Tag: business maths

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Questions Related to business maths

Multiple choice business maths pair of straight lines condition for perpendicular and coincident lines and bisectors of angles pair of straight lines through origin analytical geometry

Equation of line in the place $P=\equiv 2x-y+z-4=0$ which is perpendicular to the line I whose equation is $\dfrac{x-2}{1}=\dfrac{y-2}{-1}=\dfrac{z-3}{-2}$ and which passes through point of intersection of I and P is

  1. $\dfrac{x-2}{3}=\dfrac{y-1}{5}=\dfrac{z-1}{-1}$
  2. $\dfrac{x-1}{3}=\dfrac{y-3}{5}=\dfrac{z-5}{-1}$
  3. $\dfrac{x+2}{2}=\dfrac{y+1}{-1}=\dfrac{z+1}{1}$
  4. $\dfrac{x-2}{2}=\dfrac{y-1}{-1}=\dfrac{z-1}{1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The line must pass through the intersection of the given line and plane, and be perpendicular to the given line. Solving the system of equations for the intersection point and applying the cross product of the normal vector of the plane and the direction vector of the line yields the direction ratios.

Multiple choice business maths definite integral and it applications to areas fundamental theorem of calculus interpretation of define integral as an area fundamental theorem of integral calculus

The value of $\displaystyle \int _0^1\tan^{-1}\left (\frac {2x-1}{1+x-x^2}\right )dx$ is

  1. $1$
  2. $0$
  3. $-1$
  4. $\dfrac {\pi}{4}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let $I=\int _0^1\tan^{-1}\left (\dfrac {2x-1}{1+x-x^2}\right )dx$
$\Rightarrow I=\int _0^1 \tan^{-1}\left (\dfrac {x-(1-x)}{1+x(1-x)}\right )dx$
$\Rightarrow I=\int _0^1[\tan^{-1}x-\tan^{-1}(1-x)]dx$ ................ (1)
$\Rightarrow I=\int _0^1[\tan^{-1}(1-x)-\tan^{-1}(1-1+x)]dx$
$\Rightarrow I=\int _0^1[\tan^{-1}(1-x)-\tan^{-1}(x)]dx$
$\Rightarrow I=\int _0^1[\tan^{-1}(1-x)-\tan^{-1}(x)]dx$ ........... (2)
Adding (1) and (2), we obtain
$2I=\int _0^1(\tan^{-1}x+\tan^{-1}(1-x)-\tan^{-1}(1-x)-\tan^{-1}x)dx=0$
$\Rightarrow I=0$
Hence, the correct Answer is B.

Multiple choice business maths definite integral and it applications to areas fundamental theorem of calculus interpretation of define integral as an area fundamental theorem of integral calculus

$\int _{0}^{\pi /2}sin2xtan^{-1}\left ( sinx \right )dx=$

  1. $\dfrac{\pi }{2}$-1
  2. $\dfrac{\pi }{2}$+1
  3. $\dfrac{3\pi }{2}$+1
  4. $\dfrac{3\pi }{2}$-1
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We have,

$I=\int _{0}^{\dfrac{\pi }{2}}{\sin 2x{{\tan }^{-1}}\left( \sin x \right)dx}$

$=\int _{0}^{\dfrac{\pi }{2}}{2\sin x\cos x{{\tan }^{-1}}\left( \sin x \right)dx}$

Let

$ \sin x=t $

$ \cos xdx=dt $

Change limit

$ \sin 0=t $

$ t=0 $

And,

$ \sin \dfrac{\pi }{2}=t $

$ t=1 $

Then,

$ \int _{0}^{1}{2t{{\tan }^{-1}}t\cos xdx} $

$ =\int _{0}^{1}{2t{{\tan }^{-1}}tdt} $

$ =2\int _{0}^{1}{t{{\tan }^{-1}}tdt} $

On integrating and we get,

$ 2\left[ {{\tan }^{-1}}t\int _{0}^{1}{t}dt-\int _{0}^{1}{\left( \dfrac{d\left( {{\tan }^{-1}}t \right)}{dt}\int _{0}^{1}{tdt} \right)}dt \right] $

$ =2\left[ {{\tan }^{-1}}t\left[ {{\left( \dfrac{{{t}^{2}}}{2} \right)} _{0}}^{1} \right]-\int _{0}^{1}{\dfrac{1}{1+{{t}^{2}}}}\dfrac{{{t}^{2}}}{2}dt \right] $

$ =2{{\tan }^{-1}}t{{\left( \dfrac{{{t}^{2}}}{2} \right)} _{0}}^{1}-\int _{0}^{1}{\dfrac{{{t}^{2}}}{1+{{t}^{2}}}}dt $

$ =2{{\tan }^{-1}}t{{\left( \dfrac{{{t}^{2}}}{2} \right)} _{0}}^{1}-\int _{0}^{1}{\dfrac{{{t}^{2}}+1-1}{1+{{t}^{2}}}}dt $

$ =2{{\tan }^{-1}}t{{\left( \dfrac{{{t}^{2}}}{2} \right)} _{0}}^{1}-\int _{0}^{1}{\dfrac{{{t}^{2}}+1}{1+{{t}^{2}}}}dt+\int _{0}^{1}{\dfrac{1}{1+{{t}^{2}}}}dt $

