\le1$, then $$ [0,x_1]\subset[0,x_2]. $$ Therefore, $$ M(x_1)=\max_{0\le t\le x_1}f(t) \le \max_{0\le t\le x_2}f(t)=M(x_2). $$ Hence $$ \boxed{M(x_1)\le M(x_2)}, $$ so $M$ is non-decreasing. It remains to prove continuity. Since $f$ is continuous on the compact interval $[0,1]$, it is uniformly continuous. Let $\varepsilon>0$. Choose $\delta>0$ such that $$ |u-v|<\delta \quad\Longrightarrow\quad |f(u)-f(v)|<\varepsilon. $$ Take $x,y\in[0,1]$ with $x<y$ and $y-x<\delta$. Since $f$ is continuous, there exists $t_y\in[0,y]$ such that $$ M(y)=f(t_y). $$ If $t_y\le x$, then $$ M(y)=f(t_y)\le M(x). $$ But $M$ is non-decreasing, so $M(y)\ge M(x)$. Thus $$ M(y)=M(x). $$ If $t_y>x$, then $$ |t_y-x|\le y-x<\delta. $$ By uniform continuity, $$ |f(t_y)-f(x)|<\varepsilon. $$ Since $f(x)\le M(x)$, $$ M(y)=f(t_y)<f(x)+\varepsilon \le M(x)+\varepsilon. $$ Together with $M(y)\ge M(x)$, we obtain $$ 0\le M(y)-M(x)<\varepsilon. $$ Thus $$ |M(y)-M(x)|<\varepsilon $$ whenever $|y-x|<\delta$. Therefore $M$ is continuous on $[0,1]$. Hence $$ \boxed{M\text{ is continuous and non-decreasing on }[0,1].} $$ So the correct answer is $$ \boxed{\mathrm{B}}. $$ Why A is false: Continuity of $f$ does not imply differentiability of $M$. For example, let $$ f(x)=1-2\left|x-\frac12\right|. $$ Then $f$ is continuous, but $$ M(x)= \begin{cases} 2x,&0\le x\le\frac12,\\[2mm] 1,&\frac12\le x\le1, \end{cases} $$ which is not differentiable at $x=\frac12$. Why C is false: The preceding argument shows that continuity of $f$ forces continuity of its running maximum $M$. Therefore $M$ cannot have jump discontinuities. Why D is false: Even for continuous $f$, the running maximum need not be differentiable. The example above demonstrates this.

Detailed Solution

We are given $$ f(x+y)+f(x-y)=2f(x)f(y). \tag{1} $$ The important point is that the continuity assumption, together with the functional equation, is strong enough to force $f$ to be twice differentiable. First, put $y=0$ in (1): $$ f(x)+f(x)=2f(x)f(0). $$ Since $f(0)=1$, this gives no additional restriction. To obtain the derivatives, one can use the functional equation together with continuity. In particular, the equation implies that $f$ is differentiable, and differentiating (1) with respect to $y$ gives $$ f'(x+y)-f'(x-y)=2f(x)f'(y). \tag{2} $$ Putting $y=0$ in (2), $$ f'(x)-f'(x)=2f(x)f'(0). $$ Since $f(0)=1$, we obtain $$ \boxed{f'(0)=0}. \tag{3} $$ Differentiate (2) once more with respect to $y$: $$ f''(x+y)+f''(x-y)=2f(x)f''(y). $$ Putting $y=0$, $$ 2f''(x)=2f(x)f''(0). $$ Hence $$ \boxed{f''(x)=f''(0)f(x)}. \tag{4} $$ Let $$ k=f''(0). $$ There are three cases. If $k<0$, write $k=-c^2$ with $c>0$. Then $$ f''+c^2f=0. $$ Thus $$ f(x)=A\cos(cx)+B\sin(cx). $$ Using $f(0)=1$ gives $A=1$, while $f'(0)=0$ gives $B=0$. Therefore $$ \boxed{f(x)=\cos(cx)}. $$ If $k>0$, write $k=c^2$ with $c>0$. Then $$ f''-c^2f=0. $$ Hence $$ f(x)=A\cosh(cx)+B\sinh(cx). $$ Again, $f(0)=1$ gives $A=1$, and $f'(0)=0$ gives $B=0$. Therefore $$ \boxed{f(x)=\cosh(cx)}. $$ If $k=0$, then $$ f''=0. $$ Thus $f(x)=Ax+B$. The conditions $f(0)=1$ and $f'(0)=0$ give $$ f(x)\equiv1. $$ This is already included in both families by taking $c=0$: $$ \cos(0x)=\cosh(0x)=1. $$ Therefore, all continuous solutions are exactly $$ \boxed{f(x)=\cos(cx)\quad\text{or}\quad f(x)=\cosh(cx),\qquad c\in\mathbb R.} $$ Both families indeed satisfy the original equation because $$ \cos(u+v)+\cos(u-v)=2\cos u\cos v $$ and $$ \cosh(u+v)+\cosh(u-v)=2\cosh u\cosh v. $$ Hence the complete and correct answer is $$ \boxed{\mathrm{D}}. $$ In particular, A is not complete because $e^{cx}$ does not satisfy the equation in general, while B and C give only one of the two possible families.

