Tag: derivatives - introduction and interpretation

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

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

  1. $2$
  2. $\sqrt{2}$
  3. $1$
  4. $-1$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $a=-1$.

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

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

  1. 1

  2. 3

  3. 4

  4. 5
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $5.$

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

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 $(a, b)$.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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

  1. $0$
  2. $1$
  3. $2$
  4. $3$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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

  1. $1+[g(x)]^{3}$
  2. $\dfrac{1}{1+[g(x)]^{3}}$
  3. $[g(x)]^{3}$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $g'(x)=\dfrac{1}{f'[g(x)]}$

$\therefore$  $g'(x)=1+[g(x)]^3$

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

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?

  1. $\{-2, 2, 1, 3\}$
  2. $\{-2, 0, 3\}$
  3. $\{-2, 2, 0\}$
  4. $\{1, 3\}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

 

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

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

  1. $\cfrac { x-y }{ x+y } $
  2. $\cfrac { x+y }{ x-y } $
  3. $\cfrac { y-x }{ y+x } $
  4. $\cfrac { x-y }{ 2x+y } $
Reveal answer Fill a bubble to check yourself
B Correct answer
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 } $