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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} $

                                                 $=\dfrac{0.8642}{0.2562}$

                                                $=3.371459$

                                                $=0.3371459 \times 10^1$

                                                $=0.33715 : E : 01$

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

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

The second approximation of roots of $x^3-5x-7=0$ in the interval $(2,3)$ by the method of false position is?

  1. $1.735$
  2. $2.375$
  3. $3.735$
  4. $2.735$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Here, $x^3-5x-7=0$

Let $f(x)=x^3-5x-7$
First Iteration:
Here, $f(2)=-9<0$ and $f(3)=5>0$
Now, Root lies between $x _0=2$ and $x _1=3$
$x _2=x _0-f(x _0)\times \dfrac{x _1-x _0}{f(x _1)-f(x _0)}=2-(-9)\times \dfrac{3-2}{5-(-9)}=2.64286$
Second Iteration:
Here, $f(2.64286)=-1.75474$ and $f(3)=5>0$

Now, Root lies between $x _0=2.64286$ and $x _1=3$
$x _3=x _0-f(x _0)\times \dfrac{x _1-x _0}{f(x _1)-f(x _0)}=2.64-(-1.75)\times \dfrac{3-2.64}{5-(-1.75)}=2.73564$

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

The third approximation of root of $x^3-x^2-1=0$ in the interval $(1,2)$ using successive bisection method is?

  1. $1.475$
  2. $1.375$
  3. $2.213$
  4. $1.564$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We have to find the third approximation of root of the equation $x^3-x^2-1=0$ 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 $a _1=a _0,b _1=c _0$

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

$c _1=\dfrac{a _1+b _1}{2}=\dfrac{1+1.5}{2}=1.25$

Since $f(c _1)f(a _1)=f(1.25)f(1)>0$

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

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

$c _2=\dfrac{a _2+b _2}{2}=\dfrac{1.25+1.5}{2}=1.375$

Thus the third approximation of the root is $1.375$ respectively.

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

By successive bisection method, the cube root of $2$ between the interval (1,1.5)_is?

  1. $1.2813$
  2. $1.2121$
  3. $1.013$
  4. $1.475$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Function can be written as $f(x)=x^3-2$


First Iteration:
$f(1)=-1<0$ and $f(1.5)=1.375>0$
Now, root lies between $1$ and $1.5$
So,
$x _0=\dfrac{1+1.5}{2}=1.25$
$f(x _0)=-0.04688<0$

Second Iteration:

$f(1.25)=-0.04688<0$ and $f(1.5)=1.375>0$
Now, root lies between $1.25$ and $1.5$
So,
$x _1=\dfrac{1.25+1.5}{2}=1.375$

$f(x _1)=0.59961>0$

Third Iteration:

$f(1.25)=-0.04688<0$ and $f(1.375)=0.59961>0$
Now, root lies between $1.25$ and $1.375$
So,
$x _2=\dfrac{1.25+1.375}{2}=1.3125$
$f(x _2)=0.26099>0$

Fourth Iteration:
$f(1.25)=-0.04688<0$ and $f(1.3125)=0.26099>0$
Now, root lies between $1.25$ and $1.3125$
So,
$x _2=\dfrac{1.25+1.3125}{2}=1.28125\approx 1.2813$
Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

The value of $\cdot4267\ E\ 10 \div \cdot2437\ E -02.$ is?

  1. $\cdot1751\ E\ 03$
  2. $\cdot1752\ E\ 13$
  3. $\cdot1751\ E\ 13$
  4. $\cdot1762\ E\ 13$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$0.4267:E:10\div0.2437 : E :-02=?$

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 $1.23 \times 10^{10}$

$0.4267:E:10\div0.2437 : E :-02=\dfrac{0.4267 \times 10^{10}}{0.2437 \times 10^{-2}}$

                                                       $= 1.75092 \times 10^{12}$

                                                       $= 1.751 \times 10^{12}$

                                                       $= .1751 \times 10^{13}$

                                                       $= 0.1751 : E :13$

Hence $0.4267:E:10\div0.2437 : E :-02=0.1751 : E :13$

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

The third approximation of roots of $x^3-9x+1=0$ in the interval $(2,4)$ by the method of false position is?

  1. $8.23$
  2. $1.25$
  3. $2.85$
  4. $2.12$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Here, $x^3-9x+1=0$

Let $f(x)=x^3-9x+1$
First Iteration:
Here, $f(2)=-9<0$ and $f(4)=29>0$
Now, Root lies between $x _0=2$ and $x _1=4$
$x _2=x _0-f(x _0)\times \dfrac{x _1-x _0}{f(x _1)-f(x _0)}=2-(-9)\times \dfrac{4-2}{29-(-9)}=2.47368$
Second Iteration:
Here, $f(2.47368)=-6.1264$ and $f(2)=29>0$

Now, Root lies between $x _0=2.47368$ and $x _1=4$
$x _3=x _0-f(x _0)\times \dfrac{x _1-x _0}{f(x _1)-f(x _0)}=2.47-(-6.13)\times \dfrac{4-2.47}{29-(-6.13)}=2.73989$

Third Iteration:

Here, $f(2.73989)=-3.09067$ and $f(4)=29>0$
Now, Root lies between $x _0=2.73989$ and $x _1=4$
$x _4=x _0-f(x _0)\times \dfrac{x _1-x _0}{f(x _1)-f(x _0)}=2.74-(-3.09)\times \dfrac{4-2.74}{29-(-3.09)}=2.86125$