Question

If
$a = \sqrt{21 } - \sqrt{20} \: \: and \\ b = \sqrt{18} - \sqrt{17} \: \: then$
A) a = b
B) a + b = 0
C) a > b
D) a < b

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otherwise I will report your answer​

Answer 1

Answer:

D) a< b

Step-by-step explanation:

a= sqrt(21)-sqrt(20)= 4.58 - 4.47= 0.11

b= sqrt(18)- sqrt(17)= 4.24 - 4.12 = 0.12

so a<b

Answer 2

Answer:

D) a< b

Step-by-step explanation:

a= sqrt(21)-sqrt(20)= 4.58 - 4.47= 0.11

b= sqrt(18)- sqrt(17)= 4.24 - 4.12 = 0.12

so a<b

Topic :- This is your punishment for spamming in my question.

Differentiation

Given :-

$\sf{y=\dfrac{x+1}{(x+2)^2}}$

To Find :-

$\sf {\dfrac{dy}{dx}\:for\:given\:'y'.}$

Solution :-

$\sf{y=\dfrac{x+1}{(x+2)^2}}$

$\sf {\dfrac{dy}{dx}=\dfrac{d}{dx}\left[\dfrac{x+1}{(x+2)^2}\right]}$

$\sf{\dfrac{(x+2)^2\cdot\dfrac{d(x+1)}{dx}-(x+1)\cdot\dfrac{d(x+2)^2}{dx}}{\{(x+2)^2\}^2}}$

$\sf {\left( \because \dfrac{d}{dx}\left(\dfrac{f}{g}\right)=\dfrac{g\cdot\dfrac{df}{dx}-f\cdot\dfrac{dg}{dx}}{g^2}\right)}$

$\sf{\dfrac{(x+2)^2\cdot\left(\dfrac{dx}{dx}+\dfrac{d(1)}{dx}\right)-(x+1)\cdot\dfrac{d(x+2)^2}{dx}}{(x+2)^4}}$

$\sf {\left(\because \dfrac{d(f\pm g)}{dx}=\dfrac{df}{dx}\pm\dfrac{dg}{dx}\right)}$

$\sf{\dfrac{(x+2)^2\cdot\left(1+0\right)-(x+1)\cdot\dfrac{d(x+2)^2}{dx}}{(x+2)^4}}$

$\sf {\left(\because \dfrac{dx}{dx}=1 \right)}$

$\sf {\left(\because \dfrac{dk}{dx}=0,when\:k\:is\:constant. \right)}$

$\sf{\dfrac{(x+2)^2\cdot1-(x+1)\cdot2(x+2)^{2-1}\cdot\dfrac{d(x+2)}{dx}}{(x+2)^4}}$

$\sf {\left(\because \dfrac{d(x+k)^n}{dx}=n(x+k)^{n-1}\cdot \dfrac{d(x+k)}{dx} \right)}$

$\sf{\dfrac{(x+2)^2-2(x+1)(x+2)\cdot\left(\dfrac{dx}{dx}+\dfrac{d(2)}{dx}\right)}{(x+2)^4}}$

$\sf {\left(\because \dfrac{d(f\pm g)}{dx}=\dfrac{df}{dx}\pm\dfrac{dg}{dx}\right)}$

$\sf{\dfrac{(x+2)^2-2(x+1)(x+2)\cdot\left(1+0\right)}{(x+2)^4}}$

$\sf {\left(\because \dfrac{dx}{dx}=1 \right)}$

$\sf {\left(\because \dfrac{dk}{dx}=0,when\:k\:is\:constant. \right)}$

$\sf{\dfrac{(x+2)^2-2(x+1)(x+2)}{(x+2)^4}}$

$\sf{\dfrac{x^2+4x+4-2(x^2+3x+2)}{(x+2)^4}}$

$\sf{\dfrac{x^2+4x+4-2x^2-6x-4}{(x+2)^4}}$

$\sf{\dfrac{x^2-2x^2+4x-6x+4-4}{(x+2)^4}}$

$\sf{\dfrac{-x^2-2x}{(x+2)^4}}$

$\sf{\dfrac{-x(x+2)}{(x+2)^4}}$

$\sf{\dfrac{-x}{(x+2)^3}}$

Answer :-

$\underline{\boxed{\sf{\dfrac{d}{dx}\left[\dfrac{x+1}{(x+2)^2}\right]=\dfrac{-x}{(x+2)^3}}}}$

