Question
Download Solution PDF\({\left( {\sqrt {\rm{x}} + \frac{1}{{3{{\rm{x}}^2}}}} \right)^{10}}\) के विस्तार में (x से स्वतंत्र) स्थिर पद का मान क्या है?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFधारणा:
सामान्य पद: (x + y)n के विस्तार में सामान्य पद निम्न द्वारा दिया जाता है
- \({{\rm{T}}_{\left( {{\rm{r\;}} + {\rm{\;}}1} \right)}} = {\rm{\;}}{{\rm{\;}}^{\rm{n}}}{{\rm{C}}_{\rm{r}}} \times {{\rm{x}}^{{\rm{n}} - {\rm{r}}}} \times {{\rm{y}}^{\rm{r}}}\)
गणना:
दिया गया विस्तार \({\left( {\sqrt {\rm{x}} + \frac{1}{{3{{\rm{x}}^2}}}} \right)^{10}}\) है
सामान्य पद = \({{\rm{T}}_{\left( {{\rm{r\;}} + {\rm{\;}}1} \right)}} = {\rm{\;}}{{\rm{\;}}^{10}}{{\rm{C}}_{\rm{r}}} \times {{\rm{x}}^{\frac{{10{\rm{\;}} - {\rm{\;r}}}}{2}}} \times {\left( {\frac{1}{{3{{\rm{x}}^2}}}} \right)^{\rm{r}}} = {\rm{\;}}{{\rm{\;}}^{10}}{{\rm{C}}_{\rm{r}}} \times {3^{ - {\rm{r}}}} \times {{\rm{x}}^{\frac{{10{\rm{\;}} - 5{\rm{\;r}}}}{2}}}\)
x से स्वतंत्र पद के लिए x की घात शून्य होनी चाहिए
यानी \(\frac{{10{\rm{\;}} - 5{\rm{\;r}}}}{2} = 0\)
⇒ r = 2
∴ आवश्यक पद \({{\rm{T}}_{\left( {2{\rm{\;}} + {\rm{\;}}1} \right)}} = {{\rm{\;}}^{10}}{{\rm{C}}_2} \times {3^{ - 2}} = 5\)हैLast updated on May 30, 2025
->UPSC has released UPSC NDA 2 Notification on 28th May 2025 announcing the NDA 2 vacancies.
-> A total of 406 vacancies have been announced for NDA 2 Exam 2025.
->The NDA exam date 2025 has been announced for cycle 2. The written examination will be held on 14th September 2025.
-> Earlier, the UPSC NDA 1 Exam Result has been released on the official website.
-> 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.