On the unit circle the constraint gradient $(2x,2y)$ never vanishes. The Lagrange multiplier equations areA nonzero solution of the first two equations requiresso $\lambda =9/2$ or $\lambda =-1/2$. Multiplying the first equation by $x$, the second by $y$, and adding shows $2f=2\lambda (x^2+y^2)=2\lambda$. Thus each candidate value of $f$ equals its multiplier.

For $\lambda =9/2$, $y=-x/3$, giving $(x,y)=\pm (3,-1)/\sqrt{10}$. For $\lambda =-1/2$, $y=3x$, giving $(x,y)=\pm (1,3)/\sqrt{10}$. Since a continuous function attains its extrema on the compact set given by the unit circle and these exhaust the Lagrange multiplier candidates,Each $\pm$ changes both coordinates together.

The gradient vanishes when $4x+3y-1=0$ and $3x+2y-1=0$. Subtracting these equations gives $x+y=0$, so the unique critical point is $\boxed{(x,y)=(1,-1)}$, with $f(1,-1)=0$.

The Hessian matrix isIt is indefinite, so the critical point is a saddle point. Explicitly, around that point the change is exactly $2u^2+3uv+v^2$: it is positive for $v=0,u\ne 0$ and negative for $v=-3u/2,u\ne 0$. Thus arbitrarily close points have both larger and smaller values.

The characteristic equation of a constant-coefficient differential equation here is $r^2-3r+2=(r-1)(r-2)=0$. Its homogeneous solution is $C_1{e}^x+C_2{e}^{2x}$. Because the forcing $e^x$ is already a homogeneous mode, the method of undetermined coefficients requires $y_p=Ax{e}^x$ rather than $A{e}^x$.

Here $y_p^{\prime }=A(x+1)e^x$ and $y_p^{\prime \prime }=A(x+2)e^x$, so $y_p^{\prime \prime }-3y_p^{\prime }+2y_p=-A{e}^x$. Thus $A=-1$, and the general solution is

Use partial fraction decomposition:Integrating each term gives the antiderivativeThis holds on any interval avoiding $-2,0,2$; the constant may differ between these intervals. Differentiation of the logarithm recovers $1/[4(x^2-4)]$, providing a direct check.

Assume $n\ge 1$ and list the elements of $X$ as $a_1,\dots ,a_n$. Define the partial sums $s_j=a_1+\cdots +a_j$ for $1\le j\le n$. If one is divisible by $n$, its corresponding nonempty prefix is a suitable $Y$.

Otherwise the $n$ partial sums have remainders in the $n-1$ nonzero residue classes. The pigeonhole principle gives $i<j$ with $s_i\equiv s_j(\mathrm{mod}n)$. Thus $s_j-s_i=a_{i+1}+\cdots +a_j$ is divisible by $n$, giving a nonempty subset $Y=\{a_{i+1},\dots ,a_j\}$.

Therefore a nonempty $Y$ exists. This strengthens the literal assertion, which would otherwise allow the trivial empty set with sum zero.

The fifty disjoint sets $\{1,100\},\{2,99\},\dots ,\{50,51\}$ form a partition of a set $\{1,\dots ,100\}$. If a chosen subset met each pair at most once, it would contain at most fifty elements. The pigeonhole principle therefore forces a $51$-element subset to contain both elements of at least one pair. Those two distinct elements sum to $101$.

For a proof by contradiction, suppose $\sqrt{3}=a/b$ for positive integers $a,b$ in lowest terms, so $\gcd (a,b)=1$. Squaring gives $a^2=3b^2$, and part (a) implies $3\mid a$. Write $a=3c$; then $9c^2=3b^2$, hence $b^2=3c^2$, and part (a) again implies $3\mid b$. This contradicts the coprimality of $a,b$. Therefore $\sqrt{3}$ is an irrational number.

We prove the contrapositive. If $3\nmid a$, Euclidean division gives $a=3q+1$ or $a=3q+2$ for an integer $q$. Squaring gives respectively $a^2=3(3q^2+2q)+1$ or $a^2=3(3q^2+4q+1)+1$. Both have remainder $1$ upon division by $3$, so $3\nmid a^2$. Taking the contrapositive yields $3\mid a^2\Rightarrow 3\mid a$.

Put $H_n={\sum }_{k=0}^{\lfloor n/2\rfloor }\left(\genfrac{}{}{0ex}{}{n-k}{k}\right)$; the other summands in the printed sum vanish. The initial values are $H_0=1=F_1$ and $H_1=1=F_2$. For $n\ge 2$, Pascal's identity givesFor the first sum, when $n$ is even its final term is zero; otherwise its range already agrees with $H_{n-1}$. In the second sum put $j=k-1$, obtaining $\left(\genfrac{}{}{0ex}{}{n-2-j}{j}\right)$ for $0\le j\le \lfloor (n-2)/2\rfloor$. Thus both range identifications hold without negative upper arguments in a binomial coefficient.

Now strong induction gives $H_n=F_n+F_{n-1}=F_{n+1}$ from the two preceding cases and the Fibonacci recurrence. Hence