e^2x+lambda+|2kappa+lambda-1|+e^k-lambda-1+|kappa-lambda-2|=4 का समाधान करें

$$e ^ { 2 x + \lambda } + | 2 \kappa + \lambda - 1 | + e ^ { k - \lambda - 1 } + | \kappa - \lambda - 2 | = 4$$

k के लिए हल करें

$$\left\{\begin{matrix}k=\ln(7-3\kappa -e^{\lambda +2x})+\lambda +1\text{, }&\kappa <\frac{7-e^{\lambda +2x}}{3}\text{ and }\lambda \leq \kappa -2\text{ and }\lambda \geq 1-2\kappa \\k=\ln(3-2\lambda -\kappa -e^{\lambda +2x})+\lambda +1\text{, }&\kappa <3-2\lambda -e^{\lambda +2x}\text{ and }\lambda \geq \kappa -2\text{ and }\lambda \geq 1-2\kappa \\k=\ln(5+2\lambda +\kappa -e^{\lambda +2x})+\lambda +1\text{, }&\kappa >-\left(5+2\lambda -e^{\lambda +2x}\right)\text{ and }\lambda \leq \kappa -2\text{ and }\lambda \leq 1-2\kappa \\k=\ln(1+3\kappa -e^{\lambda +2x})+\lambda +1\text{, }&\kappa >\frac{e^{\lambda +2x}-1}{3}\text{ and }\lambda \geq \kappa -2\text{ and }\lambda \leq 1-2\kappa \end{matrix}\right.$$

x के लिए हल करें

$$\left\{\begin{matrix}x=\frac{\ln(7-e^{k-\lambda -1}-3\kappa )-\lambda }{2}\text{, }&\kappa <\frac{7-e^{k-\lambda -1}}{3}\text{ and }\lambda \leq \kappa -2\text{ and }\lambda \geq 1-2\kappa \\x=\frac{\ln(3-e^{k-\lambda -1}-2\lambda -\kappa )-\lambda }{2}\text{, }&\kappa <3-e^{k-\lambda -1}-2\lambda \text{ and }\lambda \geq \kappa -2\text{ and }\lambda \geq 1-2\kappa \\x=\frac{\ln(\kappa +2\lambda -e^{k-\lambda -1}+5)-\lambda }{2}\text{, }&\kappa >-2\lambda +e^{k-\lambda -1}-5\text{ and }\lambda \leq \kappa -2\text{ and }\lambda \leq 1-2\kappa \\x=\frac{\ln(3\kappa -e^{k-\lambda -1}+1)-\lambda }{2}\text{, }&\kappa >\frac{e^{k-\lambda -1}-1}{3}\text{ and }\lambda \geq \kappa -2\text{ and }\lambda \leq 1-2\kappa \end{matrix}\right.$$