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If $\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{xy}{x^{2}+y^{2}};y(1)=1;$ then a value of $x$ satisfying $y(x)=e$ is :
Option: 1 $\sqrt{3}\: e$

Option: 2 $\frac{1}{2}\sqrt{3}\: e$

Option: 3 $\sqrt{2}\: e$

Option: 4 $\frac{e}{\sqrt{2}}$

√3e

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Nalla mahalakshmi

Given : $f(x)=\left\{\begin{matrix} x, &0\leq x< \frac{1}{2} \\ \frac{1}{2}, &x=\frac{1}{2} \\ 1-x, &\frac{1}{2} < x\leq 1 \end{matrix}\right.$ and $g(x)=\left ( x-\frac{1}{2} \right )^{2},\: x\epsilon \textbf{R}.$ Then the area (in sq. units) of the region bounded b the curves, $y=f(x)$ and $y=g(x)$ between the lines, $2x=1\: \: \: and\: \: \: 2x=\sqrt{3},$ is :
Option: 1 $\frac{\sqrt{3}}{4}-\frac{1}{3}$

Option: 2 $\frac{1}{3}+\frac{\sqrt{3}}{4}$

Option: 3 $\frac{1}{2}+\frac{\sqrt{3}}{4}$

Option: 4 $\frac{1}{2}-\frac{\sqrt{3}}{4}$

Area Bounded by Curves When Intersects at More Than One Point -

Area bounded by the curves y = f(x), y = g(x) and intersect each other in the interval [a, b]

First find the point of intersection of these curves y = f(x) and y = g(x) , let the point of intersection be x = c

Area of the shaded region

$=\int_{a}^{c}\{f(x)-g(x)\} d x+\int_{c}^{b}\{g(x)-f(x)\} d x$

When two curves intersects more than one point

rea bounded by the curves y=f(x), y=g(x) and intersect each other at three points at x = a, x = b amd x = c.

To find the point of intersection, solve f(x) = g(x).

For x ∈ (a, c), f(x) > g(x) and for x ∈ (c, b), g(x) > f(x).

Area bounded by curves,

$\\\mathrm{A=} \int_{a}^{b}\left |f(x)-g(x) \right |dx\\\\\mathrm{\;\;\;\;=} \int_{a}^{c}\left ( f(x)-g(x) \right )dx+\int_{c}^{b}\left ( g(x)-f(x) \right )dx$

Required area = Area of trapezium ABCD - $\int_{1/2}^{\sqrt3/2}\left ( x-\frac{1}{2} \right )^2dx$

$\\=\frac{1}{2}\left(\frac{\sqrt{3}-1}{2}\right)\left(\frac{1}{2}+1-\frac{\sqrt{3}}{2}\right)-\frac{1}{3}\left(\left(x-\frac{1}{2}\right)^{3}\right)_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}\\=\frac{\sqrt3}{4}-\frac{1}{3}$

Correct Option (1)

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avinash.dongre

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Let a function $f:\left [ 0,5 \right ]\rightarrow \textbf{R}$ be continuous, $f\left ( 1 \right )=3$ and F be defined as : $F(x)=\int_{1}^{x}t^{2}g\left ( t \right )dt$ , where $g(t)=\int_{1}^{t}f\left ( u \right )du.$ Then for the function F, the point $x=1$ is :
Option: 1 a point of infection.
Option: 2 a point of local maxima.
Option: 3 a point of local minima.
Option: 4 not a critical point.

Integration as Reverse Process of Differentiation -

Integration is the reverse process of differentiation. In integration, we find the function whose differential coefficient is given.

For example,

$\\\mathrm{\frac{d}{dx}(\sin x)=\cos x}\\\\\mathrm{\frac{d}{dx}\left ( x^2 \right )=2x}\\\\\mathrm{\frac{d}{dx}\left ( e^x \right )=e^x}$

In the above example, the function cos x is the derived function of sin x. We say that sin x is an anti derivative (or an integral) of cos x. Similarly, x² and e^x are the anti derivatives (or integrals) of 2x and e^x respectively.

Also note that the derivative of a constant (C) is zero. So we can write the above examples as:

$\\\mathrm{\frac{d}{dx}(\sin x+c)=\cos x}\\\\\mathrm{\frac{d}{dx}\left ( x^2+c \right )=2x}\\\\\mathrm{\frac{d}{dx}\left ( e^x +c\right )=e^x}$

Thus, anti derivatives (or integrals) of the above functions are not unique. Actually, there exist infinitely many anti derivatives of each of these functions which can be obtained by selecting C arbitrarily from the set of real numbers.

