Question

1. What is sustainable development ?

2. The three pillars of sustainable development are environmental, social and __________

3. The four dimensions of sustainable development are : ______ , _______ , ________ and ________ .

4. How many SDGs are included in Agenda for sustainable development 2030 by General Assembly?

5. List the 17 SDGs included in agenda 2030 for sustainable development?

6. What are the measures being taken for implementing the SDGs in India ?​

Answer 1

Explanation:

The Sustainable Development Goals (SDGs), also known as the Global Goals, were adopted by all United Nations Member States in 2015 as a universal call to action to end poverty, protect the planet and ensure that all people enjoy peace and prosperity by 2030.

Answer 2

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Topic :-

Differentiation

To Differentiate :-

$f(x)=\cot x \cdot \ln \sec x$

Solution :-

$f(x)=\cot x \cdot \ln \sec x$

$\dfrac{d(f(x))}{dx}=\dfrac{d(\cot x \cdot \ln \sec x)}{dx}$

$\dfrac{d(f(x))}{dx}=\ln \sec x\cdot\dfrac{d(\cot x )}{dx}+\cot x\cdot\dfrac{d(\ln\sec x )}{dx}$

$\left(\because \dfrac{d(fg)}{dx}=g\cdot \dfrac{d(f)}{dx}+f\cdot\dfrac{d(g)}{dx} \right)$

$\dfrac{d(f(x))}{dx}=\ln \sec x\cdot(-\csc^2x)+\cot x\cdot\dfrac{d(\ln\sec x )}{dx}$

$\left(\because \dfrac{d(\cot x)}{dx}=-\csc^2x \right)$

$\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+\cot x\cdot\dfrac{1}{\sec x}\cdot\dfrac{d(\sec x )}{dx}$

$\left(\because \dfrac{d(\ln t)}{dx}=\dfrac{1}{t}\cdot\dfrac{dt}{dx} \right)$

$\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+\cot x\cdot\dfrac{1}{\sec x}\cdot \sec x\cdot \tan x$

$\left(\because \dfrac{d(\sec x)}{dx}=\sec x\cdot \tan x \right)$

$\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+(\cot x\cdot\tan x)\cdot\left (\dfrac{1}{\cancel{\sec x}}\cdot \cancel{\sec x}\right)$

$\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+1$

$(\because \cot x \cdot \tan x = 1)$

Answer :-

$\underline{\boxed{\dfrac{d(f(x))}{dx}=1-\csc^2x \cdot\ln \sec x}}$

Note : csc x = cosec x

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Correct Expression :-

$\mathtt{\dfrac{x^3+12x}{6x^2+8}=\dfrac{y^3+27y}{9y^2+27}}$

To Find :-

$\mathtt{x:y\:\:by\:using\:Componendo\:and\:Dividendo.}$

Concept Used :-

$\mathtt{If\:\dfrac{a}{b}=\dfrac{c}{d},then\:using\:Componendo\:and\:Dividendo}$

$\mathtt{\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}}$

Solution :-

$\mathtt{\dfrac{x^3+12x}{6x^2+8}=\dfrac{y^3+27y}{9y^2+27}}$

Using Componendo and Dividendo,

$\mathtt{\dfrac{x^3+12x+(6x^2+8)}{x^3+12x-(6x^2+8)}=\dfrac{y^3+27y+(9y^2+27)}{y^3+27y-(9y^2+27)}}$

Opening brackets,

$\mathtt{\dfrac{x^3+12x+6x^2+8}{x^3+12x-6x^2-8}=\dfrac{y^3+27y+9y^2+27}{y^3+27y-9y^2-27}}$

Rewriting it,

$\mathtt{\dfrac{x^3+3(2)^2(x)+3(2)(x^2)+2^3}{x^3+3(2^2)x-3(2)(x^2)-2^3}=\dfrac{y^3+3(3^2)(y)+3(3)(y^2)+3^3}{y^3+3(3^2)(y)-3(3)(y^2)-3^3}}$

We know that,

$\mathtt{(a+b)^3=a^3+3ab^2+3a^2b+b^3}$

$\mathtt{(a-b)^3=a^3+3ab^2-3a^2b-b^3}$

Using Identities,

$\mathtt{\dfrac{(x+2)^3}{(x-2)^3}=\dfrac{(y+3)^3}{(y-3)^3}}$

$\mathtt{\left(\dfrac{x+2}{x-2}\right)^3=\left(\dfrac{y+3}{y-3}\right)^3}$

Taking Cube Root on both sides,

$\mathtt{\sqrt[3]{\left(\dfrac{x+2}{x-2}\right)^3}=\sqrt[3]{\left(\dfrac{y+3}{y-3}\right)^3}}$

$\mathtt{\dfrac{x+2}{x-2}=\dfrac{y+3}{y-3}}$

Using Componendo and Dividendo again,

$\mathtt{\dfrac{x}{2}=\dfrac{y}{3}}$

We can write it as,

$\mathtt{\dfrac{x}{y}=\dfrac{2}{3}}$

Answer :-

Hence, x : y is equivalent to 2 : 3.

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Question :-

Solve for 'x'.

$\dfrac{x-1}{x-2}+\dfrac{x-3}{x-4}=3\dfrac{1}{3};\:x\neq2,4$

Solution :-

$\dfrac{x-1}{x-2}+\dfrac{x-3}{x-4}=3\dfrac{1}{3}$

$\dfrac{x-1}{x-2}+\dfrac{x-3}{x-4}=\dfrac{(3\times3)+1}{3}$

$\dfrac{(x-1)(x-4)+(x-3)(x-2)}{(x-2)(x-4)}=\dfrac{9+1}{3}$

$\dfrac{x^2-x-4x+4+x^2-3x-2x+6}{x^2-2x-4x+8}=\dfrac{10}{3}$

$\dfrac{x^2+x^2-x-4x-3x-2x+6+4}{x^2-2x-4x+8}=\dfrac{10}{3}$

$\dfrac{2x^2-10x+10}{x^2-6x+8}=\dfrac{10}{3}$

Cross Multiplying,

$3(2x^2-10x+10)=10(x^2-6x+8)$

$6x^2-30x+30=10x^2-60x+80$

$10x^2-6x^2-60x+30x+80-30=0$

$4x^2-30x+50=0$

$2(2x^2-15x+25)=0$

$2x^2-15x+25=0$

Factorising it using Splitting method,

$2x^2-10x-5x+25=0$

$2x(x-5)-5(x-5)=0$

$(x-5)(2x-5)=0$

So,

$x-5=0$

$x=5$

and

$2x-5=0$

$2x=5$

$x=\dfrac{5}{2}$

Answer :-

$So,value\:of\:\bold{x=5, \dfrac{5}{2}}.$