JEE PYQ: Inverse Trigonometric Functions Question 20
Question 20 - 2019 (11 Jan Shift 2)
All $x$ satisfying the inequality $(\cot^{-1} x)^2 - 7(\cot^{-1} x) + 10 > 0$ lie in the interval:
(1) $(-\infty, \cot 5) \cup (\cot 4, \cot 2)$
(2) $(\cot 2, \infty)$
(3) $(-\infty, \cot 5) \cup (\cot 2, \infty)$
(4) $(\cot 5, \cot 4)$
Show Answer
Answer: (2)
Solution
Let $t = \cot^{-1} x$. Then $t^2 - 7t + 10 > 0 \Rightarrow (t-5)(t-2) > 0 \Rightarrow t \in (-\infty, 2) \cup (5, \infty)$. But $\cot^{-1} x \in (0, \pi)$, so $t \in (0, 2)$. Since $\cot^{-1} x < 2 \Rightarrow x > \cot 2$. Hence $x \in (\cot 2, \infty)$.
BETA