For this reason C is referred to as arbitrary constant. In fact, C is the parameter by varying which one gets different anti derivatives (or integrals) of the given function.

If the function F(x) is an antiderivative of f(x), then the expression F(x) + C is the indefinite integral of the function f(x) and is denoted by the symbol ∫ f(x) dx.

By definition,

$\\\mathrm{\int f(x)dx=F(x)+c,\;\;\;where\;\;F'(x)=f(x)\;\;and\;\;'c' \;is\;constant.}$

Maxima and Minima of a Function -

Maxima and Minima of a Function

Let y = f(x) be a real function defined at x = a. Then the function f(x) is said to have a maximum value at x = a, if f(x) ≤ f(a) ∀ a ≥∈ R.

And also the function f(x) is said to have a minimum value at x = a, if f(x) ≥ f(a) ∀ a ∈ R

Concept of Local Maxima and Local Minima

The function f(x) is said to have a maximum (or we say that f(x) attains a maximum) at a point ‘a’ if the value of f(x) at ‘a’ is greater than its values for all x in a small neighborhood of ‘a’ .

In other words, f(x) has a maximum at x = ‘a’, if f(a + h) ≤ f(a) and f(a - h) ≤ f(a), where h ≥ 0 (very small quantity).

The function f(x) is said to have a minimum (or we say that f(x) attains a minimum) at a point ‘b’ if the value of f(x) at ‘b’ is less than its values for all x in a small neighborhood of ‘b’ .

In other words, f(x) has a maximum at x = ‘b’, if f(a + h) ≥ f(a) and f(a - h) ≥ f(a), where h ≥ 0 (very small quantity).

$\\ \begin{aligned} &\mathrm{F}^{\prime}(\mathrm{x})=\mathrm{x}^{2} \mathrm{g}(\mathrm{x})\\ &\Rightarrow \mathrm{F}^{\prime}(1)=1.\;\;g(1)=0\;\;\;\;\;\ldots(1)\;\;\;\;\;\;\;\;(\because\;g(1)=0)\\ &\text { Now }(\mathrm{x})=2 \mathrm{xg}(\mathrm{x})+\mathrm{x}^{2} \mathrm{g}^{\prime}(\mathrm{x}) \end{aligned}$

$\\ \begin{aligned} &\Rightarrow F^{\prime \prime}(x)=2 x g(x)+x^{2} f(x)\;\;\;\;(\because\;g'(x)=f(x))\\ &\Rightarrow F^{\prime \prime}(1)=0+1 \times 3\\ &\Rightarrow F^{\prime \prime}(1)=3\;\;\;\;\;\;\;\;\;\;\ldots (2) \end{aligned}\\\text{From (1) and (2) F(x) has local minimum at x = 1 }$

Correct Option (1)

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avinash.dongre

The value of $\int_{0}^{2x}\frac{x\sin ^{8}x}{\sin ^{8}x+\cos ^{8}x}dx$ is equal to :
Option: 1 $2\pi$

Option: 2 $4\pi$

Option: 3 $2\pi ^{2}$

Option: 4 $\pi ^{2}$

Properties of the Definite Integral (Part 2) - King's Property -

Property 4 (King's Property)

This is one of the most important properties of definite integration.

$\\\mathbf{\int_{a}^{b}f(x)\;dx=\int_{a}^{b}f(a+b-x)\;dx}$

Application of Periodic Properties in Definite Integration -

Property 9

If f(x) is a periodic function with period T, then the area under f(x) for n periods would be n times the area under f(x) for one period, i.e.

$\mathbf{\int_{0}^{n T} f(x) d x=n \int_{0}^{T} f(x) d x}$

$\\I=\int _0^{2\pi }\frac{x\sin ^8x\:}{\sin ^8x+\cos ^8x\:\:}dx\\I=\int _0^{2\pi }\frac{(2\pi-x)\sin ^8(2\pi-x)\:}{\sin ^8(2\pi-x)+\cos ^8(2\pi-x)\:\:}dx\\2I=\int _0^{2\pi }\frac{2\pi \sin ^8x\:}{\sin ^8x+\cos ^8x\:\:}dx\\I=\int _0^{2\pi }\frac{\pi \sin ^8x\:}{\sin ^8x+\cos ^8x\:\:}dx$

$\\I=\int _0^{2\pi }\frac{\pi \sin ^8x\:}{\sin ^8x+\cos ^8x\:\:}dx\\I=4\int _0^{\pi/2 }\frac{\pi \sin ^8x\:}{\sin ^8x+\cos ^8x\:\:}dx\\I=4\int _0^{\pi/2 }\frac{\pi \sin ^8(\pi/2-x)\:}{\sin ^8(\pi/2-x)+\cos ^8(\pi/2-x)\:\:}dx\\I=2\int_{0}^{\pi/2}\pi dx\\I=\pi^2$

