Q103 / 180 · COMEDK 2023 (Morning Shift) · Mathematics
If $2 f\left(x^2\right)+3 f\left(\dfrac{1}{x^2}\right)=x^2-1, \forall x \in R-\{0\}$, then $f\left(x^8\right)$ is equal to
- A$\dfrac{\left(1+x^8\right)\left(2 x^8-3\right)}{5 x^8}$
- B$\dfrac{\left(1-x^8\right)\left(2 x^8-3\right)}{5 x^8}$
- Correct answerC$\dfrac{\left(1-x^8\right)\left(2 x^8+3\right)}{5 x^8}$
- DNone of these
Solution
Given the equation $2 f(x^2) + 3 f\left(\dfrac{1}{x^2}\right) = x^2 - 1$.
Let $t = x^2$. Then the equation becomes $2 f(t) + 3 f\left(\dfrac{1}{t}\right) = t - 1$ for all $t > 0$.
Replacing $t$ with $\dfrac{1}{t}$, we get $2 f\left(\dfrac{1}{t}\right) + 3 f(t) = \dfrac{1}{t} - 1$.
We have a system of two linear equations in $f(t)$ and $f\left(\dfrac{1}{t}\right)$:
1) $2 f(t) + 3 f\left(\dfrac{1}{t}\right) = t - 1$
2) $3 f(t) + 2 f\left(\dfrac{1}{t}\right) = \dfrac{1}{t} - 1$
Multiply equation (1) by 2 and equation (2) by 3:
$4 f(t) + 6 f\left(\dfrac{1}{t}\right) = 2t - 2$
$9 f(t) + 6 f\left(\dfrac{1}{t}\right) = \dfrac{3}{t} - 3$
Subtracting the first from the second:
$5 f(t) = \dfrac{3}{t} - 3 - (2t - 2) = \dfrac{3}{t} - 2t - 1 = \dfrac{3 - 2t^2 - t}{t} = -\dfrac{2t^2 + t - 3}{t} = -\dfrac{(2t + 3)(t - 1)}{t} = \dfrac{(1 - t)(2t + 3)}{t}$.
Therefore, $f(t) = \dfrac{(1 - t)(2t + 3)}{5t}$.
Substituting $t = x^8$ into the expression for $f(t)$:
$f(x^8) = \dfrac{(1 - x^8)(2x^8 + 3)}{5x^8}$.
Answer: $\dfrac{\left(1-x^8\right)\left(2x^8+3\right)}{5x^8}$
Hence, the correct answer is option: 'C'.