QJEEM24Ma0e60: JEE Main 2024 Mathematics - Class 12 JEEM24Ma0e60 Solution: Let be a linear function and , is continuous at . If , then the …

Let $g(x)=a x+b$

Now function $\mathrm{f}(\mathrm{x})$ in continuous at $\mathrm{x}=0$

$$\begin{array}{rl}& \therefore \underset{\mathrm{x}\to 0^{+}}{lim}\mathrm{f}(\mathrm{x})=\mathrm{f}(0)\\ & \Rightarrow \underset{\mathrm{x}\to 0}{lim}{\left(\frac{1+\mathrm{x}}{2+\mathrm{x}}\right)}^{\frac{1}{\mathrm{x}}}=\mathrm{b}\\ & \Rightarrow 0=\mathrm{b}\\ & \therefore \mathrm{g}(\mathrm{x})=\mathrm{ax}\end{array}$$

Now, for \mathrm{x}>0

$\begin{aligned}& \mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\mathrm{x}} \cdot\left(\frac{1+\mathrm{x}}{2+\mathrm{x}}\right)^{\frac{1}{\mathrm{x}}-1} \cdot \frac{1}{(2+\mathrm{x})^2} \\& +\left(\frac{1+\mathrm{x}}{2+\mathrm{x}}\right)^{\frac{1}{\mathrm{x}}} \cdot \ln \left(\frac{1+\mathrm{x}}{2+\mathrm{x}}\right) \cdot\left(-\frac{1}{\mathrm{x}^2}\right) \\& \therefore \mathrm{f}^{\prime}(1)=\frac{1}{9}-\frac{2}{3} \cdot \ln \left(\frac{2}{3}\right) \\& \text { And } \mathrm{f}(-1)=\mathrm{g}(-1)=-\mathrm{a} \\& \therefore \mathrm{a}=\frac{2}{3} \ln \left(\frac{2}{3}\right)-\frac{1}{9} \\& \therefore \mathrm{g}(3)=2 \ln \left(\frac{2}{3}\right)-\frac{1}{3} \\& =\ln \left(\frac{4}{9 \cdot \mathrm{e}^{1 / 3}}\right)\end{aligned}$