Question:
In the expansion of \(\left(\sqrt{x} + \frac{1}{3x^2}\right)^{10}\), the value of the constant term (independent of \(x\)) is:
\(\left(\sqrt{x} + \frac{1}{3x^2}\right)^{10}\) के विस्तार में अचर पद (\(x\) से स्वतंत्र पद) का मान है:
Options:
- 5
- 8
- 45
- 90
- 5
- 8
- 45
- 90
Correct Answer: a)
Explanation:
The general term \(T_{r+1}\) in the expansion of \((a + b)^{10}\) is:
\[T_{r+1} = {}^{10}C_r (x^{1/2})^{10-r} \left(\frac{1}{3x^2}\right)^r = {}^{10}C_r \cdot \frac{1}{3^r} \cdot x^{\frac{10-r}{2} – 2r} = {}^{10}C_r \cdot \frac{1}{3^r} \cdot x^{\frac{10-5r}{2}}\]
For the term to be independent of \(x\), the power of \(x\) must be 0:
\[\frac{10-5r}{2} = 0 \implies 10 – 5r = 0 \implies r = 2\]
Substituting \(r = 2\):
\[T_3 = {}^{10}C_2 \cdot \frac{1}{3^2} = \frac{10 \times 9}{2} \cdot \frac{1}{9} = 45 \times \frac{1}{9} = 5\]
Hence, the correct option is (a).
द्विपद प्रसार में व्यापक पद:
\[T_{r+1} = {}^{10}C_r (x^{1/2})^{10-r} \left(\frac{1}{3x^2}\right)^r = {}^{10}C_r \cdot 3^{-r} \cdot x^{\frac{10-5r}{2}}\]
\(x\) से स्वतंत्र पद के लिए घातांक शून्य होना चाहिए:
\[\frac{10-5r}{2} = 0 \implies r = 2\]
मान रखने पर:
\[T_3 = {}^{10}C_2 \cdot \frac{1}{9} = \frac{45}{9} = 5\]
अतः सही विकल्प (a) है।