Set 2

If the tangent line to the curve $y = x^3 - 3x^2 - 9x$ at point $(a, b)$ has a gradient of $15$ , then the possible value of $a + b$ is....

Given $x^2 + 2xy + 4x = -3$ and $9y^2 + 4xy + 12y = -1$ . The value of $x + 3y$ is....

If an integer $p$ is a root of $f(x) = 0$ with $f(x) = px^2 - 3x - p - 3$ , then the gradient of the tangent line to the curve $y = f(x)$ at the point with abscissa $x = p$ is....

If $(p, q)$ is the vertex point of the graph of function $f(x) = ax^2 + 2ax + a + 1$ , with $f(a) = 19$ , then $p + 2q + 3a = ....$

Given a straight line passing through $(0, -2)$ and $\left(\frac{3}{2}, 0\right)$ . The distance from the parabola $y = x^2 - 1$ to that line is....

Explanation

Create the equation of the line from the points it passes through, namely $(0, -2)$ and $\left(\frac{3}{2}, 0\right)$

\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}

\frac{y - (-2)}{0 - (-2)} = \frac{x - 0}{\frac{3}{2} - 0}

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

3y + 6 = 4x

-4x + 3y + 6 = 0

We cannot directly find the distance from a point to the line. Find the point whose position is closest to the line. Assume the point is $(a, b)$ so that

y = x^2 - 1

b = a^2 - 1

The point becomes $(a, a^2 - 1)$ as shown in the following illustration

Parabola and Line Visualization

The graph shows the parabola

y = x^2 - 1

and the line

-4x + 3y + 6 = 0

with the closest point

\left(\frac{2}{3}, -\frac{5}{9}\right)

We need to determine the distance from point $(a, a^2 - 1)$ to the line $-4x + 3y + 6 = 0$

distance = \frac{|-4x + 3y + 6|}{\sqrt{(-4)^2 + 3^2}}

f(a) = \frac{|-4a + 3(a^2 - 1) + 6|}{5}

f(a) = \frac{|3a^2 - 4a + 3|}{5}

f(a) = \frac{1}{5}(3a^2 - 4a + 3)

The condition for minimum value is $f'(x) = 0$

f'(a) = \frac{1}{5}(6a - 4)

f'(a) = 0

\frac{1}{5}(6a - 4) = 0

6a = 4

a = \frac{4}{6} = \frac{2}{3}

Distance between line and parabola when $a = \frac{2}{3}$

f(a) = \frac{1}{5}(3a^2 - 4a + 3)

f(a) = \frac{1}{5}\left(3\left(\frac{2}{3}\right)^2 - 4\left(\frac{2}{3}\right) + 3\right)

f(a) = \frac{1}{5}\left(\frac{4}{3} - \frac{8}{3} + \frac{9}{3}\right)

f(a) = \frac{1}{5}\left(\frac{5}{3}\right) = \frac{1}{3}

Given a sequence $-\frac{1}{2}, \frac{3}{4}, -\frac{1}{8}, \frac{3}{16}, ...$ , the $12$ th term of this sequence is....

$\frac{1}{2^{11}}$

$\frac{1}{2^{12}}$

$\frac{3}{2^{11}}$

$\frac{3}{2^{12}}$

$\frac{1}{2^{11}} + \frac{1}{3^{11}}$

Given a sequence $0, \frac{3}{4}, \frac{3}{16}, \frac{9}{64}, ...$ , then the $12$ th term of this sequence is....

$\frac{1}{2^{11}} + \frac{1}{2^{12}}$

$\frac{1}{2^{11}} + \frac{1}{2^{22}}$

$\frac{1}{2^{11}}$

$\frac{1}{2^{22}}$

$\frac{3}{2^{12}}$

Given a sequence $0, \frac{5}{6}, \frac{5}{36}, \frac{35}{216}, ...$ , the $12$ th term of this sequence is....

$\frac{1}{2^{11}} - \frac{1}{3^{11}}$

$\frac{1}{2^{11}} - \frac{2}{3^{11}}$

$\frac{3}{2^{11}} - \frac{1}{3^{11}}$

$\frac{1}{2^{11}} + \frac{1}{3^{11}}$

$\frac{2}{2^{11}} + \frac{3}{3^{11}}$

A geometric sequence has $3$ first terms $a, b, b^2$ . If $a$ and $b$ are roots of the quadratic equation $2x^2 + kx + 6 = 0$ , then the fourth term of the sequence and the value of $k$ respectively are....