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avinash.dongre

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$\lim_{n\rightarrow \infty }\left ( \frac{(n+1)(n+2).....3n}{n^{2n}} \right )^{\frac{1}{n}}$ is equal to
Option: 1 $\frac{18}{e^{4}}$
Option: 2 $\frac{27}{e^{2}}$
Option: 3 $3\log ^{3-2}$
Option: 4 $3\log ^{3-2}$

As we learnt

Walli's Method -

Definite integral by first principle

$\int_{a}^{b}f(x)dx= \left ( b-a \right )\lim_{n \to \infty }\frac{1}{n}\left [ f(a) +f(a+h)+f(a+2h)....\right ]$

where

$h=\frac{b-a}{n}$

- wherein

$y=\lim_{n\rightarrow \infty }\left [ \frac{(n+1)(n+2)...(n+2n)}{n^{2n}} \right ]^{1/n}$

$=\lim_{n\rightarrow \infty }\left [ \left ( 1+\frac{1}{n} \right )\left ( 1+\frac{2}{n} \right ) ...\left ( 1+\frac{2n}{n} \right )\right ]^{1/n}$

$\Rightarrow \log y=\lim_{n\rightarrow \infty }\frac{1}{n}\left [ \left ( 1+\frac{1}{n} \right )\left ( 1+\frac{2}{n} \right ) ...\left ( 1+\frac{2n}{n} \right )\right ]$

$=\lim_{n\rightarrow \infty }\frac{1}{n}\sum_{r=1}^{2n}\log\left ( 1+\frac{r}{n} \right )$

$=\int_{0}^{2}\log\left ( 1+x \right )dx$

$=\left [ \log(1+x).x \right ]-\int_{0}^{2}\frac{1}{1+x}.xdx$

$= \left [ x\log(1+x)-x +\log(1+x)\right ]^{2}_{0}$

$=2\log 3-2+\log 2$

$\Rightarrow y=\frac{27}{e^{2}}$

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avinash.dongre

If $\int \frac{\cos \: x\: dx}{\sin ^{3}x(1+\sin ^{6}x)^{2/3}}= f(x)(1+ \sin ^{6}x)^{1/\lambda }+c$ where c is a constant of integration, then $\lambda f\left ( \frac{\pi }{3} \right )$ is equal to :
Option: 1 $-\frac{9}{8}$
Option: 2 $\frac{9}{8}$
Option: 3 $2$
Option: 4 $-2$

Integration Using Substitution -

The method of substitution is one of the basic methods for calculating indefinite integrals.

Substitution - change of variable

$\\\mathrm{To\;solve\;the\;integrate\;of\;the\;form}\\\\\mathrm{I=\int f\left ( g(x) \right )\cdot g'(x)\;dx,\;}\\\\\mathrm{\;where\;g(x)\;is\;continuously\;differentiable\;function.}\\\mathrm{put\;\;g(x)=t,\;\;g'(x)\;dx=dt}\\\mathrm{After\;substitution,\;we\;get\;\;\int f(t)\;dt.}\\\text{Evalute this integration and substitute back the value of }t.$

$\\\sin x=t \quad\Rightarrow \cos x d x=d t\\\\\int \frac{d t}{t^{2}\left(1+t^{6}\right)^{2 / 3}}=\int \frac{d t}{t^{3} t^{4}\left(\frac{1}{t^6}+1\right)^{2/3}}\\\\u=\frac{1}{t^{6}}+1 \Rightarrow d u=-\frac{6}{t^{7}} d t\\\\\frac{d u}{-6}=\frac{d t}{t^{7}}\\$

$\\=\int\frac{du}{-6u^{2/3}}=-\frac{1}{2 }u^{1/3}+C\\\\=-\frac{1}{2}\left ( \frac{1}{t^6}+1 \right )^{1/3}+C\\\\=-\frac{1}{2}\left ( \frac{(1+\sin^6x)^{1/3}}{\sin^2x} \right )\\\\f(x)=-\frac{1}{2}\frac{1}{\sin^2x}\;\;\;\lambda=3\\\\\lambda f(\pi/3)=-2$

Correct option (4)

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Kuldeep Maurya

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Let, $f(x)=\left ( \sin (\tan ^{-1}x)+\sin (\cot ^{-1}x) \right )^{2}-1,$ $\left | x \right |>1.\; If\frac{dy}{dx}=\frac{1}{2}\frac{d}{dx}\left ( \sin ^{-1}(f(x)) \right )$ and $y(\sqrt{3})=\frac{\pi }{6},$ then $y(-\sqrt{3})$ is equal to :
Option: 1 $\frac{\pi }{3}$
Option: 2 $\frac{2\pi }{3}$
Option: 3 $-\frac{\pi }{6}$
Option: 4 $\frac{5\pi }{6}$

Integration as Reverse Process of Differentiation -

Integration is the reverse process of differentiation. In integration, we find the function whose differential coefficient is given.

