JEE Main 2024 — Mathematics Calculus

Let, $g(x)=ax+b$g(x)=ax+b

Now function $f(x)$f(x) in continuous at $x=0$x=0

$$\Rightarrow \underset{x\rightarrow {0}^{+}}{\mathrm{lim}}f(x)=f(0)$$⇒x→0+lim​f(x)=f(0)

$$\Rightarrow \underset{x\rightarrow 0}{\mathrm{lim}}{(\frac{1+x}{2+x})}^{\frac{1}{x}}=b$$⇒x→0lim​(2+x1+x​)x1​=b

$$\Rightarrow 0=b$$⇒0=b

$$\Rightarrow g(x)=ax$$⇒g(x)=ax

Now, for $x>0$x>0

$$f(x)={(\frac{1+x}{2+x})}^{\frac{1}{x}}$$f(x)=(2+x1+x​)x1​

$$\Rightarrow \mathrm{log}f(x)=\mathrm{log}{(\frac{1+x}{2+x})}^{\frac{1}{x}}$$⇒logf(x)=log(2+x1+x​)x1​

$$\Rightarrow \mathrm{log}f(x)=\frac{\mathrm{log}(1+x)-\mathrm{log}(2+x)}{x}$$⇒logf(x)=xlog(1+x)−log(2+x)​

$$\Rightarrow \frac{1}{f(x)}\times {f}^{'}(x)=\frac{x(\frac{1}{1+x}-\frac{1}{2+x})-[\mathrm{log}(1+x)-\mathrm{log}(2+x)]}{{x}^{2}}$$⇒f(x)1​×f′(x)=x2x(1+x1​−2+x1​)−[log(1+x)−log(2+x)]​

$$\Rightarrow {f}^{'}(x)=f(x){\frac{x(\frac{1}{1+x}-\frac{1}{2+x})-[\mathrm{log}(1+x)-\mathrm{log}(2+x)]}{{x}^{2}}}$$⇒f′(x)=f(x)x2x(1+x1​−2+x1​)−[log(1+x)−log(2+x)]​

$$\Rightarrow {f}^{'}(1)=f(1){\frac{(\frac{1}{2}-\frac{1}{3})-[\mathrm{log}(2)-\mathrm{log}(3)]}{{1}^{2}}}$$⇒f′(1)=f(1)12(21​−31​)−[log(2)−log(3)]​

$$\Rightarrow {f}^{'}(1)=\frac{2}{3}{\frac{1}{6}-\mathrm{log}(\frac{2}{3})}$$⇒f′(1)=32​61​−log(32​)

And, $f(-1)=g(-1)=-a$f(−1)=g(−1)=−a

$$\Rightarrow a=\frac{2}{3}\mathrm{log}(\frac{2}{3})-\frac{1}{9}$$⇒a=32​log(32​)−91​

Now, finding $g(3)=3a+0$g(3)=3a+0

$$\Rightarrow g(3)=2\mathrm{log}(\frac{2}{3})-\frac{1}{3}$$⇒g(3)=2log(32​)−31​

$$\Rightarrow g(3)=\mathrm{log}(\frac{4}{9\cdot {e}^{\frac{1}{3}}})$$⇒g(3)=log(9⋅e31​4​)