Question

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

Answer 1

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:

Answer 2

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

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

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

$\bf :\longmapsto\:f(2)$

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

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

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

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

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

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

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

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