Tag: maths

Questions Related to maths

$\dfrac {d}{dx}(x^{\ell n x})$ is equal to

maths differencial calculus - differenciability and methods of differnciation differentiation by substitution methods of differentiation derivative of a function

  1. $2x^{\ell n x-1}\ell n x$

  2. $x^{\ell n x-1}$

  3. $2/3(\ell n x)$

  4. $x^{\ell n x-1}.\ell n x$

Show answer
Correct Option: A
Explanation:

Let $lnx=u$

$\therefore\ x={e}^{u}.$
$\frac { d }{ dx } \left( { x }^{ lmx } \right) =\frac { d }{ dx } \left( { e }^{ u.lnu } \right) $
$=\frac { d }{ dx } \left( { e }^{ { lnu }^{ 2 } } \right) =\frac { d }{ dx } \left( { u }^{ 2 } \right) $
$=2u.\frac { du }{ dx } $
$=2u.\frac { d }{ dx } \left( lnx \right) =\boxed{2lnx\quad { x }^{ lnx-1 }}$

If $y=(tan \, x)^{(tan\, x)^{tan\,x}}, $ then at $x=\dfrac{\pi}{4}, \dfrac{dy}{dx}$ is equal to 

maths differencial calculus - differenciability and methods of differnciation differentiation by substitution methods of differentiation derivative of a function

  1. 0

  2. 3

  3. 2

  4. None of these

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Correct Option: C
Explanation:
$y={ \left( tanx \right)  }^{ { \left( tanx \right)  }^{ \left( tanx \right)  } }$
We have to find $\dfrac { dy }{ dx } $ at $x=\dfrac { \Pi  }{ 4 } $.
Now consider:
$y={ f\left( x \right)  }^{ g\left( x \right)  }$
$ln\left( y \right) =g\left( x \right) lnf\left( x \right) $
$\Rightarrow \dfrac { 1 }{ y } \dfrac { dy }{ dx } =g\left( x \right) \dfrac { { f }^{ 1 }\left( x \right)  }{ f\left( x \right)  } +{ g }^{ 1 }\left( x \right) lnf\left( x \right) $
$\Rightarrow \dfrac { dy }{ dx } =y\left[ g\left( x \right) \dfrac { { f }^{ 1 }\left( x \right)  }{ f\left( x \right)  } +{ g }^{ 1 }\left( x \right) lnf\left( x \right)  \right] $
$\Rightarrow \dfrac { dy }{ dx } ={ f\left( x \right)  }^{ g\left( x \right)  }\left[ { g }^{ 1 }\left( x \right) lnf\left( x \right) +g\left( x \right) \dfrac { { f }^{ 1 }\left( x \right)  }{ f\left( x \right)  }  \right] $
So, We have
$y={ \left( tanx \right)  }^{ { \left( tanx \right)  }^{ tanx } }$
Where  $f\left( x \right) =tan\left( x \right) $
${ f }^{ 1 }\left( x \right) ={ \sec }^{ 2 }x$
$g\left( x \right) ={ \left( tanx \right)  }^{ tanx }$
${ g }^{ 1 }\left( x \right) ={ \left( tanx \right)  }^{ tanx\left[ { \sec }^{ 2 }xln\left( tan\left( x \right)  \right) +\dfrac { \tan\left( x \right) { \sec }^{ 2 }x }{ tanx }  \right]  }$
Now, $\tan\left( \dfrac { \Pi  }{ 4 }  \right) =1$ and $\sec\left( \Pi /4 \right) =\sqrt { 2 } $.
$f\left( \Pi /4 \right) =1$
${ f }^{ 1 }\left( \Pi /4 \right) =2$
$g\left( \Pi /4 \right) =1$
${ g }^{ 1 }\left( \Pi /4 \right) =1\left( 2ln\left( 1 \right) +2 \right) =2$
$\dfrac { dy }{ dx } $ at $\Pi /4$ is
$\dfrac { dy }{ dx } ={ f\left( \Pi /4 \right)  }^{ g\left( \Pi /4 \right) \left[ { g }^{ 1 }\left( \Pi /4 \right) ln\left( f\left( \Pi /4 \right)  \right) +g\left( \Pi /4 \right) \dfrac { { f }^{ 1 }\left( \Pi /4 \right)  }{ f\left( \Pi /4 \right)  }  \right]  }$
$\Rightarrow \dfrac { dy }{ dx } =1\left[ 2ln\left( 1 \right) +1\times 2 \right] =2$
$\Rightarrow \dfrac { dy }{ dx } =2$.

