Let $F$ be a vector subspace of a symplectic space $(E , \omega)$, and let $F^{\omega} = \{ x \in E \mid \forall y \in F , \omega ( x , y ) = 0 \}$. Is the subspace $F ^ { \omega }$ necessarily in direct sum with $F$?

The functions $p _ { \alpha }$ are defined by $p_\alpha : t \mapsto t^\alpha$ for $\alpha \in \mathbb{R}_+^*$, and the inner product on $E$ is $\langle f \mid g \rangle = \int _ { 0 } ^ { + \infty } f ( t ) g ( t ) \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t$. Is the family $\left( p _ { n } \right) _ { n \in \mathbb { N } ^ { * } }$ an orthogonal family of $E$?

If $\varphi : \mathbb{R} \rightarrow \mathbb{R}$ is a continuous function, the support of $\varphi$ is defined by: $$\operatorname{Supp}(\varphi) = \overline{\{x \in \mathbb{R} : \varphi(x) \neq 0\}}$$ We say that $\varphi$ has compact support if $\operatorname{Supp}(\varphi)$ is a bounded subset of $\mathbb{R}$. We denote by $\mathcal{C}_{c}(\mathbb{R})$ the set of continuous functions with compact support on $\mathbb{R}$. If $\varphi \in \mathcal{C}_{c}(\mathbb{R})$, we set $$\|\varphi\|_{\infty} = \sup_{x \in \mathbb{R}} |\varphi(x)| \text{ and } \|\varphi\|_{1} = \int_{-\infty}^{+\infty} |\varphi(t)| dt$$
Are the norms $\|\cdot\|_{\infty}$ and $\|\cdot\|_{1}$ equivalent?

By setting for every integer $k \geqslant 1$, $a_k = \frac{1}{k} - \frac{1}{2^{k+1}}$ and $b_k = \frac{1}{k} + \frac{1}{2^{k+1}}$, show that we can choose an integer $k_0 \geqslant 1$ such that: $$\forall k \geqslant k_0, \quad b_{k+1} < a_k.$$ Deduce that the function $f : ]0,1[ \longrightarrow \mathbf{R}$ defined by: $$f : t \longmapsto \begin{cases} k^2 \cdot 2^{k+1} \cdot (t - a_k), & \text{if there exists an integer } k \geqslant k_0 \text{ such that } t \in \left[a_k, a_k + \frac{1}{2^{k+1}}\right] \\ k^2 \cdot 2^{k+1} \cdot (b_k - t), & \text{if there exists an integer } k \geqslant k_0 \text{ such that } t \in \left[a_k + \frac{1}{2^{k+1}}, b_k\right] \\ 0, & \text{otherwise} \end{cases}$$ is a well-defined and continuous function on $]0,1[$, integrable on $]0,1[$ and that this function $f$ does not belong to the set $\mathscr{D}_{0,1}$.

Show that $\zeta ( 2 )$ is an irrational number.

Consider the function $f ( x ) = x ^ { n } ( 1 - x ) ^ { n } / n !$, where $n \geq 1$ is a fixed integer. Let $f ^ { ( k ) }$ denote the $k$-th derivative of $f$. Which of the following is true for all $k \geq 1$?
(a) $f ^ { ( k ) } ( 0 )$ and $f ^ { ( k ) } ( 1 )$ are integers.
(b) $f ^ { ( k ) } ( 0 )$ is an integer, but not $f ^ { ( k ) } ( 1 )$
(c) $f ^ { ( k ) } ( 1 )$ is an integer, but not $f ^ { ( k ) } ( 0 )$
(d) Neither $f ^ { ( k ) } ( 1 )$ nor $f ^ { ( k ) } ( 0 )$ is an integer.

Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function such that for any two real numbers $x$ and $y$, $$|f(x) - f(y)| \leq 7|x - y|^{201}$$ Then,
(A) $f(101) = f(202) + 8$
(B) $f(101) = f(201) + 1$
(C) $f(101) = f(200) + 2$
(D) None of the above.

Let $f : \mathbb { R } \rightarrow \mathbb { R }$ and $g : \mathbb { R } \rightarrow \mathbb { R }$ be two functions. Consider the following two statements: $\mathbf { P ( 1 ) }$: If $\lim _ { x \rightarrow 0 } f ( x )$ exists and $\lim _ { x \rightarrow 0 } f ( x ) g ( x )$ exists, then $\lim _ { x \rightarrow 0 } g ( x )$ must exist. $\mathbf { P ( 2 ) }$: If $f , g$ are differentiable with $f ( x ) < g ( x )$ for every real number $x$, then $f ^ { \prime } ( x ) < g ^ { \prime } ( x )$ for all $x$. Then, which one of the following is a correct statement?
(A) Both $\mathrm { P } ( 1 )$ and $\mathrm { P } ( 2 )$ are true.
(B) Both $P ( 1 )$ and $P ( 2 )$ are false.
(C) $\mathrm { P } ( 1 )$ is true and $\mathrm { P } ( 2 )$ is false.
(D) $\mathrm { P } ( 1 )$ is false and $\mathrm { P } ( 2 )$ is true.

