Let $F$ be a vector subspace of a symplectic space $(E , \omega)$, with $\omega$-orthogonal $F^{\omega} = \{ x \in E \mid \forall y \in F , \omega ( x , y ) = 0 \}$. 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 $p \to \sim q$

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) $( ( P \Rightarrow Q ) \wedge \sim Q ) \Rightarrow ( P \wedge Q )$

Consider the statement "The match will be played only if the weather is good and ground is not wet". Select the correct negation from the following:
(1) The match will not be played and weather is not good and ground is wet.
(2) If the match will not be played, then either weather is not good or ground is wet.
(3) The match will be played and weather is not good or ground is wet.
(4) The match will not be played or weather is good and ground is not wet.