Q198 / 269 · NDA 2025 (Phase 1) · Mathematics
If $k$ is a root of $x^2 - 4x + 1 = 0$, then what is $\tan^{-1}k + \tan^{-1}\dfrac{1}{k}$ equal to?
- A$-\dfrac{\pi}{2}$
- B$0$
- C$\dfrac{\pi}{4}$
- Correct answerD$\dfrac{\pi}{2}$
Solution
The given equation is $x^2 - 4x + 1 = 0$.
The roots of the equation are given by $x = \dfrac{4 \pm \sqrt{16 - 4}}{2} = 2 \pm \sqrt{3}$.
Since $\sqrt{3} \approx 1.732$, both roots $2 + \sqrt{3}$ and $2 - \sqrt{3}$ are positive. Therefore, $k > 0$.
Using the standard inverse trigonometric identity for $x > 0$:
$\tan^{-1}x + \tan^{-1}\left(\dfrac{1}{x}\right) = \dfrac{\pi}{2}$
Substituting $x = k$, we get:
$\tan^{-1}k + \tan^{-1}\left(\dfrac{1}{k}\right) = \dfrac{\pi}{2}$
Answer: $\dfrac{\pi}{2}$
The roots of the equation are given by $x = \dfrac{4 \pm \sqrt{16 - 4}}{2} = 2 \pm \sqrt{3}$.
Since $\sqrt{3} \approx 1.732$, both roots $2 + \sqrt{3}$ and $2 - \sqrt{3}$ are positive. Therefore, $k > 0$.
Using the standard inverse trigonometric identity for $x > 0$:
$\tan^{-1}x + \tan^{-1}\left(\dfrac{1}{x}\right) = \dfrac{\pi}{2}$
Substituting $x = k$, we get:
$\tan^{-1}k + \tan^{-1}\left(\dfrac{1}{k}\right) = \dfrac{\pi}{2}$
Answer: $\dfrac{\pi}{2}$
Hence, the correct answer is option: 'D'.