Jee Main 2024 27 01 2024 Shift 1 - Question10
Question 10
If $f(x)=\left[\begin{array}{ccc}\cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right]$
Then
S-I : $f(x) f(y)=f(x+y)$.
S-II : $f(-x)=0$ is invertible.
(1) S-I True, S-II False
(2) S-I True, S-II True
(3) S-I False, S-II True
(4) S-I False, S-II False
Show Answer
Answer: (2)
Solution:
$f(x)=\left[\begin{array}{ccc}\cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right]$
S-I:
$$ \begin{aligned} & f(x) \cdot f(y)=\left[\begin{array}{ccc} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{ccc} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{array}\right] \\ & =\left[\begin{array}{ccc} \cos (x+y) & -\sin (x+y) & 0 \\ \sin (x+y) & \cos (x+y) & 0 \\ 0 & 0 & 1 \end{array}\right]=f(x+y) \end{aligned} $$
$\therefore S-I$ is true Now,
$f(-x)=\left[\begin{array}{ccc}\cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right]$
$\operatorname{det}(f(-x))=\cos ^{2} x+\sin ^{2} x=1$
$\Rightarrow|f(-x)| \neq 0$
$\therefore$ Non-singular
$\therefore S$-II is true
BETA
