Question

pita putra ke liye kalam lata hai in Sanskrit
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Answer 1

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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.

$\Huge{\textbf{\textsf{{\purple{Ans}}{\pink{wer}}{\color{pink}{:}}}}}$

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 :-

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