Tag: physics

Questions Related to physics

The domain of the derivative of the function
$\displaystyle f\left ( x \right )=\begin{cases}
\tan^{-1}x & \text{ if } \left | x \right |\leq 1 \
\frac{1}{2}\left ( \left | x \right |-1 \right ) & \text{ if } \left | x \right |> 1
\end{cases}$

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

  1. $R\sim \left { 0 \right }$

  2. $R\sim \left { 1 \right }$

  3. $R\sim \left { -1 \right }$

  4. $R\sim \left { -1, 1 \right }$

Show answer
Correct Option: D
Explanation:

We have
$f\left ( x \right )=\begin{cases}
\left ( 1/2 \right )\left ( -x-1 \right ) & \text{ if } x< -1 \
\tan^{-1}x & \text{ if } -1\leq x\leq 1 \
\left ( 1/2 \right )\left ( x-1 \right ) & \text{ if } x> 1
\end{cases}$
Since $\displaystyle f\left ( -1 \right )=-\frac{\pi }{4}$, $\displaystyle f\left ( 1 \right )=\frac{\pi }{4}$ and $\lim _{x\rightarrow 1-}f\left ( x \right )=-1$, $\lim _{x\rightarrow 1+}f\left ( x \right )=0$
so f is not continuous at -1, 1, hence not differentiable at -1, 1. Also
$\displaystyle {f}'\left ( x \right )=\begin{cases}
-1/2 & \text{ if } x< -1 \
\frac{1}{1+x^{2}} & \text{ if } -1< x< 1 \
1/2 & \text{ if } x> 1
\end{cases}$
Thus the domain of ${f}'$ is $R - \left { -1, 1 \right }$.

Let $f(x)=ax^2+bx+c$ such that $f(1)=f(-1)$ and a, b, c are in Arithmetic Progression.

Then find what kind of progression $f'(a), f'(b), f'(c)$ form.

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

  1. A.P.

  2. G.P.

  3. H.P.

  4. Arithmetico-geometric progression

Show answer
Correct Option: A
Explanation:

The given equation is:


$ y= f(x) = ax^2 + bx + c$

Differentiating w.r.t to x we get,

$\Rightarrow y' = 2ax + b$

Since a, b, c are in A.P. we have, $2b = a + c$

$\Rightarrow b -a = c -b$

Finding the respective derivative values we get,

$y'(a) = 2a^2 + b$

$y'(b) = 2ab + b$

$y'(c) = 2ac + b$

Finding the common difference between the terms we get,

$y'(b) - y'(a) = (2ab + b) - (2a^2 + b)$

$\Rightarrow y'(b) - y'(a) = 2ab - 2a^2$

$\Rightarrow y'(b) - y'(a) = 2a(b -a)$

$\Rightarrow y'(b) - y'(a) = 2a(c -b)$

$y'(c) - y'(b) = (2ac + b) - (2ab + b)$

$\Rightarrow y'(c) - y'(b) = 2ac -2ab$

$\Rightarrow y'(c) - y'(b) = 2a(c-b)$


We got the same common difference between the three terms hence the derivatives are in A.P.         .....Answer

If $y=\displaystyle\dfrac{1}{a-z}$, then $\displaystyle\dfrac{dz}{dy}$ is:

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

  1. $(a-z)^2$

  2. $-(z-a)^2$

  3. $(z+a)^2$

  4. $-(z+a)^2$

Show answer
Correct Option: A
Explanation:
$y=\dfrac{1}{a-z}$
$\dfrac{{d} y}{{d} z}=\dfrac{-1}{(a-z)^{2}}(-1)$   (By differentiating w.r.t z)
$\dfrac{{d} y}{{d} z}=\dfrac{1}{(a-z)^{2}}$
   $\therefore \dfrac{{d} z}{{d} y}=(a-z)^{2}$

Which of the following given statements is/are correct?

