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
- A \(\left( { - \infty ,\cot \,5} \right) \cup \left( {\cot \,4,\cot \,2} \right)\)
- B \(\left( {\cot \,2,\infty } \right)\)
- C \(\left( { - \infty ,\cot \,5} \right) \cup \left( {\cot \,2,\infty } \right)\)
- D \(\left( {\cot \,5,\cot \,4} \right)\)
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…
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