The positive value of lambda for which the co | vedclass

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(A) The given expression is $x^2 \left( x^{1/2} + \lambda x^{-2} \right)^{10}$.The general term in the expansion of $\left( x^{1/2} + \lambda x^{-2} \

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$4$

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(A) The given expression is $x^2 \left( x^{1/2} + \lambda x^{-2} ight)^{10}$.The general term in the expansion of $\left( x^{1/2} + \lambda x^{-2} ight)^{10}$ is given by $T_{r+1} = {}^{10}C_r (x^{1/2})^{10-r} (\lambda x^{-2})^r$.$T_{r+1} = {}^{10}C_r \lambda^r x^{(10-r)/2} x^{-2r} = {}^{10}C_r \lambda^r x^{(10-5r)/2}$.Multiplying by the external $x^2$,the general term of the full expression is $x^2 \cdot T_{r+1} = {}^{10}C_r \lambda^r x^{(10-5r)/2 + 2}$.To find the coefficient of $x^2$,we set the exponent equal to $2$:$\frac{10-5r}{2} + 2 = 2$ $\Rightarrow 10-5r = 0$ $\Rightarrow r = 2$.Substituting $r=2$ into the coefficient expression:${}^{10}C_2 \lambda^2 = 720$.Since ${}^{10}C_2 = \frac{10 	imes 9}{2} = 45$,we have $45 \lambda^2 = 720$.$\lambda^2 = \frac{720}{45} = 16$.Since $\lambda$ is positive,$\lambda = 4$.

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

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