$27 \text{ and } -8$

$27 \text{ and } 8$

$24 \text{ and } -8$

$24 \text{ and } -4$

$24 \text{ and } 4$

Suppose $x_1$ and $x_2$ are integers that are roots of the quadratic equation $x^2 - (2k + 4)x + (3k + 4) = 0$ . If $x_1, k, x_2$ form the first three terms of a geometric series, then the formula for the $n$ th term of the series is....

\lim_{x \to \infty} (5^x + 5^{3x})^{\frac{1}{x}} = ....

Given $f(x) = \sin^2 x$ . If $f'(x)$ represents the first derivative of $f(x)$ , then

\lim_{h \to \infty} h \left[f'\left(x + \frac{1}{h}\right) - f'(x)\right] = ....

Given $f(x) = \sqrt{1 + x}$ . The value of $\lim_{h \to 0} \frac{f(3 + 2h^2) - f(3 - 3h^2)}{h^2}$ is....

\lim_{x \to 0} \frac{\cos x \sin x - \tan x}{x^2 \sin x} = ....

If $\lim_{x \to -3} \frac{\frac{1}{ax} + \frac{1}{3}}{bx^3 + 27} = -\frac{1}{3^5}$ , the value of $a + b$ for $a$ and $b$ positive integers is....

If $\log_{a^2}(3^a - 8)^{-4} \cdot \log_3 \sqrt{a} = a - 2$ , then $\log_a\left(\frac{1}{8}\right) = ....$

If $(\log_2 x)^2 - (\log_2 y)^2 = \log_2 256$ and $\log_2 x^2 - \log_2 y^2 = \log_2 16$ . Then the value of $\log_2 x^6 y^{-2}$ is....

If $2 \log_4 x - \log_4(4x + 3) = -1$ , then $\log_2 x = ....$

$\log_2 3 - 1$

$\log_2 3$

$1 - \log_2 3$

$-1 - \log_2 3$

$\log_2 3 + \log_3 2$

If $a$ satisfies the equation $\log_2 2x + \log_3 3x = \log_4 4x^2$ , then the value of $\log_a 3 = ....$

If $\alpha$ and $\beta$ are roots of the equation $\log_3 x - \log_x\left(2x - 4 + \frac{4}{x}\right) = 1$ , then $\alpha + \beta = ....$

If $b > a$ , the value of $x$ that satisfies $|x - 2a| + a \leq b$ is....

$3a \leq x \leq 2b + a$

$x \geq -b + 3a$

$x \leq b + a$

$b - 3a \leq x \leq -b + a$

$-b + 3a \leq x \leq b + a$

The solution set of $9 - x^2 \geq |x + 3|$ is....

$\{x \in R: -3 \leq x \leq 3\}$

$\{x \in R: -3 \leq x \leq 2\}$

$\{x \in R: x \leq -3 \text{ or } x \geq 2\}$

$\{x \in R: 0 \leq x \leq 2\}$

$R$

The solution set of $16 - x^2 \leq |x + 4|$ is....

$\{x \in R: -4 \leq x \leq 4\}$

$\{x \in R: -4 \leq x \leq 3\}$

$\{x \in R: x \leq -4 \text{ or } x \geq 4\}$

$\{x \in R: 0 \leq x \leq 3\}$

$\{x \in R: x \leq -4 \text{ or } x \geq 3\}$

The solution set of inequality $\log|x + 1| \geq \log 3 + \log|2x - 1|$ is....

$\left\{\frac{2}{7} < x < \frac{4}{5}, x \neq \frac{1}{2}\right\}$

$\left\{\frac{2}{7} \leq x \leq \frac{4}{5}, x \neq \frac{1}{2}\right\}$

$\left\{\frac{2}{5} \leq x \leq \frac{4}{5}, x \neq \frac{1}{3}\right\}$

$\left\{\frac{1}{5} \leq x < \frac{4}{5}, x \neq \frac{1}{7}\right\}$

$\left\{x \leq \frac{4}{5} \vee x \geq \frac{1}{5}, x \neq \frac{3}{5}\right\}$

The number of real numbers $x$ that satisfy the equation $|x^2 - 4| = x + |x - 2|$ is....

