Tag: physics

Questions Related to physics

State whether the given statement is True or False.
A function is said to be differentiable in an interval $(a, b)$, if it is differentiable at every point of $(a, b)$.

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 function as $f(x)$.
A function $f(x)$ is differentiable on an interval if the derivative exists for each point in that interval.
Hence, a function is said to be differentiable in an interval $(a,b)$, if it is differentiable at every point of $(a,b)$.
Therefore, the given statement is true.

Let $f(x) = \left{\begin{matrix} 2 + 1,& x \leq 1\ x^{2} + 2, & 1 < x \leq 2\ 4x - 2, & x > 2\end{matrix}\right.$ then the number of points where $f(x)$ is non-differentiable, is equal to

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

  1. $0$

  2. $1$

  3. $2$

  4. $3$

Show answer
Correct Option: B
Explanation:
$f'(1^-)=0$ and 
$f'(1^+)=2x=2$
Hence $f(x)$ is not derivable at $x=1$

$f'(2^-)=2x=4$ and 
$f'(2^+)=4$
Hence $f(x)$ is derivable at $x=2$


So, $f(x)$ is non-differentiable at only $1$ point.

$f(x)=|\cos x|$ is not differentiable for the points given by $x=?$

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

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

  2. $(2n+1)\pi, \forall n\in I$

  3. $(2n+1)\dfrac{\pi}{2}\forall n\in I$

  4. $0$

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Correct Option: A

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

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

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

  2. $\dfrac{1}{1+[g(x)]^{3}}$

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

  4. $None\ of\ these$

Show answer
Correct Option: A
Explanation:
Here, $f'(x)=\dfrac{1}{1+x^3}$
$\Rightarrow$  $g$ is inverse of function $f$, then
$f[g(x)]=x$
Differentiating w.r.t $x,$
$f'[g(x)]\times g'(x)=1$

So $g'(x)=\dfrac{1}{f'[g(x)]}$

$\therefore$  $g'(x)=1+[g(x)]^3$

If $f(x)=(x^2-4)\left|x^3-6x^2+11x-6\right|+\dfrac{x}{1+|x|}$, then the set of points at which the function $f(x)$ is not differentiable is?

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

  1. ${-2, 2, 1, 3}$

  2. ${-2, 0, 3}$

  3. ${-2, 2, 0}$

  4. ${1, 3}$

Show answer
Correct Option: D
Explanation:


$\\f(x)=(x^2-4)|(x-1)(x^2-5x+6)|+(\frac{x}{1+|x|})\\=(x^2-4)|(x-1)(x-2)(x-3)|+(\frac{x}{1+|x|})$

|x| functions are not differentiable, at points where their value becomes zero because at these points it has sharp corners.

so at x=1, 2, 3 are possible points where it might not be differentiable but at x=2, it is multiplied by term $(x^2-4)$ which gives value zero and hence will neutralise the effect, so only at x={1, 3}, f(x) is not differentiable.

 

If $\sqrt { { x }^{ 2 }+{ y }^{ 2 } } ={ e }^{ t }$ where $t=\sin ^{ -1 }{ \left( \cfrac { y }{ \sqrt { { x }^{ 2 }+{ y }^{ 2 } }  }  \right)  } $ then $\cfrac { dy }{ dx } $ is equal to

