Set 3

Point $(a, b)$ on the curve $y = x^2 + 2$ has the closest distance to the line $y = x$ . The value of $a + b$ that satisfies is....

If the quadratic equation $ax^2 + bx + c = 0$ has no real roots, then the graph of the function $y = ax^2 + bx + c$ is tangent to the line $y = -x$ when....

$b < -\frac{1}{2}$

$-\frac{1}{2} < b < 0$

$b > -\frac{1}{2}$

$0 < b < \frac{1}{2}$

$b > 0$

The equation of the tangent line to the parabola $y = \sqrt{x} + 1$ passing through the point $(-8, 0)$ is....

$4y - x - 8 = 0$

$4y + x - 8 = 0$

$4y + 3x - 2 = 0$

$4y - x + 2 = 0$

$4y - 3x - 2 = 0$

Given the function $mx^2 - 2x^2 + 2mx + m - 3$ . For the function to always be below the $x$ -axis, the possible value of $m$ is....

If a function $y = \sqrt{x^2 - 7}$ , then

$y = \frac{4}{3}x - \frac{7}{3}$ is the equation of the tangent line at $x = 4$ .
The curve is a circle centered at $(0, 0)$ .
The line $y = -\frac{3}{4}x + 6$ intersects perpendicularly the tangent line at $x = 4$ .
$y = \frac{4}{3}x - \frac{25}{3}$ is the tangent line of the curve at $(4, -3)$ .

Given $a, \frac{1}{a}, \frac{1}{a^2 + 2a}$ , $a \neq 0$ , are respectively the 3rd, 4th, and 5th terms of a geometric sequence with ratio $r \neq 1$ . The product of the first five terms of the geometric sequence is....

Let $u_n$ denote the $n$ -th term of an arithmetic sequence. Given $u_1 \times u_2 = 10$ and $u_1 \times u_3 = 16$ . If the terms of the arithmetic sequence are positive numbers, then $u_{10} = ....$

The roots of the equation $x^3 - 7x^2 + px + q = 0$ form a geometric sequence with ratio $2$ . The value of $p + q$ is....

The sum of an infinite geometric series has a value of $\frac{9}{4}$ . With the first term $u_1 = a$ and ratio $r = -\frac{1}{a}$ . If $a > 0$ , determine the value of $3u_6 - u_5$ .

If the roots of the polynomial equation $x^3 - 12x^2 + (p + 4)x - (p + 8) = 0$ form an arithmetic sequence with common difference $2$ , then $p - 36 = ....$

The value of $\lim_{x \to \frac{\pi}{2}} \frac{\sec 2x + 2}{\tan 2x}$ is....

If $p > 0$ and $\lim_{x \to p} \frac{x^3 + px^2 + qx}{x - p} = 12$ , then the value of $p - q$ is....

The value of $\lim_{x \to -y} \frac{\tan x + \tan y}{\frac{x^2 - y^2}{-2y^2} \cdot (1 - \tan x \tan y)}$ is....

If $b, c \neq 0$ and $\lim_{x \to a} \frac{(x - a) \tan(b(a - x))}{\cos(c(x - a)) - 1} = d$ , then $b = ....$

The value of $\lim_{x \to 3} \frac{(x + 6) \tan(2x - 6)}{x^2 - x - 6}$ is....

The inequality $\log_2(x^2 - x) \leq 1$ has a solution....

$x < 0$ or $x > 1$

$-1 < x < 2; x \neq 1; x \neq 0$

$-1 \leq x < 0$ or $1 < x \leq 2$

$-1 \leq x \leq 0$ or $1 \leq x \leq 2$

$-1 < x < 0$ or $1 \leq x < 2$

If $x > y \geq 1$ and $\log(x^2 + y^2 + 2xy) = 2 \log(x^2 - y^2)$ , then $\log_x(1 + y) = ....$

If $\log_3 x + \log_4 y^2 = 5$ , then the maximum value of $\log_3 x \cdot \log_2 y$ is....

