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JEE Mathematics 2024 Question with Solution

Given f(x)=[cos⁡x−sin⁡x0sin⁡xcos⁡x0001]f(x)=\begin{bmatrix}\cos x & -\sin x & 0\\ \sin x & \cos x & 0\\ 0 & 0 & 1\end{bmatrix} Evaluate: Statement I: f(−x)f(-x) is the inverse of f(x)f(x). Statement II: f(x)⋅f(y)=f(x+y)f(x)\cdot f(y)=f(x+y).

  • A

    I is false, II is true

  • B

    Both I and II are false

  • C

    I is true, II is false

  • D

    Both I and II are true

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given:

f(x)=[cos⁡x−sin⁡x0sin⁡xcos⁡x0001]f(x)=\begin{bmatrix}\cos x & -\sin x & 0\\ \sin x & \cos x & 0\\ 0 & 0 & 1\end{bmatrix}

Find: Whether Statement I and Statement II are true.

For Statement I, compute f(−x)f(-x):

f(−x)=[cos⁡(−x)−sin⁡(−x)0sin⁡(−x)cos⁡(−x)0001]=[cos⁡xsin⁡x0−sin⁡xcos⁡x0001]f(-x)=\begin{bmatrix}\cos(-x) & -\sin(-x) & 0\\ \sin(-x) & \cos(-x) & 0\\ 0 & 0 & 1\end{bmatrix}=\begin{bmatrix}\cos x & \sin x & 0\\ -\sin x & \cos x & 0\\ 0 & 0 & 1\end{bmatrix}

Also,

f(x)T=[cos⁡xsin⁡x0−sin⁡xcos⁡x0001]f(x)^T=\begin{bmatrix}\cos x & \sin x & 0\\ -\sin x & \cos x & 0\\ 0 & 0 & 1\end{bmatrix}

Thus f(−x)=f(x)Tf(-x)=f(x)^T. Since this is a rotation matrix, its transpose is its inverse. Therefore, Statement I is true.

For Statement II, multiply the matrices:

f(x)f(y)=[cos⁡x−sin⁡x0sin⁡xcos⁡x0001][cos⁡y−sin⁡y0sin⁡ycos⁡y0001]f(x)f(y)=\begin{bmatrix}\cos x & -\sin x & 0\\ \sin x & \cos x & 0\\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}\cos y & -\sin y & 0\\ \sin y & \cos y & 0\\ 0 & 0 & 1\end{bmatrix}

This gives

f(x)f(y)=[cos⁡xcos⁡y−sin⁡xsin⁡y−(cos⁡xsin⁡y+sin⁡xcos⁡y)0sin⁡xcos⁡y+cos⁡xsin⁡ycos⁡xcos⁡y−sin⁡xsin⁡y0001]f(x)f(y)=\begin{bmatrix}\cos x\cos y-\sin x\sin y & -(\cos x\sin y+\sin x\cos y) & 0\\ \sin x\cos y+\cos x\sin y & \cos x\cos y-\sin x\sin y & 0\\ 0 & 0 & 1\end{bmatrix}

Using

cos⁡(x+y)=cos⁡xcos⁡y−sin⁡xsin⁡y,\cos(x+y)=\cos x\cos y-\sin x\sin y,

and

sin⁡(x+y)=sin⁡xcos⁡y+cos⁡xsin⁡y,\sin(x+y)=\sin x\cos y+\cos x\sin y,

we get

f(x)f(y)=[cos⁡(x+y)−sin⁡(x+y)0sin⁡(x+y)cos⁡(x+y)0001]=f(x+y)f(x)f(y)=\begin{bmatrix}\cos(x+y) & -\sin(x+y) & 0\\ \sin(x+y) & \cos(x+y) & 0\\ 0 & 0 & 1\end{bmatrix}=f(x+y)

Hence Statement II is also true.

Therefore, both Statement I and Statement II are true. The correct option is D.

Verification by Identity Matrix

Given:

f(x)=[cos⁡x−sin⁡x0sin⁡xcos⁡x0001]f(x)=\begin{bmatrix}\cos x & -\sin x & 0\\ \sin x & \cos x & 0\\ 0 & 0 & 1\end{bmatrix}

Find: Verify both statements directly.

To check Statement I, verify whether f(x)⋅f(−x)=If(x)\cdot f(-x)=I.

f(−x)=[cos⁡xsin⁡x0−sin⁡xcos⁡x0001]f(-x)=\begin{bmatrix}\cos x & \sin x & 0\\ -\sin x & \cos x & 0\\ 0 & 0 & 1\end{bmatrix}

Now,

f(x)⋅f(−x)=[cos⁡x−sin⁡x0sin⁡xcos⁡x0001][cos⁡xsin⁡x0−sin⁡xcos⁡x0001]=[100010001]=If(x)\cdot f(-x)=\begin{bmatrix}\cos x & -\sin x & 0\\ \sin x & \cos x & 0\\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}\cos x & \sin x & 0\\ -\sin x & \cos x & 0\\ 0 & 0 & 1\end{bmatrix}=\begin{bmatrix}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}=I

So Statement I is true.

To check Statement II,

f(x)⋅f(y)=[cos⁡x−sin⁡x0sin⁡xcos⁡x0001][cos⁡y−sin⁡y0sin⁡ycos⁡y0001]=[cos⁡(x+y)−sin⁡(x+y)0sin⁡(x+y)cos⁡(x+y)0001]=f(x+y)f(x)\cdot f(y)=\begin{bmatrix}\cos x & -\sin x & 0\\ \sin x & \cos x & 0\\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}\cos y & -\sin y & 0\\ \sin y & \cos y & 0\\ 0 & 0 & 1\end{bmatrix}=\begin{bmatrix}\cos(x+y) & -\sin(x+y) & 0\\ \sin(x+y) & \cos(x+y) & 0\\ 0 & 0 & 1\end{bmatrix}=f(x+y)

So Statement II is true.

Hence, the correct option is D.

Common mistakes

  • Assuming f(−x)f(-x) is obtained by changing only cos⁡x\cos x and not the sine terms is incorrect. Use the identities cos⁡(−x)=cos⁡x\cos(-x)=\cos x and sin⁡(−x)=−sin⁡x\sin(-x)=-\sin x carefully before comparing with the inverse.

  • Treating matrix multiplication as ordinary scalar multiplication can lead to an incorrect check of Statement II. Multiply corresponding rows and columns completely, then apply angle addition identities.

  • Concluding that transpose and inverse are unrelated here is wrong. This matrix is an orthogonal rotation matrix, so f(x)−1=f(x)Tf(x)^{-1}=f(x)^T, which is exactly what proves Statement I.

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