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

  1. $\cfrac { x-y }{ x+y } $

  2. $\cfrac { x+y }{ x-y } $

  3. $\cfrac { y-x }{ y+x } $

  4. $\cfrac { x-y }{ 2x+y } $

Show answer
Correct Option: B
Explanation:
$t=\sin ^{ -1 }{ \left( \cfrac { y }{ \sqrt { { x }^{ 2 }+{ y }^{ 2 } }  }  \right)  } $, differentiate on both sides.
$\cfrac { dt }{ dx } =\cfrac { 1 }{ \sqrt { 1-{ \left( \cfrac { y }{ \sqrt { { x }^{ 2 }+{ y }^{ 2 } }  }  \right)  }^{ 2 } }  } \cfrac { d\left( \cfrac { y }{ \sqrt { { x }^{ 2 }+{ y }^{ 2 } }  }  \right)  }{ dx } \quad \left( \because \cfrac { d\left( \sin ^{ -1 }{ x }  \right)  }{ dx } =\cfrac { 1 }{ \sqrt { 1-{ x }^{ 2 } }  }  \right) $
$=\left( \cfrac { \sqrt { { x }^{ 2 }+{ y }^{ 2 } }  }{ x }  \right) \left[ \cfrac { \left( \sqrt { { x }^{ 2 }+{ y }^{ 2 } }  \right) \cfrac { dy }{ dx } -\left( \cfrac { 2x+xy\cfrac { dy }{ dx }  }{ x\sqrt { { x }^{ 2 }+{ y }^{ 2 } }  }  \right) y }{ \left( { x }^{ 2 }+{ y }^{ 2 } \right)  }  \right] $
$=\left( \cfrac { \sqrt { { x }^{ 2 }+{ y }^{ 2 } }  }{ x }  \right) \left[ \cfrac { \left( { x }^{ 2 }+{ y }^{ 2 } \right) \cfrac { dy }{ dx } -\left( xy+{ y }^{ 2 }\cfrac { dy }{ dx }  \right)  }{ \left( { x }^{ 2 }+{ y }^{ 2 } \right) \sqrt { { x }^{ 2 }+{ y }^{ 2 } }  }  \right] $
$\therefore \cfrac { dt }{ dx } =\cfrac { { x }^{ 2 }\cfrac { dy }{ dx } -xy }{ x\left( { x }^{ 2 }+{ y }^{ 2 } \right)  } =\cfrac { x\cfrac { dy }{ dx } -y }{ { x }^{ 2 }+{ y }^{ 2 } } $
Given that $\sqrt { { x }^{ 2 }+{ y }^{ 2 } } ={ e }^{ t }$
${ x }^{ 2 }+{ y }^{ 2 }={ e }^{ 2t }$ differentiate on both sides
$2xx+2y\cfrac { dy }{ dx } ={ e }^{ 2t }\left( 2 \right) \cfrac { dt }{ dx } $
$x+y\cfrac { dy }{ dx } =\left( \quad { x }^{ 2 }+{ y }^{ 2 } \right) \left( \cfrac { x\cfrac { dy }{ dx } -y }{ { x }^{ 2 }+{ y }^{ 2 } }  \right) $
$x+y=(x-y)\cfrac { dy }{ dx } $
$\therefore \cfrac { dy }{ dx } =\cfrac { x+y }{ x-y } $

Value of c is :-
$\dfrac{d}{dx}(c\ ^{f(x)}) = f' (x)e^{f(x)}$

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

  1. $e$

  2. $e^2$

  3. $1$

  4. $ln{x}$

Show answer
Correct Option: A
Explanation:

$\dfrac{d}{dx}(C^{f(x)}) = f(x) . e^{f(x)}$

$C^{f(x)} . ln C. f(x) = f(x) . e^{f(x)}$
$C^{f(x)} , ln (C) = e^{f(x)}$
$ln\ C = 1$
$C = e$

The condition that the line $\dfrac{x}{a}+\dfrac{y}{b}=1$ is tangent to the curve $x^{2/3}+y^{2/3}=1$ is

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

  1. $a^{2}+b^{2}=2$

  2. $a^{2}+b^{2}=1$

  3. $\dfrac{1}{a^{2}}+\dfrac{1}{b^{2}}=1$

  4. $a^{2}$

Show answer
Correct Option: A

If line $PQ$, whose equation is $y = 2x + k,$  is a normal to the parabola whose vertex is   $(-2,3)$ and the axis parallel to the $x$-axis with latus rectum equal to $2$, then the possible value of k is

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

  1. $\dfrac{{58}}{8}$

  2. $\dfrac{{50}}{8}$

  3. $1$

  4. $-1$

Show answer
Correct Option: C
Explanation:
Normal to parabola $\rightarrow y=2x+k$
parabola $\rightarrow (y-3)^{2}=4a(x+2)$
Latus rectum$=2$
$4a=2$
$a=\dfrac{1}{2}$
parabola $\rightarrow (y-3)^{2}=2(x+2)$
$y=3+\sqrt{2}\sqrt{x+2}$
slope of normal =$\dfrac{1}{-y'}$
$y'=\dfrac{\sqrt{2}}{2\sqrt{x+2}}=\dfrac{1}{\sqrt{2x+4}}$
$m=-\sqrt{-2x+4}$
if $2=-\sqrt{2x+4}$
then $x=0, y=5, 1$
$(0,5$) should also lies on $y=2x+k$
then
$k=5,1$
$C$ is correct

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 $\dfrac { d y } { d x }$ at $x = e ^ { m ^ { n p } }$ is equal to

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

  1. $e ^ { m n p }$

  2. $e ^ { m n / p }$

  3. $e ^ { n p / m }$

  4. $0$

Show answer
Correct Option: D
Explanation:
$\\y=(\dfrac{1}{1+(\dfrac{x^n}{x^m})+(\dfrac{x^p}{x^m})})+(\dfrac{1}{1+(\dfrac{x^m}{x^n})+(\dfrac{x^p}{x^n})})+(\dfrac{1}{1+(\dfrac{x^m}{x^p})+(\dfrac{x^n}{x^p})})$

$\\=(\dfrac{x^m}{x^m+x^n+x^p})+(\dfrac{x^n}{x^n+x^m+x^p})+(\dfrac{x^p}{x^p+x^m+x^n})$

$\\=(\dfrac{x^m+x^n+x^p}{x^m+x^n+x^p})$

$\\=1$
$\\\therefore\>(\dfrac{dy}{dx})=0$