State the difference between "physical change" and "chemical change" and provide examples for each.

Chemistry
State the difference between "physical change" and "chemical change" and provide examples for each.

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Answer

\dfrac{3\pi}{16}

(1) (ক)

Step 1: sin⁡4x=(sin⁡2x)2=(1−cos⁡2x2)2\sin^4 x = (\sin^2 x)^2 = \left( \frac{1 - \cos 2x}{2} \right)^2

sin⁡4x=(1−cos⁡2x2)2=1−2cos⁡2x+cos⁡22x4\sin^4 x = \left( \frac{1 - \cos 2x}{2} \right)^2 = \frac{1 - 2\cos 2x + \cos^2 2x}{4}

Step 2: cos⁡22x=1+cos⁡4x2\cos^2 2x = \frac{1 + \cos 4x}{2}

sin⁡4x=1−2cos⁡2x+1+cos⁡4x24=1−2cos⁡2x+12+cos⁡4x24=32−2cos⁡2x+cos⁡4x24\sin^4 x = \frac{1 - 2\cos 2x + \frac{1 + \cos 4x}{2}}{4} = \frac{1 - 2\cos 2x + \frac{1}{2} + \frac{\cos 4x}{2}}{4} = \frac{\frac{3}{2} - 2\cos 2x + \frac{\cos 4x}{2}}{4} =38−12cos⁡2x+18cos⁡4x= \frac{3}{8} - \frac{1}{2} \cos 2x + \frac{1}{8} \cos 4x

Step 3: Integrate term by term from 00 to π2\frac{\pi}{2}

∫0π/2sin⁡4x dx=∫0π/2(38−12cos⁡2x+18cos⁡4x)dx\int_0^{\pi/2} \sin^4 x \, dx = \int_0^{\pi/2} \left( \frac{3}{8} - \frac{1}{2} \cos 2x + \frac{1}{8} \cos 4x \right) dx = \left[ \frac{3}{8} x - \frac{1}{4} \sin 2x + \frac{1}{32} \sin 4x \right]_0^{\pi/2} = \frac{3}{8} \cdot \frac{\pi}{2} - 0 = \frac{3\pi}{16} ** \dfrac{3\pi{16}} (1) (খ) Step 1: $\cos^5 x = \cos^4 x \cos x = (1 - \sin^2 x)^2 \cos x$ Let $u = \sin x$, $du = \cos x \, dx$

\int \cos^5 x , dx = \int (1 - u^2)^2 , du

Step2:Expand Step 2: Expand

(1 - u^2)^2 = 1 - 2u^2 + u^4

\int (1 - 2u^2 + u^4) , du = u - \frac{2}{3} u^3 + \frac{1}{5} u^5 + C

Step3:Substituteback Step 3: Substitute back

\sin x - \frac{2}{3} \sin^3 x + \frac{1}{5} \sin^5 x + C **

\sin x - \dfrac{2{3} \sin^{3} x + \dfrac{1}{5} \sin^{5} x + C}

(1) (গ)

Step 1: Complete the square in denominator

x2+4x+3=(x+2)2−4+3=(x+2)2−1x^2 + 4x + 3 = (x+2)^2 - 4 + 3 = (x+2)^2 - 1 ∫dx(x+2)2−1\int \frac{dx}{\sqrt{(x+2)^2 - 1}}

Step 2: Let u=x+2u = x+2, du=dxdu = dx

\int \frac{du}{\sqrt{u^2 - 1}} = \cosh^{-1} u + C \quad or \quad \ln |u + \sqrt{u^2 - 1}| + C ** \ln \left| x + 2 + \sqrt{x^{2 + 4x + 3} \right| + C} (2) (ক) (i) Step 1: $y = \frac{\sin(x + a)}{\cos(x - a)}$ Use quotient rule: $\frac{dy}{dx} = \frac{ u' v - u v' }{v^2}$ where $u = \sin(x+a)$, $v = \cos(x-a)$ $u' = \cos(x+a)$, $v' = \sin(x-a)$ Step 2: Numerator: $\cos(x+a) \cos(x-a) - \sin(x+a) (-\sin(x-a))$

= \cos(x+a)\cos(x-a) + \sin(x+a)\sin(x-a) = \cos( (x+a) - (x-a) ) = \cos 2a

Step 3: Denominator: $\cos^2 (x-a)$

\frac{dy}{dx} = \frac{\cos 2a}{\cos^2 (x - a)} **

\dfrac{\cos 2a{\cos^{2}(x - a)}}

(2) (ক) (ii)

Step 1: y=xsin⁡xy = x^{\sin x}

Take ln⁡\ln: ln⁡y=sin⁡xln⁡x\ln y = \sin x \ln x

Step 2: Differentiate: 1yy′=cos⁡xln⁡x+sin⁡x⋅1x\frac{1}{y} y' = \cos x \ln x + \sin x \cdot \frac{1}{x}

