Math, asked by Anonymous, 3 months ago


Q14. If 2f(x) – \sf{3f\bigg(\dfrac{1}{x}\bigg)} = x², x ≠ 0, then find f(2).​

Answers

Answered by brainlychallenger99
0

Answer:

-7/4

Step-by-step explanation:

Given: 2f(x) - 3f (1/x) = x^2

Let x=2. Then, 2f(2) - 3f(1/2) = 4

Let x=1/2. Then, 2f(1/2) - 3f (1/2) = 1/4

From the second equation, f(1/2) = 1/8 + (3/2).f(2)

Substitute this into the first:

2f(2) - 3/8 - (9/2).f(2) = 4

==> (-5/2).f(2) = 35/8

==> f(2) = -7/4

Step-by-step explanation:

Answered by mathdude500
3

\large\underline{\sf{Given- }}

\rm :\longmapsto\:2f(x) - 3f\bigg(\dfrac{1}{x}\bigg) =  {x}^{2}, \:  \: x \ne \: 0

\large\underline{\sf{To\:Find - }}

\bf :\longmapsto\:f(2)

 \purple{\large\underline{\sf{Solution-}}}

Given that

\rm :\longmapsto\:2f(x) - 3f\bigg(\dfrac{1}{x}\bigg) =  {x}^{2}  -  - (1)

Now, Replace

 \red{\rm :\longmapsto\:x \: by \: \dfrac{1}{x} }

So, we get

\rm :\longmapsto\: 2f\bigg(\dfrac{1}{x}\bigg) - 3f(x) =  \dfrac{1}{ {x}^{2} }   -  - (2)

Now multiply equation (1) by 2 and equation (2) by 3, we get

\rm :\longmapsto\:4f(x) - 6f\bigg(\dfrac{1}{x}\bigg) =  2{x}^{2}  -  - (3)

and

\rm :\longmapsto\: 6f\bigg(\dfrac{1}{x}\bigg) - 9f(x) =  \dfrac{3}{ {x}^{2} }   -  - (4)

On adding equation (3) and (4), we get

\rm :\longmapsto\: - 5f(x) =  {2x}^{2}  + \dfrac{3}{ {x}^{2} }

Now, to find f(2), Substitute x = 2, we get

\rm :\longmapsto\: - 5f(2) =  {2(2)}^{2}  + \dfrac{3}{ {(2)}^{2} }

\rm :\longmapsto\: - 5f(2) =  2 \times 4  + \dfrac{3}{4}

\rm :\longmapsto\: - 5f(2) = 8  + \dfrac{3}{4}

\rm :\longmapsto\: - 5f(2) =  \dfrac{32 + 3}{4}

\rm :\longmapsto\: - 5f(2) =  \dfrac{35}{4}

\bf :\longmapsto\: f(2)  \: = \:  -  \:   \dfrac{7}{4}

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