The solution set of the inequality $\log_{\frac{1}{2}}(2x - 1) + \log_{\frac{1}{2}}(2 - x) \geq 2 \log_{\frac{1}{2}} x$ is....

$\frac{2}{3} \leq x \leq 1$

$x \leq \frac{2}{3}$ or $x \geq 1$

$\frac{1}{2} < x \leq \frac{2}{3}$ or $1 \leq x < 2$

$\frac{1}{2} \leq x \leq \frac{2}{3}$ or $1 \leq x \leq 2$

$x \leq \frac{1}{2}$ or $x > 2$

If $x_1$ and $x_2$ satisfy the equation $(2 \log x - 1) \cdot \frac{1}{\log_x 10} = \log 10$ , then $x_1x_2 = ....$

The solution set of the inequality $|x - 5|^2 - 3|x - 5| + 2 < 0$ is....

$(3,4) \cup [6,7)$

$(3,4) \cup (6,7)$

$(1,2) \cup (3,4]$

$(-\infty, 1) \cup [6,\infty)$

$(-\infty, 2) \cup (3,7)$

All values of $x$ that satisfy $|x| + |x - 2| > 3$ are....

$x < -1$ or $x > \frac{5}{2}$

$x < -\frac{1}{2}$ or $x > 3$

$x < -\frac{1}{2}$ or $x > \frac{5}{2}$

$x < -1$ or $x > 3$

$x < -\frac{3}{2}$ or $x > \frac{5}{2}$

Explanation

The definition of absolute value $|x - 2|$ is

|x - 2| = \begin{cases} x - 2, & \text{for } x \geq 2 \\ -(x - 2), & \text{for } x < 2 \end{cases}

To facilitate finding the solution, we divide it into 3 regions based on the definition of absolute value above

Region Division

Three regions based on absolute value definition:

x < 0

0 \leq x < 2

, and

x \geq 2

\text{Region I}

\text{Region II}

\text{Region III}

0

2

\text{Region I:}

(x < 0)

|x| = -x

|x - 2| = -(x - 2) = -x + 2

|x| + |x - 2| > 3

-x + (-x + 2) > 3

-2x + 2 > 3

-2x > 1

x < -\frac{1}{2}

So the solution set is $\{x < 0\} \cap \{x < -\frac{1}{2}\} = \{x < -\frac{1}{2}\}$ .

\text{Region II:}

(0 \leq x < 2)

|x| = x

|x - 2| = -(x - 2) = -x + 2

|x| + |x - 2| > 3

x + (-x + 2) > 3

2 > 3

This means there is no value of $x$ that satisfies region II.

\text{Region III:}

(x \geq 2)

|x| = x

|x - 2| = x - 2

|x| + |x - 2| > 3

x + (x - 2) > 3

2x > 5

x > \frac{5}{2}

So the solution set is $\{x \geq 2\} \cap \{x > \frac{5}{2}\} = \{x > \frac{5}{2}\}$ .

So the total solution is the union of the three regions

\{x < -\frac{1}{2}\} \cup \{x > \frac{5}{2}\}

So the solution is $\{x < -\frac{1}{2} \cup x > \frac{5}{2}\}$ .

All values of $x$ that satisfy $|x + 1| > x + 3$ and $|x + 2| < 3$ are....

All real numbers $x$ that satisfy $|2x + 1| < 5 - |2x|$ are....

$-\frac{3}{2} < x < 1$

$-\frac{5}{2} < x < 3$

$-\frac{7}{2} < x < 5$

$x < -\frac{3}{2} \cup x > 2$

$x < -\frac{5}{2} \cup x > 3$

Explanation

Define each absolute value

|2x + 1| = \begin{cases} 2x + 1, & \text{for } 2x + 1 \geq 0 \text{ or } x \geq -\frac{1}{2} \\ -(2x + 1), & \text{for } 2x + 1 < 0 \text{ or } x < -\frac{1}{2} \end{cases}

|2x| = \begin{cases} 2x, & \text{for } 2x \geq 0 \text{ or } x \geq 0 \\ -2x, & \text{for } 2x < 0 \text{ or } x < 0 \end{cases}

It is seen that the boundary values of $x$ from both absolute values are $x = -\frac{1}{2}$ and $x = 0$ . This means the boundaries will form three regions, namely $\text{Region I } (x < -\frac{1}{2})$ , $\text{Region II } (-\frac{1}{2} \leq x < 0)$ , and $\text{Region III } (x \geq 0)$ . Based on the regions and their definitions, the problem can be solved.