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

  1. If L.H.D $\neq $ R.H.D, then $f(x)$ is not differentiable at $x= c$

  2. If a function is differentiable at a point, it is necessarily continuous at the point.

  3. If a function is differentiable at each $x \in R$ then it is said to be every where differentiable.

  4. $\dfrac{d}{dx}(c f (x)) =c \dfrac{d}{dx} (f (x))$, where $c$ is a constant.

Show answer
Correct Option: A,B,C,D
Explanation:

Using concept 1 we can say that Option A is correct.

It is mandatory that  at  a point where a function is differentiable, Left hand limit = Right Hand Limit. Thus, it is continuous at that point. therefore, Option B  is correct.
Option C is correct.
Using concept 2, we can  say  that option  D is  correct.

The value of $\displaystyle \frac{d}{dx} (|x-1|+ |x-5|) $ at x = 3 is

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

  1. -2

  2. 0

  3. 2

  4. 4

Show answer
Correct Option: B
Explanation:

$\dfrac{d}{dx} |x| = \dfrac{|x|}{x}$


Applying the above formula ,

$\dfrac{d}{dx} ( |x-1|+|x-5|) = \dfrac{d}{dx}|x-1|+\dfrac{d}{dx}|x-5|= \dfrac{|x-1|}{x-1}+\dfrac{|x-5|}{x-5}$

Substituting $x=3$,

$\dfrac{|3-1|}{3-1}+\dfrac{|3-5|}{3-5} = \dfrac{|2|}{2} +\dfrac{|-2|}{-2} = \dfrac{2}{2} +\dfrac{2}{-2} =1-1 =0$

Let $f : R \rightarrow R$ be a function defined by $f(x)= \max\left { x,  x^3 \right }$. The set of all points where $f(x)$ is NOT differentiable is:

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

  1. ${-1, 1}$

  2. ${-1, 0}$

  3. ${0, 1}$

  4. ${-1, 0, 1}$

Show answer
Correct Option: D
Explanation:

$ f(x)=m\left{ x, { x }^{ 3 } \right} $

$ =x;x<-1$ and
$ ={ x }^{ 3 };-1\le x\le 0$
$ \Rightarrow  f(x)=x;0\le x\le 1$ and
$={ x }^{ 3 };x\ge 1$
$ \therefore  f(x)=1;x<-1$
$ \therefore  f'(x)=3{ x }^{ 2 };-1\le x\le 0$ and $ =1$
$0<x<1$
Hence, option D is correct.

If $f(x) = \left{\begin{matrix}e^x+ax & x< 0 \ b(x-1)^2 & x \geq 0 \end{matrix}\right.$ is differentiable at $x= 0$, then $(a, b)$ is

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

  1. $(-3, -1)$

  2. $(-3, 1)$

  3. $(3, 1)$

  4. $(3, -1)$

Show answer
Correct Option: B
Explanation:

Given that $f(x)$ is differentiable at $x=0$ , which implies $f(x)$ is continuous at $x=0$

$\Rightarrow b=1$
By using differentiable condition , we get ${e}^{0}+a=2b(0-1)$
$\Rightarrow 1+a=-2b=-2$
$\Rightarrow a=-3$
Therefore the correct option is $B$

If $f(x) =x[x \sqrt{x}-\sqrt{x+1}]$, then:

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

  1. $f(x)$ is continuous but not differentiable at $x= 0$

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

  3. $f(x)$ is differentiable at $x= 0$

  4. none of these

Show answer
Correct Option: C
Explanation:

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

$f^{ 1 }\left( x \right) =2x\sqrt { x } +\dfrac{x^2}{2\sqrt x} -\sqrt { x+1 } -\dfrac { x }{ 2\sqrt { x+1 }  } =\frac { 3 }{ 2 } x\sqrt { x } -\sqrt { x+1 } -\dfrac { x }{ 2\sqrt { x+1 }  } $
We get $f^{ 1 }\left( 0^{+} \right)=f^{ 1 }\left( 0^{-} \right) =-1$  
Therefore $f(x)$ is differentiable at $x=0$

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. $0$

Show answer
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 $1,-1$.