Continuity And Differentiability - Class 12 Engineering Maths - Extra Questions

If,  is continuous at $$x = 2$$, find the value of $$k$$.
$$f(x) = \begin{cases}  \dfrac{x^3 + x^2 - 16x + 20}{(x-2)^2},& x \neq 2\\ k,&x = 2 \end{cases}$$

Since the function $$f(x)$$ is continuous $$x=2$$ then $$\lim\limits_{x\to 2}f(x)=f(2)$$......(1).
Now,
$$\lim\limits_{x\to 2}f(x)$$
$$=\lim\limits_{x\to 2}\dfrac{x^3+x^2-16x+20}{(x-2)^2}$$ $$\dfrac{0}{0}$$ form
Using L'Hospital's rule we get,
$$=\lim\limits_{x\to 2}\dfrac{3x^2+2x-16}{2(x-2)}$$$$\dfrac{0}{0}$$ form
Again using L'Hospital's rule we get,
$$=\lim\limits_{x\to 2}\dfrac{6x+2}{2}=7$$.
From (1) we get, $$k=7$$.


Check the continuity of the function $$f ( x ) = \left\{ \begin{array} { l l } { \frac { | x | } { x } , } & { x \neq 0 } \\ { 0 , } & { x = 0 } \end{array} \right. \text { at } x = 0$$

$$f ( x ) = \left\{ \begin{array} { l l } { \frac { | x | } { x } , } & { x \neq 0 } \\ { 0 , } & { x = 0 } \end{array} \right. \text { at } x = 0$$
$$\displaystyle\lim_{x\to 0^{+}} f(x)=\lim_{x\to 0^+}\dfrac{|x|}{x}=\lim_{x\to 0^{+}} \dfrac{x}{x}=1$$
$$\displaystyle\lim_{x\to 0^{-}} f(x)=\lim_{x\to 0^-}\dfrac{|x|}{x}=\lim_{x\to 0^{-}} \dfrac{-x}{x}=-1$$
Here $$\displaystyle\lim_{x\to 0^+} f(x)\ne \lim_{x\to 0^-} f(x)\ne f(0)$$
So $$f(x)$$ is not continuous at $$x=0$$


Discuss the continuity of the function $$f$$, where $$f$$ is defined by 
$$f(x) =$$ $$\begin{cases} 3,\ if\ 0 \le x \le 1 \\ 4,\ if\ 1< x < 3 \\ 5,\ if\ 3 \le x \le 10 \end{cases}$$ 

The given function is f(x) = $$\begin{cases} 3,\ if\ 0 \le x \le 1 \\ 4,\ if\ 1< x < 3 \\ 5,\ if\ 3 \le x \le 10 \end{cases}$$ 
The given function is defined at all points of the interval $$[0, 10]$$.
Let c be a point in the interval $$[0, 10]$$. 

Case I :
If 0 $$\le$$ c < 1, then f(c) = 3 and $$\underset{x \rightarrow c}{\lim}$$ f(x) = $$\underset{x \rightarrow c}{\lim}$$ (3) = 3
$$\therefore$$ $$\underset{x \rightarrow c}{\lim}$$ f(x) = f(c)
Therefore, f is continuous in the interval [0, 1].

Case II :
If $$c = 1,$$ then $$f(3) = 3$$
The left hand limit of f at $$x =- 1$$ is 
$$\underset{x \rightarrow 1}{\lim}$$ f(x) = $$\underset{x \rightarrow 1}{\lim}$$ (3) = 3
The right hand limit of f at $$x = 1$$ is, 
$$\underset{x \rightarrow 1}{\lim}$$ f(x) = $$\underset{x \rightarrow 1}{\lim}$$ (4) = 4
It is observed that the left and right hand limit of f at x = 1 do not coincide.
Therefore, f is not continuous at x = 1

Case III :
If 1 < c < 3, then f(c) = 4 and $$\underset{x \rightarrow c}{\lim}$$ f(x) = $$\underset{x \rightarrow c}{\lim}$$ (4) = 4
$$\therefore$$ $$\underset{x \rightarrow c}{\lim}$$ $$f(x) = f(c)$$
Therefore, f is continuous at all points of the interval $$(1, 3).$$

Case IV :
If $$c = 3$$, then $$f(c) = 5$$
The left hand limit of f at x = 3 is,
$$\underset{x \rightarrow 3}{\lim}$$ f(x) = $$\underset{x \rightarrow 3}{\lim}$$ (4) = 4
The right hand limit of f at x = 3 is, 
$$\underset{x \rightarrow 3}{\lim}$$ f(x) = $$\underset{x \rightarrow 3}{\lim}$$( (5) = 5
It is observed that left and right hand limit of fat x = 3 do not coincide.
Therefore, f is not continuous at x = 3.

Case V:
If 3 < c $$\le$$ 10, then $$f(c) = 5$$ and $$\underset{x \rightarrow c}{lim}$$ f(x) = $$\underset{x \rightarrow c}{\lim}$$ (5) = 5
$$\underset{x \rightarrow c}{\lim}$$ f(x) = f(c)
Therefore, f is continuous at all points of the interval $$(3, 10]$$.
Hence, f is not continuous at x = 1 and x = 3


Find the values of $$a$$ and $$b$$ so that the function, $$f(x)=\begin{cases} \cfrac { 1-\sin ^{ 2 }{ x }  }{ 3\cos ^{ 2 }{ x }  } ,\quad x<\pi /2 \\ a,\quad \quad x=\pi /2 \\ \cfrac { b\left( 1-\sin { x }  \right)  }{ { \left( \pi -2x \right)  }^{ 2 } } ,\quad \quad x>\pi /2 \end{cases}$$ is continuous.