\text{Region I:}

(x < -\frac{1}{2})

|2x + 1| = -(2x + 1)

|2x| = -2x

|2x + 1| < 5 - |2x|

-(2x + 1) < 5 - (-2x)

-2x - 1 < 5 + 2x

-4x < 6

x > -\frac{3}{2}

The solution for Region I is $\{x < -\frac{1}{2}\} \cap \{x > -\frac{3}{2}\} = \{-\frac{3}{2} < x < -\frac{1}{2}\}$ .

\text{Region II:}

(-\frac{1}{2} \leq x < 0)

|2x + 1| = 2x + 1

|2x| = -2x

|2x + 1| < 5 - |2x|

2x + 1 < 5 - (-2x)

2x + 1 < 5 + 2x

1 < 5

All $x$ in Region II are true, so the solution is $\{-\frac{1}{2} \leq x < 0\}$ .

\text{Region III:}

(x \geq 0)

|2x + 1| = 2x + 1

|2x| = 2x

|2x + 1| < 5 - |2x|

2x + 1 < 5 - 2x

4x < 4

x < 1

The solution is $\{x \geq 0\} \cap \{x < 1\} = \{0 \leq x < 1\}$ .

So the overall solution is the union

\left\{-\frac{3}{2} < x < -\frac{1}{2}\right\} \cup \left\{-\frac{1}{2} \leq x < 0\right\} \cup \{0 \leq x < 1\} = \left\{-\frac{3}{2} < x < 1\right\}

The solution set of $\left|\frac{x}{3} - \frac{1}{4}\right| = \frac{1}{12}$ is....

$\{-\frac{1}{2}, 2\}$

$\{2, 3\}$

$\{1, 2\}$

$\{-\frac{3}{2}, \frac{1}{3}\}$

$\{\frac{1}{2}, 1\}$

Given the system of equations

x + y^2 = y^3

y + x^2 = x^3

The number of real pairs $(x, y)$ that satisfy the system of equations above is....

Explanation

Multiply the first equation by $x^2$ and the second equation by $y^2$

(x + y^2 = y^3) \times x^2 \Rightarrow x^3 + x^2y^2 = x^2y^3

(y + x^2 = x^3) \times y^2 \Rightarrow y^3 + x^2y^2 = x^3y^2

Subtract the second equation from the first equation

\begin{aligned} (x^3 + x^2y^2) - (y^3 + x^2y^2) &= x^2y^3 - x^3y^2 \\ x^3 - y^3 &= x^2y^3 - x^3y^2 \end{aligned}

Factor both sides

x^3 - y^3 = (x - y)(x^2 + xy + y^2)

x^2y^3 - x^3y^2 = x^2y^2(y - x) = -x^2y^2(x - y)

Substitute into the equation

(x - y)(x^2 + xy + y^2) = -x^2y^2(x - y)

(x - y)(x^2 + xy + y^2) + x^2y^2(x - y) = 0

(x - y)(x^2 + xy + y^2 + x^2y^2) = 0

So there are two cases

x - y = 0 \quad \text{or} \quad x^2 + xy + y^2 + x^2y^2 = 0

For the first case, $x = y$ . Substitute $y = x$ into the first equation

x + y^2 = y^3

x + x^2 = x^3

x^3 - x^2 - x = 0

x(x^2 - x - 1) = 0

So $x = 0$ or $x^2 - x - 1 = 0$ .

For $x^2 - x - 1 = 0$ , the discriminant value is

D = b^2 - 4ac = (-1)^2 - 4(1)(-1) = 1 + 4 = 5

The discriminant value $D = 5 > 0$ , which means the quadratic equation has 2 distinct real roots.