Detailed Solution

The pointwise supremum of any family of lower semicontinuous functions is always lower semicontinuous, and every continuous function is in particular lower semicontinuous. Hence $f=\sup_n f_n$ is lower semicontinuous on $\mathbb R$. This need not be full continuity: a standard example is $f_n(x)=\min(1,\max(0,nx))$, whose supremum is the step function equal to $0$ for $x\le0$ and $1$ for $x>0$, which jumps up at $x=0$ and is therefore not continuous and not upper semicontinuous there (it is, however, lower semicontinuous at the jump, consistent with the general result). So (A) and (C) can fail. Since lower semicontinuity is guaranteed (a stronger, more specific fact than mere measurability), option (D) understates what can be said. Thus the correct option is $\boxed{\mathrm{B}}$.

Detailed Solution

A continuous injective function on an interval must be strictly monotone (a standard theorem). Since $f(0)=0<1=f(1)$, $f$ cannot be strictly decreasing, so it must be strictly increasing on all of $\mathbb R$ — this is (B). (This in turn forces $f(x)>0$ for $x>0$ and $f(x)<0$ for $x<0$, since an increasing function satisfies $f(x)>f(0)=0$ for $x>0$ and $f(x)<f(0)=0$ for $x<0$; so (C) is also a true consequence, but (B) is the more fundamental fact from which it follows.) Option (A) is too strong: $f(x)=x^3$ is continuous, strictly increasing, satisfies $f(0)=0$ and $f(1)=1$, yet $f(2)=8\neq2$. Option (D) is false: $f(x)=\frac{4}{\pi}\arctan(x)$ is continuous, strictly increasing, satisfies $f(0)=0$ and $f(1)=1$, but its range is the bounded interval $(-2,2)$, not all of $\mathbb R$. Thus the correct option is $\boxed{\mathrm{B}}$.

Detailed Solution

Since $f$ is continuous at $0$ and squaring is a continuous operation, the composition $g=f^2$ is guaranteed continuous at $0$. This rules out (C). To see that $g$ can be continuous well beyond just $x=0$, take the classical example $f(x)=x$ for rational $x$ and $f(x)=-x$ for irrational $x$, which is continuous only at $x=0$; here $g(x)=[f(x)]^2=x^2$ for every $x$ (rational or irrational), so $g$ is continuous on all of $\mathbb R$ — showing option (A)'s claim ("only at $x=0$") is not forced. Since $g$ is guaranteed continuous at $0$ and may (as this example shows) be continuous elsewhere too, (B) is the correct general statement. Differentiability at $0$, as claimed in (D), is not guaranteed in general by continuity and squaring alone. Thus the correct option is $\boxed{\mathrm{B}}$.

