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If  $2f(x)-3f(\frac{1}{x})=x^2 \: (x \neq0)$,  then $f(2)$ is

  1. $\frac{2}{3}$
  2. $ – \frac{3}{2}$
  3. $ – \frac{7}{4}$
  4. $\frac{5}{4}$
in Calculus
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Answer: $\mathbf C$

Explanation:

$2f(x) - 3f(\frac{1}{x}) = x^2\tag{1}\;\;\text{[Given]}$ 

Substitute, $x = \frac{1}{x}$, we get:

$2f(\frac{1}{x}) - 3f(x) = \frac{1}{x^2}\tag{2}$

Multiplying Equation $(1)$ by $3$, we get:

$6f(x) - 9f(\frac{1}{x}) = 3x^2\tag{3}$

Multiplying Equation $(2)$ by $2$, we get:

$4f(\frac{1}{x}) - 6f(x) = \frac{2}{x^2}\implies -6f(x) + 4f(\frac{1}{x}) = \frac{2}{x^2}\tag{4}$

Adding Equations $(3)$ and $(4)$, we get:

$f(\frac{1}{x}) = -\frac{1}{5}\left(3x^2 + \frac{2}{x^2}\right)$

Now, on putting $x = \frac{1}{2}$ in above equation, we get:

$x = -\frac{7}{4}$


$\therefore \mathbf C$ is the correct option.


edited by

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