So from the first case we obtain:

$x = 0$ , then $y = 0$ , so the pair $(0, 0)$
$x^2 - x - 1 = 0$ has 2 distinct real roots, so 2 pairs $(x, y)$ with $y = x$

For the second case $x^2 + xy + y^2 + x^2y^2 = 0$ , since all terms are non-negative and $x^2y^2 \geq 0$ , this equation is only satisfied if $x = 0$ and $y = 0$ , which is already included in the first case.

So the system of equations has 3 distinct pairs.

Let $\alpha$ and $\beta$ be the roots of the equation $x^2 - bx + 6 = 0$ . If $\alpha + \beta$ and $\alpha - \beta$ are the roots of the equation $x^2 - 4x + c = 0$ , the equation whose roots are $b$ and $c$ is....

$x^2 - 5x + 5 = 0$

$(x - 5)^2 = 0$

$x^2 - 5^2 = 0$

$(x + 5)^2 = 0$

$x^2 + 5x + 5 = 0$

What is the value of $a$ such that the solution $(x, y)$ of the system of equations

-2x + y = a^2 - 1

3x + 2y = 2a^2 + 7a + 5

satisfies $x\sqrt{y} + 3 > 0$ ?

$a > -1 + \sqrt{3}$

$a < -1 - \sqrt{3}$

$a < 1 - \sqrt{3}$

$a > 1 + \sqrt{3}$

$a < -\sqrt{3}$

Explanation

This system of linear equations can be written in matrix form $AX = B$ , where

A = \begin{pmatrix} -2 & 1 \\ 3 & 2 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \end{pmatrix}, \quad B = \begin{pmatrix} a^2 - 1 \\ 2a^2 + 7a + 5 \end{pmatrix}

Cramer's rule can be used because the determinant of the coefficient matrix $\det(A) = -4 - 3 = -7 \neq 0$ (determinant is nonzero). Cramer's rule states that for the system $AX = B$ with $\det(A) \neq 0$ , the solution is

x = \frac{\det(A_x)}{\det(A)}, \quad y = \frac{\det(A_y)}{\det(A)}

where $A_x$ is the matrix $A$ with the first column replaced by $B$ , and $A_y$ is the matrix $A$ with the second column replaced by $B$ .

By using Cramer's rule, we obtain

x = \frac{\begin{vmatrix} a^2 - 1 & 1 \\ 2a^2 + 7a + 5 & 2 \end{vmatrix}}{\begin{vmatrix} -2 & 1 \\ 3 & 2 \end{vmatrix}} = \frac{2(a^2 - 1) - (2a^2 + 7a + 5)}{-4 - 3} = \frac{-7a - 7}{-7} = a + 1

y = \frac{\begin{vmatrix} -2 & a^2 - 1 \\ 3 & 2a^2 + 7a + 5 \end{vmatrix}}{\begin{vmatrix} -2 & 1 \\ 3 & 2 \end{vmatrix}} = \frac{-2(2a^2 + 7a + 5) - 3(a^2 - 1)}{-7} = \frac{-7a^2 - 14a - 7}{-7} = a^2 + 2a + 1 = (a + 1)^2

Next, substitute the values of $x$ and $y$ into $x\sqrt{y} + 3 > 0$ to obtain

x\sqrt{y} + 3 > 0

(a + 1)\sqrt{(a + 1)^2} + 3 > 0

To simplify $\sqrt{(a + 1)^2}$ , note that $\sqrt{(a + 1)^2} = |a + 1|$ . For $\sqrt{y}$ to be defined, we must have $y \geq 0$ , that is $(a + 1)^2 \geq 0$ , which is always true.

