Tag: approximation

Questions Related to approximation

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

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

  1. $2.884$

  2. $3.256$

  3. $5.231$

  4. $4.526$

Show answer
Correct Option: A
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 $2.884$

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

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

  1. $ 2.375,2.135 \  & \ \ 2.2815$

  2. $1.25,1.375 \ \ & \ \ 1.4375$

  3. $4.23,3.214 \ \  & \ \  2.135$

  4. $2.4475,2.175 \ \ & \ \ 3.2815$

Show answer
Correct Option: A
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.

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

maths the trapezium rule approximation errors and approximations the need for approximation
  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 $x _3 = 2.07$
Show answer
Correct Option: A
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:

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

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

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

  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. $x _2=7.856,x _3=1.8561$

Show answer
Correct Option: C
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:


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

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

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

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 $(1,2)$

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

  1. $2.75,13.875 , 1.8125$

  2. $1.75,1.875 , 1.8125$

  3. $1.725,1.5 , 1.8125$

  4. $2.75,1.875 , 1.8125$

Show answer
Correct Option: B
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.

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

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

  1. $1.78049$

  2. $1.276$

  3. $2.123$

  4. $0.726$

Show answer
Correct Option: A
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 $x _1=2$

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

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

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

  1. $0.75,1.875,0.8125$

  2. $1.75,0.875,0.8125$

  3. $0.75,0.875,0.8125$

  4. $0.75,0.875,1.8125$

Show answer
Correct Option: C
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.

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

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

  1. $2.430$

  2. $1.340$

  3. $1.430$

  4. $1.230$

Show answer
Correct Option: C
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 $x _1=2$
$x _4=x _0-f(x _0)\times \dfrac{x _1-x _0}{f(x _1)-f(x _0)}=1.38-(-0.29)\times \dfrac{2-1.38}{3-(-0.29)}=1.43093$

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

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

  1. $1.011$

  2. $2.265$

  3. $1.255$

  4. $1.294$

Show answer
Correct Option: D
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 $x _1=2$
$x _4=x _0-f(x _0)\times \dfrac{x _1-x _0}{f(x _1)-f(x _0)}=1.25-(-0.29)\times \dfrac{2-1.25}{5-(-0.29)}=1.29344$

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

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

  1. $\cdot12057\ E\ 09$

  2. $\cdot33715\ E\ 01$

  3. $\cdot33715\ E\ 05$

  4. $\cdot33725\ E\ 01$

Show answer
Correct Option: B
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} $

                                                 $=\dfrac{0.8642}{0.2562}$

                                                $=3.371459$

                                                $=0.3371459 \times 10^1$

                                                $=0.33715 : E : 01$

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