Math / Algebra

QuestionPure Mathematics (2) The domain of the function $f(x)=\sqrt{x^{2}-25}$ is (a) $\mathbb{R}$ - $\{5\}$ (b) $\mathbb{R}-]-5,5[$ (c) $]-5,5[$ (3) Which of the following functions is an even function $\qquad$ (d) $[-5,5]$ (a) $f(x)=x^{2}+\sin ^{2} x$ (b) $f(x)=x+\cos x$ (c) $f(x)=x^{3}$ (d) $f(x)=x^{4}+\sin x$ (4) The opposite figure represents the curve of the function $f(x)$ which of the following is true? (a) range of $f$ is $]-\infty, \infty[$ (b) domain of $f$ is $] 0, \infty[$ (c) $f$ is increasing on $] 0, \infty[$ (d) $f$ is not one-to-one function (5) The solution set of the equation: $3^{2 x}+12 \times 3^{x}+27=0$ in $\mathbb{R}$ is $\qquad$ (a) $\{-9,-3\}$ (b) $\{1,2\}$ (c) $\{-1,-2\}$ (d) $\varnothing$ (6) The solution set of the equation : $|2 x+6|=\sqrt{x^{2}}$ in $\mathbb{R}$ is $\qquad$ (a) $\{-6,-2\}$ (b) $\{6,2\}$ (c) $\{-1,3\}$ (d) $\varnothing$ (7) The value of : $\log _{6} \log _{\sqrt{3}} 27=$ $\qquad$ (a) -6 (b) -3 (c) 3 (d) 1 (8) $\sqrt[4]{81 x^{8} y^{12}}=$ $\qquad$ (a) $3\left|x^{2}\right| y^{3}$ (b) $|3| x^{2} y^{3}$ (c) $3 x^{2}\left|y^{3}\right|$ (d) $3 x^{2} y^{3}$ (9) If $f(x)=\frac{1}{2} x-6$ , then $f^{-1}(12)=$ $\qquad$ (a) 6 (b) 12 (c) 24 (d) 36 (10) If $\log \mathrm{A}=x, \log \mathrm{~B}=\mathrm{y}$ where $\mathrm{A}>0, \mathrm{~B}>0$ , then $\log \mathrm{AB}=$ $\qquad$ (a) $A+B$ (b) $x+y$ (c) $x_{y}$ (d) $(X)^{y}$ (11) The intersection point of the curve of the function $f(X)=3^{x}+5$ with $y$ -axis is (a) $(1,0)$ (b) $(0,5)$ (c) $(0,6)$ (d) $(1,8)$ (12) If $\log _{2} 2 x+\log _{2} x=5$ , then $x=$ (a) 5 (b) $\pm 5$ (c) 4 (d) $\pm 4$ (13) The solution set of equation : $(4)^{x-3}=(8)^{2 x-6}$ is $\qquad$ (a) $\{3\}$ (b) $\{0\}$ (c) $\{4\}$ (d) $\{5\}$

Studdy Solution

STEP 1

What is this asking? We need to solve a bunch of different math problems covering functions, equations, logarithms, and exponents. Watch out! It's easy to make small mistakes with signs or exponents, so let's be super careful with every step!

STEP 2

1. Domain of a Square Root Function
2. Even Function Identification
3. Function Properties from Graph
4. Exponential Equation
5. Absolute Value Equation
6. Logarithm Value
7. Simplifying a Radical Expression
8. Inverse Function
9. Logarithm of a Product
10. Y-Intercept of an Exponential Function
11. Logarithmic Equation
12. Exponential Equation with Different Bases

STEP 3

We want to find the domain of $f(x) = \sqrt{x^2 - 25}$ .
The expression inside a square root must be greater than or equal to zero.

STEP 4

So, we need $x^2 - 25 \ge 0$ .
This means $x^2 \ge 25$ .

STEP 5

Taking the square root of both sides, we get $|x| \ge 5$ .
This means $x \ge 5$ or $x \le -5$ .

