If $\cot^{-1}(3x + 5) > \frac{\pi}{4}$, then find the range of the value of $x$.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $x < -\frac{4}{3}$ ##
$\cot^{-1}(3x + 5) > \frac{\pi}{4}$
$⇒\cot^{-1}(3x+5) > \cot^{-1} 1$
$⇒3x + 5 < 1$
$\text{(As } \cot^{-1}x \text{ is strictly } \text{decreasing function in its domain)}$
$⇒3x < -4$
$⇒x < -\frac{4}{3}$
$∴x \in \left( -\infty, -\frac{4}{3} \right)$