For f(x) to be continuous $$\displaystyle \lim_{x\to {\pi/2}^{-}}f(x)=f(\pi/2)=\lim_{x\to {\pi/2}^{+}}f(x)$$
$$\displaystyle \lim_{x\to {\pi/2}^{-}}f(x)=\lim_{x\to {\pi/2}^{-}}\dfrac{1-\sin^2 x}{3\cos^2 x}$$
$$\displaystyle \lim_{x\to {\pi/2}^{-}}\dfrac{\cos^2 x}{3\cos^2 x}=\dfrac{1}{3}$$
$$\therefore a=\dfrac{1}{3}$$
$$\displaystyle \lim_{x\to {\pi/2}^{+}}f(x)=\lim_{x\to {\pi/2}^{+}}\dfrac{b(1-\sin x)}{(\pi-2x)^2}$$
Multiply and divide by $$(1+\sin x )$$, then put $$(1-\sin x)(1+\sin x)=1 - \sin ^2 x=\cos ^2 x$$
$$\Rightarrow \displaystyle \lim_{x\to {\pi/2}^{+}}\dfrac{b(\cos^2 x)}{(\pi-2x)^2(1+\sin x)}$$
$$\displaystyle =\lim_{x\to {\pi/2}^{+}}\dfrac{b(\sin^2 (\pi/2-x))}{(\pi-2x)^2(1+\sin x)}$$
$$\displaystyle =\lim_{x\to {\pi/2}^{+}}\dfrac{b(\sin^2 (\pi/2-x))}{4(\pi/2-x)^2(1+\sin x)}$$
Putting limit for $$(1+\sin x)$$ because it's value is not affecting the limit,
$$=\displaystyle \lim_{x\to {\pi/2}^{+}}\dfrac{b(\sin^2 (\pi/2-x))}{4(\pi/2-x)^2(2)}$$
$$=\displaystyle \dfrac{1}{8}\lim_{x\to {\pi/2}^{+}}\dfrac{b(\sin^2 (\pi/2-x))}{(\pi/2-x)^2}$$
Put $$h=\pi/2-x$$, we get
$$=\displaystyle \dfrac{1}{8}\lim_{h\to {0}^{+}}\dfrac{b(\sin^2 (h))}{(h)^2}$$
$$=\dfrac{b}{8}$$
$$\dfrac{b}{8}=a=\dfrac{1}{3}$$
$$\Rightarrow b=\dfrac{8}{3}$$
Hence, $$a=\dfrac{1}{3},b=\dfrac{8}{3}$$


Differentiate $$\sin^{-1}\left [\dfrac {2^{x + 1}.3^{x}}{1 + (36)^{x}}\right ]$$ w.r.t $$x$$.

We have $$y=\sin^{-1}[\dfrac{2^{x+1}.3^{x}}{1+(36^x)}]$$
$$\Rightarrow y=\sin^{-1}[\dfrac{2.6^{x}}{1+(6^{2x})}]$$
Put $$6^{x}=\tan\theta$$
$$\Rightarrow \theta=\tan^{-1}(6^{x})$$
Now,$$y=\sin^{-1}\left [\dfrac{2\tan\theta}{1+\tan^{2}\theta}\right]$$
$$\Rightarrow y=\sin^{-1}[\sin(2\theta)]$$
$$\Rightarrow y=2\theta$$
$$\Rightarrow y=2\tan^{-1}(6^{x})$$
We know$$ \dfrac{d(\tan^{-1}x)}{dx}=\dfrac{1}{{1+x^{2}}}$$
Also $$\dfrac{d(a^{x})}{dx}=a^{x}\ln(a)$$
Therefore, $$\dfrac{dy}{dx}=\dfrac{2}{1+6^{2x}}\times\dfrac{d}{dx}(6^{x})$$
$$\dfrac{dy}{dx}=\dfrac{2}{1+6^{2x}}\times6^{x}\ln 6$$


Find the relationship between $$a$$ and $$b$$ so that the function $$f$$ defined by
$$f(x) = \left\{\begin{matrix} ax + 1,&if\ x\leq 3 \\bx + 3,  & if\ x > 3\end{matrix}\right.$$
is continuous at $$x = 3$$.

Given:

$$f(x) = \left\{\begin{matrix} ax + 1,&if\ x\leq 3 \\bx + 3,  & if\ x > 3\end{matrix}\right.$$
We have $$f(3) = 3a + 1$$

$$\therefore \displaystyle \lim_{x\rightarrow 3^{-}} f(x) = \displaystyle \lim_{x\rightarrow 3^{+}} f(x) = f(3)$$

$$LHL = \displaystyle \lim_{x\rightarrow 3^{-}} f(x) = \displaystyle \lim_{x\rightarrow 3^{-}} (ax + 1) = 3a + 1$$

$$RHL = \displaystyle \lim_{x\rightarrow 3^{+}} = \displaystyle \lim_{x\rightarrow 3^{+}} (bx + 3) = 3b + 3$$

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


Using the definition, show that the function.
$$\displaystyle\, f(x) = x\sin\, (1/x)$$ if $$x \neq 0,  \, 0$$ if $$x = 0$$ is continuous at the point $$x = 0$$

$$f(x) = x\sin \left (\dfrac {1}{x}\right )\ for \  x\neq 0$$
$$and \ f(x)=0\ \ for \  x = 0$$
For a continuous function $$\underset {x\rightarrow a}{L.H.L} = \underset {x\rightarrow a}{R.H.L.} = f(a)$$
Let $$L.H.L.$$
$$= \displaystyle \lim_{x\rightarrow 0^{-}} x\sin \dfrac {1}{x}$$
$$= \displaystyle \lim_{n\rightarrow 0} (0 - 4) \sin \left (\dfrac {1}{-n}\right )$$
$$\Rightarrow \displaystyle \lim_{n \rightarrow 0} - n \sin \left (\dfrac {-1}{n}\right )$$
$$= \displaystyle \lim_{n\rightarrow 0} n \sin \dfrac {1}{n} [\because \sin (-\theta) = -\sin \theta]$$
$$= 0\sin (\infty)$$
$$= 0$$
R.H.L,
$$= \displaystyle \lim_{x \rightarrow 0^{+}} x\sin \dfrac {1}{x}$$
$$= \displaystyle \lim_{n\rightarrow 0} (0 + n) \sin \dfrac {1}{0 + n}$$
$$= \displaystyle \lim_{n\rightarrow 0} n\sin \dfrac {1}{n} = 0. \sin \infty$$
$$= 0$$
So, here
$$L.H.L. = R.H.L. = f(0)$$
$$0 = 0 = 0$$
So function is continuous.


