Question:
Coefficient of \(x\) in the expansion of \(\left(x^2 + \frac{a}{x}\right)^5\) is:
\(\left(x^2 + \frac{a}{x}\right)^5\) के विस्तार में \(x\) का गुणांक है:
Options:
- \(9a^2\)
- \(10a^3\)
- \(10a^2\)
- \(10a\)
- \(9a^2\)
- \(10a^3\)
- \(10a^2\)
- \(10a\)
Correct Answer: b)
Explanation:
The general term in the expansion is:
\[T_{r+1} = {}^5C_r (x^2)^{5-r} \left(\frac{a}{x}\right)^r = {}^5C_r a^r x^{10 – 2r – r} = {}^5C_r a^r x^{10 – 3r}\]
To find the coefficient of \(x^1\), equate the exponent of \(x\) to 1:
\[10 – 3r = 1 \implies 3r = 9 \implies r = 3\]
Substituting \(r = 3\):
\[\text{Coefficient} = {}^5C_3 a^3 = \frac{5 \times 4}{2} a^3 = 10a^3\]
Hence, the correct option is (b).
द्विपद प्रसार में व्यापक पद:
\[T_{r+1} = {}^5C_r a^r x^{10 – 3r}\]
\(x\) के गुणांक हेतु घातांक को 1 के बराबर रखने पर:
\[10 – 3r = 1 \implies r = 3\]
गुणांक:
\[{}^5C_3 a^3 = 10a^3\]
अतः सही विकल्प (b) है।