Topic :-

Indefinite Integration

To Solve :-

$\sf{I=\displaystyle \int e^{2x}\sin 3x\:dx}$

Concept Used :-

Integration by Parts

$\sf{\displaystyle \int uv \:dx = u\int v\:dx - \int \left(\dfrac{du}{dx}\int v\:dx \right)dx}$

where u and v are differentiable functions.

u and v are generally chosen as per the order of the letters in ILATE, where

I represents Inverse Trigonometric Function

L represents Logarithmic Function

A represents Algebraic Function

T represents Trigonometric Function and

E represents Exponential Function

Solution :-

$\sf{I=\displaystyle \int e^{2x}\sin 3x\:dx}$

$\sf{Taking\:u=\sin3x \:and\:v=e^{2x}\:as\:per\:ILATE.}$

Applying Integration by parts,

$\sf{I=\displaystyle\sin3x\int e^{2x}\:dx - \int \left(\dfrac{d(\sin 3x)}{dx}\int e^{2x}\:dx \right)dx}$

$\sf{I=\displaystyle\sin3x\cdot\dfrac{e^{2x}}{2} - \int \left(\dfrac{d(\sin 3x)}{dx}\cdot\dfrac{e^{2x}}{2}\right)dx}$

$\sf{\left(\because\displaystyle \int e^{ax}\:dx=\dfrac{e^{ax}}{a}+C \right)}$

$\sf{\displaystyle I=\sin3x\cdot\dfrac{e^{2x}}{2} - \int \left(\cos3x\cdot\dfrac{d(3x)}{dx}\cdot\dfrac{e^{2x}}{2}\right)dx}$

$\sf{\left(\because \dfrac{d(\sin t)}{dx}=\cos t\cdot\dfrac{dt}{dx} \right)}$

$\sf{I = \displaystyle\sin3x\cdot\dfrac{e^{2x}}{2} - \int \left(3\cos3x\cdot\dfrac{e^{2x}}{2}\right)dx}$

$\sf{\left(\because \dfrac{d(kx)}{dx}=k,when\:k\:is\:a\:constant. \right)}$

$\sf{I=\displaystyle\sin3x\cdot\dfrac{e^{2x}}{2} - \dfrac{3}{2}\int \cos3x\cdot e^{2x}\:dx}$

$\sf{\left(\because \displaystyle \int k\cdot f(x)\:dx=k\int f(x)\:dx,when\:k\:is\:a\:constant. \right)}$

$\sf{Take\:\displaystyle\int \cos3x\cdot e^{2x}\:dx=J}$

$\sf{I=\displaystyle \sin3x\cdot\dfrac{e^{2x}}{2} - \dfrac{3}{2}J}$

Solving J,

$\sf{J=\displaystyle \int \cos3x\cdot e^{2x}\:dx}$

$\sf{Taking\:u=\cos3x \:and\:v=e^{2x}\:as\:per\:ILATE.}$

Applying Integration by parts,

$\sf {J=\displaystyle\cos3x\int e^{2x}\:dx - \int \left(\dfrac{d(\cos 3x)}{dx}\int e^{2x}\:dx \right)dx}$

$\sf{J=\displaystyle \cos3x\cdot\dfrac{e^{2x}}{2} - \int \left(\dfrac{d(\cos 3x)}{dx}\cdot\dfrac{e^{2x}}{2}\right)dx}$

$\sf{J=\displaystyle \cos3x\cdot\dfrac{e^{2x}}{2} - \int \left(-\sin3x\cdot\dfrac{d(3x)}{dx}\cdot\dfrac{e^{2x}}{2}\right)dx}$

$\sf{\left(\because \dfrac{d(\cos t)}{dx}=-\sin t\cdot\dfrac{dt}{dx} \right)}$

$\sf{J=\displaystyle \cos3x\cdot\dfrac{e^{2x}}{2} +\int \left(3\sin3x\cdot\dfrac{e^{2x}}{2}\right)dx}$