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If (f:\mathbb{R}\to[-\frac{\pi}{2},\frac{\pi}{2}]), (f(x)=\tan^{-1}x), why is (f) not onto?यदि (f:\mathbb{R}\to[-\frac{\pi}{2},\frac{\pi}{2}]), (f(x)=\tan^{-1}x) है, तो (f) आच्छादी क्यों नहीं है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: Because (-\frac{\pi}{2}) and (\frac{\pi}{2}) are not obtainedक्योंकि (-\frac{\pi}{2}) और (\frac{\pi}{2}) नहीं मिलते
Step 1: The range of (\tan^{-1}x) is the open interval ((-\frac{\pi}{2},\frac{\pi}{2})). Step 2: The codomain includes both endpoint values, but the function never attains them. Step 3: The difference between open and closed endpoints is very important in onto questions.चरण 1: (\tan^{-1}x) का परास खुला अंतराल ((-\frac{\pi}{2},\frac{\pi}{2})) है। चरण 2: सहप्रांत में दोनों अंतिम मान शामिल हैं, लेकिन फलन उन्हें कभी प्राप्त नहीं करता। चरण 3: खुले और बंद सिरों का अंतर आच्छादीपन में बहुत महत्वपूर्ण है।
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