MaxWong

g(0) is not well-defined since f^{-1}(0) does not exist in [0, \infty), i.e., there is no x \in [0, \infty) such that f(x) = 0.

Proof:

Since \(x\in [0, \infty)\), \(x^3, x^2, x\) are all nonnegative.

Then \(f(x) = x^3 + x^2 + x + 2 \geq 0 + 0 + 0 + 2 = 2 > 0\) for any \(x\in [0, \infty)\).

Then we have \(f(x) > 0\) for any \(x\in [0, \infty)\).

This implies there are no \(x\in [0, \infty)\) such that \(f(x) = 0\).

Therefore, we cannot find the value of \(g(0) + g(3)\).

Let \(h\) be the original height and suppose a decrease of \(k\%\) is required.

Let \(r\) be the original radius,

Now,

\(\pi r^2 h = \pi (r(1 + 36\%))^2(h(1 - k\%))\\ (1.36)^2 (1 - k\%) = 1\\ k\% = 1 - \dfrac1{1.36^2}\\ k\% = \dfrac{531}{1156}\\ k = \dfrac{13275}{289}\)

Therefore, the height needs to be decreased by \(\dfrac{13275}{289}\%\), which is approximately \(45.9\%\).

We divide the problem into cases.

Case 1: f(x) > 0.

Then,

\(\begin{array}{rcl} f(f(x)) &=& 4\\ 2f(x) - 15 &=& 4\\ f(x) &=& \dfrac{19}2 \end{array}\)

Case 2: f(x) <= 0.

Then,

\(\begin{array}{rcl} f(f(x)) &=& 4\\ -f(x) + 3&=& 4\\ f(x) &=& -1 \end{array}\)

Then f(x) = 19/2 or f(x) = -1.

We split the problem into 2 cases again.

Case A: x > 0.

Then,

\(2x - 15 = \dfrac{19}2\text{ or }2x - 15 = -1\\ x = \dfrac{49}4\text{ or }x = 7\)

Case B: x <= 0.

Then,

\(-x+3= \dfrac{19}2\text{ or } -x+3 = -1\\ x = -\dfrac{13}2\text{ or }x = 4\text{ (rejected)}\)

Then, solution set = \(\left\{\dfrac{49}4, 7, -\dfrac{13}2\right\}\). So there are a total of 3 solutions.

The main diagonal of a cube is \(\sqrt 3\) times longer than its side length.

Therefore, the side length of cube = \(\dfrac2{\sqrt 3}\).

Now, we can calculate the surface area by the formula \(\text{surface area of cube} = 6(\text{side length})^2\).

\(\text{surface area of cube} = 6\left(\dfrac2{\sqrt 3}\right)^2 = 8\)

Since the amount of paint used is exactly the same as the amount of paint to paint the surface of a sphere, the surface area of the sphere is the same as that of the cube.

\(\text{surface area of sphere} = 8\).

Let r be the radius of the sphere. Then,

\(4\pi r^2 = 8\\ r^2 = \dfrac2\pi\\ r = \sqrt{\dfrac2\pi}\)

Therefore, radius of sphere = \(\sqrt{\dfrac{2}\pi}\).

Happy birthday, Melody!

We calculate the area of the triangle.

Area of triangle JKL = \(\dfrac12 \cdot 10 \cdot \sqrt{25^2 - \left(\dfrac{10}2\right)^2} = 50\sqrt 6\)

Then, we use the formula \(\text{circumradius} = \dfrac{\text{product of 3 sides}}{4(\text{area})}\)

\(\text{circumradius} = \dfrac{25(25)(10)}{4(50\sqrt 6)} = \dfrac{125}{24}\sqrt6\)

We visualize the equation as a circle on the coordinate plane.

We first need to find out an explicit form of g(2x).

\(\begin{array}{rcl} g(x) &=& \sqrt{\dfrac{x + 3}{4}} \end{array}\)

Replace x by 2x:

\(\begin{array}{rcl} g(2x) &=& \sqrt{\dfrac{2x+ 3}{4}} \end{array}\)

Now we know what g(2x) and g(x) are, we proceed to solve the equation.

\(\begin{array}{rcl} g(2x) &=& 2g(x)\\ \sqrt{\dfrac{2x + 3}4} &=& 2\sqrt{\dfrac{x + 3}{4}}\\ \dfrac{\sqrt{2x + 3}}2 &=& \sqrt{x + 3}\\ \left(\dfrac{\sqrt{2x + 3}}2\right)^2 &=& \left(\sqrt{x + 3}\right)^2\\ \dfrac{2x + 3}4&=&x + 3\\ 2x+3&=&4(x+3)\\ 2x+3&=&4x+12\\ -9&=&2x\\ x &=& -\dfrac92 \end{array}\)

However, when we substitute \(x =-\dfrac92\) into the equation, we see that \(x = -\dfrac92\) is not in the domain of g(x).

Hence, it is an extraneous root and needs to be rejected.

There is no such value of x that satisfies g(2x) = 2g(x).

Note that \(168|a\) and \(88|a\) by the property of gcd.

Take \(a = \operatorname{lcm}(168, 88) = 1848\).

Then \(b = 168\) and \(c = 88\) satisfies the conditions. You can check that gcd(b, c) is minimum in this case because for any positive integers \(k,l\), \(\gcd(kb, lc) = \gcd(k, l)\gcd(b, c)\gcd\left(\dfrac{k}{\gcd(k,l)}, \dfrac{c}{\gcd(b, c)}\right)\gcd\left(\dfrac{b}{\gcd(b, c)},\dfrac{l}{\gcd(k,l)}\right)\geq \gcd(b, c)\), and b = 168k for some positive integer k and c = 88l for some positive integer l.

\(\min\gcd(b, c)=\gcd(168, 88) = 8\).

Here are the first 6 primes: \(\{2,3,5,7,11,13\}\)

We can take a look at all the possibilities by making a table. 

\(\begin{matrix}&\color{blue}2&\color{blue}3&\color{blue}5&\color{blue}7&\color{blue}11&\color{blue}13\\\color{red}2&4&5&7&\color{magenta}9&13&\color{magenta}15\\\color{red}3&5&\color{magenta}6&8&10&14&16\\\color{red}5&7&8&10&\color{magenta}12&16&\color{magenta}18\\\color{red}7&\color{magenta}9&10&\color{magenta}12&14&\color{magenta}18&20\\\color{red}11&13&14&16&\color{magenta}18&22&\color{magenta}24\\\color{red}13&\color{magenta}15&16&\color{magenta}18&20&\color{magenta}24&26\end{matrix}\)

Blue numbers and red numbers represent the number that Paul chose and the number that Jesse chose respectively.

Pink numbers are those divisible by 3.

By trial-and-error, out of all 36 possible cases, 13 cases has a sum divisible by 3.

Therefore, 

\(\text{Prob}(\text{sum divisible by 3}) = \dfrac{13}{36}\)