STEP 6

In interval notation, the domain is $(-\infty, -5] \cup [5, \infty)$ , which is the same as $\mathbb{R} - ]-5, 5[$ .
So the answer is (b).

STEP 7

An even function satisfies $f(-x) = f(x)$ .
Let's check each option.

STEP 8

For (a), $f(-x) = (-x)^2 + \sin^2(-x) = x^2 + (-\sin x)^2 = x^2 + \sin^2 x = f(x)$ .
So, (a) is even.

STEP 9

For (b), $f(-x) = -x + \cos(-x) = -x + \cos x \ne f(x)$ .
So (b) is not even.

STEP 10

For (c), $f(-x) = (-x)^3 = -x^3 \ne f(x)$ .
So (c) is not even.

STEP 11

For (d), $f(-x) = (-x)^4 + \sin(-x) = x^4 - \sin x \ne f(x)$ .
So (d) is not even.

STEP 12

Therefore, the answer is (a).

STEP 13

Looking at the graph, we see the function is defined for all $x > 0$ , so the domain is $]0, \infty[$ .

STEP 14

The function is increasing as $x$ increases, so it's increasing on its domain $]0, \infty[$ .

STEP 15

The graph passes the horizontal line test, meaning it's a one-to-one function.

STEP 16

Therefore, the correct answer is (c).

STEP 17

Let $y = 3^x$ .
Then the equation becomes $y^2 + 12y + 27 = 0$ .

STEP 18

Factoring, we get $(y+3)(y+9) = 0$ , so $y = -3$ or $y = -9$ .

STEP 19

Since $3^x$ is always positive, there's no real solution for $x$ .
The solution set is $\varnothing$ , so the answer is (d).

STEP 20

We have $|2x + 6| = \sqrt{x^2} = |x|$ .

STEP 21

Case 1: $2x + 6 = x$ .
Then $x = -6$ .

STEP 22

Case 2: $2x + 6 = -x$ .
Then $3x = -6$ , so $x = -2$ .

STEP 23

The solution set is $\{-6, -2\}$ , so the answer is (a).

STEP 24

We have $\log_6 \log_{\sqrt{3}} 27$ .

STEP 25

First, $\log_{\sqrt{3}} 27 = \log_{\sqrt{3}} (\sqrt{3})^6 = 6$ .

STEP 26

Then, $\log_6 6 = 1$ .
So the answer is (d).

STEP 27

We have $\sqrt[4]{81x^8y^{12}}$ .

STEP 28

This simplifies to $3|x^2||y^3| = 3x^2|y^3|$ .
The answer is (c).

STEP 29

Given $f(x) = \frac{1}{2}x - 6$ , we want to find $f^{-1}(12)$ .

STEP 30

Let $y = f(x)$ .
Then $y = \frac{1}{2}x - 6$ .

STEP 31

To find the inverse, swap $x$ and $y$ : $x = \frac{1}{2}y - 6$ .

STEP 32

Solving for $y$ , we get $2x = y - 12$ , so $y = 2x + 12$ .
Thus, $f^{-1}(x) = 2x + 12$ .

STEP 33

$f^{-1}(12) = 2(12) + 12 = 36$ .
The answer is (d).

STEP 34

Given $\log A = x$ and $\log B = y$ , we want to find $\log(AB)$ .

STEP 35

Using the logarithm property $\log(AB) = \log A + \log B$ , we have $\log(AB) = x + y$ .
The answer is (b).

STEP 36

The y-intercept occurs when $x = 0$ .

STEP 37

$f(0) = 3^0 + 5 = 1 + 5 = 6$ .
So the y-intercept is $(0, 6)$ .
The answer is (c).

STEP 38

We have $\log_2(2x) + \log_2(x) = 5$ .

STEP 39

Using the logarithm property $\log_a(b) + \log_a(c) = \log_a(bc)$ , we get $\log_2(2x \cdot x) = \log_2(2x^2) = 5$ .

STEP 40

This means $2x^2 = 2^5 = 32$ , so $x^2 = 16$ .
Thus, $x = \pm 4$ .

