Tag: physics

Questions Related to physics

The derivative of a differentiable even function is always an even function.

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

  1. True

  2. False

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Correct Option: B
Explanation:

The statement is false.

For example take the function,
$f(x)=\sin^2 x$, $x\in \mathrm{R}$.
This function is even as $f(-x)=f(x)\forall x\in\mathrm{R}$.
But the derivative of $f(x)$, $f'(x)=-2\sin x.\cos x$, $x\in \mathrm{R}$..
This function $f'(x)$ is not even.
As $f'(-x)=-f'(x)$, $\forall x\in \mathrm{R}$.

If $g$ is the inverse of $f$ and $\displaystyle f'\left( x \right) =\frac { 1 }{ 1+{ x }^{ 3 } } $, then $g'\left( x \right) $ is equal to

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

  1. $1+{ \left[ g\left( x \right) \right]  }^{ 3 }$

  2. $\displaystyle \frac { 1 }{ 1+{ \left[ g\left( x \right) \right]  }^{ 3 } } $

  3. ${ \left[ g\left( x \right)  \right]  }^{ 3 }$

  4. None of these

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Correct Option: A
Explanation:

We have, $g=$inverse of $f={ f }^{ -1 }$

$\Rightarrow g\left( x \right) ={ f }^{ -1 }\left( x \right) \Rightarrow f\left[ g\left( x \right)  \right] =x.$
Differentiating w.r.t. $x,$ we get
$f'\left[ g\left( x \right)  \right] .g'\left( x \right) =1$
$\displaystyle \therefore g'\left( x \right) =\frac { 1 }{ f'\left[ g\left( x \right)  \right]  } =1+{ \left[ g\left( x \right)  \right]  }^{ 3 }$
$\displaystyle \left[ \because f'\left( x \right) =\frac { 1 }{ 1+{ x }^{ 3 } } ,\therefore f'\left[ g\left( x \right)  \right] =\frac { 1 }{ 1+{ \left[ g\left( x \right)  \right]  }^{ 3 } }  \right] $

If $y=mx+c$ is the normal at a point $(8,8)$  on the parabola ${ y }^{ 2 }=8x$ Find $m$

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

  1. $-2 $

  2. $8 $

  3. $10 $

  4. $16 $

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Correct Option: A
Explanation:

Given equation $y^2=8x $

Slope of tangent is given as $2y\dfrac{dy}{dx}=8\\dfrac{dy}{dx}=\dfrac{4}{y}\\left.\dfrac{dy}{dx}\right| _{(8,8)}=\dfrac{4}{8}=\dfrac 1{2}$
Slope of normal is $\dfrac{-1}{\dfrac{1}{2}}=-2$

Function $ f(x)= \sin^{-1} (3x-4x^3) $ is-

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

  1. Always differentiable

  2. Not differentiable at 2 points

  3. Not continuous at 2 points

  4. Not differentiable at 3 points

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Correct Option: B
Explanation:

Given the function 

$f(x)=\sin^{-1}(3x-4x^3)=3\sin^{-1}x$.
We have $f'(x)$ does not exists at two points and they are $x=\pm 1$ as $f'(x)=\dfrac{1}{\sqrt{1-x^2}}$ for $x\in (-1,1)$.
So option (B) is correct.

If $\log \sqrt{x^2+y^2}=\tan^{-1}\left(\dfrac{y}{x}\right)$ , then $\dfrac{dy}{dx}$ is:

