Gaokao Mathematics

1. There are nine increases between the first and tenth terms.

   $$a_{10}=3+9\cdot2=21$$

2. Pair the first and last terms. Their average is the average of all ten terms.

   $$\frac{a_1+a_{10}}{2}=\frac{3+21}{2}=12$$

3. Multiply that average by the number of terms.

   $$S_{10}=10\cdot12=120$$

Common mistake: Do not use ten increases to find the tenth term. The first term is already counted.

1. Perpendicular nonzero vectors have a dot product of zero.

   $$\mathbf a\cdot\mathbf b=0$$

2. Multiply matching coordinates and add the products.

   $$2x+3(-2)=0$$

3. Solve the resulting linear equation.

   $$2x=6\quad\Rightarrow\quad x=3$$

Common mistake: The dot product is a single number. Do not multiply the coordinates across diagonally.

1. Use the multiplication rule with the given conditional probability.

   $$P(A\cap B)=P(A)P(B\mid A)$$

2. Substitute the two fractions.

   $$P(A\cap B)=\frac34\cdot\frac23$$

3. Simplify the result. No independence assumption is needed.

   $$P(A\cap B)=\frac{6}{12}=\frac12$$

Common mistake: Do not replace P(B given A) with P(B). Those probabilities need not be equal.

1. Compare the equation with the standard form to identify the parameter p.

   $$y^2=2px\quad\Rightarrow\quad p=4$$

2. The focus has x-coordinate p/2, and the directrix is the vertical line x = -p/2.

   $$F=(2,0),\qquad x=-2$$

3. Subtract the two horizontal coordinates to get the distance.

   $$d=\lvert 2-(-2)\rvert=4$$

Common mistake: The distance from the vertex to the focus is 2. The requested distance goes all the way from the focus to the directrix.

1. Differentiate and factor the result.

   $$\begin{aligned}f'(x)&=3x^2-3\\&=3(x-1)(x+1)\end{aligned}$$

2. The derivative is positive when its two factors have the same sign.

   $$x<-1\quad\lor\quad x>1$$

3. Write the two intervals separately. The derivative is negative between the critical points.

   $$(-\infty,-1),\quad(1,\infty)$$

Common mistake: A positive derivative describes an increase in the function, not a positive value of the function itself.

1. Differentiate using the product rule.

   $$f'(x)=e^{-x}(1-x)$$

2. The exponential factor is always positive. The derivative changes from positive to negative at x = 1.

   $$f'(x)\begin{cases}>0&x<1\\=0&x=1\\<0&x>1\end{cases}$$

3. The function increases before 1 and decreases after 1. Substitute 1 to obtain its maximum value.

   $$\max_{x\in\mathbb R} f(x)=f(1)=\frac1e$$

Common mistake: The x-value of the maximum is 1, but the maximum value of the function is 1/e.