Trigonometry & Inverse Trigonometry Question - None

Question

If the equation $5 \arctan(x^2 + x + k) + 3 \text{arccot}(x^2 + x + k) = 2\pi$ has two distinct solutions, then the range of $k$ is

Accepted Answer

Step 1: We use the property $\arctan a + ext{arccot} a = \frac{\pi}{2}$ for all $a \in \mathbb{R}$.

Step 2: Let $u = x^2 + x + k$. Then:

$5 \arctan u + 3 ext{arccot} u = 2\pi$

Step 3: Write $ ext{arccot} u = \frac{\pi}{2} - \arctan u$, so:

$5 \arctan u + 3\left(\frac{\pi}{2} - \arctan u ight) = 2\pi$

$2 \arctan u + \frac{3\pi}{2} = 2\pi$

$\arctan u = \frac{\pi}{4}$

Step 4: Thus $u = an\frac{\pi}{4} = 1$, so $x^2 + x + k = 1$.

Step 5: Rearranging: $x^2 + x + (k-1) = 0$

Step 6: For two distinct real solutions, we need discriminant $\Delta > 0$:

$1 - 4(k-1) > 0$

$5 - 4k > 0$

$k < \frac{5}{4}$

∴ Answer is (B).

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