Let $f(x) = 2 + \cos x$ for all real $x$. STATEMENT-1: For each real $t$, there exists a point $c$ in $[t, t+\pi]$ such that $f'(c) = 0$. because STATEMENT-2: $f(t) = f(t+2\pi)$ for each real $t$.
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement-1 is True, Statement-2 is False
(D) Statement-1 is False, Statement-2 is True

The following statement $(p \to q) \to [ ( \sim p \to q ) \to q ]$ is:
(1) a fallacy
(2) a tautology
(3) equivalent to $\sim p \to q$
(4) equivalent to $p \to \sim q$

The statement $(p \rightarrow q) \rightarrow (\sim p \rightarrow q) \rightarrow q$ is
(1) A tautology
(2) Equivalent to $\sim p \rightarrow q$
(3) Equivalent to $p \rightarrow \sim q$
(4) A fallacy

If $p \rightarrow ( \sim p \vee \sim q )$ is false, then the truth values of $p$ and $q$ are, respectively
(1) $F , F$
(2) $T , T$
(3) $F , T$
(4) $T , F$

Consider the statement: ``$P ( n ) : n ^ { 2 } - n + 41$ is prime''. Then which one of the following is true?
(1) $P ( 3 )$ is false but $P ( 5 )$ is true
(2) Both $P ( 3 )$ and $P ( 5 )$ are false
(3) Both $P ( 3 )$ and $P ( 5 )$ are true
(4) $P ( 5 )$ is false but $P ( 3 )$ is true

Which of the following statement is a tautology?
(1) $p \vee (\sim q) \rightarrow p \wedge q$
(2) $\sim(p \wedge \sim q) \rightarrow p \vee q$
(3) $\sim(p \vee \sim q) \rightarrow p \wedge q$
(4) $\sim(p \vee \sim q) \rightarrow p \vee q$

If $p \rightarrow ( p \wedge \sim q )$ is false, then the truth values of $p$ and $q$ are respectively
(1) $F , F$
(2) $T , F$
(3) $T , T$
(4) $F , T$

Which of the following is a tautology?
(1) $( \sim p ) \wedge ( p \vee q ) \rightarrow q$
(2) $( q \rightarrow p ) \vee \sim ( p \rightarrow q )$
(3) $( \sim q ) \vee ( p \wedge q ) \rightarrow q$
(4) $( p \rightarrow q ) \wedge ( q \rightarrow p )$

Let $p , q , r$ be three statements such that the truth value of $( p \wedge q ) \rightarrow ( \sim q \vee r )$ is $F$. Then the truth values of $p , q , r$ are respectively :
(1) $T , T , F$
(2) $T , T , T$
(3) $T , F , T$
(4) $F , T , F$

Given the following two statements: $\left( \mathrm { S } _ { 1 } \right) : ( \mathrm { q } \vee \mathrm { p } ) \rightarrow ( \mathrm { p } \leftrightarrow \sim \mathrm { q } )$ is a tautology $\left( \mathrm { S } _ { 2 } \right) : \sim \mathrm { q } \wedge ( \sim \mathrm { p } \leftrightarrow \mathrm { q } )$ is a fallacy. Then :
(1) both ( $S _ { 1 }$ ) and ( $S _ { 2 }$ ) are not correct.
(2) only ( $S _ { 1 }$ ) is correct.
(3) only ( $S _ { 2 }$ ) is correct.
(4) both $\left( S _ { 1 } \right)$ and $\left( S _ { 2 } \right)$ are correct.

The statement $(p \rightarrow (q \rightarrow p)) \rightarrow (p \rightarrow (p \vee q))$ is:
(1) equivalent to $(p \wedge q) \vee (\sim q)$
(2) a contradiction
(3) equivalent to $(p \vee q) \wedge (\sim p)$
(4) a tautology

The negation of the statement $\sim p \wedge ( p \vee q )$ is:
(1) $\sim p \vee q$
(2) $\sim p \wedge q$
(3) $p \vee \sim q$
(4) $p \wedge \sim q$

For the statements $p$ and $q$, consider the following compound statements: $( a ) ( \sim q \wedge ( p \rightarrow q ) ) \rightarrow \sim p$
(b) $( ( p \vee q ) \wedge \sim p ) \rightarrow q$ Then which of the following statements is correct?
(1) (b) is a tautology but not (a).
(2) (a) and (b) both are tautologies.
(3) (a) and (b) both are not tautologies.
(4) (a) is a tautology but not (b).

The statement among the following that is a tautology is:
(1) $A \vee A \wedge B$
(2) $A \wedge A \vee B$
(3) $B \rightarrow A \wedge A \rightarrow B$
(4) $A \wedge A \rightarrow B \rightarrow B$

Let $F _ { 1 } ( A , B , C ) = ( A \wedge \sim B ) \vee [ \sim C \wedge ( A \vee B ) ] \vee \sim A$ and $F _ { 2 } ( A , B ) = ( A \vee B ) \vee ( B \rightarrow \sim A )$ be two logical expressions. Then :
(1) $F _ { 1 }$ is a tautology but $F _ { 2 }$ is not a tautology
(2) $F _ { 1 }$ is not a tautology but $F _ { 2 }$ is a tautology
(3) Both $F _ { 1 }$ and $F _ { 2 }$ are not tautologies
(4) $F _ { 1 }$ and $F _ { 2 }$ both are tautologies

If the Boolean expression $( p \Rightarrow q ) \Leftrightarrow \left( q ^ { * } ( \sim p ) \right)$ is a tautology, then the Boolean expression $p ^ { * } ( \sim q )$ is equivalent to:
(1) $q \Rightarrow p$
(2) $\sim q \Rightarrow p$
(3) $p \Rightarrow \sim q$
(4) $p \Rightarrow q$

If $P$ and $Q$ are two statements, then which of the following compound statement is a tautology?
(1) $( ( P \Rightarrow Q ) \wedge \sim Q ) \Rightarrow Q$
(2) $( ( P \Rightarrow Q ) \wedge \sim Q ) \Rightarrow \sim P$
(3) $( ( P \Rightarrow Q ) \wedge \sim Q ) \Rightarrow P$
(4) $( ( P \Rightarrow Q ) \wedge \sim Q ) \Rightarrow ( P \wedge Q )$