For example,

$\\\mathrm{\frac{d}{dx}(\sin x)=\cos x}\\\\\mathrm{\frac{d}{dx}\left ( x^2 \right )=2x}\\\\\mathrm{\frac{d}{dx}\left ( e^x \right )=e^x}$

Also note that the derivative of a constant (C) is zero. So we can write the above examples as:

$\\\mathrm{\frac{d}{dx}(\sin x+c)=\cos x}\\\\\mathrm{\frac{d}{dx}\left ( x^2+c \right )=2x}\\\\\mathrm{\frac{d}{dx}\left ( e^x +c\right )=e^x}$

For this reason C is referred to as arbitrary constant. In fact, C is the parameter by varying which one gets different anti derivatives (or integrals) of the given function.

If the function F(x) is an antiderivative of f(x), then the expression F(x) + C is the indefinite integral of the function f(x) and is denoted by the symbol ∫ f(x) dx.

By definition,

$\\\mathrm{\int f(x)dx=F(x)+c,\;\;\;where\;\;F'(x)=f(x)\;\;and\;\;'c' \;is\;constant.}$

Trigonometric Identities -

Trigonometric Identities-

These identities are the equations that hold true regardless of the angle being chosen.

$\\\mathrm{\sin^2\mathit{t}+\cos^2\mathit{t}=1}\\\mathrm{1+\tan^2\mathit{t}=\sec^2\mathit{t}}\\\mathrm1+{\cot^2\mathit{t}=\csc^2\mathit{t}}\\\mathrm{\tan \mathit{t}=\frac{\sin \mathit{t}}{\cos \mathit{t}},\;\;\cot \mathit{t}=\frac{\cos\mathit{t}}{\sin\mathit{t}}}$

Principal Value of function f-1 (f (x)) -

Principal Value of function f^-1 (f (x))

$\begin{array} {l}\mathrm{1.\;\;\sin^{-1}(\sin (\theta))=\theta} \quad\quad\quad \;\mathrm{for\;all\;\theta\in[-\pi/2,\pi/2] }\\\mathrm{2.\;\;\cos^{-1}(\cos(\theta))=\theta} \quad\quad\quad \mathrm{for\;all\;\theta\in[0,\pi]}\\\mathrm{3.\;\;\tan^{-1}(\tan(\theta))=\theta} \;\;\;\quad\quad \mathrm{for\;all\;\theta\in(-\pi/2,\pi/2)}\\\mathrm{4.\;\;\cot^{-1}(\cot(\theta))=\theta} \quad\quad\quad \mathrm{for\;all\;\theta\in(0,\pi)} \\\mathrm{5.\;\;\sec^{-1}(\sec(\theta))=\theta} \quad\quad\quad \mathrm{for\;all\;\theta\in[0,\pi]-\left \{ \pi/2 \right \}}\\\mathrm{6.\;\;\csc^{-1}(\csc(\theta))=\theta} \quad\quad\quad \mathrm{for\;all\;\theta\in[-\pi/2,\pi/2]-\left \{ 0 \right \}}\end{array}$

$\\y=\frac{1}{2} \sin ^{-1}(f(x))+c\\\\\begin{array}{l}{\text {Let } \theta=\tan ^{-1} x \Rightarrow \cot ^{-1} x=\frac{\pi}{2}-\theta} \\\\ {f(x)=\left\{\sin \theta+\sin \left(\frac{\pi}{2}-\theta\right)\right\}^{2}-1}\end{array}\\\\\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta-1=\sin2\theta$

$\\y=\frac{1}{2} \sin ^{-1}\left(\sin \left(2 \tan ^{-1} x\right)\right)+C\\\\\frac{\pi}{6}=\frac{1}{2} \sin ^{-1}\left(\sin \left(2 \frac{\pi}{3}\right)\right)+C\\\\\frac{\pi}{6}=\frac{1}{2} \frac{\pi}{3}+C\Rightarrow C=0\\\\y=\frac{1}{2} \sin ^{-1}\left(\sin \left(-2 \frac{\pi}{3}\right)\right)=-\frac{\pi}{6}$

Correct Option (4)

