$(a)\;\lambda_{C}=\lambda_{\infty}+(B)C\qquad(b)\;\lambda_{C}=\lambda_{\infty}-(B)C\qquad(c)\;\lambda_{C}=\lambda_{\infty}-(B)\sqrt{C}\qquad(d)\;\lambda_{C}=\lambda_{\infty}+(B)\sqrt{C}$

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