Tag: proving properties of curves

Questions Related to proving properties of curves

If $y = \dfrac {1}{1 + x^{n - m} + x^{p - m}} + \dfrac {1}{1 + x^{m - n} + x^{p - n}} + \dfrac {1}{1 + x^{m - p} +x^{n - p}}$ then $\dfrac {dy}{dx}$ at $e^{m^{n^{p}}}$ is equal to

physics parametric equations proving properties of curves derivatives - introduction and interpretation introduction to calculus - differentiation

  1. $e^{mnp}$

  2. $e^{mn/p}$

  3. $e^{np/m}$

  4. $0$

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Correct Option: D
Explanation:

$\quad \quad y=\cfrac { 1 }{ 1+{ x }^{ n-m }+{ x }^{ p-m } } +\cfrac { 1 }{ 1+{ x }^{ m-n }+{ x }^{ p-n } } +\cfrac { 1 }{ 1+{ x }^{ m-p }+{ x }^{ n-p } } \\ \Rightarrow y=\cfrac { 1 }{ 1+\cfrac { { x }^{ n } }{ { x }^{ m } } +\cfrac { { x }^{ p } }{ { x }^{ m } }  } +\cfrac { 1 }{ 1+\cfrac { { x }^{ m } }{ { x }^{ n } } +\cfrac { { x }^{ p } }{ { x }^{ n } }  } +\cfrac { 1 }{ 1+\cfrac { { x }^{ m } }{ { x }^{ p } } +\cfrac { { x }^{ n } }{ { x }^{ p } }  } $

$\Rightarrow y=\cfrac { { x }^{ m } }{ { x }^{ m }+{ x }^{ n }+{ x }^{ p } } +\cfrac { { x }^{ n } }{ { x }^{ m }+{ x }^{ n }+{ x }^{ p } } +\cfrac { { x }^{ p } }{ { x }^{ m }+{ x }^{ n }+{ x }^{ p } } $

$\quad \quad \quad =\cfrac { 1 }{ \left( { x }^{ m }+{ x }^{ n }+{ x }^{ p } \right)  } \left( { x }^{ m }+{ x }^{ n }+{ x }^{ p } \right) $

$\Rightarrow y=1\\ \Rightarrow \cfrac { dy }{ dx } =0\quad $ at all $x$

$\therefore \dfrac { dy }{ dx } $ at ${ e }^{  m ^ { n ^{ p } } }=0$

D answer

 

If the distance between a tangent to the parabola $y^{2} = 4x$ and a parallel normal to the same parabola is $2\sqrt{2}$, then possible values of gradient of either of them are:

physics parametric equations proving properties of curves derivatives - introduction and interpretation introduction to calculus - differentiation

  1. $-1$

  2. $+1$

  3. $-\sqrt{\sqrt{5} - 2}$

  4. $+\sqrt{\sqrt{5} - 2}$

Show answer
Correct Option: A,B
Explanation:

$y=\frac { x }{ t } +at\ y=-xs+2as+a{ s }^{ 3 }\ s=-\frac { 1 }{ t } $

For parallel condition.
$ts=-1$

$\left| \dfrac { \left( 2as+a{ s }^{ 3 } \right) -\left( at \right)  }{ \sqrt { 1+{ s }^{ 2 } }  }  \right| =2\sqrt { 2 } \ \ \left| \dfrac { \left( 2as+a{ s }^{ 3 } \right) +\frac { a }{ s }  }{ \sqrt { 1+{ s }^{ 2 } }  }  \right| =\left| \dfrac { a\left( { s }^{ 4 }+2{ s }^{ 2 }+1 \right)  }{ s\sqrt { 1+{ s }^{ 2 } }  }  \right| =\left| \dfrac { a{ \left( { s }^{ 2 }+1 \right)  }^{ 2 } }{ s\sqrt { 1+{ s }^{ 2 } }  }  \right| =\left| \dfrac { a{ \left( { s }^{ 2 }+1 \right)  }^{ \frac { 3 }{ 2 }  } }{ s }  \right| $
As $a=1$
$\left| \dfrac { { \left( { s }^{ 2 }+1 \right)  }^{ \frac { 3 }{ 2 }  } }{ s }  \right| =2\sqrt { 2 } ={ 2 }^{ \frac { 3 }{ 2 }  }$
Putting $s$ as $\pm 1$ the above equation satisfies.
Hence, the answer is $1,-1$.