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

  1. $1$

  2. $2$

  3. $\dfrac{2x}{\sqrt{x^2+y^2}}$

  4. $\dfrac{x+y}{x-y}$

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Correct Option: D
Explanation:
Given, 
$\log\sqrt{x^2+y^2}=\tan ^{-1}\left(\dfrac{y}{x}\right)$
Now, differentiating both sides w.r.to $x$ we get,
or, $\dfrac{1}{2}\dfrac{2x+2y\dfrac{dy}{dx}}{x^2+y^2}=\dfrac{x^2}{x^2+y^2}.\left(-\dfrac{y}{x^2}+\dfrac{\dfrac{dy}{dx}}{x}\right)$
or, $x+y\dfrac{dy}{dx}=-y+x\dfrac{dy}{dx}$
or, $(x-y)\dfrac{dy}{dx}=(x+y)$
or, $\dfrac{dy}{dx}=\dfrac{x+y}{x-y}$

State whether the given statement is True or False.
The derivative of a function is defined at $a$, provided  $lim _{h\rightarrow 0} \displaystyle \frac{f(a+h)-f(a)}{h}$ exists.

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

  1. True

  2. False

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Correct Option: A
Explanation:

Consider the given function as $f(x)$
Then, its derivative at a point $a$ is $f'(x)=lim _{h\rightarrow 0} \displaystyle \frac{f(a+h)-f(a)}{h}$ wherever the limit exists.
Hence, the derivative of a function is defined provided the limit exists.
Therefore, the given statement is true.

Average Speed and Instantaneous speed are the same things.

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

  1. True

  2. False

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Correct Option: B
Explanation:

False.
An average speed tells how much distance a body covers during a certain time span, but it does not tell much about the actual motion that occurred. Instantaneous speed is the exact speed that a body is moving at, at a given instant in time. It is a true measure of the body's motion for that point in time.

If $y = \left| {\cos x} \right| + \left| {\sin x} \right|$ , then ${\dfrac{dy} {dx}}$ at $x = {\dfrac {2\pi } 3}$ is

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

  1. ${1 \over 2}\left( {\sqrt 3 + 1} \right)$

  2. $2\left( {\sqrt 3 - 1} \right)$

  3. ${1 \over 2}\left( {\sqrt 3 - 1} \right)$

  4. none of these

Show answer
Correct Option: A
Explanation:

$y=|\cos x|+|\sin x|$


at  $x=\dfrac{2\pi}{3},$

$y=-\cos x+\sin x$

$now, \dfrac{d{y}}{d{x}}=\sin x+\cos x$

$\dfrac{d{y}}{d{x}}=\sin\dfrac{2\pi}{3} +\cos \dfrac{2\pi}{3}$

$ \dfrac{d{y}}{d{x}}=\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}$

$ \dfrac{d{y}}{d{x}}=\dfrac{1}{2}(\sqrt{3}+1)$

The line $y =\sqrt{2}x + 4\sqrt{2}$ is a normal to $y^{2} =4ax$ then a = 

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

  1. $2$

  2. $\sqrt{2}$

  3. $1$

  4. $-1$

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Correct Option: D
Explanation:

The general form of equation of normal to the parabola $y^2=4ax$ is $y=mx-2am-am^3$.....(1) where $m$ is the slope of the normal.

According to the problem, $y =\sqrt{2}x + 4\sqrt{2}$.....(2) is the normal to the parabola.
Equation (1) and (2) are identical.
Then $m=\sqrt{2}$ and $-2am-am^3=4\sqrt{2}$ or, $a(-2\sqrt{2}-2\sqrt{2})=4\sqrt{2}$ or, $a=-1$ [ Using the value of $m$].
So $a=-1$.

The  number of points where $f(x) = \mid | x |^2 - 5| x | + 6 \mid $ is non-derivable is/are

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

  1. 1

  2. 3

  3. 4

  4. 5

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Correct Option: D
Explanation:

Let $\mid x \mid = t$

$f(x) = \mid t^2 -5t +6 \mid = \mid t(t-3) -2(t - 3)\mid = \mid (t - 3)(t - 2)\mid$
$\Rightarrow f(x) = \mid (\mid x\mid - 3)(\mid x\mid -2)\mid$
Non-derivable at all $x$ where $f(x) = 0$
$\pm 3, 0, \pm 2$
Hence $5.$