STEP 41

Since we must have $x > 0$ for the logarithm to be defined, the solution is $x = 4$ .
The answer is (c).

STEP 42

We have $4^{x-3} = 8^{2x-6}$ .

STEP 43

Rewrite both sides with base 2: $(2^2)^{x-3} = (2^3)^{2x-6}$ .

STEP 44

This simplifies to $2^{2x-6} = 2^{6x-18}$ .

STEP 45

Therefore, $2x - 6 = 6x - 18$ , so $4x = 12$ , and $x = 3$ .
The answer is (a).

STEP 46

(2) b, (3) a, (4) c, (5) d, (6) a, (7) d, (8) c, (9) d, (10) b, (11) c, (12) c, (13) a

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Studdy Solution

STEP 1

What is this asking? We need to solve a bunch of different math problems covering functions, equations, logarithms, and exponents. Watch out! It's easy to make small mistakes with signs or exponents, so let's be super careful with every step!

STEP 2

STEP 3

We want to find the **domain** of f(x)=x2−25f(x) = \sqrt{x^2 - 25}f(x)=x2−25​.The expression inside a square root must be greater than or equal to zero.

STEP 4

So, we need x2−25≥0x^2 - 25 \ge 0x2−25≥0.This means x2≥25x^2 \ge 25x2≥25.

STEP 5

Taking the square root of both sides, we get ∣x∣≥5|x| \ge 5∣x∣≥5.This means x≥5x \ge 5x≥5 or x≤−5x \le -5x≤−5.

STEP 6

In interval notation, the domain is (−∞,−5]∪[5,∞)(-\infty, -5] \cup [5, \infty)(−∞,−5]∪[5,∞), which is the same as R−]−5,5[\mathbb{R} - ]-5, 5[R−]−5,5[.So the answer is (b).

STEP 7

An **even function** satisfies f(−x)=f(x)f(-x) = f(x)f(−x)=f(x).Let's check each option.

STEP 8

For (a), f(−x)=(−x)2+sin⁡2(−x)=x2+(−sin⁡x)2=x2+sin⁡2x=f(x)f(-x) = (-x)^2 + \sin^2(-x) = x^2 + (-\sin x)^2 = x^2 + \sin^2 x = f(x)f(−x)=(−x)2+sin2(−x)=x2+(−sinx)2=x2+sin2x=f(x).So, (a) is even.

STEP 9

For (b), f(−x)=−x+cos⁡(−x)=−x+cos⁡x≠f(x)f(-x) = -x + \cos(-x) = -x + \cos x \ne f(x)f(−x)=−x+cos(−x)=−x+cosx=f(x).So (b) is not even.

STEP 10

For (c), f(−x)=(−x)3=−x3≠f(x)f(-x) = (-x)^3 = -x^3 \ne f(x)f(−x)=(−x)3=−x3=f(x).So (c) is not even.

STEP 11

For (d), f(−x)=(−x)4+sin⁡(−x)=x4−sin⁡x≠f(x)f(-x) = (-x)^4 + \sin(-x) = x^4 - \sin x \ne f(x)f(−x)=(−x)4+sin(−x)=x4−sinx=f(x).So (d) is not even.

STEP 12

Therefore, the answer is (a).

STEP 13

Looking at the graph, we see the function is defined for all x>0x > 0x>0, so the domain is ]0,∞[]0, \infty[]0,∞[.

STEP 14

The function is increasing as xxx increases, so it's increasing on its domain ]0,∞[]0, \infty[]0,∞[.

STEP 15

The graph passes the horizontal line test, meaning it's a one-to-one function.

STEP 16

Therefore, the correct answer is (c).

STEP 17

Let y=3xy = 3^xy=3x.Then the equation becomes y2+12y+27=0y^2 + 12y + 27 = 0y2+12y+27=0.

STEP 18

Factoring, we get (y+3)(y+9)=0(y+3)(y+9) = 0(y+3)(y+9)=0, so y=−3y = -3y=−3 or y=−9y = -9y=−9.