Consider the function $f(x)=\begin{cases} x^2 \sin \dfrac{1}{x};x\neq 0 \ 0 ; otherwise  \end{cases}$
then,

physics parametric equations proving properties of curves derivatives - introduction and interpretation introduction to calculus - differentiation

  1. $f$ is derivable at $x=0$

  2. $f $ is not derivable at $x=0$

  3. $f$ is derivable at $x=0$ and $f'(0)=0$

  4. $f$ is derivable at $x=0$ and $f'(0)\neq0$

Show answer
Correct Option: A
Explanation:
Consider the function 
$f(x)=\begin{cases} x^{2}\sin\dfrac{1}{x}\ \ \ x\neq 0 \\ 0\ \ \ \ otherwise \end{cases}$
To check derivability at $x=0$
Concept : A function is derivable at $x=a$ if
$LHD \, at(x=a)=RHD\, at\, (x=a)$
$RHD \, at\, x=$
$=\underset{h \rightarrow 0}{\lim}\dfrac{f(0+h)-f(0)}{h}$
$=\underset{h \rightarrow 0}{\lim}\dfrac{f(h)-f(0)}{h}$
$=\underset{h \rightarrow 0}{\lim}\dfrac{h^2\sin \left(\dfrac{1}{h}\right)-0}{h}$
$=\underset{h \rightarrow 0}{\lim} h \sin \left(\dfrac{1}{h}\right)=0\times (value \, between\, -1\&1)$
$=0$
Now 
LHD at $x=0$.
$=\underset{h \rightarrow 0}{\lim}\dfrac{f(0-h)-f(0)}{-h}$
$=\underset{h \rightarrow 0}{\lim}\dfrac{f(-h)-f(0)}{-h}$
$\underset{h \rightarrow 0}{\lim} \dfrac{(-h)^2\sin \left(-\dfrac{1}{h}\right)-0}{-h}$
$\underset{h \rightarrow 0}{\lim} -h \sin \left(-\dfrac{1}{h}\right)=\underset{h \rightarrow 0}{\lim} h \sin \dfrac{1}{h}$
$=0\times [-1,1]$
$=0$
Here $\because LHD =RHD =0$ at $x=0$
Hence $f$ is derivable at $x=0$
Important Concept : Left Hand derivative $ (LHD)=\underset{h \rightarrow 0}{\lim} \dfrac{f(a-h)-f(a)}{-h}$
$RHD$ at $(x=a)=\underset{h \rightarrow 0}{\lim}\dfrac{f(a+h)-f(a)}{h}$

Consider the following statements:
$1.$ Derivative of $f(x)$ may not exist at some point.
$2.$ Derivative of $f(x)$ may exist finitely at some point.
$3.$ Derivative of $f(x)$ may be infinite (geometrically) at some point.
Which of the above statements are correct?

physics parametric equations proving properties of curves derivatives - introduction and interpretation introduction to calculus - differentiation

  1. $1$ and $2$ only

  2. $2$ and $3$ only

  3. $1$ and $3$ only

  4. $1, 2$ and $3$

Show answer
Correct Option: D
Explanation:

Lets check each statement one by one

1. This statement is true. Take example $y=|x|$ derivative does not exist at $x=0$
2. This statement is true. Derivative may  be finite may be not. Take example of $y=\tan x$
3. Derivative of a function can be infinite(undefined) 
Hence, all three statements are true.

$f(x)= \left\{\begin{matrix}x^2+3x+a & \text{for}\, x \leq 1 \\ bx+2 & \text{for}\, x > 1 \end{matrix}\right.$ 
is everywhere differentiable. Then value of constant $b$ is
physics parametric equations proving properties of curves derivatives - introduction and interpretation introduction to calculus - differentiation

  1. $5$

  2. $3$

  3. $\dfrac{1}{5}$

  4. None of these

Show answer
Correct Option: A
Explanation:

Given it is differentiable everywhere,
Therefore, 

$f'(x)=\begin{cases} 2x+3 &, x \leq 1 \ b &, x>1 \end{cases}$

At $ x=1, f'(1)=2(1)+3=5$

When $x>1$

As $f$ is differentiable at every point, both left-hand and right-hand derivatives should be equal.

$\therefore f'(1)=b$

$\Rightarrow 5=b$

Hence, $b=5$.

A function is defined in $(0, \infty)$ by
$f(x) = \left{\begin{matrix}1 - x^{2} & for & 0 < x  \leq 1\ \ln\ x & for & 1 < x \leq 2\ \ln\ 2 - 1 + 0.5x & for & 2 < x < \infty\end{matrix}\right.$
Which one of the following is correct in respect of the derivative of the function, i.e., $f'(x)$?

physics parametric equations proving properties of curves derivatives - introduction and interpretation introduction to calculus - differentiation

  1. $f'(x) = 2x$  for  $0 < x \leq 1$

  2. $f'(x) = -2x$  for  $0 < x \leq 1$

  3. $f'(x) = -2x$  for  $0 < x < 1$

  4. $f'(x) = 0$  for  $0 < x < \infty$

Show answer
Correct Option: B
Explanation:
$f(x)$ is continuous at $x=1$ and at $x=2$.
Differentiating w.r.t $x$
$f'(x)=\left\lbrace\begin{matrix}-2x & \text{for} & 0<x\leq 1 \\ \dfrac{1}{x} & \text{for} & 1<x\leq2 \\ 0.5 & \text{for} & 2<x<\infty\end{matrix}\right\rbrace$

