Question

If the function \(f(x) = \frac{4\sqrt{2}(sin3x+sinx)}{2sin2xsin\frac{3x}{2}+cos\frac{5x}{2}-cos\frac{3x}{2}}\) for \(x\neq \frac{π}{2}\) is continuous at \(x = \frac{π}{2}\), then the value of \(f(\frac{π}{2})\) is equal to

• A. \((2)^2\)
• B. \((3)^2\)
• C. \(4\sqrt{2}\)
• D. \(2\sqrt{2}\)

Hint:

The (3,2) element of the inverse is the cofactor of the (2,3) element divided by the determinant.

Answer 1

The Correct Option is A

Step 1: Solve a linear system instead:
The second column of $A^{-1}$ is the solution $x$ of $Ax = e_2 = (0, 1, 0)^T$. The third entry of that solution is what we need.

Step 2: Solve:
$x_1 + 3x_2 + 3x_3 = 0$, $3x_1 + x_2 + 3x_3 = 1$, $3x_1 + 3x_2 + 4x_3 = 0$.
Subtract the third from the second: $-2x_2 - x_3 = 1$, so $x_3 = -1 - 2x_2$. Subtract the first from the third: $2x_1 + x_3 = 0$, so $x_1 = -\frac{x_3}{2}$.
Put these in the first: $-\frac{x_3}{2} + 3x_2 + 3x_3 = 0$, so $3x_2 + \frac52 x_3 = 0$, $x_2 = -\frac56 x_3$. Then $x_3 = -1 + \frac53 x_3$, so $-\frac23 x_3 = -1$ and $x_3 = \frac32$. Option (A).

Final Answer:
$\frac{3}{2}$. \[ \boxed{\frac{3}{2}} \]