Tag: approximation

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Questions Related to approximation

Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

Using Newton-Raphson method, the cube root of $24$ is?

  1. $2.884$
  2. $3.256$
  3. $5.231$
  4. $4.526$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To find the cube root of $24$ using Newton - Raphson method,
we need to solve $f(x)=x^3-24$.

$ \Rightarrow f'(x)=3x^2$

Notice $3^3=27$

Therefore the cube root of $24$ is slightly less than $3$.

We have $f(x)=x^3-24, f'(x)=3x^2$

Let us start estimating the root $x$

Let the first estimation be $a=2.9$ (slightly less than 3)

Hence the subsequent estimates will be $b=a-\dfrac{f(a)}{f'(a)},c=b-\dfrac{f(b)}{f'(b)}$.

$f(a)=f(2.9)=(2.9)^3-24=0.389$ and $f'(a)=f'(2.9)=3(2.9)^2=25.23$

Therefore $b=2.9-\dfrac{0.389}{25.23}\approx 2.88458$

Now $c=2.88458-\dfrac{f(2.88458)}{f'(2.88458)}=2.88449$

Hence the cube root of $24$ is $2.884$

Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

Using successive Bisection method find the second, third and fourth approximation of root of the  equation $x^3-3x-5=0$ in the interval $(2,2.5)$

  1. $ 2.375,2.135 \ \& \ \ 2.2815$
  2. $1.25,1.375 \ \ \& \ \ 1.4375$
  3. $4.23,3.214 \ \ \& \ \ 2.135$
  4. $2.4475,2.175 \ \ \& \ \ 3.2815$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We have to find the second,third and fourth approximation of root of the equation $x^3-3x-5=0$ in the interval $(2,2.5)$ using successive Bisection method.

$\textbf{Iteration 1: k=0}$

$c _0=\dfrac{a _0+b _0}{2}=\dfrac{2+2.5}{2}=2.25$

Since $f(c _0)f(a _0)=f(2.25)f(2)>0$

Therefore set $a _1=2.25,b _1=b _0$

$\textbf{Iteration 2: k=1}$

$c _1=\dfrac{a _1+b _1}{2}=\dfrac{2.25+2.5}{2}=2.375$

Since $f(c _1)f(a _1)=f(2.375)f(2.25)<0$

Therefore set $a _2=a _1,b _2=c _1$

$\textbf{Iteration 3: k=2}$

$c _2=\dfrac{a _2+b _2}{2}=\dfrac{2.25+2.375}{2}=2.3125$

Since $f(c _2)f(a _2)=f(2.3125)f(2.25)<0$

Therefore set $a _3=a _2,b _3=c _2$

$\textbf{Iteration 4: k=3}$

$c _3=\dfrac{a _3+b _3}{2}=\dfrac{2.25+2.3125}{2}=2.28125$

Thus the second,third and fourth approximations are $2.375,2.3125,2.28125$ respectively.

Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

The second and third approximation of $x^3-2x-5=0$ in the interval $(2,3)$ is?

  1. $x _2 = 2.0946$ and $x _3 = 2.0947$
  2. $x _2 = 1.636 $ and $x _3 = 2.98$
  3. $x _2 = 4.0946 $ and $x _3 = 5.0947$
  4. $x _2 = 2.946$ and $x _3 = 2.07$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Here, $x^3-2x-5=0$

Let $f(x)=x^3-2x-5$
$f'(x)=3x^2-2$
Here $f(2)=-1<0$ and $f(3)=16>0$
Root lies between $2$ and $3$.
$x _0=\dfrac{2+3}{2}=2.5$
First Iteration:
$f(x _0)=f(2.5)=5.625$
$f'(x _0)=f'(2.5)=16.75$
$x _1=x _0-\dfrac{f(x _0)}{f'(x _0)}=2.5-\dfrac{5.625}{16.75}=2.16418$
Second Iteration:

$f(x _1)=f(2.16418)=0.80795$
$f'(x _1)=f'(2.16418)=12.05101$
$x _2=x _1-\dfrac{f(x _1)}{f'(x _1)}=2.16418-\dfrac{0.80795}{12.05101}=2.09714$

Third Iteration:

$f(x _2)=f(2.09714)=0.02888$
$f'(x _2)=f'(2.09714)=11.19393$
$x _3=x _2-\dfrac{f(x _2)}{f'(x _2)}=2.09714-\dfrac{0.02888}{11.19393}=2.09456$

Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

The second and third approximation to the roots of $x^4-x-10=0$ in the interval $(1,2)$ is?