Explanation

Define the first absolute value for $|x - 2|$ .

$x - 2$ is positive for $x - 2 \geq 0 \rightarrow x \geq 0$ .

$x - 2$ is negative for $x - 2 < 0 \rightarrow x < 2$ .

So the definition is

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

For $|x^2 - 4|$

$x^2 - 4$ is positive for $x^2 - 4 \geq 0 \rightarrow x \leq -2 \vee x \geq 0$ .

$x^2 - 4$ is negative for $x^2 - 4 < 0 \rightarrow -2 < x < 2$ .

So the definition is

|x^2 - 4| = \begin{cases} x^2 - 4; \text{for } x \leq -2 \vee x \geq 2 \\ -(x^2 - 4); \text{for } -2 < x < 2 \end{cases}

Based on the definitions above, the absolute value is bounded by $x = -2$ and $x = 2$ . This means there are $3$ possible regions/values of $x$ .

Solve the problem based on regions.

Region I when $x < -2$

|x^2 - 4| = x + |x - 2|

x^2 - 4 = x + (-x + 2)

x^2 = 6

x = \pm\sqrt{6}

Since region I is negative, the root that satisfies is $x_1 = -\sqrt{6}$ .

Region II when $-2 \leq x < 2$

|x^2 - 4| = x + |x - 2|

-x^2 + 4 = x + (-x + 2)

x^2 = 2

x = \pm\sqrt{2}

Since region II includes both positive and negative, all roots satisfy region II.

Region III when $x \geq 2$

|x^2 - 4| = x + |x - 2|

x^2 - 4 = x + x - 2

x^2 - 2x - 2 = 0

Determine the discriminant value to find the type of roots.

D = b^2 - 4ac

D = (-2)^2 - 4(1)(-2)

D = 4 + 8

D = 12

Since $D > 0$ , the roots are distinct, one positive root satisfies region III.

Therefore the total number of solutions is $4$ solutions.

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 $x, y, z$ satisfy the system of equations

3x + 2y - z = 3

2x + y - 3z = 4

x - y + 2z = -1

Then the value of $2x + 2y - 3z =$ ....

Explanation

Given the system of equations

3x + 2y - z = 3 \quad ...(1)

2x + y - 3z = 4 \quad ...(2)

x - y + 2z = -1 \quad ...(3)

Eliminate the first and second equations

Multiply the first equation by $3$ and the second equation by $1$ , then subtract.

3x + 2y - z = 3 \quad |\times 3| \quad 9x + 6y - 3z = 9

2x + y - 3z = 4 \quad |\times 1| \quad 2x + y - 3z = 4 \quad -

7x + 5y = 5 \quad ...(4)

Eliminate the first and third equations

Multiply the first equation by $2$ and the third equation by $1$ , then subtract.

3x + 2y - z = 3 \quad |\times 2| \quad 6x + 4y - 2z = 6

x - y + 2z = -1 \quad |\times 1| \quad x - y + 2z = -1 \quad -

7x + 3y = 5 \quad ...(5)

Eliminate the fourth and fifth equations

Subtract the fifth equation from the fourth equation to find the value of $y$ .

7x + 5y = 5

7x + 3y = 5 \quad -

2y = 0

y = 0

Substitute the value of y

Substitute $y = 0$ into the fourth equation to find the value of $x$ .

7x + 5y = 5

7x + 5(0) = 5

7x = 5

x = \frac{5}{7}

Substitute the values of x and y

Substitute $x = \frac{5}{7}$ and $y = 0$ into the third equation to find the value of $z$ .

x - y + 2z = -1

\frac{5}{7} - 0 + 2z = -1

2z = -\frac{12}{7}

z = -\frac{6}{7}

Calculate the requested value

Substitute the values $x = \frac{5}{7}$ , $y = 0$ , and $z = -\frac{6}{7}$ into $2x + 2y - 3z$ .

2x + 2y - 3z = 2\left(\frac{5}{7}\right) + 2(0) - 3\left(-\frac{6}{7}\right)

2x + 2y - 3z = \frac{10}{7} - 0 + \frac{18}{7}

2x + 2y - 3z = \frac{28}{7}

2x + 2y - 3z = 4

Therefore, the value of $2x + 2y - 3z$ is $4$ .

