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 and give an explanation or a counter example.

Answer to Problem 1RE

The given statement is true.

Explanation of Solution

Given:

The given statement is “A function could have a property that $f\left(-x\right)=f\left(x\right)$ for all $x$”.

Calculation:

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

Now, check whether the above mentioned function satisfies the property $f\left(-x\right)=f\left(x\right)$ or not.

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

Therefore, the statement is true.

To determine

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

Answer to Problem 1RE

The statement is false.

Explanation of Solution

Given:

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

Calculation:

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

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

Similarly, calculate the right hand side of the identity as follows.

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

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:

The given statement is “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.

$f\left(1+1\right)=f\left(1\right)+f\left(1\right)$        (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 and give an explanation or a counter example.

Answer to Problem 1RE

The statement is true.

Explanation of Solution

Given:

The given statement is “The function $f\left(x\right)=1-x$ has the property that $f\left(f\left(x\right)\right)=x$”.

Calculation:

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

Calculate $f\left(f\left(x\right)\right)=x$ as shown below.

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

Therefore, the statement is true.

To determine

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

Answer to Problem 1RE

The statement is false.

Explanation of Solution

Given:

The given statement is “The set $\{x:|x+3|>4\}$ can be drawn on the number line without lifting pencil”.

Calculation:

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

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

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:

The given statement is “The equation, ${\log }_{10}\left(xy\right)=\left({\log }_{10}x\right)\left({\log }_{10}y\right)$.”

Calculation:

Take, $x=10$ and $y=10$ ,and compute both the sides of the above mentioned equation as shown below.

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

Similarly, calculate the right hand side of the equation as shown below.

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

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 and give an explanation or a counter example.

Answer to Problem 1RE

The statement is true.

Explanation of Solution

Given:

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

Calculation:

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.