Mathematics I

Euler's theorem. If $f(x,y)$ is a homogeneous function of degree $n$ with continuous first-order partial derivatives, then

$x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}=n\,f.$

Verification for $u=\log(x^2+y^2)$. Here $u$ itself is not homogeneous, but $f=e^u=x^2+y^2$ is homogeneous of degree $n=2$. Applying Euler's theorem to $f$:

$x f_x+y f_y=2f\ \Rightarrow\ x\,e^u u_x+y\,e^u u_y=2e^u\ \Rightarrow\ x u_x+y u_y=2.$

Direct check.

$u_x=\frac{2x}{x^2+y^2},\qquad u_y=\frac{2y}{x^2+y^2}$

$x u_x+y u_y=\frac{2x^2+2y^2}{x^2+y^2}=2$

This agrees with the value $n=2$ predicted by Euler's theorem, so the theorem is verified. $\blacksquare$

Euler's theorem: if $f(x,y)$ is homogeneous of degree $n$, then $x f_x+y f_y=n f$.

For $x\ne y$, $f(x,y)=\frac{x^3-y^3}{x-y}=x^2+xy+y^2,$ which is homogeneous of degree $n=2$ (check: $f(tx,ty)=t^2f(x,y)$).

Partial derivatives: $f_x=2x+y,\qquad f_y=x+2y.$ $xf_x+yf_y=x(2x+y)+y(x+2y)=2x^2+2xy+2y^2=2(x^2+xy+y^2)=2f.$ Thus $xf_x+yf_y=2f$, and Euler's theorem is verified.

Euler's theorem. If $f(x,y)$ is a homogeneous function of degree $n$, then $x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}=n\,f.$

Verification. $f=4x^2+8xy+4y^2$ is homogeneous of degree $2$, since every term has degree 2 (indeed $f=4(x+y)^2$). $f_x=8x+8y,\qquad f_y=8x+8y.$ $xf_x+yf_y=x(8x+8y)+y(8x+8y)=8x^2+16xy+8y^2=2(4x^2+8xy+4y^2)=2f.$ So $xf_x+yf_y=2f$, and the theorem is verified.

Euler's theorem: If $u=f(x,y)$ is a homogeneous function of degree $n$, then $x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=nu.$

Verification. For $u=ax^2+2hxy+by^2$, every term has degree $2$, so $u$ is homogeneous of degree $n=2$.

$\frac{\partial u}{\partial x}=2ax+2hy,\qquad\frac{\partial u}{\partial y}=2hx+2by.$

$x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=2ax^2+2hxy+2hxy+2by^2=2(ax^2+2hxy+by^2)=2u.$

This equals $nu$ with $n=2$, so Euler's theorem is verified.

Euler's theorem: If $u=f(x,y)$ is a homogeneous function of degree $n$, then $x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=nu.$

Verification. In $u=x^3+y^3+3xy^2$ every term has degree $3$, so $u$ is homogeneous of degree $n=3$.

$\frac{\partial u}{\partial x}=3x^2+3y^2,\qquad\frac{\partial u}{\partial y}=3y^2+6xy.$

$x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=3x^3+3xy^2+3y^3+6xy^2=3x^3+3y^3+9xy^2=3(x^3+y^3+3xy^2)=3u.$

Hence Euler's theorem is verified.

Given $p=\dfrac{2x+5}{x^2+5x+5}$.

Price falls as $x\to\infty$. Divide numerator and denominator by $x^2$:

$\lim_{x\to\infty}p=\lim_{x\to\infty}\frac{\frac{2}{x}+\frac{5}{x^2}}{1+\frac{5}{x}+\frac{5}{x^2}}=\frac{0+0}{1+0+0}=0$

So the price tends to zero as the quantity demanded becomes infinitely large.