Differentiate the following functions w.r.t $$x$$
$$(i) \cos^{-1} ( \dfrac{2x}{1+x^2})$$
$$(ii) \sin^{-1} (2x\sqrt{1-x^2})$$

$$(i)$$ Let $$y=\cos ^{-1}\left ( \cfrac{2x}{1+x^{2}} \right )\rightarrow (1)$$
  Let $$x=\tan \theta $$
  $$\mathrm{d} x=sec^{2}\theta \mathrm{d}\theta\Rightarrow \cfrac{\mathrm{d} \theta}{\mathrm{d} x} =\cfrac{1}{\sec^{2}\theta}=\cfrac{1}{1+x^{2}}$$
  Differentiate (1) w.r.t x
 $$\cfrac{\mathrm{d} y}{\mathrm{d} x}=\cfrac{\mathrm{d} }{\mathrm{d} x}\left ( \cos ^{-1}\left ( \cfrac{2\tan \theta }{1+\tan ^{2}\theta } \right ) \right )$$
   $$=\cfrac{\mathrm{d} }{\mathrm{d} x}(\cos ^{-1}(\sin 2\theta))$$
   $$=\cfrac{\mathrm{d} \left ( \cfrac{\pi }{2}\pm 2\theta  \right )}{\mathrm{d} x}$$
     $$=\pm 2\cfrac{\mathrm{d} \theta }{\mathrm{d} x}$$
$$\therefore \cfrac{\mathrm{d} y}{\mathrm{d} x}=\cfrac{2}{1+x^{2}}$$    $$x \in (-\infty ,-1)\cup (1,\infty )$$
    $$\quad \   = -\cfrac{2}{1+x^{2}}$$   $$x \in[-1,1]$$

$$(ii)y=\sin ^{-1}\left ( 2x\sqrt{1-x^{2}} \right )$$
  Let $$x=\sin \theta $$
  $$\mathrm{d} x=\cos \theta \mathrm{d}\theta\Rightarrow \cfrac{\mathrm{d} \theta}{\mathrm{d} x} =\cfrac{1}{\sqrt{1-x^{2}}}$$
  $$y=\sin ^{-1}(2\sin \theta\sqrt{1-\sin ^{2}\theta})$$
      $$=\sin ^{-1}(\sin 2\theta)$$
  $$\therefore \cfrac{\mathrm{d} y}{\mathrm{d} x}=2\cfrac{\mathrm{d} \theta}{\mathrm{d} x}(-1)^{n}=\cfrac{2(-1)^{n}}{\sqrt{1-x^{2}}}$$
    $$\therefore \cfrac{\mathrm{d} y}{\mathrm{d} x}=\cfrac{2}{\sqrt{1-x^{2}}}$$ $$\forall x \in \left [ \cfrac{-1}{\sqrt{2}},\cfrac{1}{\sqrt{2}} \right ]$$
           $$=\cfrac{-2}{\sqrt{1-x^{2}}}$$   $$\forall x \in R - \left [  \cfrac{-1}{\sqrt{2}},\cfrac{1}{\sqrt{2}} \right ]$$


Show that the function defined by $$f(x)=|\cos x|$$ is a continuous function.


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If $$y=\tan x$$, prove that $$y_{2}=2yy_{1}$$.

$$y =\tan x$$

$$\therefore \dfrac{dy}{dx} = y_1 = \sec^2x$$

$$\dfrac{d}{dx}(y_1) = y_2 = 2\sec x\dfrac{d}{dx}(\sec x)$$

$$y_2= 2\sec^2x\tan x$$

$$y_2 = 2yy_1$$


If $$y=\tan^{-1}x$$, find $$\dfrac{d^2y}{dx^2}$$ in terms of y alone.

$$y=\tan^{-1}x$$

$$x=\tan y$$

Differentiating w.r.t.y, we get

$$\dfrac{dx}{dy}=\sec^{2}y$$

Diff both sides w.r.t x, 

$$\dfrac{d^{2}y}{dx^{2}}=\dfrac{d}{dx}(\cos^{2}y)$$

$$\dfrac{d^{2}y}{dx^{2}}=-2\cos y\sin y\dfrac{dy}{dx}$$

$$\dfrac{d^{2}y}{dx^{2}}=-2\cos y\sin y\times \cos^{2}y$$

$$\dfrac{d^{2}y}{dx^{2}}=-2\sin y\cos^{3}y$$


$$f(x)$$ = $$\Bigg [\begin{array}   d5, & x\leq 2 \\ ax+b, & 2 < x < 10 \\ 21, & x > 10  \end{array}$$
If $$f(3)=5\\f(5)=3$$
Find $$a,b$$

$$f(3)=5\\3a+b=5\cdots(1)\\f(5)=3\\5a+b=3\cdots(2)\\(2)-(1)\\5a+b-3a-b=3-5\\2a=-2\\a=-1\implies b-3=5\\b=8$$


Differentiate: $$y = 8 \sin x \cos x $$ w.r.t $$x$$

$$\textbf{Step 1: Simplify and differentiating.}$$

                $$y = 8. sin x. cos x$$

                $$\text{On differentiating w.r.t } x,$$

                $$\Rightarrow$$$$y = 4 \times  2 . sin x . cos x$$      $$\boldsymbol[\because\textbf{sin2x = 2 sinx.cosx}]$$

                $$\Rightarrow$$$$y = 4 sin 2x $$

                $$\Rightarrow$$$$\dfrac{dy}{dx} = 4 \dfrac{d}{dx}(sin 2x)$$

                $$\Rightarrow$$$$\dfrac{dy}{dx} = 4 \times \text{cos 2x}\dfrac{d}{dx}(2x)$$

                $$\Rightarrow$$$$\dfrac{dy}{dx} = 8 cos 2x$$

$$\textbf{Hence, on differentiating we get, }$$$$\boldsymbol{\dfrac{dy}{dx} = 8 cos 2x}$$


If $$y=x^x,$$ prove that $$\dfrac{d^2y}{dx^2}-\dfrac{1}{y}(\dfrac{dy}{dx})^2-\dfrac{y}{x}=0$$


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Write the value of the derivative of $$f(x)=|x-1|+|x-3|$$ at $$x=2$$.

