Tag: maths

Questions Related to maths

The value of $\cdot3214\ E\ - 02 \times \cdot3781\ E\ 05.$ is?

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

  1. $\cdot699045\ E\ 05$

  2. $\cdot699045\ E\ 02$

  3. $\cdot699055\ E\ 02$

  4. $\cdot698045\ E\ 02$

Show answer
Correct Option: B

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

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

  1. $11$

  2. $22$

  3. $10$

  4. $44$

Show answer
Correct Option: B
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 $22\%$.

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

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

  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

Show answer
Correct Option: B
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

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

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

  1. $0.005$

  2. $-0.005$

  3. $0.003$

  4. $-0.003$

Show answer
Correct Option: B
Explanation:

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

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

$\Delta T=-0.005T$

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

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

  1. $0.05$ radians

  2. $0.01$ radians

  3. $0.05$ degree

  4. $0.01$ degree

Show answer
Correct Option: A
Explanation:

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

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.