Choose $a < -1$ so that the form $\sqrt{(a + 1)^2} = -(a + 1)$ , thus we obtain

(a + 1)[-(a + 1)] + 3 > 0

-(a + 1)^2 + 3 > 0

(a + 1)^2 < 3

a^2 + 2a + 1 < 3

a^2 + 2a - 2 < 0

To find the value of $a$ , we can use the quadratic formula. The quadratic equation $a^2 + 2a - 2 = 0$ has roots

a_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

a_{1,2} = \frac{-2 \pm \sqrt{2^2 - 4(1)(-2)}}{2(1)}

a_{1,2} = \frac{-2 \pm \sqrt{4 + 8}}{2}

a_{1,2} = \frac{-2 \pm \sqrt{12}}{2}

a_{1,2} = \frac{-2 \pm 2\sqrt{3}}{2}

a_{1,2} = -1 \pm \sqrt{3}

For the inequality $a^2 + 2a - 2 < 0$ , since the coefficient of $a^2$ is positive, the inequality is satisfied for values of $a$ between its two roots, that is $-1 - \sqrt{3} < a < -1 + \sqrt{3}$ .

Since we chose $a < -1$ , the intersection is $-1 - \sqrt{3} < a < -1$ .

However, since $a = -1 - \sqrt{3}$ is a root of the equation $a^2 + 2a - 2 = 0$ , and for $a < -1$ we require $(a + 1)^2 < 3$ , the values of $a$ that satisfy are $a < -1 - \sqrt{3}$ .

Therefore, the values of $a$ that satisfy are $a < -1 - \sqrt{3}$ .

Given the quadratic equation $x^2 - 4(k + 1)x + k^2 - k + 7 = 0$ where one root is three times the other root and all roots are greater than $2$ . The set of all values of $k$ that satisfy is....

No value satisfies

$\left\{k \in \mathbb{R} \mid k < \frac{-3 - \sqrt{13}}{2} \text{ or } k > \frac{-3 + \sqrt{13}}{2}\right\}$

$\{k \in \mathbb{R} \mid k > -1\}$

$\{-4, \frac{1}{2}\}$

$\left\{\frac{1}{2}\right\}$

Both roots of the quadratic equation $(m + 2)x^2 - (2m - 1)x + m + 1 = 0$ are negative. The range of values of $m$ that satisfies this is....

$m < -2 \text{ or } m > -1$

$-2 < m < -1$

$-2 < m < -\frac{1}{2}$

$-2 < m \leq -\frac{7}{16}$

$-1 < m \leq -\frac{7}{16}$

Explanation

Given the quadratic equation $(m + 2)x^2 - (2m - 1)x + m + 1 = 0$ with the condition that $x_1 < 0$ and $x_2 < 0$ .

For the equation to be a quadratic equation, the coefficient of $x^2$ must not be zero, that is $m + 2 \neq 0$ or $m \neq -2$ .

The discriminant condition is $D \geq 0$ , then we obtain

D = b^2 - 4ac \geq 0

[-(2m - 1)]^2 - 4(m + 2)(m + 1) \geq 0

(2m - 1)^2 - 4(m + 2)(m + 1) \geq 0

4m^2 - 4m + 1 - 4(m^2 + 3m + 2) \geq 0

4m^2 - 4m + 1 - 4m^2 - 12m - 8 \geq 0

-16m - 7 \geq 0

-16m \geq 7

m \leq -\frac{7}{16}

Condition for sum of roots: $x_1 + x_2 < 0$

x_1 + x_2 = -\frac{b}{a} = -\frac{-(2m - 1)}{m + 2} = \frac{2m - 1}{m + 2}

\frac{2m - 1}{m + 2} < 0

The inequality $\frac{2m - 1}{m + 2} < 0$ is satisfied if the numerator and denominator have different signs. Then

(2m - 1)(m + 2) < 0

This inequality is satisfied for $-2 < m < \frac{1}{2}$ .

Condition for product of roots: $x_1x_2 > 0$

x_1x_2 = \frac{c}{a} = \frac{m + 1}{m + 2}

\frac{m + 1}{m + 2} > 0

The inequality $\frac{m + 1}{m + 2} > 0$ is satisfied if the numerator and denominator have the same sign. Then

(m + 1)(m + 2) > 0

This inequality is satisfied for $m < -2$ or $m > -1$ .

