Question

ꇙꄲ꒒꒦ꏂ ꓄ꁝ꒐ꇙ ꆰ꒤ꏂꇙ꓄꒐ꄲꋊ ꒯꒤꒯ꏂ ​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

GIVEN

$\displaystyle \: \: f(x) = \frac{2020x + 131}{3x - 2020}$

TO DETERMINE

The least value of

$\displaystyle \: f(f( x ))+ f(f( \frac{4}{x} ))$

CALCULATION

Here

$f(f(x))$

$= \displaystyle \: \: f( \frac{2020x + 131}{3x - 2020} )$

$= \displaystyle \: \:\frac{2020(\frac{2020x + 131}{3x - 2020}) + 131}{3(\frac{2020x + 131}{3x - 2020} )- 2020}$

$\displaystyle \: = \frac{ {2020}^{2} x + 131 \times 2020 + 131 \times 3x - 131 \times 2020}{3 \times 2020x + 3 \times 131 - 3 \times 2020x + {2020}^{2} }$

$\displaystyle \: = \frac{ {2020}^{2} x + 131 \times 3x }{ 3 \times 131 + {2020}^{2} }$

$\displaystyle \: = x \times \frac{ {2020}^{2} + 131 \times 3 }{ 3 \times 131 + {2020}^{2} }$

$= x$

So

$f(f(x)) = x$

Hence

$\displaystyle \: f(f( \frac{4}{x} )) = \frac{4}{x}$

Let

$y = \displaystyle \: f(f( x ))+ f(f( \frac{4}{x} ))$

$\implies \: y = \displaystyle \: x+ \frac{4}{x}$

Now Differentiating both sides with respect to x two times

$\displaystyle \: \frac{dy}{dx} = 1 - \frac{4}{ {x}^{2} }$

$\displaystyle \: \frac{ {d}^{2}y }{d {x}^{2} } = \frac{8}{ {x}^{3} }$

For extremum value of y

$\displaystyle \: \frac{dy}{dx} = 0$

$\implies \: \displaystyle \: 1 - \frac{4}{ {x}^{2} } = 0$

$\implies \: {x}^{2} = 4$

So

$x = \pm \: 2$

Since

$x > 0$

So

$x = 2$

Now at x = 2

$\displaystyle \: \frac{ {d}^{2}y }{d {x}^{2} } = \frac{8}{ {2}^{3} } = 1 > 0$

So y has least value at x = 2

And the required least value is

$\displaystyle \: 2 + \frac{4}{2} = 2 + 2 = 4$

Answer 2

Given,

$\longrightarrow f(x)=\dfrac{2020x+131}{3x-2020}$

Then,

$\longrightarrow f(f(x))=\dfrac{2020f(x)+131}{3f(x)-2020}$

$\longrightarrow f(f(x))=\dfrac{2020\left(\dfrac{2020x+131}{3x-2020}\right)+131}{3\left(\dfrac{2020x+131}{3x-2020}\right)-2020}$

$\longrightarrow f(f(x))=\dfrac{2020(2020x+131)+131(3x-2020)}{3(2020x+131)-2020(3x-2020)}$

$\longrightarrow f(f(x))=\dfrac{2020^2x+2020\times131+3\times131x-2020\times131}{3\times2020x+3\times131-3\times2020x+2020^2}$

$\longrightarrow f(f(x))=\dfrac{2020^2x+3\times131x}{3\times131+2020^2}$

$\longrightarrow f(f(x))=\dfrac{(2020^2+3\times131)x}{2020^2+3\times131}$

$\longrightarrow f(f(x))=x$

Therefore,

$\longrightarrow f(f(x))+f\left(f\left(\dfrac{4}{x}\right)\right)=x+\dfrac{4}{x}$

For least value,

$\longrightarrow \dfrac{d}{dx}\left[x+\dfrac{4}{x}\right]=0$

$\longrightarrow1-\dfrac{4}{x^2}=0$

$\longrightarrow x=\pm2\quad\quad\dots(1)$

And,

$\longrightarrow \dfrac{d^2}{dx^2}\left[x+\dfrac{4}{x}\right]>0$

$\longrightarrow \dfrac{8}{x^3}>0$

$\Longrightarrow x>0\quad\quad\dots(2)$

From (1) and (2) we get,

$\longrightarrow x=2$

Hence,

$\longrightarrow f(f(x))+f\left(f\left(\dfrac{4}{x}\right)\right)=2+\dfrac{4}{2}$

$\longrightarrow\underline{\underline{f(f(x))+f\left(f\left(\dfrac{4}{x}\right)\right)=4}}$

Hence 4 is the answer.