If the roots of the equation $x^2 - ax + b = 0$ satisfy the equation $2x^2 - (a + 3)x + (3b - 2) = 0$ , then....

$a = 3$
$b = 2$
$2a - 2ab + 3b = 0$
$ab = 5$

ONLY $(1), (2), (3)$ are correct

ONLY $(1)$ and $(3)$ are correct

ONLY $(2)$ and $(4)$ are correct

ONLY $(4)$ is correct

ALL statements are correct

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 with the tangent line at $x = 4$
$y = \frac{4}{3}x - \frac{25}{3}$ is the tangent line to the curve at $(4, -3)$

ONLY $(1), (2), (3)$ are correct

ONLY $(1)$ and $(3)$ are correct

ONLY $(2)$ and $(4)$ are correct

ONLY $(4)$ is correct

ALL statements are correct

Explanation

Given the function $y = \sqrt{x^2 - 7}$ . The derivative of this function is

y' = \frac{x}{\sqrt{x^2 - 7}}

Testing the first statement

The ordinate of the point on the curve with abscissa $4$ is

f(4) = \sqrt{4^2 - 7} = \sqrt{9} = 3

The gradient of the tangent line at point $(4,3)$ can be found by substituting $x = 4$ into the derivative.

y' = f'(x) = \frac{x}{\sqrt{x^2 - 7}}

f'(4) = \frac{4}{\sqrt{4^2 - 7}} = \frac{4}{\sqrt{9}} = \frac{4}{3}

With gradient $m = \frac{4}{3}$ and point $(4,3)$ , the equation of the tangent line is

y - y_1 = m(x - x_1)

y - 3 = \frac{4}{3}(x - 4)

y - 3 = \frac{4}{3}x - \frac{16}{3}

y = \frac{4}{3}x - \frac{7}{3}

The first statement is correct.

Testing the second statement

For values of $x$ and $y$ that satisfy the condition, square both sides.

y = \sqrt{x^2 - 7}

y^2 = \left(\sqrt{x^2 - 7}\right)^2

y^2 = x^2 - 7

x^2 - y^2 = 7

The form $x^2 - y^2 = 7$ is not the equation of a circle.

The second statement is incorrect.

Testing the third statement

The line $y = -\frac{3}{4}x + 6$ has gradient $m_1 = -\frac{3}{4}$ .

From the first statement, the gradient of the tangent line at $x = 4$ is $m_2 = \frac{4}{3}$ .

Check if both lines are perpendicular by multiplying both gradients.

m_1 \cdot m_2 = -\frac{3}{4} \cdot \frac{4}{3} = -1

Since $m_1 \cdot m_2 = -1$ , both lines are perpendicular.

The third statement is correct.

Testing the fourth statement

Check if the curve passes through point $(4, -3)$ by substituting the value $x = 4$ .

y = \sqrt{4^2 - 7} = \sqrt{9} = 3

The curve does not pass through point $(4, -3)$ , but passes through point $(4,3)$ .

The fourth statement is incorrect.

Therefore, the correct statements are the first and third.

If $k$ is the smallest positive integer such that two quadratic functions $f(x) = (k - 1)x^2 + kx - 1$ and $g(x) = (k - 2)x^2 + x + 2k$ intersect at two different points $(x_1, y_1)$ and $(x_2, y_2)$ , then the quadratic equation with roots $x_1 + x_2$ and $y_1 + y_2$ is....

$x^2 - 1 = 0$

$x^2 + 4x - 5 = 0$

$x^2 - 10x = 0$

$x^2 - 6x - 7 = 0$

$x^2 - 26x - 56 = 0$

Explanation

Given two quadratic functions

f(x) = (k - 1)x^2 + kx - 1, \quad k \neq 1

g(x) = (k - 2)x^2 + x + 2k, \quad k \neq 2

Finding the condition for intersection

Eliminate $y$ from both functions by equating $f(x) = g(x)$ .

(k - 1)x^2 + kx - 1 = (k - 2)x^2 + x + 2k

x^2 + (k - 1)x - (2k + 1) = 0 \quad ...(1)

The condition for both functions to intersect at two different points is that the discriminant must be positive, namely $D > 0$ .

