Class 12 Engineering Maths - Continuity And Differentiability

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

$\displaystyle\lim_{x\to 0^+} f(x)\ne \lim_{x\to 0^-} f(x)\ne f(0)$

$f(x)$ 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.

1309158_1254090_ans_d6c05f075ef54e5588b4fc65b1610f07.JPG

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

1294351_1504189_ans_6b2a16ad88e644829b5c365cea38f081.jpg

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 $1+a^2 =b$

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

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

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

$x=0\Rightarrow\,\dfrac{dy}{dx}=1$ (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 $(x^2-1)\dfrac{d^2y}{dx^2}+x\dfrac{dy}{dx}- n^2 y = 0.$

2001852_1661974_ans_2fb1d02014f14e53b2a590fc0e64eebb.jpg

Continuity And Differentiability - Class 12 Engineering Maths - Extra Questions

If, is continuous at x=2x = 2, find the value of kk.f(x)={x3+x2−16x+20(x−2)2,x≠2k,x=2f(x) = \begin{cases} \dfrac{x^3 + x^2 - 16x + 20}{(x-2)^2},& x \neq 2\\ k,&x = 2 \end{cases}

Check the continuity of the function f(x)={|x|x,x≠00,x=0 at x=0f ( x ) = \left\{ \begin{array} { l l } { \frac { | x | } { x } , } & { x \neq 0 } \\ { 0 , } & { x = 0 } \end{array} \right. \text { at } x = 0

Discuss the continuity of the function ff, where ff is defined by f(x)=f(x) = {3, if 0≤x≤14, if 1<x<35, if 3≤x≤10\begin{cases} 3,\ if\ 0 \le x \le 1 \\ 4,\ if\ 1< x < 3 \\ 5,\ if\ 3 \le x \le 10 \end{cases}

Differentiate sin−1[2x+1.3x1+(36)x]\sin^{-1}\left [\dfrac {2^{x + 1}.3^{x}}{1 + (36)^{x}}\right ] w.r.t xx.

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

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

Differentiate the following functions w.r.t xx(i)cos−1(2x1+x2)(i) \cos^{-1} ( \dfrac{2x}{1+x^2})(ii)sin−1(2x√1−x2)(ii) \sin^{-1} (2x\sqrt{1-x^2})

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

If y=tanxy=\tan x, prove that y2=2yy1y_{2}=2yy_{1}.

If y=tan−1xy=\tan^{-1}x, find d2ydx2\dfrac{d^2y}{dx^2} in terms of y alone.

f(x)f(x) = [5,x≤2ax+b,2<x<1021,x>10\Bigg [\begin{array} d5, & x\leq 2 \\ ax+b, & 2 < x < 10 \\ 21, & x > 10 \end{array}If f(3)=5f(5)=3f(3)=5\\f(5)=3Find a,ba,b

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

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

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

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

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

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

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

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

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

If the derivatives of \tan^{-1}(a+bx )\tan^{-1}(a+bx ) takes the value 11 at x=0x=0, prove that 1+a^2 =b1+a^2 =b

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

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

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

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

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

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

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

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

Find \dfrac{d^{2}y}{dx^{2}}\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 ).

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

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

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

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

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

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

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

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

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

Class 12 Engineering Maths Extra Questions

Support mcqexams.com by disabling your adblocker.

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

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

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

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

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 $x = 3$ .

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 $x = 0$

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

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

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

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

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

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

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

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

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

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

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

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

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

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 $\dfrac {dy}{dx}$ .

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

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

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

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

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

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

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

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

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

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

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

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 $y_{2}-4y_{1}+4y=0$ .

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

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

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

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

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

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