  1. $x _2=2.856,x _3=3.8561$
  2. $x _2=1.7756,x _3=1.061$
  3. $x _2=1.87409, x _3=1.85587$
  4. $x _2=7.856,x _3=1.8561$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Here, $x^4-x-10=0$


Let $f(x)=x^4-x-10$

$f'(x)=4x^3-1$

Here $f(1)=-10<0$ and $f(2)=4>0$

Root lies between $1$ and $2$

$x _0=\dfrac{1+2}{2}=1.5$

First Iteration:

$f(x _0)=f(1.5)=-6.4375$

$f'(x _0)=f'(1.5)=12.5$

$x _1=x _0-\dfrac{f(x _0)}{f'(x _0)}=1.5-\dfrac{-6.4375}{12.5}=2.015$

Second Iteration:


$f(x _1)=f(2.015)=4.47043$

$f'(x _1)=f'(2.015)=31.72541$

$x _2=x _1-\dfrac{f(x _1)}{f'(x _1)}=2.015-\dfrac{4.47043}{31.72541}=1.87409$


Third Iteration:


$f(x _2)=f(1.87409)=0.46155$

$f'(x _2)=f'(1.87409)=25.32882$

$x _3=x _2-\dfrac{f(x _2)}{f'(x _2)}=1.87409-\dfrac{0.46155}{25.32882}=1.85587$

Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

Using successive Bisection method find the second, third and fourth approximation of root of the given  equation $x^3-x-4=0$ in the interval $(1,2)$

  1. $2.75,13.875 , 1.8125$
  2. $1.75,1.875 , 1.8125$
  3. $1.725,1.5 , 1.8125$
  4. $2.75,1.875 , 1.8125$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We have to find the second,third and fourth approximation of root of the equation $x^3-x-4=0$ in the interval $(1,2)$ using successive Bisection method.

$\textbf{Iteration 1: k=0}$

$c _0=\dfrac{a _0+b _0}{2}=\dfrac{1+2}{2}=1.5$

Since $f(c _0)f(a _0)=f(1.5)f(1)>0$

Therefore set $a _1=1.5,b _1=b _0$

$\textbf{Iteration 2: k=1}$

$c _1=\dfrac{a _1+b _1}{2}=\dfrac{1.5+2}{2}=1.75$

Since $f(c _1)f(a _1)=f(1.75)f(1.5)>0$

Therefore set $a _2=c _1,b _2=b _1$

$\textbf{Iteration 3: k=2}$

$c _2=\dfrac{a _2+b _2}{2}=\dfrac{1.75+2}{2}=1.875$

Since $f(c _2)f(a _2)=f(1.875)f(1.75)<0$

Therefore set $a _3=a _2,b _3=c _2$

$\textbf{Iteration 4: k=3}$

$c _3=\dfrac{a _3+b _3}{2}=\dfrac{1.75+1.875}{2}=1.8125$

Thus the second,third and fourth approximations are $1.75,1.875,1.8125$ respectively.

Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

The second approximation of roots of $x^3-x-4=0$ in the interval $(1,2)$ by the method of false position is?

  1. $1.78049$
  2. $1.276$
  3. $2.123$
  4. $0.726$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Here, $f(x)=x^3-x-4=0$


first iteration:

]Here $f(1)=-4<0$ and $f(2)=2>0$

Now, root lies between $x _0=1$ and $x _1=2$

$x _2=x _0-f(x _0) \times \dfrac{x _1-x _0}{f(x _1)-f(x _0)}=1-(-4)\times \dfrac{2-1}{2-(-4)}=1.66667$

$f(x _2)=f(1.66667)=-1.03704<0$

2nd iteration:

Here $f(1.66667)=-1.03704<0$ and $f(2)=2>0$

Now, root lies between $x _0=1.66667$ and $x _1=2$

$x _3=x _0-f(x _0) \times \dfrac{x _1-x _0}{f(x _1)-f(x _0)}=1.67-(-1.04)\times \dfrac{2-1.67}{2-(-1.04)}=1.78049$

Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

Using successive Bisection method find the second, third and fourth approximation of root of the equation $x^3+x^2-1$ in the interval $(0,1)$

  1. $0.75,1.875,0.8125$
  2. $1.75,0.875,0.8125$
  3. $0.75,0.875,0.8125$
  4. $0.75,0.875,1.8125$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We have to find the second,third and fourth approximation of root of the equation $x^3+x^2-1=0$ in the interval $(0,1)$ using successive Bisection method.