D > 0

(k - 1)^2 - 4(1)(-2k - 1) > 0

k^2 - 2k + 1 + 8k + 4 > 0

k^2 + 6k + 5 > 0

(k + 5)(k + 1) > 0

From the inequality $(k + 5)(k + 1) > 0$ , we get $k < -5$ or $k > -1$ .

Since $k \neq 1$ and $k \neq 2$ , the smallest positive integer that satisfies the condition is $k = 3$ .

Substituting the value of k

Substitute the value $k = 3$ into the first equation to find the sum of roots $x$ .

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

x^2 + (3 - 1)x - (2(3) + 1) = 0

x^2 + 2x - 7 = 0

From this quadratic equation, the sum of its roots is

x_1 + x_2 = -2

Finding the value of y

Substitute the value $k = 3$ into functions $f(x)$ and $g(x)$ .

For $f(x)$

f(x) = (k - 1)x^2 + kx - 1

f(3) = (3 - 1)x^2 + 3x - 1

f(3) = 2x^2 + 3x - 1

For $g(x)$

g(x) = (k - 2)x^2 + x + 2k

g(3) = (3 - 2)x^2 + x + 2(3)

g(3) = x^2 + x + 6

Then find the value of $x$ by eliminating $x^2$ .

y = 2x^2 + 3x - 1 \quad |\times 1|

y = x^2 + x + 6 \quad |\times 2|

y = 2x^2 + 3x - 1

2y = 2x^2 + 2x + 12 \quad -

-y = x - 13

x = 13 - y

Substitute $x = 13 - y$ into $g(x)$ .

y = x^2 + x + 6

y = (13 - y)^2 + (13 - y) + 6

y = 169 - 26y + y^2 + 13 - y + 6

y^2 - 28y + 176 = 0

From this quadratic equation, the sum of its roots is

y_1 + y_2 = 28

Forming the new quadratic equation

The new quadratic equation with roots $x_1 + x_2$ and $y_1 + y_2$ is

x^2 - ((x_1 + x_2) + (y_1 + y_2))x + ((x_1 x_2)(y_1 y_2)) = 0

x^2 - (-2 + 28)x + (-2)(28) = 0

x^2 - 26x - 56 = 0

Therefore, the quadratic equation with roots $x_1 + x_2$ and $y_1 + y_2$ is $x^2 - 26x - 56 = 0$ .

If the polynomial $f(x)$ is divisible by $(x - 1)$ , then the remainder when $f(x)$ is divided by $(x - 1)(x + 1)$ is....

$\frac{-f(-1)}{2}(1 + x)$

$\frac{-f(-1)}{2}(1 - x)$

$\frac{f(-1)}{2}(1 + x)$

$\frac{f(-1)}{2}(1 - x)$

$\frac{f(-1)}{2}(x - 1)$

If the polynomial $ax^3 + 2x^2 + 5x + b$ is divided by $(x^2 - 1)$ and gives a remainder of $(6x + 5)$ , then $a + 3b$ equals....

Given that $p(x)$ and $g(x)$ are two different polynomials, with $p(10) = m$ and $g(10) = n$ . If $p(x)h(x) = \left(\frac{p(x)}{g(x)} - 1\right)(p(x) + g(x))$ , $h(10) = -\frac{16}{15}$ , then the maximum value of $|m + n| =$ ....

Given that the polynomial $f(x)$ divided by $2x^2 - x - 1$ has a remainder of $4ax - b$ and divided by $2x^2 + 3x + 1$ has a remainder of $-2bx + a - 11$ . If $f(x - 2)$ is divisible by $x - 3$ , then $a + 2b + 6 =$ ....

Explanation

Factor the first divisor

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

x = -\frac{1}{2} \cup x = 1

The remainder is $s(x) = 4ax - b$ . The roots of the divisor are $-\frac{1}{2}$ and $1$ , so

For $x = -\frac{1}{2}$

f\left(-\frac{1}{2}\right) = s\left(-\frac{1}{2}\right)

f\left(-\frac{1}{2}\right) = 4a\left(-\frac{1}{2}\right) - b

f\left(-\frac{1}{2}\right) = -2a - b \quad ...(1)

For $x = 1$

f(1) = s(1)

f(1) = 4a - b \quad ...(2)

Factor the second divisor

Given the polynomial $f(x)$ with divisor $2x^2 + 3x + 1 = (2x + 1)(x + 1) = 0$ , then $x = -\frac{1}{2} \cup x = -1$ .