$ =2{{\tan }^{-1}}t{{\left( \dfrac{{{t}^{2}}}{2} \right)} _{0}}^{1}-\int _{0}^{1}{1}dt+\int _{0}^{1}{\dfrac{1}{1+{{t}^{2}}}}dt $

$ =2{{\tan }^{-1}}t{{\left( \dfrac{{{t}^{2}}}{2} \right)} _{0}}^{1}-{{\left[ t \right]} _{0}}^{1}+{{\left[ {{\tan }^{-1}}t \right]} _{0}}^{1}+C $

$ =2\left[ {{\tan }^{-1}}1-{{\tan }^{-1}}0 \right]\left[ \dfrac{{{1}^{2}}}{2}-\dfrac{{{0}^{2}}}{2} \right]-\left[ 1-0 \right]+\left[ {{\tan }^{-1}}1-{{\tan }^{-1}}0 \right]+C $

$ =2\left[ \dfrac{\pi }{4}-0 \right]\left[ \dfrac{1}{2} \right]-1+\left[ \dfrac{\pi }{4}-0 \right] $

$ =\dfrac{\pi }{4}+\dfrac{\pi }{4}-1 $

$ =\dfrac{2\pi }{4}-1 $

$ =\dfrac{\pi }{2}-1 $

Hence, this is the answer.

Multiple choice business maths definite integral and it applications to areas fundamental theorem of calculus interpretation of define integral as an area fundamental theorem of integral calculus

Evaluate: $\displaystyle \int _{0}^{\sqrt{3}}[x^{3} -1] dx$

  1. $\dfrac{1}{4}-\sqrt3$
  2. $\dfrac{1}{4}-\sqrt2$
  3. $\dfrac{9}{4}-\sqrt3$
  4. $\dfrac{9}{4}-\sqrt2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Consider, $\displaystyle I= \int _{0}^{\sqrt{3}}[x^{3} -1] dx$


$\Rightarrow I=\left [\dfrac{x^4}{4}-x\right]^{\sqrt3} _{0}$

$I=\dfrac94-\sqrt3$

Multiple choice business maths definite integral and it applications to areas fundamental theorem of calculus interpretation of define integral as an area fundamental theorem of integral calculus

$\displaystyle\int^{100} _0[\tan^{-1}x]dx$.

  1. $100+\tan 1$
  2. $100-\tan 1$
  3. $\tan 1$
  4. $99+\tan 1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The integral is the sum of integrals from n to n+1. For x in [tan(n), tan(n+1)], [tan^-1(x)] = n. The sum is 0*integral(0 to tan(1)) + 1*integral(tan(1) to tan(2)) + ... + 99*integral(tan(99) to tan(100)). This simplifies to 100 - tan(1).

Multiple choice business maths definite integral and it applications to areas fundamental theorem of calculus interpretation of define integral as an area fundamental theorem of integral calculus

Solve $\displaystyle\int^{100} _0e^{x-[x]}dx=?$ where $[x]$ is greatest integer function.

  1. $100e$
  2. $100(e-1)$
  3. $100(e+1)$
  4. $100(1-e)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Consider, $I=\displaystyle\int^{100} _0e^{x-[x]}dx$

$I=100\displaystyle\int^{1} _0e^{x}dx$

$I=100(e^x) _0^1$

$I=100(e^1-e^0)$

$I=100(e-1)$
Multiple choice business maths definite integral and it applications to areas fundamental theorem of calculus interpretation of define integral as an area fundamental theorem of integral calculus

If $I _1 = \displaystyle \int^{2\pi /3} _{\pi / 2}\left|cos\dfrac{x}{2}cosx\right|dx,I _2=\left|\displaystyle \int _{\pi/2}^{2\pi/3} cos\dfrac{x}{2}cosxdx\right|$ then $I _1 - I _2$ equals 

  1. $\dfrac{1}{3}(\sqrt{32}-\sqrt{27})$
  2. $\dfrac{1}{3}(\sqrt{32}-\sqrt{25})$
  3. $\dfrac{1}{3}(\sqrt{27}-\sqrt{25})$

  4. None
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice business maths definite integral and it applications to areas fundamental theorem of calculus interpretation of define integral as an area fundamental theorem of integral calculus

The value of the definite integral, $\displaystyle \int _0^{\pi/2} \dfrac{sin5x}{sinx}dx$ is 

  1. 0
  2. $\dfrac{\pi}{2}$
  3. $\pi$
  4. $2\pi$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$\displaystyle = \int _{0}^{\dfrac{\pi}{2}} \dfrac{\sin 5x}{\sin x} dx$
We are going to use a important property if.
$\displaystyle \int _{0}^{\dfrac{\pi}{2}} \dfrac{\sin nx}{\sin x} = \begin{cases} \dfrac{\pi}{2} & ,\ if\ n\ is\ odd \\ o & ,\ if\ is\ even \end{cases}$
So, less $n=s (odd)$
$\displaystyle \int _{0}^{\dfrac{\pi}{2}} \dfrac{\sin 5x}{\sin x} =\dfrac{\pi}{2}$