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

The percentage error in the surface area of a cube with edge x cm, when the edge is increased by $11\%$ is _________.

  1. $11$
  2. $22$
  3. $10$
  4. $44$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Surface area of cube $=$ S $=6x^2$
$\therefore \dfrac{dS}{dt}=12x\dfrac{dx}{dt}$
Side of cube increase $11\%$.
$\therefore \dfrac{dx}{dt}=11\%$ increment in side
$=\dfrac{11x}{100}$
$\therefore \dfrac{dx}{dt}=12\left(\dfrac{11x}{100}\right)$
$=6\left(\dfrac{22}{100}\right)x^2$
$=6x^2\left(\dfrac{22}{100}\right)$
$=22\%$ in surface area
$\therefore$ Surface area increase $22\%$.

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

The focal length of a mirror is given by $\dfrac {1}{v}-\dfrac {1}{u}=\dfrac {2}{f}$. If equal errors ($\alpha$) are made in measuring $u$ and $v$, then the relative error in $f$ is

  1. $\dfrac {2}{\alpha}$
  2. $\alpha \left (\dfrac {1}{u}+\dfrac {1}{v}\right )$
  3. $\alpha \left (\dfrac {1}{u}-\dfrac {1}{v}\right )$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given, $\displaystyle \dfrac {1}{v}-\dfrac {1}{u}=\dfrac {2}{f}$
$\Rightarrow \displaystyle -\dfrac {\Delta v}{v^2}+\dfrac {\Delta u}{u^2}=-\dfrac {2\Delta f}{f^2}$
Since, given equal errors in measuring u and v i.e.$\Delta u=\Delta v=\alpha$
$\Rightarrow {\alpha}\left(\dfrac{1}{u}-\dfrac{1}{v}\right)\left(\dfrac{1}{u}+\dfrac{1}{v}\right)=-\dfrac {2\Delta f}{f^2}$
But, $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{2}{f}$
$\Rightarrow\displaystyle \frac{ \Delta f}{f}={\alpha}\left(\dfrac{1}{u}+\dfrac{1}{v}\right)$
Hence, correct option is B

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

The period of oscillation $T$ of a pendulum of length $l$ at a place of acceleration due to gravity $g$ is given by $T=2\pi \sqrt {\dfrac {l}{g}}$. If the calculated length is $0.992$ times the actual length and if the value assumed for $g$ is $1.002$ times its actual value, the relative error in the computed value of $T$ is

  1. $0.005$
  2. $-0.005$
  3. $0.003$
  4. $-0.003$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Relative error will calculate by
$\dfrac{\Delta T}{T}=\dfrac{1}{2}\left[\dfrac{\Delta L}{L}-\dfrac{\Delta g}{g}\right]$

$\dfrac{\Delta T}{T}=\dfrac{1}{2}\left[\dfrac{0.992L}{L}-\dfrac{1.002 g}{g}\right]$

$\Delta T=-0.005T$

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

The area of a triangle is computed using the formula $S=\dfrac {1}{2}$ bc sin A. If the relative errors made in measuring b, c and calculating S are respectively $0.02$, $0.01$ and $0.13$ the approximate error in A when $A=\pi /6$ is

  1. $0.05$ radians
  2. $0.01$ radians
  3. $0.05$ degree
  4. $0.01$ degree
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Error formula for given equation is
$\dfrac{\Delta s}{s}=\dfrac{\Delta b}{b}+\dfrac{\Delta c}{c}+\dfrac{\Delta sinx}{sinx}$
$0.13=0.02+0.01+\dfrac{\Delta sinx}{1/2}$
$\Delta sinx=0.05$

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.