The remainder is $s(x) = -2bx + a - 11$ .

For $x = -\frac{1}{2}$

f\left(-\frac{1}{2}\right) = s\left(-\frac{1}{2}\right)

f\left(-\frac{1}{2}\right) = -2b\left(-\frac{1}{2}\right) + a - 11

f\left(-\frac{1}{2}\right) = a + b - 11 \quad ...(3)

Use the divisibility condition

$f(x - 2)$ is divisible by $x - 3$ , so the divisor $x - 3 = 0 \rightarrow x = 3$ .

The remainder is $s(x) = 0$ (because it is divisible).

x = 3 \rightarrow f(x - 2) = 0

f(3 - 2) = 0

f(1) = 0 \quad ...(4)

Solve the system of equations

From the second and fourth equations

4a - b = 0

b = 4a \quad ...(5)

From the first and third equations, with $b = 4a$

f\left(-\frac{1}{2}\right) = f\left(-\frac{1}{2}\right)

a + b - 11 = -2a - b

3a + 2b = 11

3a + 2(4a) = 11

3a + 8a = 11

11a = 11

a = 1

Substitute the value $a = 1$ into the fifth equation

b = 4a = 4(1) = 4

Therefore, the value of $a + 2b + 6 = 1 + 2(4) + 6 = 15$ .

Given that the polynomial $f(x)$ divided by $x^2 + 3x + 2$ has a remainder of $3bx + a - 2$ and divided by $x^2 - 2x - 3$ has a remainder of $ax - 2b$ . If $f(3) + f(-2) = 6$ , then $a + b =$ ....

Explanation

Factor the first divisor

The divisor is $x^2 + 3x + 2 = (x + 1)(x + 2) \rightarrow x = -1 \vee x = -2$ .

The remainder is $s(x) = 3bx + a - 2$ .

For $x = -1$

f(-1) = s(-1)

f(-1) = 3b(-1) + a - 2

f(-1) = -3b + a - 2 \quad ...(1)

For $x = -2$

f(-2) = s(-2)

f(-2) = 3b(-2) + a - 2

f(-2) = -6b + a - 2 \quad ...(2)

Factor the second divisor

For the second polynomial, the divisor is $x^2 - 2x - 3 = (x + 1)(x - 3) \rightarrow x = -1 \vee x = 3$ .

The remainder is $s(x) = ax - 2b$ .

For $x = -1$

f(-1) = s(-1)

f(-1) = a(-1) - 2b

f(-1) = -a - 2b \quad ...(3)

For $x = 3$

f(3) = s(3)

f(3) = a(3) - 2b

f(3) = 3a - 2b \quad ...(4)

Solve the system of equations

From the first and third equations, we get

f(-1) = f(-1)

-3b + a - 2 = -a - 2b

b = 2a - 2 \quad ...(5)

Then combine equations two, four, and five into

f(3) + f(-2) = 6

3a - 2b + (-6b + a - 2) = 6

4a - 8b = 8

a - 2b = 2

a - 2(2a - 2) = 2

-3a = -2

a = \frac{2}{3}

Then the value of $b = 2a - 2 = 2\left(\frac{2}{3}\right) - 2 = -\frac{2}{3}$ .

Therefore, $a + b = \frac{2}{3} + \left(-\frac{2}{3}\right) = 0$ .

If angles $A$ and $B$ satisfy the system of equations

2 \tan A + \tan B = 4

\tan A - 3 \tan B = -\frac{17}{2}

Then $\tan(2A + B)$ equals....

Explanation

Recall the trigonometric concepts for angle addition

\tan(x + y) = \frac{\tan x + \tan y}{1 - \tan x \cdot \tan y}

\tan(2x) = \frac{2 \tan x}{1 - \tan^2 x}

Solve the system of equations

Get the first equation by transforming to find $\tan B$ .

2 \tan A + \tan B = 4 \rightarrow \tan B = 4 - 2 \tan A

Then substitute into the second equation

\tan A - 3 \tan B = -\frac{17}{2}

\tan A - 3(4 - 2 \tan A) = -\frac{17}{2}

\tan A - 12 + 6 \tan A = -\frac{17}{2}

7 \tan A = -\frac{17}{2} + 12

7 \tan A = \frac{7}{2}

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

Therefore, the value of $\tan B = 4 - 2 \tan A = 4 - 2\left(\frac{1}{2}\right) = 3$ .

