Let `f(x)={{:(,3sinx+a^(2)-10a+30,x in Q),(,4 cos x,x in Q):}"which"`one of the following statements is correct?

Options

A. f(x) is continuous for all x when a=5
B. f(x) must be continuous for all, x when a=5
C. f(x) is continuous for all x, `=2pix-tan^(-1)((3)/(4)),n in Z,"when a=5"`
D. f(x) is continuous for all `x=2pix-tan^(-1)((4)/(3)),n in Z` when a=5

Explanation

To determine the correct statement regarding the function \( f(x) \), we need to analyze its continuity. The function is defined as follows:

\[ f(x) = \begin{cases} 3\sin x + a^2 - 10a + 30 & \text{if } x \in \mathbb{Q} \\ 4\cos x & \text{if } x \in \mathbb{R} \setminus \mathbb{Q} \end{cases} \]

### Step 1: Set the condition for continuity
For \( f(x) \) to be continuous at a point \( x = k \), the left-hand limit (LHL), right-hand limit (RHL), and the function value at that point must be equal:

\[ \lim_{x \to k} f(x) = f(k) \]

This means:

\[ \lim_{x \to k} (3\sin x + a^2 - 10a + 30) = \lim_{x \to k} (4\cos x) \]

### Step 2: Evaluate the limits
We need to equate the two expressions:

\[ 3\sin k + a^2 - 10a + 30 = 4\cos k \]

### Step 3: Substitute \( a = 5 \)
We will check the continuity condition by substituting \( a = 5 \):

\[ 3\sin k + 5^2 - 10 \cdot 5 + 30 = 4\cos k \]

Calculating the constants:

\[ 5^2 = 25, \quad -10 \cdot 5 = -50, \quad 30 = 30 \]

Thus, we have:

\[ 3\sin k + 25 - 50 + 30 = 4\cos k \]

This simplifies to:

\[ 3\sin k + 5 = 4\cos k \]

### Step 4: Rearranging the equation
Rearranging gives:

\[ 3\sin k = 4\cos k - 5 \]

### Step 5: Formulate as a trigonometric identity
Dividing through by 5, we can express this as:

\[ \frac{3}{5}\sin k = \frac{4}{5}\cos k - 1 \]

### Step 6: Use the right triangle relationships
Recognizing \( \frac{3}{5} \) and \( \frac{4}{5} \) as sine and cosine of an angle in a right triangle, we can let:

\[ \sin \theta = \frac{3}{5}, \quad \cos \theta = \frac{4}{5} \]

### Step 7: Solve for \( k \)
Using the identity \( \sin(k + \theta) = 1 \), we find:

\[ k + \theta = 2n\pi \quad \text{for integers } n \]

Thus, we can express \( k \) as:

\[ k = 2n\pi - \theta \]

### Conclusion
The function \( f(x) \) is continuous for \( a = 5 \) at points of the form \( k = 2n\pi - \tan^{-1}\left(\frac{3}{4}\right) \), where \( n \) is an integer.

### Final Answer
The correct statement is that \( f(x) \) is continuous for all \( x \) when \( a = 5 \).

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