If (x) is a function given by `f(x) = |{:(sinx , sin a, sin b),(cosx,cosa,cosb),(tan x , tan a, tan b):}|" where"0 lt altblt(pi)/(2)` Then the equatiion f'(x)=0

Options

A. has at least one root in (a,b)
B. has at most one root in (a,b)
C. has exactly one root in (a,b)
D. has no root in (a,b)

Answer

To solve the problem step by step, we will analyze the function \( f(x) \) given by the determinant:

\[ f(x) = \begin{vmatrix} \sin x & \sin a & \sin b \\ \cos x & \cos a & \cos b \\ \tan x & \tan a & \tan b \end{vmatrix} \]

### Step 1: Evaluate \( f(a) \) and \( f(b) \)

First, we substitute \( x = a \) into \( f(x) \):

\[ f(a) = \begin{vmatrix} \sin a & \sin a & \sin b \\ \cos a & \cos a & \cos b \\ \tan a & \tan a & \tan b \end{vmatrix} \]

Notice that the first and second columns are identical. Therefore, the determinant will be zero:

\[ f(a) = 0 \]

Now, we substitute \( x = b \):

\[ f(b) = \begin{vmatrix} \sin b & \sin a & \sin b \\ \cos b & \cos a & \cos b \\ \tan b & \tan a & \tan b \end{vmatrix} \]

Again, the first and second columns are identical, so the determinant will also be zero:

\[ f(b) = 0 \]

### Step 2: Apply Rolle's Theorem

Since we have established that:

\[ f(a) = 0 \quad \text{and} \quad f(b) = 0 \]

and given that \( a < b \) (with \( a, b \in (0, \frac{\pi}{2}) \)), we can apply Rolle's Theorem. This theorem states that if a function is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one \( c \in (a, b) \) such that:

\[ f'(c) = 0 \]

### Step 3: Conclusion

From the application of Rolle's Theorem, we conclude that there is at least one root of \( f'(x) = 0 \) in the interval \( (a, b) \).

### Final Answer

Thus, the equation \( f'(x) = 0 \) has at least one root in the interval \( (a, b) \).

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