JEE Main 201911 JanuaryEvening Shift

Exam: JEE Main
Year: 2019
Date: 11 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Inverse Trigonometric Functions

JEE Main 2019 · 11 January · Evening Shift · Mathematics · Inverse Trigonometric Functions

All x satisfying the inequality ( ^ -1 x)^ 2 -7( ^ -1 x)+10> 0 ,...

All

x

satisfying the inequality \left(\cot ^{-1} x\right)^{2}-7\left(\cot ^{-1} x\right)+10> 0 , lie in the interval : All x satisfying the inequality ( ^ -1 x)^ 2 -7( ^ -1 x)+10> 0 , lie in the interval :

All x satisfying the inequality ( ^ -1 x)^ 2 -7( ^ -1 x)+10> 0 , lie in the interval :

Options

A $(-\infty, \cot 5) \cup(\cot 4, \cot 2)$
B $(\cot 2, \infty)$
C $(-\infty, \cot 5) \cup(\cot 2, \infty)$
D $(\cot 5, \cot 4)$

Correct answer

B. $(\cot 2, \infty)$

Step-by-step solution

\left(\cot ^{-1} x\right)^{2}-7\left(\cot ^{-1} x\right)+10>0 \left(\cot ^{-1} x-5\right)\left(\cot ^{-1}-2\right)>0 \cot ^{-1} x \in(-\infty, 2) \cup(5, \infty)

But

\cot ^{-1} x

lies in

(0, \pi)

Now, from equation (1)

\cot ^{-1} x \in(0,2)

Now, it is clear from the graph

x \in(\cot 2, \infty)

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