Question

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Pls give the correct answer for the above question and explain it with details (step by step)

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Answer 1

Solution :-

Given :-

2f(x²) + 3f(1/x²) = x² - 1 ....(i)

x belongs to R - {0}

Now we will find out the f(x²) first and then we will replace x² from x⁴ from the equation.

Now :-

By replacing x² as 1/x² and 1/x² as x² in the function

$\rightarrow 2f\left(\dfrac{1}{x^2}\right) + 3f(x^2) = \dfrac{1}{x^2} - 1$ .....(ii)

Now we will eliminate the 1/x² from the equation. by 2(i) - 6(ii)

$4f\left(x^2\right) + 6f\left(\dfrac{1}{x^2}\right) = 2x^2 - 2$

$- 6f\left(\dfrac{1}{x^2}\right) - 3f(x^2) = -3\dfrac{1}{x^2} + 3$

$\rightarrow -5f(x^2) = 2x^2 - 2 + 3 - 3\dfrac{1}{x^2}$

$\rightarrow -5f(x^2) = 2x^2 + 1 - 3\dfrac{1}{x^2}$

$\rightarrow -5f(x^2) = \dfrac{2x^4 + x^2 - 3}{x^2}$

$\rightarrow f(x^2) = \dfrac{2x^4 + x^2 - 3}{-5x^2}$

$\rightarrow f(x^2) = \dfrac{3 - 2x^4 - x^2 }{5x^2}$

$\rightarrow f(x^2) = \dfrac{ 3 - 3x^2 + 2x^2 - 2x^4 }{5x^2}$

$\rightarrow f(x^2) = \dfrac{ 3(1 - x^2) + 2x^2(1 - x^4)}{5x^2}$

$\rightarrow f(x^2) = \dfrac{(1-x^2)(2x^2 + 3)}{5x^2}$

Now we will replace x² with x⁴ in the function

$\rightarrow f(x^4) = \dfrac{(1-x^4)(2x^4 + 3)}{5x^4}$

So Answer is option (a)

Answer 2

$\huge\mathfrak\red{Answer}$

$2f( {x}^{2} ) + 3f( \frac{1}{ {x}^{2} } ) = {x}^{2} - 1 \\ \\ put \: x = \frac{1}{x} \\ \\ 2f( \frac{1}{ {x}^{2} } ) + 3f( {x}^{2} ) = \: \frac{1}{ {x}^{2} } - 1 \\ \\ now \: you \: have \: two \: equations \\ and \: all \: the \: 95 \: percentile \: girls \ \\ know \: how \: to \: solve \\ \\ let \: f( \frac{1}{ {x}^{2} } ) = m \\ f( {x}^{2} ) = n \\ \\ two \: equations \\ \\ 2n + 3m = {x}^{2} - 1 \\ 3n + 2m = \frac{1}{ {x}^{2} } - 1 \\ \\ eleminate \: m \\ \\ MULTIPLY \: by \: 2 \: and \: by \: 3 \: in \\ eq(1) \: and \: eq(2) \: and \: subtract \: both \: \\ the \: equation \\ \\ 4n - 9n = 2 {x}^{2} - 2 - \frac{3}{ {x}^{2} } - 3 \\ - 5n = 2 {x}^{2} - \frac{3}{ {x}^{2} } - 5 \\ \\ 5n = - 2 {x}^{2} + \frac{3}{ {x}^{2} } + 5 \\ \\ 5n = \frac{ - 2 {x}^{4} + 3 + 5 {x}^{2} }{ {x}^{2} } \\ \\ n = \frac{ - 2 {x}^{4} + 5 {x}^{2} + 3}{5 {x}^{2} } \\ \\ n = \frac{ - 2 {x}^{4} + 3 {x}^{2} - 2 {x}^{2} + 3 }{5 {x}^{2} } \\ \\ n = \frac{ - {x}^{2}(2 {x}^{2} - 3) - 1(2 {x}^{2} - 3)}{5 {x}^{2} } \\ \\ n = \frac{ - (2 {x}^{2} - 3)( {x}^{2} - 1) }{5 {x}^{2} } \\ \\ put \: the \: original \: value \: of \: n \\ \\ f( {x}^{2} ) = \frac{ - (2 {x}^{2} - 3)( {x}^{2} - 1) }{5 {x}^{2} } \\ \\ put \: x = {x}^{2} \\ \\ f( {x}^{4} ) = \frac{(1 - {x}^{4} )(2 {x}^{4} - 3)}{5 {x}^{4} }$

PRECAUTIONS

• YOU CAN USE THIS METHOD ONLY AND ONLY WHEN THE GIVEN EQUATION HAVE SOME COFFECIENT EXCEPT 1
• IF THE EQUATION HAVE CONJUGATED PLUS AND MINUS SIGN.