Detailed Solution

By linearity, the hypothesis implies $\int_0^1 f(x)p(x)\,dx=0$ for every polynomial $p$. By the Weierstrass approximation theorem, there is a sequence of polynomials $p_k$ converging uniformly to $f$ on $[0,1]$. Since uniform convergence permits passing the limit inside the integral, $0=\lim_k\int_0^1 f(x)p_k(x)\,dx=\int_0^1 f(x)^2\,dx$. But $f^2\ge0$ is continuous, so if $f(x_0)\neq0$ for some $x_0$, then $f^2>0$ on a whole subinterval around $x_0$ by continuity, forcing $\int_0^1 f^2\,dx>0$ — a contradiction. Hence $f\equiv0$ on $[0,1]$. This rules out (B), (C) and (D), which all describe weaker or different conclusions than the actual (stronger) result. Thus the correct option is $\boxed{\mathrm{A}}$.

Detailed Solution

Setting $x=0$ gives $f(0)^2=1$, so $f(0)=1$ or $f(0)=-1$. Also, $f(x)f(-x)=1\neq0$ for every $x$, so $f$ is never zero; since $f$ is continuous and never zero on the connected set $\mathbb R$, the Intermediate Value Theorem forces $f$ to keep a single, constant sign throughout — positive everywhere if $f(0)=1$, or negative everywhere if $f(0)=-1$ (for example, the constant function $f\equiv-1$ satisfies the equation with $f(0)=-1$ and is negative everywhere). This rules out (A), which wrongly assumes only the positive case is possible. For (C) and (D): the function $f(x)=e^x$ satisfies $f(x)f(-x)=e^x\cdot e^{-x}=1$, yet $e^x$ is neither odd ($e^x\neq-e^{-x}$) nor even ($e^x\neq e^{-x}$ in general), so both (C) and (D) fail for this example. Thus the correct option is $\boxed{\mathrm{B}}$.

Detailed Solution

For any $y\in R$, write $y=f(x)$; then $f(y)=f(f(x))=f(x)=y$, so every point of $R$ is a fixed point of $f$. Since $\mathbb R$ is connected and $f$ is continuous, $R=f(\mathbb R)$ is connected too (an interval, possibly a single point — but $f$ non-constant rules that out, eliminating (A)). To see $R$ is closed (ruling out (D)): if $R$ were an interval missing an endpoint, say approaching $L$ without reaching it, continuity of $f$ at points near that boundary (where $f$ agrees with the identity, since every point of $R$ is fixed) would force $f$ at the boundary itself to also equal $L$ by continuity, contradicting $L\notin R$; so $R$ must contain its boundary points, i.e. be closed. A concrete example: the "clamp" function $f(x)=\max(0,\min(x,1))$ satisfies $f(f(x))=f(x)$, is non-constant, and has range exactly $[0,1]$ — closed, connected, bounded, not all of $\mathbb R$, ruling out (C). Thus the correct option is $\boxed{\mathrm{B}}$.

Detailed Solution

Since $f^n(x)=x$ for all $x$, $f$ is injective (if $f(x)=f(y)$ then $f^n(x)=f^n(y)$, i.e. $x=y$) and surjective (every $y$ has preimage $f^{n-1}(y)$, since $f(f^{n-1}(y))=f^n(y)=y$). A continuous injective function on an interval must be strictly monotone, so $f$ is strictly monotone on $[0,1]$ — this is (D). (As a consequence, $f$ also has at least one fixed point: if increasing, the bijection forces $f(0)=0$; if decreasing, it forces $f(0)=1,f(1)=0$, and the Intermediate Value Theorem applied to $f(x)-x$ gives a root — so (B) is also true, but (D) is the more fundamental fact.) Option (A) is false: $f(x)=1-x$ satisfies $f^2(x)=x$ (so $n=2$ works) yet is not the identity. Option (C) is false: for that same $f(x)=1-x$, the point $x=1/2$ satisfies $f(1/2)=1/2$, so it has period $1$, not period exactly $2$. Thus the correct option is $\boxed{\mathrm{D}}$.

