Question

The positive value of \(\lambda\) for which the co-efficient of \(x^2\) in the expression \(x^2\left(\sqrt{x} + \frac{\lambda}{x^2}\right)^{10}\) is 720, is -

• A. 4
• B. \(2\sqrt2\)
• C. 3
• D. \(\sqrt5\)

Hint:

The coefficient of \(x^2\) in the whole expression equals the coefficient of \(x^0\) in \(\left(\sqrt x + \frac{\lambda}{x^2}\right)^{10}\); use the general term.

Answer 1

The Correct Option is A

General term of \(\left(\sqrt x + \frac{\lambda}{x^2}\right)^{10}\): \(T_{r+1} = \binom{10}{r}(\sqrt x)^{10-r}\left(\frac{\lambda}{x^2}\right)^r = \binom{10}{r}\lambda^r x^{\frac{10-r}{2} - 2r}\).
After multiplying by \(x^2\), we need the total power of x to be 2, i.e. \(\frac{10-r}{2} - 2r = 0 \Rightarrow 10 - r = 4r \Rightarrow r = 2\).
Coefficient \(= \binom{10}{2}\lambda^2 = 45\lambda^2 = 720 \Rightarrow \lambda^2 = 16\).
Positive value: \(\lambda = 4\).
Answer: (1) 4