Consider the given function.
$$f(x)=|x-1|+|x-3|$$

Since, $$x=2\Rightarrow |x-1|=x-1$$ and $$|x-3|=-(x-3)$$

$$f(x)=(x-1)-(x-3)$$

$$f(x)=x-1-x+3=2$$

Now, differentiating with respect to $$x$$
$$f'(x)=0$$

Hence, this is the answer.


Discuss the continuity and differentiability of the function $$f(x)=|x|+|x-1|$$ in the interval $$(-1, 2)$$.

We can rewrite f(x) as ,
$$f(x) = -x - (x-1) = -2x+1$$ for $$x \in (\infty,0]$$
         $$= x-(x-1)=1$$ for $$x\in (0,1)$$
         $$=x+(x-1) = 2x-1\ for\  x\in [1,\infty)$$
 f(x) is clearly continuous and differentiable in intervals $$(-1,0),(0,1),(1,2)$$
Now let's check continuity and differentiability of f(x) at x=0 and x=1

LHL of f(x) at x=0 is 1.
RHL of f(x) at x=0 is 1.
So, f(x) is continuous at x=0.

LHD of f(x) at x=0 is $$-2$$.
RHD of f(x) at x=0 is $$0$$.
So, f(x) is not differentiable at x=0.

LHL of f(x) at x=1 is 1.
RHL of f(x) at x=1 is 1.
So, f(x) is continuous at x=1.

LHD of f(x) at x=1 is $$0$$.
RHD of f(x) at x=1 is $$2$$.
So, f(x) is not differentiable at x=1.


Prove that the greatest integer function by $$f(x)=[x], 0<x<3$$ is not differentiable at $$x=1$$.

The given function is $$f(x)=[x], 0<x<3$$.

A function f is differentiable at a point c in its domain if both 

$$\lim_{h\rightarrow 0^+} \dfrac{f(c+h)-f(c)}{h}$$ and $$\lim_{h\rightarrow 0^-} \dfrac{f(c+h)-f(c)}{h}$$ are finite and equal.

To check the differentiability of the given function at $$x=1$$, consider the left hand limit of $$f$$ at $$x=1$$.

$$\lim_{h\rightarrow 0^-}\dfrac{[1+h]-[1]}{h}=lim_{h\rightarrow 0^-}\dfrac{1}{h}=-\infty$$

Consider the right hand limit of f at $$x=1$$.

$$\lim_{h\rightarrow 0^+}\dfrac{[1+h]-[1]}{h}=\lim_{h\rightarrow 0^+}0=0$$

Since left and right hand limits of f at $$x=1$$ are not equal, f is not differentiable at $$x=1$$.


Prove that the greatest integer function by $$f(x)=[x], 0<x<3$$ is not differentiable at $$x=2$$.

The given function is $$f(x)=[x], 0<x<3$$.
A function f is differentiable at a point c in its domain if both 
$$\displaystyle \lim_{h\rightarrow 0^+} \dfrac{f(c+h)-f(c)}{h}$$ and $$\lim_{h\rightarrow 0^-} \dfrac{f(c-h)-f(c)}{h}$$ are finite and equal.

To check the differentiability of the given function at $$x=2$$, consider the left hand limit of $$f$$ at $$x=2$$.
$$\displaystyle \lim_{h\rightarrow 0^-}\dfrac{[2-h]-[2]}{h}=lim_{h\rightarrow 0^-}-\dfrac{1}{h}=\infty$$

Consider the right hand limit of f at $$x=2$$.
$$\displaystyle \lim_{h\rightarrow 0^+}\dfrac{[2+h]-[2]}{h}=\lim_{h\rightarrow 0^+}0=0$$

Since left and right hand limits of f at $$x=2$$ are not equal, f is not differentiable at $$x=2$$.


Show that the function $$f(x)=x-[x]$$, where $$[\cdot]$$ denotes the greatest integer function is discontinuous at all integral points.

The given function is $$f(x)=x-[x]$$.

f is defined at all integral points.
Let n be an integer.

Then,

$$f(n)=n-[n]=n-n=0$$

The left hand limit of f at x=n is $$n-(n-1)=1$$

The right hand limit of f at $$x=n$$ is $$n-n=0$$

Since, left and right hand limits are not equal. f is not contnuous at $$x=n$$.

Thus, f is discontinuous at all integral points.


Show that $$f(x)=|x-3|$$ is continuous but not differentiable at $$x=3$$.

$${\textbf{Step - 1: Defining the function piece wise}}{\text{.}}$$

                   $${\text{The given function is f}}\left( {\text{x}} \right) = \left| {{\text{x}} - \left. 3 \right|} \right.$$

                   $$ \Rightarrow {\text{f}}\left( {\text{x}} \right) = {\text{x}} - 3{\text{ if x}} \geqslant {\text{3}}{\text{.}}$$

                                  $$ = 3 - {\text{x if x < 3}}$$

$${\textbf{Step - 2: Checking the continuity at x = 3}} $$

                   $${\text{Lhl}} = \mathop {\lim }\limits_{x \to 3} {\text{f}}\left( {\text{x}} \right) = -\left( {{\text{x}} - 3} \right) = 3 - 3 = 0.$$

                   $${\text{Rhl}} = \mathop {\lim }\limits_{x \to 3} {\text{f}}\left( {\text{x}} \right) = \left( {{\text{x}} - 3} \right) = 3 - 3 = 0.$$

                   $${\text{So , we can say at x}} = 3{\text{ , LHL}} = {\text{RHL}} = {\text{f}}\left( 3 \right).$$

$${\textbf{Step - 3: Checking the differentiability at x = 3}} $$

                  $${\text{At x}} = 3{\text{ L}}{\text{.H}}{\text{.D}} = \mathop {\lim }\limits_{x \to 3} \dfrac{{{\text{f}}\left( {\text{x}} \right) - {\text{f}}\left( 3 \right)}}{{{\text{x}} - 3}}$$

