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JEE Mains · Maths · STD 12 - 2. inverse trigonometric function

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

  1. A \(\left( { - \infty ,\cot \,5} \right) \cup \left( {\cot \,4,\cot \,2} \right)\)
  2. B \(\left( {\cot \,2,\infty } \right)\)
  3. C \(\left( { - \infty ,\cot \,5} \right) \cup \left( {\cot \,2,\infty } \right)\)
  4. D \(\left( {\cot \,5,\cot \,4} \right)\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\left( {\cot \,2,\infty } \right)\)

Step-by-step Solution

Detailed explanation

\(\left( {{{\cot }^{ - 1}}\left( x \right) - \left( 5 \right)} \right)\left( {{{\cot }^{ - 1}}\left( x \right) - 2} \right) > 0\) \( \Rightarrow {\cot ^{ - 1}}\left( x \right) \in \left( { - \infty ,2} \right) \cup \left( {5,\infty } \right)\) Put…