Now we combine all conditions:

$m \neq -2$ (for the equation to be quadratic)
$m \leq -\frac{7}{16}$ (from discriminant)
$-2 < m < \frac{1}{2}$ (from sum of roots)
$m < -2$ or $m > -1$ (from product of roots)

The intersection of all these conditions is $-1 < m \leq -\frac{7}{16}$ .

Therefore, the values of $m$ that satisfy are $-1 < m \leq -\frac{7}{16}$ .

Given that the polynomial $f(x)$ when divided by $x^2 + x - 2$ leaves a remainder of $ax + b$ , and when divided by $x^2 - 4x + 3$ leaves a remainder of $2bx + a - 1$ . If $f(-2) = 7$ , then $a^2 + b^2$ = ....

If $h(x)$ is the remainder of dividing $f(x) = 5x^4 - 2x^2 + 7x + 9$ by $x^2 - 5$ . The value of $h(1)$ is....

Function $f(x)$ divided by $(x - 1)$ has a remainder of $3$ , whereas if divided by $(x - 2)$ has a remainder of $4$ . If $f(x)$ is divided by $x^2 - 3x + 2$ , then the remainder is....

A third-degree polynomial $P(x) = x^3 + 2x^2 + mx + n$ divided by $x^2 - 4x + 3$ has a remainder of $3x + 2$ . Then the value of $n$ = ....

Given the polynomial $g(x) = x^3 + x^2 - x + b$ is divisible by $(x - 1)$ . If $g(x)$ is divided by $(x^2 - 1)$ , then the remainder is....

The maximum value of the function $y = 4 \sin x \sin(x - 60°)$ is achieved when the value of $x$ = ....

$x = 30^\circ + k \cdot 180^\circ$ , $k$ is an integer

$x = 60^\circ + k \cdot 180^\circ$ , $k$ is an integer

$x = 90^\circ + k \cdot 180^\circ$ , $k$ is an integer

$x = 120^\circ + k \cdot 180^\circ$ , $k$ is an integer

$x = 150^\circ + k \cdot 180^\circ$ , $k$ is an integer

Explanation

Determine the derivative of the function

y = 4 \sin x \sin(x - 60°)

y' = 4[\cos x \cdot \sin(x - 60°) + \sin x \cdot \cos(x - 60°)]

Use trigonometric identities $\sin A \cos B + \cos A \sin B = \sin(A + B)$ and $\cos A \cos B - \sin A \sin B = \cos(A + B)$

y' = 4[\sin(x - 60°) \cos x + \cos(x - 60°) \sin x]

y' = 4 \sin[(x - 60°) + x]

y' = 4 \sin(2x - 60°)

Or using the identity $\sin A \sin B = \frac{1}{2}[\cos(A - B) - \cos(A + B)]$

y = 4 \sin x \sin(x - 60°)

y = 4 \cdot \frac{1}{2}[\cos(x - (x - 60°)) - \cos(x + (x - 60°))]

y = 2[\cos(60°) - \cos(2x - 60°)]

y = 2 \cos 60° - 2 \cos(2x - 60°)

y' = -2 \cdot (-\sin(2x - 60°)) \cdot 2

y' = 4 \sin(2x - 60°)

The maximum value occurs when $y' = 0$ or when $\sin(2x - 60°)$ reaches its maximum value, which is $\sin(2x - 60°) = 1$ .

However, since $y = 2 \cos 60° - 2 \cos(2x - 60°) = 1 - 2 \cos(2x - 60°)$ , the maximum value occurs when $\cos(2x - 60°)$ is minimum, which is $\cos(2x - 60°) = -1$ .

\cos(2x - 60°) = -1

2x - 60° = 180° + k \cdot 360°

2x = 240° + k \cdot 360°

x = 120° + k \cdot 180°

With $k$ an integer.

The values of $x$ , for $0° \leq x \leq 360°$ that satisfy $\sin x + \sin 2x > \sin 3x$ are....