The solution of the differential equation, $y\,dx + \left( {x + {x^2}y} \right)dy = 0$ is

maths differencial calculus - differenciability and methods of differnciation differentiation by substitution methods of differentiation derivative of a function

  1. $\log y = cx$

  2. $ \log y- \dfrac{1}{{xy}}  = c$

  3. $\dfrac{1}{{xy}} - \log y = c$

  4. $\dfrac{1}{{xy}} + \log y = c$

Show answer
Correct Option: B
Explanation:

We have,

$ydx+\left( { x+{ x^{ 2 } }y } \right) dy=0 \ ydx=-\left( { x+{ x^{ 2 } }y } \right) dy \ ydx+xdy=-{ x^{ 2 } }ydy \ \dfrac { { ydx+xdy } }{ { { { \left( { xy } \right)  }^{ 2 } } } } =-\dfrac { { dy } }{ y }  \ \dfrac { { d\left( { xy } \right)  } }{ { { { \left( { xy } \right)  }^{ 2 } } } } =-\dfrac { { dy } }{ y } $

On taking integrating both sides

$\begin{array}{l} \dfrac { { -1 } }{ { xy } } =-\log  y+c \\ \log  y-\dfrac { 1 }{ { xy } } =c \end{array}$


Hence, this is the answer.

If $x=a\sin \theta$ and $y=b\cos\theta$, then $\displaystyle\frac{d^2y}{dx^2}$ is 

maths differencial calculus - differenciability and methods of differnciation differentiation by substitution methods of differentiation derivative of a function

  1. $\displaystyle\frac{a}{b^2}\sec^2\theta$

  2. $\displaystyle\frac{-b}{a}\sec^2\theta$

  3. $\displaystyle\frac{b}{a^2}\sec^3\theta$

  4. $\displaystyle\frac{-b}{a^2}\sec^3\theta$

Show answer
Correct Option: D
Explanation:

Given, $x=a\sin\theta$ and $y=b\cos\theta$
On differentiating w.r.t.$\theta$, we get
$\displaystyle\frac{dx}{d\theta}=a\cos\theta$
and $\displaystyle\frac{dy}{d\theta}=-b\sin \theta$
$\Rightarrow \displaystyle\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=-\frac{b}{a}\tan\theta$
Again, differentiating w.r.t. $x$, we get
$\displaystyle\frac{d^2y}{dx^2}=-\frac{b}{a}sec^2\theta\cdot\frac{d\theta}{dx}$
$\Rightarrow \displaystyle\frac{d^2y}{dx^2}=-\frac{b}{a}sec^2\theta\cdot\frac{1}{a\cos\theta}$
$=-\displaystyle\frac{b}{a^2}sec^3\theta$

The set of all points, where the function $f(x) = \sqrt {1 - e^{-x^{2}}}$ is differentiable, is

maths differencial calculus - differenciability and methods of differnciation differentiation by substitution methods of differentiation derivative of a function

  1. $(0, \infty)$

  2. $(-\infty, \infty)$

  3. $(-\infty, 0) \cup (0, \infty)$

  4. $(-1, \infty)$

Show answer
Correct Option: C
Explanation:

Given that

$ f(x) = \sqrt{1 - e^{-x^2}}$

By chain rule of differentiation, 

$ f'(x) = \dfrac{1}{2\sqrt{1 - e^{-x^2}} } \; \dfrac{d}{dx}(1 - e^{-x^2})$
           $  = \dfrac{1}{2\sqrt{1 - e^{-x^2}} } \; (-e^{-x^2}) (-2x)$
           $  = \dfrac{x \; e^{-x^2}}{2\sqrt{1 - e^{-x^2}} } $

f(x) is differentiable at a point x if the f'(x) exists at the point.

The denominator of f'(x) vanishes for 
$ 1 - e^{-x^2} = 0 $
i.e for
$ x = 0$

f(x) is not differentiable at x = 0.

