QJEEAdv24Maaca7: JEE Advanced 2024 Mathematics - Class 11 JEEAdv24Maaca7 Solution: Let be a function such that for all , and be a function such…

$\begin{aligned}& \mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{y}) \quad \text{... (1)}\\& \Rightarrow \quad \mathrm{f}(\mathrm{nx})=\operatorname{nf}(\mathrm{x}) \forall \mathrm{n} \in \mathrm{N} \quad \text{... (2)}\end{aligned}$

$\begin{array}{ll}\text { Now } & \text { put } \mathrm{y}=-\mathrm{x} \text { in eq.(1) } \\& \mathrm{f}(\mathrm{x})+\mathrm{f}(-\mathrm{x})=\mathrm{f}(0) \quad\{\mathrm{f}(0)=0\} \\\Rightarrow \quad & \mathrm{f}(-\mathrm{x})=-\mathrm{f}(\mathrm{x}) \\\Rightarrow \quad & \mathrm{f} \text { is odd function } \\& \text { from eq. (2) } \\& \mathrm{f}(-\mathrm{nx})=\mathrm{nf}(-\mathrm{x}) \\\Rightarrow \quad & \mathrm{f}(-\mathrm{nx})=-\mathrm{nf}(\mathrm{x}) \\\Rightarrow \quad & \mathrm{f}(\mathrm{mx})=\operatorname{mf}(\mathrm{x}) \forall \mathrm{m} \in \mathrm{Z}^{-} \quad \text{... (3)}\\& \text {from eq. (2) and eq. (3) } \\& \mathrm{f}(\mathrm{nx})=\mathrm{nf}(\mathrm{x}) \forall \mathrm{n} \in \mathrm{Z} \quad \text{... (4)}\end{array}$

$\begin{aligned}\text { Now } & \text { put } x=\frac{p}{q} \text { where } p, q \in Z, q \neq 0 \\f\left(\frac{n p}{q}\right) & =n f\left(\frac{p}{q}\right) \forall n \in Z \\\text { put } n & =q \\& f(p)=q f\left(\frac{p}{q}\right)\end{aligned}$

$\Rightarrow \quad \mathrm{pf}(1)=\mathrm{qf}\left(\frac{\mathrm{p}}{\mathrm{q}}\right) \quad \{\text{from eq.(4)}\}$

Let $f(1)=a$

$\begin{array}{ll}\text { then } \quad & \mathrm{pa}=\mathrm{qf}\left(\frac{\mathrm{p}}{\mathrm{q}}\right) \\& \mathrm{f}\left(\frac{\mathrm{p}}{\mathrm{q}}\right)=\frac{\mathrm{ap}}{\mathrm{q}} \\\Rightarrow \quad & \mathrm{f}(\mathrm{x})=\mathrm{ax} \forall \mathrm{x} \in \mathbb{Q}\end{array}$

Now, $f\left(\frac{-3}{5}\right)=a\left(\frac{-3}{5}\right)=12 \Rightarrow a=-20$

$\Rightarrow \quad \mathrm{f}(\mathrm{x})=-20 \mathrm{x} \forall \mathrm{x} \in \mathbb{Q} \quad \text{... (5)}$

From the given functional equation it is not possible to find a unique function for irrational values of ' $x$ ', there are infinitely many such functions satisfying given functional equation for irrational values of $x$ , but in this problem we finally need the function at rational values of ' $x$ ' only. So, for rational values of $x$ we are getting a unique function mentioned in (5).

Now, $g(x+y)=g(x) \cdot g(y)$

$\begin{array}{ll}\Rightarrow & \ln (\mathrm{g}(\mathrm{x}+\mathrm{y})=\ln (\mathrm{g}(\mathrm{x}))+\ln (\mathrm{g}(\mathrm{y})) \\\text { Let } & \ln (\mathrm{g}(\mathrm{x}))=\mathrm{h}(\mathrm{x}) \\\Rightarrow & \mathrm{h}(\mathrm{x}+\mathrm{y})=\mathrm{h}(\mathrm{x})+\mathrm{h}(\mathrm{y}) \\\Rightarrow & \mathrm{h}(\mathrm{x})=\mathrm{kx} \forall \mathrm{x} \in \mathbb{Q} \\\Rightarrow & \mathrm{g}(\mathrm{x})=\mathrm{e}^{\mathrm{kx}} \forall \mathrm{x} \in \mathbb{Q} \quad \text{... (6)}\end{array}$

$\begin{aligned}& \text { and } \quad \mathrm{g}\left(\frac{-1}{3}\right)=\mathrm{e}^{-\frac{\mathrm{K}}{3}}=2 \quad \Rightarrow \quad \mathrm{K}=-3 \ln 2 \\& \Rightarrow \quad \mathrm{K}=\ln \left(\frac{1}{8}\right) \\& \Rightarrow \quad \mathrm{g}(\mathrm{x})=\mathrm{e}^{\ln \left(\frac{1}{8}\right) \cdot \mathrm{x}}=\left(\frac{1}{8}\right)^x=2^{-3 \mathrm{x}} \forall \mathrm{x} \in \mathbb{Q}\end{aligned}$

Now, $f\left(\frac{1}{4}\right)=-5, g(-2)=2^6=64$

$\mathrm{g}(0)=1$

$\begin{aligned}\text{So} \quad & \left(\mathrm{f}\left(\frac{1}{4}\right)+\mathrm{g}(-2)-(8) \mathrm{g}(0)\right) \\& =(-5+64-8)(1)=51\end{aligned}$