                  $$ = \mathop {\lim }\limits_{x \to 3} \dfrac{{\left( {3 - {\text{x}}} \right) - 0}}{{{\text{x}} - 3}}$$

                  $$ = \mathop {\lim }\limits_{x \to 3} \left( { - 1} \right)$$

                  $$ = \left( { - 1} \right).$$

                  $${\text{At the same point RHD}} = \dfrac{{{\text{f}}\left( {\text{x}} \right) - {\text{f}}\left( 3 \right)}}{{{\text{x}} - 3}}$$   

                  $$ = \mathop {\lim }\limits_{x \to 3} \dfrac{{\left( {{\text{x}} - 3} \right) - 0}}{{{\text{x}} - 3}}$$

                  $$ = 1$$

                  $${\text{So at x}} = 3, {\text{ LHD}} \ne {\text{RHD}}{\text{.}}$$

$${\textbf{Hence, it is proved that f}}\left( {\textbf{x}} \right){\textbf{ is continuous but not differentiable at x}} = \mathbf{3}.$$



Differentiate the following functions with respect to $$x$$:
If $$y=\cos^{-1}(2x)+2\cos^{-1} \sqrt {1-4x^2},0 < x <\dfrac {1}{2}$$, find $$\dfrac {dy}{dx}$$.

$$y=\cos^{-1}(2x)+2\cos^{-1}\sqrt{1-4x^2}$$

Putting $$2x=\cos \theta$$, we get

$$y=\theta +2\cos^{-1}(\sin \theta )$$

$$=\theta +2\cos^{-1}|\cos (\pi /2 -\theta)|$$

$$=\theta +2(\pi /2 -\theta )=\pi -\theta =\pi -\cos^{-1}(2x)$$

Thus,

$$\dfrac{dy}{dx}=\dfrac{2}{\sqrt{1-4x^2}}$$


Differentiate the following functions with respect to $$x$$:
If $$y=\sin^{-1}\left (\dfrac {2x}{1+x^2}\right)+\sec^{-1}\left (\dfrac {1+x^2}{1-x^2}\right), 0 < x < 1$$, prove that $$\dfrac {dy}{dx}=\dfrac {4}{1+x^2}$$

$$y=\sin^{-1} (\dfrac{2x}{1+x^2})+\sec^{-1} (\dfrac{1+x^2}{1-x^2})$$

Putting $$x=\tan \theta $$

$$ we\ get \ y=2\tan^{-1} x+2\tan^{-1} x=4\tan^{-1}x$$

Thus,

$$\dfrac{dy}{dx}=\dfrac{4}{1+x^2}$$   $$[Hence \ proved]$$


Differentiate the following functions with respect to $$x$$:

If $$y=\cot^{-1}\left\{\dfrac {\sqrt {1+\sin x} +\sqrt {1-\sin x}}{\sqrt {1+\sin x} -\sqrt {1-\sin x}}\right\}$$, $$0<x<\dfrac{\pi}2$$,show that $$\dfrac {dy}{dx}$$ is independent of $$x$$.

$$y=cot^{-1}\{\dfrac{\sqrt{1+sinx}+\sqrt{1-sinx}}{\sqrt{1+sinx}-\sqrt{1-sinx}}\}$$
$$y=cot^{-1}(\dfrac{2+2cosx}{2sinx})=cot^{-1}(\dfrac{1+cosx}{sinx})=cot^{-1}(\dfrac{2\cos^2{\dfrac{x}{2}}}{2sin{\dfrac{x}{2}}{\cos{\dfrac{x}{2}}}})$$
$$y=cot^{-1}(cot(\dfrac{x}{2}))$$
$$y=\dfrac{x}{2}$$
So, $$\dfrac{dy}{dx}=\dfrac{1}{2}$$ which is independent of x .


If the derivatives of $$\tan^{-1}(a+bx )$$ takes the value $$1$$ at $$x=0$$, prove that $$1+a^2 =b$$

$$y={\tan}^{-1}{\left(a+bx\right)}$$

$$\Rightarrow\,\dfrac{dy}{dx}=\dfrac{1}{1+{\left(a+bx\right)}^{2}}\dfrac{d}{dx}\left(a+bx\right)$$

$$\Rightarrow\,\dfrac{dy}{dx}=\dfrac{b}{1+{\left(a+bx\right)}^{2}}$$

At$$x=0\Rightarrow\,\dfrac{dy}{dx}=1$$(given)

$$\Rightarrow\,1=\dfrac{b}{1+{\left(a+0\right)}^{2}}$$

$$\therefore\,1+{a}^{2}=b$$ is proved.


If $$y=(\sin^{-1}x)^{2}$$, then prove that $$(1-x^{2})y_{2}-xy_{1}-2=0$$.

Given: $$y = (\sin^{-1}x)^2$$
To Prove: $$(1\times x^2) y_2 - xy_1-2 = 0$$
Solution: Consider $$y = (\sin^{-1}x)^2$$
Differentiate '$$y$$' with respect to '$$x$$'
$$\dfrac{dy}{dx} = 2\sin^{-1}x . \dfrac{1}{\sqrt{1-x^2}}$$

Again differentiate $$\dfrac{dy}{dx} $$ with respect to '$$x$$'

$$\dfrac{d^2y}{dx^2} = 2 \left[\dfrac{\dfrac{1}{\sqrt{1-x^2}}. \sqrt{1-x^2} - \sin^{-1}x.\dfrac{1}{2\sqrt{1-x^2}} (-2x)}{(1-x^2)} \right]$$

$$\dfrac{d^2y}{dx^2} = \dfrac{2\left[1+\dfrac{x}{\sqrt{1-x^2}} \sin^{-1}x\right]}{(1-x^2)}$$