Hence, the set of all points at which f(x) is differentiable:
$  (-\infty, 0) \; \cup \; (0, \infty) $

The answer: option C.

If $f(x) =x^{3} + e^{x/2}$ then $g"(1)$ is equal to, (where $f(x)$ and $g(x)$ are inverse functions of each other).

maths differencial calculus - differenciability and methods of differnciation differentiation by substitution methods of differentiation derivative of a function

  1. $2$

  2. $-2$

  3. $-\dfrac {1}{2}$

  4. $\dfrac {1}{2}$

Show answer
Correct Option: B
Explanation:
$g(f(x))=x$
$g'f(x)=1/f'(X)$
$g''f(x)=\dfrac{f''(x)}{{f'(x)}^3}$
$f(x)=1 at x=0$
$f(0)=1$
$f'(x)=3x^2+e^{(x/2)}/2$
$f'(0)=1/2$
$f''(x)=6x+e^{x/2}/4$
$f''(0)=1/4$
$g''(1)=\dfrac{-1/4}{1/8}=-2$

The cartesian equation of the plane which is at a distance of 10 unite from the original and perpendicular to the vector i + 2j -2k is 

maths vectors:planes in three dimensions cartesian equation of plane general form of the equation of a plane lines in space

  1. x+2y+2z = 30

  2. x - 2y - 2z =30

  3. x - 2y + 2z = 30

  4. x+2y-2z = 30

Show answer
Correct Option: D

The equation of the plane through the point $(0, -1, -6)$ and $(-2, 9, 3)$ are perpendicular to the plane $x-4y-2z=8$ is

maths vectors:planes in three dimensions cartesian equation of plane general form of the equation of a plane lines in space

  1. $3x+3y-2z=0$

  2. $x-2y+z=2$

  3. $2x+y-z=2$

  4. $5x-3y=2z=0$

Show answer
Correct Option: A

The normal form of $2x-2y+z=5$ is

maths vectors:planes in three dimensions cartesian equation of plane general form of the equation of a plane lines in space

  1. $12x-4y+3z=39$

  2. $\displaystyle \dfrac{-6}{7}x+\dfrac{2}{7}y+\dfrac{3}{7}z=1$

  3. $\displaystyle \dfrac{12}{13}x-\dfrac{-4}{13}y+\dfrac{3}{13}z=3$

  4. $\displaystyle \dfrac{2}{3}x-\dfrac{2}{3}y+\dfrac{1}{3}z=\dfrac{5}{3}$

Show answer
Correct Option: D
Explanation:

The dr's of the normal to the plane are $(2,-2,1)$.
The dc's will be $\left ( \dfrac{2}{3} , \dfrac{-2}{3} , \dfrac{1}{3} \right)$
Hence, the equation of the plane in the normal form will be,
$ \dfrac{2x}{3} - \dfrac{2y}{3} + \dfrac{z}{3} $ = $ \dfrac{5}{3} $

If a line is given by  $\dfrac{x-2}3 = \dfrac{y+10}5 = \dfrac{z+6}2$,  then which of the following points lies on this line?

maths vectors:planes in three dimensions cartesian equation of plane general form of the equation of a plane lines in space

  1. $(5,11,0)$

  2. $(3,10,0)$

  3. $(11,5,0)$

  4. $(0,5,11)$

Show answer
Correct Option: C
Explanation:
Putting $x=11$ and $y=5$ and $z=0$ in the given equation of line, we get
$\begin{array}{l} \dfrac { { x-2 } }{ 3 } =\dfrac { { y+10 } }{ 5 } =\dfrac { { z+6 } }{ 2 }  \\ \dfrac { { 11-2 } }{ 3 } =\dfrac { { 5+10 } }{ 5 } =\dfrac { { 0+6 } }{ 2 }  \\ \dfrac { 9 }{ 3 } =\dfrac { { 15 } }{ 3 } =\dfrac { 6 }{ 2 }  \\ 3=3=3 \end{array}$
therefore $(11,5,0)$ satisfies given equation of line 
hence  $(11,5,0)$ lies on the given line

Option $C$ is the correct answer.