Question:
If the constant term in the expansion of \(\left(\sqrt{x} – \dfrac{k}{x^2}\right)^{10}\) is 405, then what can be the values of k?
यदि \(\left(\sqrt{x} – \dfrac{k}{x^2}\right)^{10}\) के प्रसार में अचर पद 405 है, तो k के संभावित मान क्या हो सकते हैं?
Options:
- \(\pm 9\)
- \(\pm 3\)
- \(\pm 5\)
- \(\pm 2\)
- \(\pm 9\)
- \(\pm 3\)
- \(\pm 5\)
- \(\pm 2\)
Correct Answer: b)
Explanation:
General term \(T_{r+1} = {}^{10}C_r (x^{1/2})^{10-r} (-k x^{-2})^r = {}^{10}C_r (-k)^r x^{5 – \frac{5r}{2}}\).
For constant term: \(5 – \dfrac{5r}{2} = 0 \implies r = 2\).
\(T_3 = {}^{10}C_2 (-k)^2 = 45 k^2\).
Given constant term is \(405\): \(45 k^2 = 405 \implies k^2 = 9 \implies k = \pm 3\).
व्यापक पद \(T_{r+1} = {}^{10}C_r (x^{1/2})^{10-r} (-k x^{-2})^r = {}^{10}C_r (-k)^r x^{5 – \frac{5r}{2}}\)।
अचर पद के लिए: \(5 – \dfrac{5r}{2} = 0 \implies r = 2\)।
\({}^{10}C_2 k^2 = 405 \implies 45 k^2 = 405 \implies k^2 = 9 \implies k = \pm 3\)।