$\textbf{Iteration 1: k=0}$

$c _0=\dfrac{a _0+b _0}{2}=\dfrac{0+1}{2}=0.5$

Since $f(c _0)f(a _0)=f(0.5)f(0)>0$

Therefore set $a _1=0.5,b _1=b _0$

$\textbf{Iteration 2: k=1}$

$c _1=\dfrac{a _1+b _1}{2}=\dfrac{0.5+1}{2}=0.75$

Since $f(c _1)f(a _1)=f(0.75)f(0.5)>0$

Therefore set $a _2=c _1,b _2=b _1$

$\textbf{Iteration 3: k=2}$

$c _2=\dfrac{a _2+b _2}{2}=\dfrac{0.75+1}{2}=0.875$

Since $f(c _2)f(a _2)=f(0.875)f(0.75)<0$

Therefore set $a _3=a _2,b _3=c _2$

$\textbf{Iteration 4: k=3}$

$c _3=\dfrac{a _3+b _3}{2}=\dfrac{0.75+0.875}{2}=0.8125$

Thus the second,third and fourth approximations are $0.75,0.875,0.8125$ respectively.

Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

The third approximation of roots of $x^3-x^2-1=0$ in the interval $(1,2)$ by the method of false position is?

  1. $2.430$
  2. $1.340$
  3. $1.430$
  4. $1.230$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Here, $x^3-x^2-1=0$

Let $f(x)=x^3-x^2-1$
First Iteration:
Here, $f(1)=-1<0$ and $f(2)=3>0$
Now, Root lies between $x _0=1$ and $x _1=2$
$x _2=x _0-f(x _0)\times \dfrac{x _1-x _0}{f(x _1)-f(x _0)}=1-(-1)\times \dfrac{2-1}{3-(-1)}=$
Second Iteration:
Here, $f(1.25)=-0.60938<0$ and $f(2)=3>0$

Now, Root lies between $x _0=1.25$ and $x _1=2$
$x _3=x _0-f(x _0)\times \dfrac{x _1-x _0}{f(x _1)-f(x _0)}=1.25-(-0.61)\times \dfrac{2-1.25}{3-(-0.61)}=1.37662$

Third Iteration:

Here, $f(1.37662)=-0.28626<0$ and $f(2)=3>0$
Now, Root lies between $x _0=1.37662$ and $x _1=2$
$x _4=x _0-f(x _0)\times \dfrac{x _1-x _0}{f(x _1)-f(x _0)}=1.38-(-0.29)\times \dfrac{2-1.38}{3-(-0.29)}=1.43093$

Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

The third approximation of roots of $x^3-x-1=0$ in the interval $(1,2)$ by the method of false position is?

  1. $1.011$
  2. $2.265$
  3. $1.255$
  4. $1.294$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Here, $x^3-x-1=0$

Let $f(x)=x^3-x-1$
First Iteration:
Here, $f(1)=-1<0$ and $f(2)=5>0$
Now, Root lies between $x _0=1$ and $x _1=2$
$x _2=x _0-f(x _0)\times \dfrac{x _1-x _0}{f(x _1)-f(x _0)}=1-(-1)\times \dfrac{2-1}{5-(-1)}=1.16667$
Second Iteration:
Here, $f(1.16667)=-0.5787$ and $f(2)=5>0$

Now, Root lies between $x _0=1.16667$ and $x _1=2$
$x _3=x _0-f(x _0)\times \dfrac{x _1-x _0}{f(x _1)-f(x _0)}=1.17-(-0.58)\times \dfrac{2-1.17}{5-(-0.58)}=1.25311$

Third Iteration:

Here, $f(1.25311)=-0.28536$ and $f(2)=5>0$
Now, Root lies between $x _0=1.25311$ and $x _1=2$
$x _4=x _0-f(x _0)\times \dfrac{x _1-x _0}{f(x _1)-f(x _0)}=1.25-(-0.29)\times \dfrac{2-1.25}{5-(-0.29)}=1.29344$

Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

The value of $\cdot8642\ E\ 02 \div \cdot2562\ E02.$ is?

  1. $\cdot12057\ E\ 09$
  2. $\cdot33715\ E\ 01$
  3. $\cdot33715\ E\ 05$
  4. $\cdot33725\ E\ 01$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$0.8642:E:02 \div 0.2562 : E: 02 = ?$

The Scientific format displays a number in exponential notation, replacing part of the number with $E+n$, where $E$ (stands for exponent) multiplies the preceding number by $10$ to the $n^{th}$ power.

That is $1.23E+10$ can be written as $1.23 \times 10^{10}$

$0.8642:E:02 \div 0.2562 : E: 02 = \dfrac{0.8642 \times 10^2}{0.2562 \times 10^2} $

                                                 $=\dfrac{0.8642}{0.2562}$

                                                $=3.371459$

                                                $=0.3371459 \times 10^1$

                                                $=0.33715 : E : 01$

$0.8642:E:02 \div 0.2562 : E: 02 =0.33715 : E : 01$