## MATH1013 Quizzes

Quiz 1: The Bisection method
Question 1 Questions
Solve the equation $3{x}^{2}=5x+2$ to 2 decimal places.
 a) $x=-1.00$, $x=-0.67$ b) $x=-0.33$, $x=2.00$ c) $x=1.00$, $x=0.67$ d) $x=0.06$, $x=-1.73$

Choice (a) is incorrect
Choice (b) is correct!
We need to solve $3{x}^{2}-5x-2=\left(3x+1\right)\left(x-2\right)=0$.
Choice (c) is incorrect
Choice (d) is incorrect
Solve the equation $2{x}^{2}=-4x+1$ to 2 decimal places.
 a) $x=-3.22$, $x=-0.78$ b) $x=1.71$, $x=-0.29$ c) $x=-2.22$, $x=0.22$ d) $x=-2.22$, $x=-0.22$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
We need to solve $2{x}^{2}+4x-1=0$ and using the quadratic formula we find
Choice (d) is incorrect
Consider the two graphs sketched below.

Which of the equations could they be used to solve ?
 a) ${e}^{-2x}=\sqrt{-2x+3}$ b) ${e}^{-x}=2x-3$ c) ${e}^{-2x}+3=-2x+3$ d) ${e}^{-2x}=-2x+3$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
The two graphs are $y={e}^{-2x}$ and $y=-2x+3$. Their points of intersection (dotted in the sketch) are the solutions of the equation ${e}^{-2x}=-2x+3$.
Consider the two graphs sketched below.
Which of the following equations could they be used to solve ?
 a) $sin2x={x}^{3}-3x+2$ b) $sin2x+{x}^{3}-3x+2=0$ c) $sinx=\frac{{x}^{3}-3x+2}{2}$ d) $sinx={x}^{2}-3x+2$

Choice (a) is correct!
The two graphs are $y=sin2x$ and $y={x}^{3}-3x+2$. Their intersection solves the equation $sin2x={x}^{3}-3x+2$.
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Which of the following functions has roots which are solutions of the equation ${e}^{2x+5}={x}^{3}-5x+6$ ?
 a) $f\left(x\right)=2x+5-ln{x}^{3}+ln5x-ln6$ b) $f\left(x\right)={e}^{2x+5}-{x}^{3}+5x-6$ c) $f\left(x\right)={e}^{2x+5}-{x}^{3}+5x+6$ d) $f\left(x\right)=2x+5+ln\left({x}^{3}+5x-6\right)$

Choice (a) is incorrect
Choice (b) is correct!
Choice (c) is incorrect
Choice (d) is incorrect
How many solutions are there of the equation $sin\left(2x-5\right)-2={\left(3x-4\right)}^{2}$ in $ℝ$ ?
 a) None b) One c) Two d) Infinitely many

Choice (a) is correct!
$sin\left(2x-5\right)-2\le -1$ but ${\left(3x-4\right)}^{2}\ge 0$, for all $x\in ℝ$ hence there are no solutions.
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
How many solutions are there to the equation $ln\left(x+1\right)={x}^{2}-2x-3$ in $ℝ$ ?
 a) None b) One c) Two d) Infinitely many

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
The easiest way to see this is by graphing $y=ln\left(x+1\right)$ and $y={x}^{2}-2x-8$ on the same set of axes.
We see that there are only two possible solutions, one near $x=-1$ and one near $x=4.3$.
Choice (d) is incorrect
How many solutions are there to the equation ${x}^{3}+x+1=cosx$ in $ℝ$ ?
 a) None b) One c) Two d) Infinitely many

Choice (a) is incorrect
Choice (b) is correct!
Consider $f\left(x\right)={x}^{3}+x+1-cosx$. $f\left(0\right)=0$ so $x=0$ is solution. Now ${f}^{\prime }\left(x\right)=3{x}^{2}+sinx+1>0$ for all $x\in ℝ$, since $-1\le sinx\le 1$ and $3{x}^{2}+1\ge 1$. Therefore the function is always increasing and so there can be only the one root, at $x=0$.
Choice (c) is incorrect
Choice (d) is incorrect
Use the bisection method three times on the function $f\left(x\right)={x}^{2}-sinx-1$ to determine where $f\left(x\right)$ changes sign on the interval $-2.
 a) $f\left(x\right)$ changes sign on the interval $-0.75\le x\le -0.5$ b) $f\left(x\right)$ changes sign on the interval $-0.25\le x\le 0$ c) $f\left(x\right)$ changes sign on the interval $-1\le x\le -0.75$ d) We cannot use this method as $f\left(x\right)$ does not change sign on this interval.

Choice (a) is correct!
The first iteration tells us that the change of sign is between -1 and 0. The second iteration tells us that the change of sign is between -1 and -0.5. The third iteration tells us that the change of sign occurs between -0.75 and -0.5.
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Use the dissection method on $f\left(x\right)={x}^{3}-cosx+1$ to determine where the function changes sign on the interval $-1.1.
 a) $f\left(x\right)$ changes sign on the interval $-0.2\le x\le -0.1$ b) $f\left(x\right)$ changes sign on the interval $-0.9\le x\le -0.8$ c) $f\left(x\right)$ changes sign on the interval $-0.4\le x\le -0.3$ d) $f\left(x\right)$ changes sign on the interval $-0.5\le x\le -0.4$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
This is the correct response since $f\left(-0.5\right)<0$ and $f\left(-0.4\right)>0$.