Now; $$LHS = (1-x^2)y_2 - xy_1-2$$

$$= (1-x^2).2 \left[\dfrac{1+\dfrac{x}{\sqrt{1-x^2}} \sin^{-1}x}{(1-x^2)}\right] - \dfrac{x(2\sin^{-1}x)}{\sqrt{1-x^2}} + (-2)$$

$$= 2 + \dfrac{2x}{\sqrt{1-x^2}} \sin^{-1}x - \dfrac{x(2\sin^{-1}x)}{\sqrt{1-x^2}} - 2$$

$$=0$$

$$=RHS$$
Hence proved


Find $$\dfrac {dy}{dx}$$   
if   $$y=(\sin x)^x +\sin^{-1}\sqrt x$$

Let $$y=(\sin x)^x +\sin^{-1}\sqrt x$$. Then

$$y=e^{x \log \sin x}+\sin^{-1} \sqrt x$$

$$\Rightarrow \ \dfrac {dy}{dx}=\dfrac {d}{dx} (e^{x \log \sin x})+\dfrac {d}{dx} (\sin^{-1} \sqrt x)$$

$$\Rightarrow \ \dfrac {dy}{dx}=e^{x \log \sin x}\dfrac {d}{dx} (x \log \sin x)+\dfrac {d}{dx} (\sin^{-1}\sqrt x)$$

$$\Rightarrow \ \dfrac {dy}{dx}=(\sin x)^x \left\{\log \sin x +x \cot x \right\}+\dfrac {1}{\sqrt {1-x}}\times \dfrac {1}{2\sqrt x}$$

$$\Rightarrow \ \dfrac {dy}{dx}=(\sin x)^x \left\{\log \sin x+x \cot x \right\} +\dfrac {1}{2\sqrt {x-x^2}}$$


If $$y= \sin (\sin\, x),$$ then prove that $$\dfrac{d^{2}y}{dx^{2}}+\tan\, x\dfrac{dy}{dx}+ y\, \cos^{2}x=0$$. 

Given: $$y = \sin (\sin x) $$         ....(1)

To prove: $$\dfrac{d^2y}{dx^2} + \tan x . \dfrac{dy}{dx} + y\cos^2x = 0$$

Solution: 
Differentiate (1) with respect to 'x'
$$\dfrac{dy}{dx} = \cos (\sin x). \cos x$$

Again Differentiate with respect to 'x'
$$\dfrac{d^2y}{dx^2} =- \left[-\sin (\sin x)\cos x \cos x + \cos(\sin x) - \sin x\right]$$

$$\dfrac{d^2y}{dx^2} = \sin(\sin x)\cos^2x - \sin x \cos (\sin x)$$

LHS:
$$ \dfrac{d^2y}{dx^2} + \tan x \dfrac{dy}{dx} + y \cos^2x$$

$$= -\sin (\sin x)\cos^2x - \sin x\cos (\sin x) +\tan x (\cos (\sin x) \cos x) + \sin (\sin x) \cos^2 x$$  

$$= -\sin  (\sin x) \cos^2x - \sin x \cos (\sin x) + \sin x(\cos (\sin x)) + \sin (\sin x) \cos^2x$$

$$=0$$

$$=RHS$$    Hence proved


If $$y=e^{\tan^{-1}x}$$, then prove that $$\left ( 1+x^{2} \right )y_{2}+(2x-1)y_{1}=0$$.

Given: $$y = e^{\tan^{-1}x}$$

To prove : 
$$(1 + x^2)y_2 + (2x-1)y_1=0$$

Solution: Consider $$y = e^{\tan{-1}x}$$

Differentiate '$$y$$' w.r.t '$$x$$'
$$\dfrac{dy}{dx} = e^{\tan^{-1}x}. \dfrac{1}{1+x^2}$$
$$y_1 = \dfrac{e^{\tan^{-1}x}}{1+x^2}$$

Again Differentiate w.r.t '$$x$$'
$$\dfrac{d^2y}{dx^2} = \dfrac{e^{\tan^{-1}x} . \left(\dfrac{1}{1+x^2}\right) (1+x^2) - e^{\tan^{-1}x} (2x)}{(1-x^2)^2}$$

$$y^2 = \dfrac{e^{\tan^{-1}x} (1-2x)}{(1+x^2)^2}$$

Now; 
$$LHS = (1+x^2) \left[\dfrac{e^{\tan^{-1}x}(1-2x)}{(1+x^2)^2}\right]+ (2x-1)\left[ \dfrac{e^{\tan^{-1}x}}{1+x^2}\right]$$ 

$$=\dfrac{e^{\tan^{-1}x}-2 x e^{\tan^{-1}x} + 2 x e^{\tan^{-1}x} - e^{\tan^{-1}x}}{1+x^2}$$

$$=0$$

$$=RHS$$
Hence proved


If $$y=3 \cos (log\, x)+4\sin (\log\, x),$$ prove that $$x^{2}y_{2}+xy_{1}+y=0$$.

We have,
$$y=3 cos(log\, x)+4 sin (log\, x)$$
$$\Rightarrow \dfrac{dy}{dx}=-\dfrac{3}{x}sin (log\, x)+\dfrac{4}{x}cos (log\,x)$$
$$\Rightarrow x\dfrac{dy}{dx}=-3sin(log\,x)+4cos (log\,x)$$
Differentiating both sides with respect to x, we get
$$x\dfrac{d^{2}y}{dx^{2}}+\dfrac{dy}{dx}=\dfrac{-3}{x}cos (log\, x)-\dfrac{4}{x}sin (log\, x)$$
$$\Rightarrow x^{2}\dfrac{d^{2}y}{dx^{2}}+x\dfrac{dy}{dx}=-\left \{ 3 cos (log\, x)+4 sin (log\, x) \right \}$$
$$\Rightarrow x^{2}\dfrac{d^{2}y}{dx^{2}}+x\dfrac{dy}{dx}=-y\Rightarrow x^{2}\dfrac{d^{2}y}{dx^{2}}+x\dfrac{dy}{dx}+y=0$$


If $$y= \text{cosec}^{-1}x, x > 1$$, then show that $$x(x^{2}-1)\dfrac{d^{2}y}{dx^{2}}+(2x^{2}-1)\dfrac{dy}{dx}=0$$.

Given,
$$y= \text{cosec}^{-1}x, x > 1$$.