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For $a>0,$ let the curves $C_{1}:y^{2}=ax$ and $C_{2}:x^{2}=ay$ intersect at origin O and a point P. Let the line $x=b(0<b<a)$ intersect the chord OP and the x-axis at points Q and R,respectively. If the line $x=b$ bisects the are bounded by the curves, $C_{1}$ and $C_{2'}$ and the area of $\Delta OQR=\frac{1}{2}$ , then 'a' satisfies the equation :

Option: 1 $x^{6}-12x^{3}+4=0$
Option: 2 $x^{6}-12x^{3}-4=0$
Option: 3 $x^{6}+6x^{3}-4=0$
Option: 4 $x^{6}-6x^{3}+4=0$

Parabola -

Parabola

A parabola is the set of all points (x, y) in a plane that are the same distance from a fixed line, called the directrix, and a fixed point (the focus) not on the directrix in the plane.

Standard equation of a parabola

Let focus of parabola is S(a, 0) and directrix be x + a = 0, and axis as x-axis

P(x, y) is any point on the parabola.

Now, from the definition of the parabola,
$\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;SP=PM}\\\mathrm{\Rightarrow \;\;\;\;\;\;\;\;\;\;\;\;SP^2=PM^2}\\\mathrm{\Rightarrow \;\;\;\;\;(x-a)^{2}+(y-0)^{2}=(x+a)^{2}}\\\mathrm{\Rightarrow \;\;\;\;\;y^2=4ax}$

which is the required equation of a standard parabola

Area Bounded by Curves When Intersects at More Than One Point -

Area bounded by the curves y = f(x), y = g(x) and intersect each other in the interval [a, b]

First find the point of intersection of these curves y = f(x) and y = g(x) , let the point of intersection be x = c

Area of the shaded region

$=\int_{a}^{c}\{f(x)-g(x)\} d x+\int_{c}^{b}\{g(x)-f(x)\} d x$

When two curves intersects more than one point

rea bounded by the curves y=f(x), y=g(x) and intersect each other at three points at x = a, x = b amd x = c.

To find the point of intersection, solve f(x) = g(x).

For x ∈ (a, c), f(x) > g(x) and for x ∈ (c, b), g(x) > f(x).

Area bounded by curves,

$\\\mathrm{A=} \int_{a}^{b}\left |f(x)-g(x) \right |dx\\\\\mathrm{\;\;\;\;=} \int_{a}^{c}\left ( f(x)-g(x) \right )dx+\int_{c}^{b}\left ( g(x)-f(x) \right )dx$

$\begin{array}{l}{\frac{1}{2} (b \times b)=\frac{1}{2}} \\ {b=1}\end{array}$

$\int_{0}^{1}\left(\sqrt{a} \sqrt{x}-\frac{x^{2}}{a}\right) d x=\frac{a^2}{6} \text{ by property of parabola}$

By solving above you will get

$a^{6}-12 a^{3}+4=0$

Correct option (1)

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Let $y=y(x)$ be a solution of the differential equation, $\sqrt{1-x^{2}}\frac{dy}{dx}+\sqrt{1-y^{2}}=0,\left | x \right |<1.$ If $y\left ( \frac{1}{2} \right )=\frac{\sqrt{3}}{2},$ then $y\left ( \frac{-1}{\sqrt{2}} \right )$ is equal to :
Option: 1 $-\frac{1}{\sqrt{2}}$
Option: 2 $-\frac{\sqrt{3}}{2}$
Option: 3 $\frac{1}{\sqrt{2}}$
Option: 4 $\frac{\sqrt{3}}{2}$

Formation of Differential Equation and Solutions of a Differential Equation -

This is the general solution of the differential equation (2), which represents the family of the parabola (when a = 1) and one member of the family of parabola is given in Eq (1).

Also, Eq (1) is a particular solution of the differential equation (2).

The solution of the differential equation is a relation between the variables of the equation not containing the derivatives, but satisfying the given differential equation.

A general solution of a differential equation is a relation between the variables (not involving the derivatives) which contains the same number of the arbitrary constants as the order of the differential equation.

Particular solution of the differential equation obtained from the general solution by assigning particular values to the arbitrary constant in the general solution.

$\sqrt{1-x^{2}} \frac{d y}{d x}=-\sqrt{1-y^{2}}$

${\frac{d y}{\sqrt{1-y^{2}}}=-\frac{d x}{\sqrt{1-x^{2}}} \Rightarrow \sin ^{-1} y=-\sin ^{-1} x+c} \\ {x=\frac{1}{2}, y=\frac{\sqrt{3}}{2} \Rightarrow \frac{\pi}{2}=-\frac{\pi}{6}-c}$