Chapter 1, Problem 1RE

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.

1. a. A function could have the property that f(−x) = f(x), for all x.
2. b. cos (a + b) = cos a + cos b, for all a and b in [0, 2π].
3. c. If f is a linear function of the form f(x) = mx + b, then f(u + v) = f(u) + f(v), for all u and v.
4. d. The function f(x) = 1 − x has the property that f(f(x)) = x.
5. e. The set {x: |x + 3| > 4} can be drawn on the number line without lifting your pencil.
6. f. log₁₀(xy) = (log₁₀ x)(log₁₀ y)
7. g. sin⁻¹ (sin (2π)) = 0

To determine

Whether the given statement is true or not and give an explanation or a counter example.

Answer to Problem 1RE

The statement is true.

Explanation of Solution

Given:

A function could have a property that $f\left(-x\right)=f\left(x\right)$ for all $x$.

Consider the function $f\left(x\right)=x^2$.

Then,

$\begin{array}{c}f\left(-x\right)={\left(-x\right)}^2\\ =x^2\\ =f\left(x\right)\end{array}$

Therefore, the statement is true.

To determine

Whether the given statement is true or not and give an explanation or a counter example.

Answer to Problem 1RE

The statement is false.

Explanation of Solution

Given:

The identity $\cos \left(a+b\right)=\cos a+\cos b$, for all $a$ and $b$ in $\left[0,2\pi \right]$.

Calculation:

Take $a=\frac{\pi }{2}$, $b=\frac{\pi }{2}$ and substitute it, in both the parts of the identity.

Then,

$\begin{array}{c}\cos \left(a+b\right)=\cos \left(\frac{\pi }{2}+\frac{\pi }{2}\right)\\ =\cos \left(\pi \right)\\ =-1\end{array}$

And,

$\begin{array}{c}\cos a+\cos b=\cos \left(\frac{\pi }{2}\right)+\cos \left(\frac{\pi }{2}\right)\\ =0+0\\ =0\end{array}$

That is, $\cos \left(\frac{\pi }{2}+\frac{\pi }{2}\right)\ne \cos \left(\frac{\pi }{2}\right)+\cos \left(\frac{\pi }{2}\right)$ implies $\cos \left(a+b\right)\ne \cos a+\cos b$.

Therefore, the statement is false.

To determine

Whether the given statement is true and give an explanation or a counter example.

Answer to Problem 1RE

The statement is false.

Explanation of Solution

Given:

If $f$ is a linear function of the form $f\left(x\right)=mx+b$, then $f\left(u+v\right)=f\left(u\right)+f\left(v\right)$, for all $u$ and $v$.

Calculation:

Take $u=1$ and $v=1$ and compute the following.

That is,

$f\left(1+1\right)=f\left(1\right)+f\left(1\right)$ (1)

The right hand side of equation $\left(1\right)$ is given by,

$\begin{array}{c}f\left(1+1\right)=f\left(2\right)\\ =2m+b\end{array}$

And, the left hand side of equation $\left(1\right)$ is given by,

$\begin{array}{c}f\left(1\right)+f\left(1\right)=\left(m+b\right)+\left(m+b\right)\\ =2m+2b\end{array}$

Thus, $f\left(1+1\right)\ne f\left(1\right)+f\left(1\right)$ that implies $f\left(u+v\right)\ne f\left(u\right)+f\left(v\right)$.

Therefore, the statement is false.

To determine

Whether the given statement is true or not and give an explanation or a counter example.

Answer to Problem 1RE

The statement is true.

Explanation of Solution

Given:

The function $f\left(x\right)=1-x$ has the property that $f\left(f\left(x\right)\right)=x$.

Calculation:

Consider the function $f\left(x\right)=1-x$.

Then,

$\begin{array}{c}f\left(f\left(x\right)\right)=f\left(1-x\right)\\ =1-\left(1-x\right)\\ =1-1+x\\ =x\end{array}$

Therefore, the statement is true.

To determine

Whether the given statement is true or not and give an explanation or a counter example.

Answer to Problem 1RE

The statement is false.

Explanation of Solution

Given:

The set $\left\{x:\left|x+3\right|>4\right\}$ can be drawn on the number line without lifting the pencil.

Calculation:

The given set $\left\{x:\left|x+3\right|>4\right\}$ can be written as the union of the disjoint intervals $\left(-\infty ,-7\right)$ and $\left(1,\infty \right)$.

Therefore, the set $\left\{x:\left|x+3\right|>4\right\}$ cannot be drawn on the number line without lifting the pencil.

Therefore, the statement is false.

To determine

Whether the given statement is true or not and give an explanation or a counter example.

Answer to Problem 1RE

The statement is false.

Explanation of Solution

Given:

The equation, ${\log }_{10}\left(xy\right)=\left({\log }_{10}x\right)\left({\log }_{10}y\right)$.

Calculation:

Take, $x=10$, $y=10$ and compute both the sides of the following equation.

Then,

$\begin{array}{c}{\log }_{10}\left(10\times 10\right)={\log }_{10}100\\ =2\end{array}$

and,

$\begin{array}{c}\left({\log }_{10}10\right)\left({\log }_{10}10\right)=1\times 1\\ =1\end{array}$

Then, ${\log }_{10}\left(10\times 10\right)\ne \left({\log }_{10}10\right)\left({\log }_{10}10\right)$ implies ${\log }_{10}\left(xy\right)\ne \left({\log }_{10}x\right)\left({\log }_{10}y\right)$.

Therefore, the statement is false.

To determine

Whether the given statement is true or not and give an explanation or a counter example.

Answer to Problem 1RE

The statement is true.

Explanation of Solution

Given:

The equation, ${\sin }^{-1}\left(\sin \left(2\pi \right)\right)=0$.

Consider ${\sin }^{-1}\left(\sin \left(2\pi \right)\right)=0$.

Then, compute the following.

$\begin{array}{c}{\sin }^{-1}\left(\sin \left(2\pi \right)\right)={\sin }^{-1}\left(0\right)\\ =0\end{array}$

Thus, ${\sin }^{-1}\left(\sin \left(2\pi \right)\right)=0$.

Therefore, the statement is true.