Calculus: Differentiability and Curve Properties - Class XI

Given,

$\displaystyle \frac{dy}{dx}-\text{cosec}2x y=\sqrt{\tan x};I.F=e^{\displaystyle \int-\text{cosec}2xdx}$

The statement is false.

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

Given equation $y^2=8x $

Given the function 

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

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.

Let $\mid x \mid = t$

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.

The function f(x) = |cos x| has sharp corners or cusps whenever cos x = 0, because the absolute value introduces a sudden change in slope. This occurs at odd multiples of pi/2, where x = (2n + 1)pi/2 for any integer n.

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

|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.

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

$\therefore f(\alpha) = f(\beta) = 0$

$\dfrac{dy}{dx}=\dfrac{1}{(\dfrac{1+x}{1-x})^{\frac{1}{4}}}\times$ $\dfrac{1}{4\times{(\dfrac{1+x}{1-x})^{\frac{3}{4}}}}\times $ $\dfrac{(1-x)(1)-(1+x)(-1)}{(1-x)^{2}}-$ $\dfrac{1}{2}\dfrac{1}{x^2+1}$ 

We have,

$f\left( x \right)=\sqrt{2{{x}^{2}}-1}$

And

$ y=f\left( {{x}^{2}} \right) $

$ y={{\left( \sqrt{2{{x}^{2}}-1} \right)}^{2}} $

$ y=2{{x}^{2}}-1 $

On differentiating and we get,

$ \dfrac{dy}{dx}=\dfrac{d}{dx}\left( 2{{x}^{2}}-1 \right) $

$ \dfrac{dy}{dx}=4x-0 $

$ \dfrac{dy}{dx}=4x $

At point $\left( x=1 \right)$

So,

$ \dfrac{dy}{dx}=4x=4\left( 1 \right) $

$ \dfrac{dy}{dx}=4 $

Hence, this is the answer.

By differentiating the given integral equation twice with respect to x using Leibniz rule, we can extract differential relations for h(x). Evaluating the resulting equations at x = 0 allows us to solve directly for h(0) and h'(0), yielding a sum of 2.

Equation of line

$ x-y=0\,\,......\,\,\left( 1 \right) $

$ x+y=0\,\,.......\,\,\left( 2 \right) $

${{x}^{2}}-{{y}^{2}}={{a}^{2}}\,\,......\,\,\left( 3 \right)$

Let the point $P(a\sec \theta, a\tan \theta)$ on the hyperbola.

Equation of tangent is,

$ x{{x} _{1}}-y{{y} _{1}}={{a}^{2}} $

$\Rightarrow a(x\sec \theta-y\tan\theta)=a^2$

$ \Rightarrow x\sec \theta -y\tan \theta =a\,\,......\,\,\left( 4 \right) $

Now,

Area of $\Delta AOB$ $=\dfrac{1}{2}$

$=\dfrac{1}{2}\left|a^2(\tan^2\theta-\sec^2\theta)-a^2(\sec^2\theta-\tan^2\theta)\right|$

$ =\dfrac{1}{2}\left| {{a}^{2}}\left( -1 \right)-{{a}^{2}}\left( 1 \right) \right| $

$ =\dfrac{1}{2}\left| -2{{a}^{2}} \right| $

$ =\left| -{{a}^{2}} \right| $

$ =\left| {{a}^{2}} \right| $

 $\text{sgn}(x) =\begin{cases}-1 & x <0\0 & x = 0\ 1 & x>0\end{cases} $
$g(x)=x(1-x^{2})$
$=x(1-x)(1+x)$
$g(x)=0$ implies, $x={-1,0,1}$.
$\text{sgn}(g(x))=\begin{cases}-1 & x\in(-\infty,-1)\cup(1,\infty) \0 & x =  \big{-1,0,1\big} \ 1 & x\in(-1,1)\end{cases} $

The function f(x) = (sqrt(x+1) - 1) / sqrt(x) is differentiable for all x > 0. At x=0, the limit of the derivative exists, but the domain of the function is typically defined for x > 0.

The derivative of given function is $4x-6 = 0$, which gives $x = \dfrac {3}{2}$
At $x$-axis , $y=0$ 

$f(x)=x ; x< 1$
$f(x)=3-x ; 1\leq x\leq 3$
If ${f}'(x) _{x\rightarrow 1^{-}}={f}'(x) _{x\rightarrow 1^{+}}={f}'(1)$
So, ${f}'(x) _{x\rightarrow 1^{-}}=1$
${f}'(x) _{x\rightarrow 1^{+}}=-1$
$\because {f}'(x) _{x\rightarrow 1^{-}}\neq {f}'(x) _{x\rightarrow 1^{+}}$
So, ${f}'(x)$ doesn't exist.

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

The given equation is:

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

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

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

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

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

$\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

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

Lets check each statement one by one

Given it is differentiable everywhere,
Therefore, 

Given :

LHD at x=0

$f'(x)=n(4+x)^{n-1},f''=n(n-1).(4+x)^{n-2}$
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 $f^{r}(0)=0 for r>n$
$\sum _{r=0}^{\infty}\frac{(f^{r}(0))}{r!}$=$5^{n}$

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

We know that, the order of the differential equation is the order of highest derivative.
So, equation $y'+y''+y=\sin { x } $ is the second order differential equation.

Given $f(x)=\left | x-1 \right |+\left | x+2 \right |+\left | x-3 \right |$