$f(x)= \left\{\begin{matrix}x^2+3x+a & \text{for}\, x \leq 1 \\ bx+2 & \text{for}\, x > 1 \end{matrix}\right.$
is everywhere differentiable. Then value of constant $a$ is
physics parametric equations proving properties of curves derivatives - introduction and interpretation introduction to calculus - differentiation

  1. $3$

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

  3. $5$

  4. None of these

Show answer
Correct Option: A
Explanation:

Given :


$f\left( x \right) =\begin{cases} { x }^{ 2 }+3x+a\quad \quad for\quad x\le 1\quad  \ bx+2\quad \quad \quad \quad \quad for\quad x>1 \end{cases}$

$f\left( x \right) =\begin{cases} { x }^{ 2 }+3x+a\quad \quad ;\quad x\le 1\quad  \ bx+2\quad \quad \quad \quad \quad ;\quad x>1 \end{cases}$

Since $\ { x }^{ 2 }+3x+a$ quadratic and $bx+2$ is linear equation.Therefore the equations are continous.
Hence,we will check differetiability of this equation as $x=1$

$f\left( { 1 }^{ - } \right) =f\left( { 1 }^{ + } \right) $ [since the equation is continous]
$1+3+a=b+2$
$a+4=b+2$
$\Rightarrow a-b=-2\rightarrow (1)$

$f^{ 1 }\left( x \right) =\begin{cases} 2x+3\quad \quad for\quad x\le 1\quad  \ b\quad \quad \quad \quad \quad for\quad x>1 \end{cases}$

$f^{ 1 }\left( { 1 }^{ - } \right) =f^{ 1 }\left( { 1 }^{ + } \right) $ [since $f(x)$ is differentiable at each point of $x$]
$2+3=b$
$b=5$
Substitute b=5 in equation (1)
$a-b=-2$
$a-5=-2$
$a=3$
Hence the correct answer is $3$.


 lf $\mathrm{f}(\mathrm{x})=\left{\begin{array}{l}1, \mathrm{x}<0\1+ \mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}, 0\leq \mathrm{x}</\pi _{2} \end{array}\right.$, then derivative of f(x) at$\mathrm{x}=0$

physics parametric equations proving properties of curves derivatives - introduction and interpretation introduction to calculus - differentiation

  1. is equal to 1

  2. is equal to 0

  3. is equal to -1

  4. does not exist

Show answer
Correct Option: D
Explanation:

LHD at x=0

=${ \left[ \frac { d }{ dx } 1 \right]  } _{ x=0 }=0$
RHD at x=0
=${ \left[ \frac { d }{ dx } 1+\sin { x }  \right]  } _{ x=0 }=cos0=1$
hence $f'$ doesn't exist

If $f(x)=(4+x)^{n}$,$n \epsilon N$ and $f^{r}(0)$ represents the $r^{th}$ derivative of f(x) at x = 0, then the value of $\sum _{r=0}^{\infty}\frac{(f^{r}(0))}{r!}$ is equal to

physics parametric equations proving properties of curves derivatives - introduction and interpretation introduction to calculus - differentiation

  1. $2^{n}$

  2. $e^{n}$

  3. $5^{n}$

  4. $4^{n}$

Show answer
Correct Option: C
Explanation:

$f'(x)=n(4+x)^{n-1},f''=n(n-1).(4+x)^{n-2}$
If we keep on differentiating, we get
$f^{r}=n(n-1)....(n-r+1).(4+x)^{n-r},r\leq 4$
$\Rightarrow f^{r}(0)=\frac{n!}{(n-r)!}.4^{n-r},r\leq n$
and $f^{r}(0)=0 for r>n$
$\sum _{r=0}^{\infty}\frac{(f^{r}(0))}{r!}$=$5^{n}$

Let $f(x)=\begin{cases}\begin{matrix} 1 & \forall &  x<0 \ 1+\sin x & \forall & 0\leq x\leq \dfrac{\pi}2\end{matrix}\end{cases}$ then what is the value of $f'(x)$ at $x=0?$

physics parametric equations proving properties of curves derivatives - introduction and interpretation introduction to calculus - differentiation

  1. $1$

  2. $-1$

  3. $\infty$

  4. Does not exist

Show answer
Correct Option: D
Explanation:

For $x<0$ , $f(x)=1$

$\Rightarrow f^{ 1 }\left( 0^{-}\right) =0$
For $x\ge 0$ , $f(x)=1+sinx$
$\Rightarrow f^{ 1 }\left( 0^{+} \right) = cos(0)=1 $
Therefore $f^{ 1 }\left( 0^{-} \right) \neq f^{ 1 }\left( 0^{+} \right)$  
So $f^{ 1 }\left( x \right) $ at $x=0$ does not exist
Therefore the correct option is $D$