Now differentiating both sides with respect to $$x$$ we get,

$$\dfrac{dy}{dx}=-\dfrac{1}{x\sqrt{x^2-1}}$$

Now squaring both sides we get,

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

Now again differentiating both sides with respect to $$x$$ we get,

$$[2x(x^2-1)+2x^3]\left(\dfrac{dy}{dx}\right)^2+x^2(x^2-1).2.\dfrac{dy}{dx}.\dfrac{d^2y}{dx^2}=0$$

or, $$[4x^3-2x]\left(\dfrac{dy}{dx}\right)^2+x^2(x^2-1).2.\dfrac{dy}{dx}.\dfrac{d^2y}{dx^2}=0$$

or, $$x(x^2-1)\dfrac{d^2y}{dx^2}+(2x^2-1)\dfrac{dy}{dx}=0$$. [ Taking common $$2x\dfrac{dy}{dx}$$ from both sides we get]


Show that $$f(x)=(x-1){e}^{x}+1$$ is an increasing function for all $$x> 0$$

We have

$$f(x)=(x-1){e}^{x}+1\Rightarrow f'(x)=x{e}^{x}> 0$$ for all $$x> 0$$

So, $$f(x)$$ is increasing for all $$x> 0$$


If $$y= (cot^{-1}x)^{2}$$, prove that $$y_{2}(x^{2}+1)^{2}+2x(x^{2}+1)y_{1}=2$$.

Given, $$y= (cot^{-1}x)^{2}$$.

Now differentiating both sides with respect to $$x$$ we get,

$$y_1=-\dfrac{2\cot^{-1}x}{1+x^2}$$

or, $$(1+x^2)y_1=-2\cot^{-1}x$$

Now again differentiating both sides with respect to $$x$$ we get,

$$(1+x^2)y_2+2xy_1=\dfrac{2}{1+x^2}$$

or, $$(1+x^2)^2y_2+2x(1+x^2)y_1=2$$.


Find $$\dfrac{d^{2}y}{dx^{2}}$$, where $$y= \log\left ( \dfrac{x^{2}}{e^{2}} \right )$$.

$$y=\log \left(\dfrac{x^2}{e^2}\right)$$

$$\dfrac{dy}{dx}=\dfrac{1}{\dfrac{x^2}{e^2}}\times \dfrac{2x}{e^2}=\dfrac{e^2}{x^2}\times \dfrac{2x}{e^2}$$

$$\dfrac{dy}{dx}=\dfrac{2}{x}$$

$$\dfrac{d^2y}{dx^2}=\dfrac{-2}{x^2}$$


If $$y=\left \{ log(x+\sqrt{x^{2}+1}) \right \}^{2}$$, then show that $$(1+x^{2})\dfrac{d^{2}y}{dx^{2}}+x\dfrac{dy}{dx}=2.$$

Given, $$y = [\log (x + \sqrt{x^2 + 1})]^2$$
On differentiate with respect to $$x, $$ we get

$$\dfrac{dy}{dx} = 2 \log (x + \sqrt{x^2 + 1}) \times \dfrac{1}{x + \sqrt{x^2 + 1}} \times \left(1 + \dfrac{1}{2 \sqrt{x^2 + 1}} \times 2x\right)$$

$$= \dfrac{2 \log (x + \sqrt{x^2 + 1})}{x + \sqrt{x^2 + 1}} \times \left(\dfrac{x + \sqrt{x^2 + 1}}{\sqrt{x^2 + 1}} \right)$$

$$= \dfrac{2 \log (x + \sqrt{x^2 + 1})}{\sqrt{x^2 + 1}}$$

Again differentiate with respect to x, we get

$$\dfrac{d^2 y}{dx^2} = \dfrac{\sqrt{x^2 + 1} \times \dfrac{2}{x + \sqrt{x^2 + 1}} \times \left(1 + \dfrac{1}{2 \sqrt{x^2 +1}} \times 2x\right) - 2 \log (x + \sqrt{x^2 + 1}) \times \left(\dfrac{1}{2 \sqrt{x^2 + 1}} \times 2x\right)}{(\sqrt{x^2 + 1})^2}$$

$$= \dfrac{\sqrt{x^2 + 1} \times \dfrac{2}{x + \sqrt{x^2 + 1}} \times \left(\dfrac{x + \sqrt{x^2 + 1}}{\sqrt{x^2 + 1}} \right) - 2 \log (x + \sqrt{x^2 + 1}) \times \dfrac{x}{\sqrt{x^2 + 1}}}{(x^2 + 1)}$$

$$= \dfrac{2 - \dfrac{2x \log (x + \sqrt{x^2 + 1})}{\sqrt{x^2 + 1}}}{(x^2 + 1)}$$

$$(1 + x^2) \dfrac{d^2y}{dx^2} + x \dfrac{dy}{dx} = 2$$


Find the differential equation of the following.
$$\tan^{-1} \left(\dfrac{1 - \cos x}{\sin x} \right)$$



If $$y=e^{2x}(ax+b)$$, show that $$y_{2}-4y_{1}+4y=0$$.

$$y=e^{2x}(ax+b)$$...........(1)
differentiate w.r.t $$x$$
$$y_1=e^{2x}a+(ax+b)2e^{2x}$$.............(2)
Again diff w.r.t.x
$$y_2=2e^{2x}a+(ax+b)4e^{2x}+2e^{2x}a$$
$$=4ae^{2x}+(ax+b)4e^{2x}$$.........(3)
Putting (1),(2) and (3) in
$$y_2-4y_1+4y$$
$$4ae^{2x}+(ax+b)4e^{2x}-4\times e^{2x}a+4\times 2e^{2x}-(ax+b)+4e^{2x}(ax+b)$$
$$4ae^{2x}-4ae^{2x}+(ax+b)[4e^{2x}-8e^{2x}+4e^{2x}]=0$$
$$\boxed{Hence\,y_2-4y_1+4y=0}$$


The left hand derivative of $$f(x)=[x] sin \pi x$$ at $$x=k$$ is?
(Where k is an integer and [x]=greatest integer $$\leq x$$)  

