If $\cot^{-1}(3x + 5) > \frac{\pi}{4}$, then find the range of the value of $x$.

$x < -\frac{4}{3}$

The correct answer is Option (3) → $x < -\frac{4}{3}$ ##

$\cot^{-1}(3x + 5) > \frac{\pi}{4}$

$⇒\cot^{-1}(3x+5) > \cot^{-1} 1$

$⇒3x + 5 < 1$

$\text{(As } \cot^{-1}x \text{ is strictly } \text{decreasing function in its domain)}$

$⇒3x < -4$

$⇒x < -\frac{4}{3}$

$∴x \in \left( -\infty, -\frac{4}{3} \right)$