y' = y \left( \cos x \ln x + \frac{\sin x}{x} \right) = x^{\sin x} \left( \cos x \ln x + \frac{\sin x}{x} \right) ** x^{\sin x \left( \cos x \ln x + \dfrac{\sin x}{x} \right)} (2) (ক) (iii) Step 1: $y = \log (\sec x + \tan x)$ $\frac{dy}{dx} = \frac{1}{\sec x + \tan x} \cdot \frac{d}{dx} (\sec x + \tan x)$ Step 2: Derivative of $\sec x + \tan x = \sec x \tan x + \sec^2 x = \sec x (\tan x + \sec x)$ Step 3: So $\frac{dy}{dx} = \frac{ \sec x (\sec x + \tan x) }{ \sec x + \tan x } = \sec x$ ** \sec x (3) (ক) Step 1: $A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}$ $|A| = 2 \cdot 4 - 3 \cdot 1 = 8 - 3 = 5$ 5 (3) (খ) Assuming inverse of same A. Step 1: $A^{-1} = \frac{1}{|A|} \begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix} = \frac{1}{5} \begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix}$ \dfrac{1{5} \begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix}} (3) (গ) Assuming product of two matrices, say P = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, Q = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix} Step 1: PQ_{11} = 1\cdot5 + 2\cdot7 = 5+14=19 PQ_{12} = 1\cdot6 + 2\cdot8 = 6+16=22 PQ_{21} = 3\cdot5 + 4\cdot7 = 15+28=43 PQ_{22} = 3\cdot6 + 4\cdot8 = 18+32=50 \begin{bmatrix 19 & 22 \\ 43 & 50 \end{bmatrix}}
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(1) (ক) Step 1: ^4 x = (^2 x)^2 = ( (1 - 2x)/(2) )^2 ^4 x = ( (1 - 2x)/(2) )^2 = (1 - 2 2x + ^2 2x)/(4) Step 2: ^2 2x = (1 + 4x)/(2) ^4 x = (1 - 2 2x + 1 + 4x)/(2)4 = (1 - 2 2x + 1)/(2) + ( 4x)/(2)4 = (3)/(2) - 2 2x + ( 4x)/(2)4 = (3)/(8) - (1)/(2) 2x…

State the difference between "physical change" and "chemical change" and provide examples for each.
Chemistry

This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.

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(1) (ক) Step 1: ^4 x = (^2 x)^2 = ( (1 - 2x)/(2) )^2 ^4 x = ( (1 - 2x)/(2) )^2 = (1 - 2 2x + ^2 2x)/(4) Step 2: ^2 2x = (1 + 4x)/(2) ^4 x = (1 - 2 2x + 1 + 4x)/(2)4 = (1 - 2 2x + 1)/(2) + ( 4x)/(2)4 = (3)/(2) - 2 2x + ( 4x)/(2)4 = (3)/(8) - (1)/(2) 2x + (1)/(8) 4x Step 3: Integrate term by term from 0 to ()/(2) _0^/2 ^4 x \, dx = _0^/2 ( (3)/(8) - (1)/(2) 2x + (1)/(8) 4x ) dx = [ (3)/(8) x - (1)/(4) 2x + (1)/(32) 4x ]_0^/2 = (3)/(8) · ()/(2) - 0 = (3)/(16) ** (3)/(16) (1) (খ) Step 1: ^5 x = ^4 x x = (1 - ^2 x)^2 x Let u = x, du = x \, dx ^5 x \, dx = (1 - u^2)^2 \, du Step 2: Expand (1 - u^2)^2 = 1 - 2u^2 + u^4 (1 - 2u^2 + u^4) \, du = u - (2)/(3) u^3 + (1)/(5) u^5 + C Step 3: Substitute back x - (2)/(3) ^3 x + (1)/(5) ^5 x + C ** x - (2)/(3) ^3 x + (1)/(5) ^5 x + C (1) (গ) Step 1: Complete the square in denominator x^2 + 4x + 3 = (x+2)^2 - 4 + 3 = (x+2)^2 - 1 (dx)/(sqrt((x+2)^2 - 1)) Step 2: Let u = x+2, du = dx (du)/(sqrt(u^2 - 1)) = ^-1 u + C or |u + sqrt(u^2 - 1)| + C ** | x + 2 + sqrt(x^2) + 4x + 3 | + C (2) (ক) (i) Step 1: y = ((x + a))/((x - a)) Use quotient rule: (dy)/(dx) = ( u' v - u v' )/(v^2) where u = (x+a), v = (x-a) u' = (x+a), v' = (x-a) Step 2: Numerator: (x+a) (x-a) - (x+a) (-(x-a)) = (x+a)(x-a) + (x+a)(x-a) = ( (x+a) - (x-a) ) = 2a Step 3: Denominator: ^2 (x-a) (dy)/(dx) = ( 2a)/(^2 (x - a)) ** ( 2a)/(^2)(x - a) (2) (ক) (ii) Step 1: y = x^ x Take : y = x x Step 2: Differentiate: (1)/(y) y' = x x + x · (1)/(x) y' = y ( x x + ( x)/(x) ) = x^ x ( x x + ( x)/(x) ) ** x^ x ( x x + ( x)/(x) ) (2) (ক) (iii) Step 1: y = ( x + x) (dy)/(dx) = (1)/( x + x) · (d)/(dx) ( x + x) Step 2: Derivative of x + x = x x + ^2 x = x ( x + x) Step 3: So (dy)/(dx) = ( x ( x + x) )/( x + x ) = x ** x (3) (ক) Step 1: A = 2 & 3 \\ 1 & 4 |A| = 2 · 4 - 3 · 1 = 8 - 3 = 5 5 (3) (খ) Assuming inverse of same A. Step 1: A^-1 = (1)/(|A|) 4 & -3 \\ -1 & 2 = (1)/(5) 4 & -3 \\ -1 & 2 (1)/(5) 4 & -3 \\ -1 & 2 (3) (গ) Assuming product of two matrices, say P = 1 & 2 \\ 3 & 4 , Q = 5 & 6 \\ 7 & 8 Step 1: PQ_11 = 1·5 + 2·7 = 5+14=19 PQ_12 = 1·6 + 2·8 = 6+16=22 PQ_21 = 3·5 + 4·7 = 15+28=43 PQ_22 = 3·6 + 4·8 = 18+32=50 19 & 22 \\ 43 & 50