Determine the value of tan 2A

\tan(2A) = \frac{2 \tan A}{1 - \tan^2 A}

\tan(2A) = \frac{2\left(\frac{1}{2}\right)}{1 - \left(\frac{1}{2}\right)^2}

\tan(2A) = \frac{1}{1 - \frac{1}{4}} = \frac{4}{3}

Determine the value of tan(2A + B)

\tan(2A + B) = \frac{\tan 2A + \tan B}{1 - \tan 2A \cdot \tan B}

\tan(2A + B) = \frac{\frac{4}{3} + 3}{1 - \frac{4}{3}(3)}

\tan(2A + B) = \frac{\frac{13}{3}}{1 - 4}

\tan(2A + B) = \frac{\frac{13}{3}}{-3}

\tan(2A + B) = -\frac{13}{9}

Therefore, the value of $\tan(2A + B)$ is $-\frac{13}{9}$ .

Given a rectangular prism $ABCD.EFGH$ where $AB = 6$ cm, $BC = 8$ cm, and $BF = 4$ cm. If $\alpha$ is the angle between $AH$ and $BD$ , then $\cos 2\alpha =$ ....

Explanation

Consider the following rectangular prism illustration.

Rectangular Prism

ABCD.EFGH

Visualization of the rectangular prism with given dimensions and relevant lines for calculating the angle.

Lines $AH$ and $BD$ do not intersect, so we need to shift one of them so they intersect, namely shift line $AH$ to line $BG$ (because $AH$ is parallel to $BG$ ), so the angle is between lines $BG$ and $BD$ .

\alpha = \angle(AH, BD) = \angle(BG, BD)

Determine segment lengths

Then determine the length of $\triangle BDG$

BD = \sqrt{AB^2 + AD^2} = \sqrt{6^2 + 8^2} = 10

DG = \sqrt{CD^2 + CG^2} = \sqrt{6^2 + 4^2} = \sqrt{52} = 2\sqrt{13}

BG = \sqrt{CB^2 + CG^2} = \sqrt{8^2 + 4^2} = \sqrt{80} = 4\sqrt{5}

Use the cosine rule

Use the cosine rule on $\triangle BDG$

\cos \angle DBG = \frac{BD^2 + BG^2 - DG^2}{2 \cdot BD \cdot BG}

\cos \alpha = \frac{10^2 + (4\sqrt{5})^2 - (2\sqrt{13})^2}{2(10)(4\sqrt{5})}

\cos \alpha = \frac{100 + 80 - 52}{80\sqrt{5}}

\cos \alpha = \frac{128}{80\sqrt{5}}

\cos \alpha = \frac{8}{5\sqrt{5}}

Then the value of $\cos 2\alpha$ can be calculated using the double angle formula

\cos 2\alpha = 2\cos^2 \alpha - 1

\cos 2\alpha = 2\left(\frac{8}{5\sqrt{5}}\right)^2 - 1

\cos 2\alpha = 2\left(\frac{64}{125}\right) - 1

\cos 2\alpha = \frac{128}{125} - \frac{125}{125}

\cos 2\alpha = \frac{3}{125}

Therefore, the value of $\cos 2\alpha$ is $\frac{3}{125}$ .

The function $f(x) = 3 \sin x + 3 \cos x$ defined on the interval $(0, 2\pi)$ reaches its maximum value at $x =$ ....

Explanation

Recall the derivative concepts for trigonometric functions

y = a \cdot \sin x \rightarrow y' = a \cdot \cos x

y = a \cdot \cos x \rightarrow y' = -a \cdot \sin x

Finding the stationary points

The condition for maximum value is when $f'(x) = 0$ .

f'(x) = 0

3 \cos x - 3 \sin x = 0

3 \cos x = 3 \sin x

\frac{\sin x}{\cos x} = 1

\tan x = 1

Then the values of $x$ that satisfy $\tan x = 1$ are $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$ .

Testing maximum and minimum values

Check first which function has maximum and minimum values.