Total revenue. $R=px=\dfrac{2x^2+5x}{x^2+5x+5}$. Dividing by $x^2$:

$\lim_{x\to\infty}R=\lim_{x\to\infty}\frac{2+\frac{5}{x}}{1+\frac{5}{x}+\frac{5}{x^2}}=\frac{2}{1}=2$

Hence total revenue approaches the finite limiting value $2$ as the quantity demanded increases indefinitely.

$f(x,y)=x^3+y^3-3x-27y+24$

Critical points.

$f_x=3x^2-3=0\Rightarrow x=\pm1,\qquad f_y=3y^2-27=0\Rightarrow y=\pm3$

The critical points are $(1,3),\ (1,-3),\ (-1,3),\ (-1,-3)$.

Second derivatives: $r=f_{xx}=6x,\ t=f_{yy}=6y,\ s=f_{xy}=0$, so $rt-s^2=36xy$.

So $f$ has a local minimum $-32$ at $(1,3)$, a local maximum $80$ at $(-1,-3)$, and saddle points at $(1,-3)$ and $(-1,3)$.

Demand: $p=\dfrac{a}{y+c}$, $a,c>0$.

Demand rises without bound as price falls. Solving for the quantity: $y=\dfrac{a}{p}-c$. As the price falls towards zero,

$\lim_{p\to0^+}y=\lim_{p\to0^+}\left(\frac{a}{p}-c\right)=\infty$

So demand becomes infinitely large as the price falls to zero.

Total revenue. $R=py=\dfrac{ay}{y+c}$. Then

$\lim_{y\to\infty}R=\lim_{y\to\infty}\frac{a}{1+\frac{c}{y}}=\frac{a}{1+0}=a$

Thus total revenue approaches the finite limiting value $a$ as the quantity demanded increases indefinitely.

Given $p=\dfrac{a}{x+b}$, with $a,b>0$.

Demand rises without bound as price falls. Solving for $x$: $x+b=\frac{a}{p}\ \Rightarrow\ x=\frac{a}{p}-b.$ As the price falls towards zero ($p\to 0^+$), $\dfrac{a}{p}\to\infty$, so $\lim_{p\to 0^+}x=\lim_{p\to0^+}\left(\frac{a}{p}-b\right)=\infty.$ Hence the quantity demanded becomes infinitely large as the price falls.

Total revenue tends to a limit. Total revenue is $R=px=\frac{ax}{x+b}.$ Dividing numerator and denominator by $x$: $\lim_{x\to\infty}R=\lim_{x\to\infty}\frac{a}{1+\frac{b}{x}}=\frac{a}{1+0}=a.$ So total revenue approaches the finite limiting value $a$ as the quantity demanded increases indefinitely.

Let $P(n):\ 2+4+6+\cdots+2n=n(n+1)$.

Basis step. For $n=1$: LHS $=2$, RHS $=1\cdot2=2$. So $P(1)$ is true.

Induction step. Assume $P(k)$ is true: $2+4+\cdots+2k=k(k+1)$. Add the next term $2(k+1)$ to both sides: $2+4+\cdots+2k+2(k+1)=k(k+1)+2(k+1)=(k+1)(k+2).$ This is $P(k+1)$, since $(k+1)\big((k+1)+1\big)=(k+1)(k+2)$.

By the principle of mathematical induction, $P(n)$ holds for all positive integers $n$.

$f(x)=x^3-6x^2+9x+4$. $f'(x)=3x^2-12x+9=3(x-1)(x-3),\qquad f''(x)=6x-12.$

Critical points: $f'(x)=0\Rightarrow x=1$ or $x=3$.

• At $x=1$: $f''(1)=-6<0$, so a local maximum, with $f(1)=1-6+9+4=8$. Point $(1,8)$.
• At $x=3$: $f''(3)=6>0$, so a local minimum, with $f(3)=27-54+27+4=4$. Point $(3,4)$.

$f(x)=8x^5-15x^4+10x^2$

First derivative: $f'(x)=40x^4-60x^3+20x=20x(2x^3-3x^2+1)=20x(x-1)^2(2x+1).$

(Since $2x^3-3x^2+1=(x-1)(2x^2-x-1)=(x-1)^2(2x+1)$.)