STEP 19

Since 3x3^x3x is always positive, there's no real solution for xxx.The solution set is ∅\varnothing∅, so the answer is (d).

STEP 20

We have ∣2x+6∣=x2=∣x∣|2x + 6| = \sqrt{x^2} = |x|∣2x+6∣=x2​=∣x∣.

STEP 21

Case 1: 2x+6=x2x + 6 = x2x+6=x.Then x=−6x = -6x=−6.

STEP 22

Case 2: 2x+6=−x2x + 6 = -x2x+6=−x.Then 3x=−63x = -63x=−6, so x=−2x = -2x=−2.

STEP 23

The solution set is {−6,−2}\{-6, -2\}{−6,−2}, so the answer is (a).

STEP 24

We have log⁡6log⁡327\log_6 \log_{\sqrt{3}} 27log6​log3​​27.

STEP 25

First, log⁡327=log⁡3(3)6=6\log_{\sqrt{3}} 27 = \log_{\sqrt{3}} (\sqrt{3})^6 = 6log3​​27=log3​​(3​)6=6.

STEP 26

Then, log⁡66=1\log_6 6 = 1log6​6=1.So the answer is (d).

STEP 27

We have 81x8y124\sqrt[4]{81x^8y^{12}}481x8y12​.

STEP 28

This simplifies to 3∣x2∣∣y3∣=3x2∣y3∣3|x^2||y^3| = 3x^2|y^3|3∣x2∣∣y3∣=3x2∣y3∣.The answer is (c).

STEP 29

Given f(x)=12x−6f(x) = \frac{1}{2}x - 6f(x)=21​x−6, we want to find f−1(12)f^{-1}(12)f−1(12).

STEP 30

Let y=f(x)y = f(x)y=f(x).Then y=12x−6y = \frac{1}{2}x - 6y=21​x−6.

STEP 31

To find the inverse, swap xxx and yyy: x=12y−6x = \frac{1}{2}y - 6x=21​y−6.

STEP 32

Solving for yyy, we get 2x=y−122x = y - 122x=y−12, so y=2x+12y = 2x + 12y=2x+12.Thus, f−1(x)=2x+12f^{-1}(x) = 2x + 12f−1(x)=2x+12.

STEP 33

f−1(12)=2(12)+12=36f^{-1}(12) = 2(12) + 12 = 36f−1(12)=2(12)+12=36.The answer is (d).

STEP 34

Given log⁡A=x\log A = xlogA=x and log⁡B=y\log B = ylogB=y, we want to find log⁡(AB)\log(AB)log(AB).

STEP 35

Using the logarithm property log⁡(AB)=log⁡A+log⁡B\log(AB) = \log A + \log Blog(AB)=logA+logB, we have log⁡(AB)=x+y\log(AB) = x + ylog(AB)=x+y.The answer is (b).

STEP 36

The y-intercept occurs when x=0x = 0x=0.

STEP 37

f(0)=30+5=1+5=6f(0) = 3^0 + 5 = 1 + 5 = 6f(0)=30+5=1+5=6.So the y-intercept is (0,6)(0, 6)(0,6).The answer is (c).

STEP 38

We have log⁡2(2x)+log⁡2(x)=5\log_2(2x) + \log_2(x) = 5log2​(2x)+log2​(x)=5.

STEP 39

Using the logarithm property log⁡a(b)+log⁡a(c)=log⁡a(bc)\log_a(b) + \log_a(c) = \log_a(bc)loga​(b)+loga​(c)=loga​(bc), we get log⁡2(2x⋅x)=log⁡2(2x2)=5\log_2(2x \cdot x) = \log_2(2x^2) = 5log2​(2x⋅x)=log2​(2x2)=5.

STEP 40

This means 2x2=25=322x^2 = 2^5 = 322x2=25=32, so x2=16x^2 = 16x2=16.Thus, x=±4x = \pm 4x=±4.

We want to find the domain of $f(x) = \sqrt{x^2 - 25}$ .
The expression inside a square root must be greater than or equal to zero.

So, we need $x^2 - 25 \ge 0$ .
This means $x^2 \ge 25$ .