For the value $x = \frac{\pi}{4}$

x = \frac{\pi}{4} \rightarrow f\left(\frac{\pi}{4}\right) = 3 \sin \frac{\pi}{4} + 3 \cos \frac{\pi}{4}

x = \frac{\pi}{4} \rightarrow f\left(\frac{\pi}{4}\right) = 3\left(\frac{1}{2}\sqrt{2}\right) + 3\left(\frac{1}{2}\sqrt{2}\right)

x = \frac{\pi}{4} \rightarrow f\left(\frac{\pi}{4}\right) = 3\sqrt{2} \text{ (max)}

For the value $x = \frac{5\pi}{4}$

x = \frac{5\pi}{4} \rightarrow f\left(\frac{5\pi}{4}\right) = 3 \sin \frac{5\pi}{4} + 3 \cos \frac{5\pi}{4}

x = \frac{5\pi}{4} \rightarrow f\left(\frac{5\pi}{4}\right) = 3\left(-\frac{1}{2}\sqrt{2}\right) + 3\left(-\frac{1}{2}\sqrt{2}\right)

x = \frac{5\pi}{4} \rightarrow f\left(\frac{5\pi}{4}\right) = -3\sqrt{2} \text{ (min)}

Therefore, the maximum value is reached when $x = \frac{\pi}{4}$ .

If $\begin{bmatrix} \tan x & 1 \\ 1 & \tan x \end{bmatrix} \begin{bmatrix} \cos^2 x \\ \sin x \cos x \end{bmatrix} = \begin{bmatrix} a \\ b \end{bmatrix} \frac{1}{2}$ where $b = 2a$ , then $0 \leq x \leq \pi$ that satisfies is....

$\frac{\pi}{6}$
$\frac{\pi}{12}$
$\frac{5\pi}{6}$
$\frac{5\pi}{12}$

ONLY $(1), (2), (3)$ are correct

ONLY $(1)$ and $(3)$ are correct

ONLY $(2)$ and $(4)$ are correct

ONLY $(4)$ is correct

ALL statements are correct

Explanation

Given the matrices

\begin{bmatrix} \tan x & 1 \\ 1 & \tan x \end{bmatrix} \begin{bmatrix} \cos^2 x \\ \sin x \cos x \end{bmatrix} = \begin{bmatrix} a \\ b \end{bmatrix}

Multiply the matrices

This is matrix multiplication, so

\begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} \tan x \cdot \cos^2 x + \sin x \cdot \cos x \\ \cos^2 x + \tan x \cdot \sin x \cos x \end{bmatrix}

Change $\tan x = \frac{\sin x}{\cos x}$ , to

\begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} \left(\frac{\sin x}{\cos x}\right) \cdot \cos^2 x + \sin x \cdot \cos x \\ \cos^2 x + \left(\frac{\sin x}{\cos x}\right) \cdot \sin x \cos x \end{bmatrix}

Simplify to

\begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} \sin x \cdot \cos x + \sin x \cdot \cos x \\ \cos^2 x + \sin^2 x \end{bmatrix}

\begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 2 \sin x \cdot \cos x \\ \sin^2 x + \cos^2 x \end{bmatrix}

We know that the value $2 \sin x \cdot \cos x = \sin 2x$ and the value $\sin^2 x + \cos^2 x = 1$ , then

\begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} \sin 2x \\ 1 \end{bmatrix}

Therefore, the value $a = \sin 2x$ and $b = 1$ .

Substitute into the given condition

Substitute $b = 1$ into $b = 2a$ to get $a = \frac{1}{2}$ .

Then, substitute $a = \frac{1}{2}$ into $a = \sin 2x$ to get $\sin 2x = \frac{1}{2}$ .

The values of $2x$ that satisfy are

2x_1 = \frac{\pi}{6} + k \cdot 2\pi \quad \text{or} \quad 2x_2 = \frac{5\pi}{6} + k \cdot 2\pi

x_1 = \frac{\pi}{12} + k\pi \quad \text{or} \quad x_2 = \frac{5\pi}{12} + k\pi

The values of $x$ that satisfy are $\frac{\pi}{12}$ or $\frac{5\pi}{12}$ .

Therefore, the correct statements are the second and fourth.

If $\cos(A + B) = \frac{2}{5}$ , $\cos A \cos B = \frac{3}{4}$ , then the value of $\tan A \tan B =$ ....

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