Critical points: $f'(x)=0\Rightarrow x=-\tfrac12,\ 0,\ 1$.

Second derivative: $f''(x)=160x^3-180x^2+20.$

• $x=-\tfrac12$: $f''=-20-45+20=-45<0$, so a local maximum, with $f(-\tfrac12)=-\tfrac14-\tfrac{15}{16}+\tfrac52=\tfrac{21}{16}$.
• $x=0$: $f''=20>0$, so a local minimum, with $f(0)=0$.
• $x=1$: $f''(1)=160-180+20=0$, so the test fails. Look at the sign of $f'=20x(x-1)^2(2x+1)$: it is positive just before $1$ and positive just after $1$ (the factor $(x-1)^2\ge0$ and $x(2x+1)>0$ there). There is no sign change, so $x=1$ is neither a maximum nor a minimum (a point of inflection), with $f(1)=3$.

Let $P(n):\ 1^3+2^3+\cdots+n^3=\left[\dfrac{n(n+1)}{2}\right]^2$.

Basis step. For $n=1$: LHS $=1^3=1$, RHS $=\left[\dfrac{1\cdot2}{2}\right]^2=1$. So $P(1)$ is true.

Induction step. Assume $P(k)$ is true for some $k\ge1$: $1^3+2^3+\cdots+k^3=\left[\frac{k(k+1)}{2}\right]^2.$ Add $(k+1)^3$ to both sides: $1^3+\cdots+k^3+(k+1)^3=\frac{k^2(k+1)^2}{4}+(k+1)^3=\frac{(k+1)^2\left[k^2+4(k+1)\right]}{4}$ $=\frac{(k+1)^2(k^2+4k+4)}{4}=\frac{(k+1)^2(k+2)^2}{4}=\left[\frac{(k+1)(k+2)}{2}\right]^2.$ This is exactly $P(k+1)$.

By the principle of mathematical induction, $P(n)$ is true for all positive integers $n$.

Claim. $P(n):\ 1^2+2^2+\cdots+n^2=\dfrac{n(n+1)(2n+1)}{6}$ for all $n\in\mathbb{N}$.

Step 1 (basis). For $n=1$: LHS $=1^2=1$ and RHS $=\dfrac{1\cdot2\cdot3}{6}=1$. So $P(1)$ is true.

Step 2 (induction hypothesis). Assume $P(k)$ is true for some $k\ge1$:

$1^2+2^2+\cdots+k^2=\frac{k(k+1)(2k+1)}{6}$

Step 3 (induction step). Add $(k+1)^2$ to both sides:

$1^2+\cdots+k^2+(k+1)^2=\frac{k(k+1)(2k+1)}{6}+(k+1)^2$

$=\frac{(k+1)\left[k(2k+1)+6(k+1)\right]}{6}=\frac{(k+1)(2k^2+7k+6)}{6}=\frac{(k+1)(k+2)(2k+3)}{6}$

This is exactly $\dfrac{(k+1)\big((k+1)+1\big)\big(2(k+1)+1\big)}{6}$, i.e. $P(k+1)$.

Conclusion. $P(1)$ is true and $P(k)\Rightarrow P(k+1)$. By the principle of mathematical induction, $P(n)$ is true for all natural numbers $n$. $\blacksquare$

Test continuity at $x=2$.

• $f(2)=2^2=4$
• Left-hand limit ($x<2$): $\lim_{x\to2^-}(3x-2)=6-2=4$
• Right-hand limit ($x>2$): $\lim_{x\to2^+}(x+2)=2+2=4$

Since $\lim_{x\to2^-}f(x)=\lim_{x\to2^+}f(x)=f(2)=4$, the function is continuous at $x=2$.

Boundary lines

• $2x+3y=12$ passes through $(6,0)$ and $(0,4)$.
• $4x+3y=12$ passes through $(3,0)$ and $(0,4)$.

