JEE Main 2024 — Mathematics Algebra

Given: $f(x)=[\begin{matrix}\mathrm{cos}x & -\mathrm{sin}x & 0 \\ \mathrm{sin}x & \mathrm{cos}x & 0 \\ 0 & 0 & 1\end{matrix}]$f(x)=[cosxsinx0​−sinxcosx0​001​]

$$\Rightarrow f(-x)=[\begin{matrix}\mathrm{cos}x & \mathrm{sin}x & 0 \\ -\mathrm{sin}x & \mathrm{cos}x & 0 \\ 0 & 0 & 1\end{matrix}]$$⇒f(−x)=[cosx−sinx0​sinxcosx0​001​]

$$\Rightarrow f(x)\times f(-x)=[\begin{matrix}\mathrm{cos}x & -\mathrm{sin}x & 0 \\ \mathrm{sin}x & \mathrm{cos}x & 0 \\ 0 & 0 & 1\end{matrix}][\begin{matrix}\mathrm{cos}x & \mathrm{sin}x & 0 \\ -\mathrm{sin}x & \mathrm{cos}x & 0 \\ 0 & 0 & 1\end{matrix}]$$⇒f(x)×f(−x)=[cosxsinx0​−sinxcosx0​001​][cosx−sinx0​sinxcosx0​001​]

$$\Rightarrow f(x)\times f(-x)=[\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}]$$⇒f(x)×f(−x)=[100​010​001​]

$$\Rightarrow f(x)\times f(-x)=I$$⇒f(x)×f(−x)=I

$$\Rightarrow f(-x)={f}^{-1}(x)$$⇒f(−x)=f−1(x)

Hence statement-I is correct

Now, checking statement II

$$\Rightarrow f(y)=[\begin{matrix}\mathrm{cos}y & -\mathrm{sin}y & 0 \\ \mathrm{sin}y & \mathrm{cos}y & 0 \\ 0 & 0 & 1\end{matrix}]$$⇒f(y)=[cosysiny0​−sinycosy0​001​]

$$\Rightarrow f(x)f(y)=[\begin{matrix}\mathrm{cos}x & -\mathrm{sin}x & 0 \\ \mathrm{sin}x & \mathrm{cos}x & 0 \\ 0 & 0 & 1\end{matrix}][\begin{matrix}\mathrm{cos}y & -\mathrm{sin}y & 0 \\ \mathrm{sin}y & \mathrm{cos}y & 0 \\ 0 & 0 & 1\end{matrix}]$$⇒f(x)f(y)=[cosxsinx0​−sinxcosx0​001​][cosysiny0​−sinycosy0​001​]

$$\Rightarrow f(x)f(y)=[\begin{matrix}\mathrm{cos}x\mathrm{cos}y-\mathrm{sin}x\mathrm{sin}y & -\mathrm{cos}x\mathrm{sin}y-\mathrm{sin}x\mathrm{cos}y & 0 \\ \mathrm{sin}x\mathrm{cos}y+\mathrm{cos}x\mathrm{sin}y & -\mathrm{sin}x\mathrm{sin}y+\mathrm{cos}x\mathrm{cos}y & 0 \\ 0 & 0 & 1\end{matrix}]$$⇒f(x)f(y)=[cosxcosy−sinxsinysinxcosy+cosxsiny0​−cosxsiny−sinxcosy−sinxsiny+cosxcosy0​001​]

$$\Rightarrow f(x)f(y)=[\begin{matrix}\mathrm{cos}(x+y) & -\mathrm{sin}(x+y) & 0 \\ \mathrm{sin}(x+y) & \mathrm{cos}(x+y) & 0 \\ 0 & 0 & 1\end{matrix}]$$⇒f(x)f(y)=[cos(x+y)sin(x+y)0​−sin(x+y)cos(x+y)0​001​]

$$\Rightarrow f(x)\cdot f(y)=f(x+y)$$⇒f(x)⋅f(y)=f(x+y)

Hence statement-II is also correct.