Taking the square root of both sides, we get $|x| \ge 5$ .
This means $x \ge 5$ or $x \le -5$ .

In interval notation, the domain is $(-\infty, -5] \cup [5, \infty)$ , which is the same as $\mathbb{R} - ]-5, 5[$ .
So the answer is (b).

An even function satisfies $f(-x) = f(x)$ .
Let's check each option.

For (a), $f(-x) = (-x)^2 + \sin^2(-x) = x^2 + (-\sin x)^2 = x^2 + \sin^2 x = f(x)$ .
So, (a) is even.

For (b), $f(-x) = -x + \cos(-x) = -x + \cos x \ne f(x)$ .
So (b) is not even.

For (c), $f(-x) = (-x)^3 = -x^3 \ne f(x)$ .
So (c) is not even.

For (d), $f(-x) = (-x)^4 + \sin(-x) = x^4 - \sin x \ne f(x)$ .
So (d) is not even.

Looking at the graph, we see the function is defined for all $x > 0$ , so the domain is $]0, \infty[$ .

The function is increasing as $x$ increases, so it's increasing on its domain $]0, \infty[$ .

Let $y = 3^x$ .
Then the equation becomes $y^2 + 12y + 27 = 0$ .

Factoring, we get $(y+3)(y+9) = 0$ , so $y = -3$ or $y = -9$ .

Since $3^x$ is always positive, there's no real solution for $x$ .
The solution set is $\varnothing$ , so the answer is (d).

We have $|2x + 6| = \sqrt{x^2} = |x|$ .

Case 1: $2x + 6 = x$ .
Then $x = -6$ .

Case 2: $2x + 6 = -x$ .
Then $3x = -6$ , so $x = -2$ .

The solution set is $\{-6, -2\}$ , so the answer is (a).

We have $\log_6 \log_{\sqrt{3}} 27$ .

First, $\log_{\sqrt{3}} 27 = \log_{\sqrt{3}} (\sqrt{3})^6 = 6$ .

Then, $\log_6 6 = 1$ .
So the answer is (d).

We have $\sqrt[4]{81x^8y^{12}}$ .

This simplifies to $3|x^2||y^3| = 3x^2|y^3|$ .
The answer is (c).

Given $f(x) = \frac{1}{2}x - 6$ , we want to find $f^{-1}(12)$ .

Let $y = f(x)$ .
Then $y = \frac{1}{2}x - 6$ .

To find the inverse, swap $x$ and $y$ : $x = \frac{1}{2}y - 6$ .

Solving for $y$ , we get $2x = y - 12$ , so $y = 2x + 12$ .
Thus, $f^{-1}(x) = 2x + 12$ .

$f^{-1}(12) = 2(12) + 12 = 36$ .
The answer is (d).

Given $\log A = x$ and $\log B = y$ , we want to find $\log(AB)$ .

Using the logarithm property $\log(AB) = \log A + \log B$ , we have $\log(AB) = x + y$ .
The answer is (b).

The y-intercept occurs when $x = 0$ .

$f(0) = 3^0 + 5 = 1 + 5 = 6$ .
So the y-intercept is $(0, 6)$ .
The answer is (c).

We have $\log_2(2x) + \log_2(x) = 5$ .

Using the logarithm property $\log_a(b) + \log_a(c) = \log_a(bc)$ , we get $\log_2(2x \cdot x) = \log_2(2x^2) = 5$ .

This means $2x^2 = 2^5 = 32$ , so $x^2 = 16$ .
Thus, $x = \pm 4$ .

Since we must have $x > 0$ for the logarithm to be defined, the solution is $x = 4$ .
The answer is (c).

We have $4^{x-3} = 8^{2x-6}$ .

Rewrite both sides with base 2: $(2^2)^{x-3} = (2^3)^{2x-6}$ .

This simplifies to $2^{2x-6} = 2^{6x-18}$ .

Therefore, $2x - 6 = 6x - 18$ , so $4x = 12$ , and $x = 3$ .
The answer is (a).