The origin satisfies both $2x+3y\le12$ and $4x+3y\le12$ ($0\le12$), so for each line shade the side containing the origin. Together with $x\ge0,\ y\ge0$ the solution lies in the first quadrant.

Intersection of the lines: subtracting, $2x=0\Rightarrow x=0,\ y=4$, so they meet at $(0,4)$ on the y-axis.

For $x\ge0$ we have $4x+3y\ge 2x+3y$, so whenever $4x+3y\le12$ the first inequality holds automatically. The line $4x+3y=12$ is therefore the binding boundary.

Feasible region (shaded): the triangle with vertices

$O(0,0),\quad A(3,0),\quad B(0,4)$

bounded by the x-axis, the y-axis and the line $4x+3y=12$, including its boundary.

Boundary lines

• $2x+3y=6$ passes through $(3,0)$ and $(0,2)$.
• $3x+4y=12$ passes through $(4,0)$ and $(0,3)$.

The origin satisfies both inequalities ($0\le6$, $0\le12$), so shade the side of each line containing the origin, within the first quadrant ($x\ge0,\ y\ge0$).

The lines meet at $x=12,\ y=-6$ (outside the first quadrant), so they do not cross within the region. Also, in the first quadrant $3x+4y=1.5(2x+3y)-0.5y\le 9-0.5y\le12$ whenever $2x+3y\le6$ (as $1.5\times6=9$), so the second inequality is automatically satisfied.

Feasible region (shaded): the triangle with vertices

$O(0,0),\quad A(3,0),\quad B(0,2)$

bounded by the axes and the line $2x+3y=6$, including the boundary.

Boundary lines

• $4x+3y=12$: passes through $(3,0)$ and $(0,4)$. Test $(0,0)$: $0\le12$ is true, so shade the side containing the origin.
• $4x+y=8$: passes through $(2,0)$ and $(0,8)$. Test $(0,0)$: $0\le8$ is true, so shade the side containing the origin.
• $x\ge0,\ y\ge0$: the first quadrant.

Intersection of the two lines. Subtracting, $2y=4$, so $y=2$ and $4x+2=8$ gives $x=\tfrac32$. The point is $\left(\tfrac32,2\right)$.

Feasible region. The common shaded region (satisfying all four inequalities jointly) is the quadrilateral with vertices $O(0,0),\quad (2,0),\quad \left(\tfrac32,\,2\right),\quad (0,4).$ Its boundary runs along the x-axis from $(0,0)$ to $(2,0)$, up the line $4x+y=8$ to $\left(\tfrac32,2\right)$, then along $4x+3y=12$ to $(0,4)$, and down the y-axis to the origin. This quadrilateral is the area to be shaded.

A function is continuous at $x=a$ if $\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a)$.

At $x=2$:

Left-hand limit: $\lim_{x\to2^-}f(x)=\lim_{x\to2^-}(x-x^2)=2-4=-2.$

Right-hand limit: $\lim_{x\to2^+}f(x)=\lim_{x\to2^+}(-2x)=-4.$

Value: $f(2)=-6$.

Since $-2\neq-4$, the limit $\lim_{x\to2}f(x)$ does not exist (and neither one-sided limit equals $f(2)=-6$).

Hence $f(x)$ is not continuous (it is discontinuous) at $x=2$.

Let M, E, S be the sets of students reading Mathematics, Economics and Statistics, and let $x$ be the number reading all three.

Given: $n(M)=30$, $n(E)=24$, $n(S)=22$, M only $=14$, E only $=8$, M and S only $=6$, M and E only $=2$, none $=8$.

Since $n(M)=14+6+2+x=30$, we get $x=8$.

Now $n(E)=8+2+x+(\text{E and S only})=24$, so E and S only $=24-8-2-8=6$.

Also $n(S)=(\text{S only})+6+6+8=22$, so S only $=2$.

Regions: M only 14, E only 8, S only 2, M∩E only 2, M∩S only 6, E∩S only 6, all three 8, none 8.

(i) Total students in the group $=14+8+2+2+6+6+8+8=\mathbf{54}$.

