Question
Download Solution PDFयदि \({\rm{f}}\left( {\rm{x}} \right) = \frac{{{{\rm{x}}^2} - 9}}{{{{\rm{x}}^2} - 2{\rm{x}} - 3}}\), x ≠ 3, x = 3 पर निरंतर है तो निम्न में से कौन सा सही है?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFधारणा:
- a2 - b2 = (a - b) (a + b)
गणना:
दिया हुआ है कि
\(\Rightarrow {\rm{f}}\left( {\rm{x}} \right) = \frac{{{{\rm{x}}^2} - 9}}{{{{\rm{x}}^2} - 2{\rm{x}} - 3}}\)
\(\Rightarrow {\rm{f}}\left( {\rm{x}} \right) = \frac{{\left( {{\rm{x}} - 3} \right)\left( {{\rm{x}} + 3} \right)}}{{{{\rm{x}}^2} - 3{\rm{x}} + {\rm{x}} - 3}}\) [∵ a2 - b2 = (a-b) (a+b)]
⇒ \({\rm{f}}\left( {\rm{x}} \right) = \frac{{\left( {{\rm{x}} - 3} \right)\left( {{\rm{x}} + 3} \right)}}{{{\rm{x}}\left( {{\rm{x}} - 3} \right) + 1\left( {{\rm{x}} - 3} \right)}}\)
\(\Rightarrow {\rm{f}}\left( {\rm{x}} \right) = \frac{{\left( {{\rm{x}} - 3} \right)\left( {{\rm{x}} + 3} \right)}}{{\left( {{\rm{x}} - 3} \right)\left( {{\rm{x}} + 1} \right)}}\)
⇒ \({\rm{f}}\left( {\rm{x}} \right) = \frac{{\left( {{\rm{x}} + 3} \right)}}{{{\rm{x}} + 1}}\)
दिया हुआ f(x), x = 3 पर निरंतर है
\(\therefore {\rm{f}}\left( 3 \right) = \mathop {\lim }\limits_{{\rm{x}} \to 3} {\rm{f}}\left( {\rm{x}} \right) = \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to 3} \frac{{\left( {{\rm{x}} + 3} \right)}}{{{\rm{x}} + 1}} = \frac{{\left( {3 + 3} \right)}}{{3 + 1}} = \frac{6}{4} = 1.5\)Last updated on Jun 18, 2025
->UPSC has extended the UPSC NDA 2 Registration Date till 20th June 2025.
-> A total of 406 vacancies have been announced for NDA 2 Exam 2025.
->The NDA exam date 2025 has been announced. The written examination will be held on 14th September 2025.
-> The selection process for the NDA exam includes a Written Exam and SSB Interview.
-> Candidates who get successful selection under UPSC NDA will get a salary range between Rs. 15,600 to Rs. 39,100.
-> Candidates must go through the NDA previous year question paper. Attempting the NDA mock test is also essential.