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: Determinant:
\[ |A| = 1(4 - 9) - 3(12 - 9) + 3(9 - 3) = -5 - 9 + 18 = 4 \]

Step 2: Needed cofactor:
The inverse is \(\frac{\text{adj}A}{|A|}\), and \(\text{adj}A\) is the transpose of the cofactor matrix. So the element in row 3, column 2 of \(A^{-1}\) is \(\frac{C_{23}}{|A|}\), where \(C_{23}\) is the cofactor of the element in row 2, column 3.

Step 3: Compute:
Delete row 2 and column 3: the minor is \(\begin{vmatrix}1 & 3\\ 3 & 3\end{vmatrix} = 3 - 9 = -6\).
\(C_{23} = (-1)^{2+3}(-6) = 6\).
\[ (A^{-1})_{32} = \frac{6}{4} = \frac{3}{2} \]
The negative value in option (C) would result from missing the sign \((-1)^{5}\).

Final Answer:
The required element is \(\frac{3}{2}\), option (A). \[ \boxed{\frac{3}{2}} \]