We need to find the left hand derivative of $$f(x)=[x]sin\: \pi x$$ at $$x=k$$.

$$\lim_{h\rightarrow 0^-} \dfrac{f(k)-f(k-h)}{h}=\lim_{h\rightarrow 0^-} \dfrac{[k]sin \: \pi k-[k-h]sin \: \pi(k-h)}{h}$$

                                               $$=\lim_{h\rightarrow 0^-} \dfrac{-(k-1)sin \: \pi(k-h)}{h}$$

We know that, $$sin\:k\pi=0 \Rightarrow sin(k\pi-\theta)=(-1)^{k-1}sin\: \theta$$

                                               $$=\lim_{h\rightarrow 0^-} \dfrac{-(k-1)(-1)^{k-1}sin\: h\pi}{h\pi} \times \pi$$

                                               $$=\pi (k-1)(-1)^k$$

$$\lim_{h\rightarrow 0^-} \dfrac{f(k)-f(k-h)}{h}=\pi (k-1)(-1)^k$$


Differentiate $$\sin { { h }^{ -1 }\left( \cfrac { 1 }{ x }  \right)  } $$ with respect to $$x(x> 0)$$

We need to find $$\dfrac{d}{dx}\left(sinh^{-1}\left(\dfrac{1}{x}\right)\right)$$

$$\dfrac{d}{dx}\left(sinh^{-1}\left(\dfrac{1}{x}\right)\right)=-\dfrac{1}{x^2}\cdot \dfrac{1}{\sqrt{\dfrac{1}{x^2}+1}}=-\dfrac{1}{x^2}\cdot \dfrac{x}{\sqrt{x^2+1}}=-\dfrac{1}{x\sqrt{1+x^2}}$$


If $$x={ acos }^{ 3 }\theta $$ and $$y={ asin }^{ 3 }\theta $$, prove that $$\dfrac { dy }{ dx } =\sqrt [ 3 ]{ \dfrac { y }{ x }  } $$

$$x=a{\cos}^{3}{\theta}$$
$$\dfrac{dx}{d\theta}=a\dfrac{d}{d\theta}\left({\cos}^{3}{\theta}\right)$$
$$=a3{\cos}^{2}{\theta}\dfrac{d}{d\theta}\left(\cos{\theta}\right)$$
$$=-3a\sin{\theta}{\cos}^{2}{\theta}$$
$$y=a{\sin}^{3}{\theta}$$
$$\dfrac{dy}{d\theta}=a\dfrac{d}{d\theta}\left({\sin}^{3}{\theta}\right)$$
$$=a3{\sin}^{2}{\theta}\dfrac{d}{d\theta}\left(\sin{\theta}\right)$$
$$=3a{\sin}^{2}{\theta}\cos{\theta}$$
$$\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}}$$
$$=\dfrac{3a{\sin}^{2}{\theta}\cos{\theta}}{-3a\sin{\theta}{\cos}^{2}{\theta}}$$
$$=\dfrac{-\sin{\theta}}{\cos{\theta}}$$
$$=-\tan{\theta}$$    .....$$(1)$$
Again $$\dfrac{y}{x}=\dfrac{a{\sin}^{3}{\theta}}{a{\cos}^{3}{\theta}}={\tan}^{3}{\theta}$$
$$\Rightarrow \tan{\theta}=\sqrt[3]{\dfrac{y}{x}}$$    .....$$(2)$$
$$\therefore \dfrac{dy}{dx}=\sqrt[3]{\dfrac{y}{x}}$$ from $$(1)$$ and $$(2)$$


If $$x=\sin t,y=\sin k t$$,then show that $$\left(1-{x}^{2}\right)\dfrac{{d}^{2}y}{d{x}^{2}}-x\dfrac{dy}{dx}-ky=0$$

Given:$$x=\sin{t}$$
$$\Rightarrow 1-{x}^{2}=1-{\sin}^{2}{t}={\cos}^{2}{t}$$
$$y=\sin{kt}\Rightarrow \dfrac{dy}{dt}=-k\cos{kt}$$
$$x=\sin{t}\Rightarrow \dfrac{dx}{dt}=\cos{t}$$
$$\therefore \dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{-k\cos{kt}}{\cos{t}}=-k$$
$$\Rightarrow \dfrac{{d}^{2}y}{d{x}^{2}}=\dfrac{d}{dx}\left(-k\right)=0$$
Substituting the above in $$\left(1-{x}^{2}\right)\dfrac{{d}^{2}y}{d{x}^{2}}-x\dfrac{dy}{dx}-ky$$
$$={\cos}^{2}{t}\left(0\right)-x\times -k-ky$$
$$=0+kx-ky$$
$$=k\sin{t}-k\sin{kt}=0$$
Hence proved.


Prove that $$f(x)=\sin x+\sqrt{3}\cos x$$ has maximum value at $$x=\dfrac{\pi}{6}$$. 

$$f(x)=\sin x+\sqrt{x}\cos x$$

$$f'(x)=\cos x-\sqrt{3}\sin x$$

$$f'(x)=0$$ for critical points

$$\Rightarrow \cos x-\sqrt{3}\sin x=0$$

$$\Rightarrow \cos x=\sqrt{3}\sin x$$

$$\Rightarrow \cot x=\sqrt{3}$$

$$\Rightarrow x=\dfrac{\pi}{6}$$

$$f''(x)=-\sin x-\sqrt{3}\cos x$$

$$f''\left(\dfrac{\pi}{6}\right)=-\sin\left(\dfrac{\pi}{6}\right)-\sqrt{3}\cos\left(\dfrac{\pi}{6}\right) < 0$$

Hence, f is maximum at the point $$x=\dfrac{\pi}{6}$$.


If $$ y=a \left \{ x+\sqrt{x^2+1} \right \}^{n}+ b \left \{ x-\sqrt{x^2+1} \right \}^{-n}$$, then prove that $$ (x^2-1)\dfrac{d^2y}{dx^2}+x\dfrac{dy}{dx}- n^2 y = 0.$$


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Class 12 Engineering Maths Extra Questions