$0° < x < 120°, 180° < x < 240°$

$0° < x < 150°, 180° < x < 270°$

$120° < x < 180°, 240° < x < 360°$

$150° < x < 180°, 270° < x < 360°$

$0° < x < 135°, 180° < x < 270°$

Explanation

Given the form of the inequality

\sin x + \sin 2x > \sin 3x

Subtract $\sin x$ from both sides

\sin 2x > \sin 3x - \sin x

Use trigonometric identities $\sin 2x = 2 \sin x \cos x$ and $\sin A - \sin B = 2 \cos \frac{A + B}{2} \sin \frac{A - B}{2}$

2 \sin x \cos x > 2 \cos \frac{3x + x}{2} \sin \frac{3x - x}{2}

2 \sin x \cos x > 2 \cos 2x \sin x

Divide both sides by $2$ and use the identity $\cos 2x = 2 \cos^2 x - 1$

\sin x \cos x > \cos 2x \sin x

\sin x \cos x > (2 \cos^2 x - 1) \sin x

\sin x \cos x - (2 \cos^2 x - 1) \sin x > 0

\sin x (\cos x - 2 \cos^2 x + 1) > 0

\sin x (2 \cos^2 x - \cos x - 1) < 0

Factor $2 \cos^2 x - \cos x - 1 = (2 \cos x + 1)(\cos x - 1)$

\sin x (2 \cos x + 1)(\cos x - 1) < 0

Thus, the zeros are

\sin x = 0 \Rightarrow x = 0°, 180°, 360°

2 \cos x + 1 = 0 \Rightarrow \cos x = -\frac{1}{2} \Rightarrow x = 120°, 240°

\cos x - 1 = 0 \Rightarrow \cos x = 1 \Rightarrow x = 0°, 360°

The critical points are $x = 0°$ , $x = 120°$ , $x = 180°$ , $x = 240°$ , and $x = 360°$ .

The number line is as follows

Number Line

Critical points:

0°

120°

180°

240°

, and

360°

\text{Interval 1}

\text{Interval 2}

\text{Interval 3}

\text{Interval 4}

0°

120°

120°

180°

180°

240°

240°

360°

By testing the sign on each interval, we obtain that the inequality $\sin x (2 \cos x + 1)(\cos x - 1) < 0$ is satisfied for $0° < x < 120°$ and $180° < x < 240°$ .

So the solution set is $0° < x < 120°, 180° < x < 240°$ .

If $\sin x - \sin y = -\frac{1}{3}$ and $\cos x - \cos y = \frac{1}{2}$ , then the value of $\sin(x + y)$ = ....

Explanation

Recall the concept of identities

\sin x - \sin y = 2 \cos \frac{x + y}{2} \sin \frac{x - y}{2}

\cos x - \cos y = -2 \sin \frac{x + y}{2} \sin \frac{x - y}{2}

Then we will get two equations from what is known, namely

2 \cos \frac{x + y}{2} \sin \frac{x - y}{2} = -\frac{1}{3} \quad \ldots \text{(1)}

-2 \sin \frac{x + y}{2} \sin \frac{x - y}{2} = \frac{1}{2} \quad \ldots \text{(2)}

Equation $\text{(2)}$ divided by equation $\text{(1)}$

\frac{\text{(2)}}{\text{(1)}}: \quad \frac{-2 \sin \frac{x + y}{2} \sin \frac{x - y}{2}}{2 \cos \frac{x + y}{2} \sin \frac{x - y}{2}} = \frac{\frac{1}{2}}{-\frac{1}{3}}

-\tan \frac{x + y}{2} = -\frac{3}{2}

\tan \frac{x + y}{2} = \frac{3}{2}

The opposite side is $3$ , and the adjacent side is $2$ . Then, using Pythagoras, the hypotenuse is obtained

\text{Hyp} = \sqrt{3^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}

Thus

\sin \frac{x + y}{2} = \frac{\text{opp}}{\text{hyp}} = \frac{3}{\sqrt{13}}

\cos \frac{x + y}{2} = \frac{\text{adj}}{\text{hyp}} = \frac{2}{\sqrt{13}}

Then the value of $\sin(x + y)$

\sin(x + y) = 2 \sin \frac{x + y}{2} \cos \frac{x + y}{2}

\sin(x + y) = 2 \cdot \frac{3}{\sqrt{13}} \cdot \frac{2}{\sqrt{13}}

\sin(x + y) = \frac{12}{13}

If $3 \cos \theta - \sin \theta$ is expressed in the form $r \sin(\theta + \alpha)$ with $r > 0$ and $0° < \alpha < 360°$ , then....