(ii) Students reading Statistics and Economics only $=\mathbf{6}$.

(iii) Students reading Statistics only $=\mathbf{2}$.

(iv) Students reading all three subjects $=\mathbf{8}$.

Using De Morgan's law and double negation:

$\sim[(\sim p)\wedge q]\equiv \sim(\sim p)\vee(\sim q)\equiv p\vee(\sim q)$

Verification by truth table

The columns for $\sim[(\sim p)\wedge q]$ and $p\vee(\sim q)$ are identical, so

$\sim[(\sim p)\wedge q]\equiv p\vee(\sim q).\ \blacksquare$

For continuity at $x=3$ we need $\lim_{x\to3^-}f(x)=\lim_{x\to3^+}f(x)=f(3)$.

• $f(3)=3(3)=9$
• Left-hand limit ($x<3$): $\lim_{x\to3^-}(4x-3)=12-3=9$
• Right-hand limit ($x>3$): $\lim_{x\to3^+}(2x+3)=6+3=9$

Since $\lim_{x\to3^-}f(x)=\lim_{x\to3^+}f(x)=f(3)=9$, the function is continuous at $x=3$.

Let $s=a+b+c$ and

$\Delta=\begin{vmatrix}a-b-c&2a&2a\\2b&b-c-a&2b\\2c&2c&c-a-b\end{vmatrix}$

Apply $R_1\to R_1+R_2+R_3$. The column sums are $a-b-c+2b+2c=s$, $2a+b-c-a+2c=s$, and $2a+2b+c-a-b=s$. So

$\Delta=\begin{vmatrix}s&s&s\\2b&b-c-a&2b\\2c&2c&c-a-b\end{vmatrix}=s\begin{vmatrix}1&1&1\\2b&b-c-a&2b\\2c&2c&c-a-b\end{vmatrix}$

Apply $C_2\to C_2-C_1$, $C_3\to C_3-C_1$:

$\Delta=s\begin{vmatrix}1&0&0\\2b&-s&0\\2c&0&-s\end{vmatrix}$

since $b-c-a-2b=-(a+b+c)=-s$ and $c-a-b-2c=-s$. Expanding along the first row:

$\Delta=s\cdot1\cdot\left[(-s)(-s)-0\right]=s^3=(a+b+c)^3.\ \blacksquare$

Let $u=\sqrt{x^2+y^2+z^2}$, so $u^2=x^2+y^2+z^2$. Differentiating with respect to $x$: $2u\,u_x=2x$, so $u_x=\dfrac{x}{u}$.

Differentiate again: $u_{xx}=\frac{u-x\,u_x}{u^2}=\frac{u-\frac{x^2}{u}}{u^2}=\frac{u^2-x^2}{u^3}.$ By symmetry, $u_{yy}=\frac{u^2-y^2}{u^3},\qquad u_{zz}=\frac{u^2-z^2}{u^3}.$ Adding: $u_{xx}+u_{yy}+u_{zz}=\frac{3u^2-(x^2+y^2+z^2)}{u^3}=\frac{3u^2-u^2}{u^3}=\frac{2u^2}{u^3}=\frac{2}{u}.$ Hence proved.

$A=[-1,3]$ and $B=(2,7)$.

$A\cap B=(2,3]$ $A-B=[-1,2]$ (the elements of $A$ not in $B$: $-1\le x\le2$, and $2\notin B$ so $2$ is included).

Logical equivalence: Two compound statements $A$ and $B$ are logically equivalent, written $A\equiv B$, if they have the same truth value for every assignment of truth values to their component statements (equivalently, $A\Leftrightarrow B$ is a tautology).

To show: $\sim(p\wedge q)\equiv(\sim p)\vee(\sim q)$

The columns for $\sim(p\wedge q)$ and $(\sim p)\vee(\sim q)$ are identical, so $\sim(p\wedge q)\equiv(\sim p)\vee(\sim q).$ This is De Morgan's law.