Tag: the need for approximation

Quizzes (0) Questions (56)

Questions Related to the need for approximation

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 $x _1=4$
$x _4=x _0-f(x _0)\times \dfrac{x _1-x _0}{f(x _1)-f(x _0)}=2.74-(-3.09)\times \dfrac{4-2.74}{29-(-3.09)}=2.86125$

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

If the error committed in measuring the radius of the circle is $0.05\%$, then the corresponding error in calculating the area is:

  1. $0.05\%$
  2. $0.025\%$
  3. $0.25\%$
  4. $0.1\%$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\dfrac { dr }{ r } =0.05\Rightarrow dr=(0.05)r$

Area of circle $=\pi r^2$
$\ A=\pi r^{ 2 }\Rightarrow \dfrac { dA }{ dr } =2\pi r\ dA=2\pi rdr\Rightarrow \dfrac { dA }{ A } =\dfrac { 2\pi rdr }{ \pi r^{ 2 } } =\dfrac { 2dr }{ r } \ \therefore \dfrac { dA }{ A } =2(0.05)^{ 2 }=0.1$
$\therefore$ Corresponding error in area $= 0.1\%$