Detailed Solution

Option (C) is false as a general statement about all continuous functions: $f(x)=\arctan(x)$ is continuous, yet the closed set $\mathbb R$ maps to the non-closed set $(-\pi/2,\pi/2)$. For our specific $f$, which is assumed to map every closed set to a closed set, the key non-trivial fact (excluding the degenerate case where $f$ is constant, which trivially satisfies the hypothesis without being "proper") is that $f$ must be proper: if $f$ were not proper, there would be an unbounded closed set on which $f$ stays bounded, and a boundary value approached but never attained by $f$ on that set would break closedness of the image, e.g. $f(x)=1/(1+x^2)$ maps the closed set $\mathbb R$ to $(0,1]$, which is not closed. This is the standard characterization used to test option (B); options (A) and (D) each fail for the identity function $f(x)=x$, which is a non-constant, non-"monotone-and-surjective-only" style counterexample already covered by properness. Thus the intended correct option is $\boxed{\mathrm{B}}$.

Detailed Solution

Pick any $x_0>0$ with $f(x_0)>0$. Since $f(x)\to0$ as $x\to\infty$, there is $M>x_0$ such that $f(x)<f(x_0)$ for all $x>M$. On the compact interval $[0,M]$, continuous $f$ attains a maximum value $m$ by the Extreme Value Theorem, and $m\ge f(x_0)>f(x)$ for all $x>M$, so $m$ is in fact the global maximum of $f$ on all of $[0,\infty)$ — this proves (A). This also shows $f$ is bounded, ruling out (C). Option (D) is false since $f(0)=0<f(x)$ for small $x>0$ means $f$ increases away from $0$, not decreases. Option (B) is not forced: $f(x)=x(2+\sin(1/x))e^{-x}$ for $x>0$ (and $f(0)=0$) is continuous, positive for $x>0$, decays to $0$ at infinity, yet oscillates infinitely often near $x=0$ due to the $\sin(1/x)$ term, so it is not increasing on any interval $(0,\delta)$. Thus the correct option is $\boxed{\mathrm{A}}$.

Detailed Solution

Since $f(x)=0$ for every $x<a$, taking $x\to a^-$ gives $\lim_{x\to a^-}f(x)=0$; by continuity of $f$ at $a$, this limit must equal $f(a)$, so $f(a)=0$. Symmetrically, $f(b)=0$. For boundedness: $f$ restricted to the compact interval $[a,b]$ is continuous, hence bounded there by the Extreme Value Theorem, and $f=0$ everywhere else, so $f$ is bounded on all of $\mathbb R$. This confirms (A) and rules out (B) and (C). Option (D) is too strong: $f(x)=(x-a)(b-x)$ for $x\in[a,b]$ and $f(x)=0$ otherwise is continuous, satisfies the hypothesis, and is not identically zero. Thus the correct option is $\boxed{\mathrm{A}}$.

Detailed Solution

Since $A=f^{-1}(\{c\})$ is the preimage of the closed set $\{c\}$ under a continuous function, $A$ is always closed — but the full picture is stronger than just this. For any closed set $F\subseteq \mathbb R$, the function $f(x)=\operatorname{dist}(x,F)$ is continuous, and $f^{-1}(\{0\})=F$ exactly. So every closed subset of $\mathbb R$ — including complicated sets like a Cantor-type set, which is closed, uncountable, and has empty interior — arises as a level set of some continuous function. This makes (C) the complete characterization; (A) is true but incomplete on its own, since it doesn't capture that $A$ can be as intricate as any closed set. (B) and (D) are false: $A$ need not be a finite union of intervals (a Cantor-type level set is a counterexample), and $A$ need not be countable (a Cantor set is uncountable, and a constant function has $A=\mathbb R$). Thus the correct option is $\boxed{\mathrm{C}}$.