$r = \sqrt{8}$ , angle $\alpha$ is in quadrant $\text{II}$ or $\text{IV}$

$r = \sqrt{8}$ , angle $\alpha$ is in quadrant $\text{II}$

$r = \sqrt{10}$ , angle $\alpha$ is in quadrant $\text{II}$ or $\text{IV}$

$r = \sqrt{10}$ , angle $\alpha$ is in quadrant $\text{II}$

$r = 3$ , angle $\alpha$ is in quadrant $\text{IV}$

If $\sin 2x + \cos 2x = -16 \cos x + 8 \sin x + \cos^2 x$ with $0 \leq x \leq \frac{\pi}{2}$ , then $\sin 2x$ = ....

Explanation

Use the following trigonometric identities

\sin 2x = 2 \sin x \cos x

\cos 2x = 2 \cos^2 x - 1

Then

\sin 2x + \cos 2x = -16 \cos x + 8 \sin x + \cos^2 x

2 \sin x \cos x + 2 \cos^2 x - 1 = -16 \cos x + 8 \sin x + \cos^2 x

2 \sin x \cos x + 2 \cos^2 x - 1 + 16 \cos x - 8 \sin x - \cos^2 x = 0

2 \sin x \cos x + \cos^2 x + 16 \cos x - 8 \sin x - 1 = 0

2 \cos x (\sin x + 8) - \sin^2 x - 8 \sin x = 0

2 \cos x (\sin x + 8) - \sin x (\sin x + 8) = 0

(2 \cos x - \sin x)(\sin x + 8) = 0

So $2 \cos x - \sin x = 0$ or $\sin x + 8 = 0$ .

Since $0 \leq x \leq \frac{\pi}{2}$ , then $\sin x \geq 0$ , so $\sin x + 8 \neq 0$ .

Therefore $2 \cos x - \sin x = 0$

\sin x = 2 \cos x

\frac{\sin x}{\cos x} = 2

\tan x = 2 = \frac{2}{1}

The opposite side is $2$ , and the adjacent side is $1$ . Then the hypotenuse is

\text{Hyp} = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}

Thus

\sin x = \frac{\text{opp}}{\text{hyp}} = \frac{2}{\sqrt{5}}

\cos x = \frac{\text{adj}}{\text{hyp}} = \frac{1}{\sqrt{5}}

Then the value of $\sin 2x$

\sin 2x = 2 \sin x \cos x

\sin 2x = 2 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}}

\sin 2x = \frac{4}{5}

Command Palette

Number 1

Explanation

Number 2

Explanation

Number 3

Explanation

Number 4

Explanation

Number 5

Explanation

Number 6

Explanation

Number 7

Explanation

Number 8

Explanation

Number 9

Explanation

Number 10

Explanation

Number 11

Explanation

Number 12

Explanation

Number 13

Explanation

Number 14

Explanation

Number 15

Explanation

Number 16

Explanation

Number 17

Explanation

Number 18

Explanation

Number 19

Explanation

Number 20

Explanation

Number 21

Explanation

Number 22

Explanation

Number 23

Explanation

Number 24

Explanation

Number 25

Explanation

Number 26

Explanation

Number 27

Explanation

Number 28

Explanation

Number 29

Explanation

Number 30

Explanation

Number 31

Explanation

Number 32

Explanation

Number 33

Explanation

Number 34

Explanation

Number 35

Explanation

Number 36

Explanation

Number 37

Explanation

Number 38

Explanation

Number 39

Explanation