If `f:[-20,20]->R` defined by `f(x)=[x^2/a]sinx+cosx` is an even fucntion, then set of values of `a` is

Answer

To solve the problem, we need to determine the set of values for \( a \) such that the function \( f(x) = \left\lfloor \frac{x^2}{a} \right\rfloor \sin x + \cos x \) is an even function.

### Step-by-Step Solution:

1. **Understanding the Even Function Condition**:
An even function satisfies the condition \( f(x) = f(-x) \) for all \( x \) in its domain.

2. **Substituting \( f(x) \) and \( f(-x) \)**:
We have:
\[ f(x) = \left\lfloor \frac{x^2}{a} \right\rfloor \sin x + \cos x \]
For \( f(-x) \):
\[ f(-x) = \left\lfloor \frac{(-x)^2}{a} \right\rfloor \sin(-x) + \cos(-x) \]
Since \( (-x)^2 = x^2 \), \( \sin(-x) = -\sin x \), and \( \cos(-x) = \cos x \), we can rewrite \( f(-x) \):
\[ f(-x) = \left\lfloor \frac{x^2}{a} \right\rfloor (-\sin x) + \cos x \]

3. **Setting Up the Equation**:
For \( f(x) \) to equal \( f(-x) \):
\[ \left\lfloor \frac{x^2}{a} \right\rfloor \sin x + \cos x = \left\lfloor \frac{x^2}{a} \right\rfloor (-\sin x) + \cos x \]

4. **Simplifying the Equation**:
Canceling \( \cos x \) from both sides gives:
\[ \left\lfloor \frac{x^2}{a} \right\rfloor \sin x + \left\lfloor \frac{x^2}{a} \right\rfloor (-\sin x) = 0 \]
This simplifies to:
\[ 2 \left\lfloor \frac{x^2}{a} \right\rfloor \sin x = 0 \]

5. **Analyzing the Equation**:
For the equation \( 2 \left\lfloor \frac{x^2}{a} \right\rfloor \sin x = 0 \) to hold for all \( x \), either \( \sin x = 0 \) (which is not true for all \( x \)) or \( \left\lfloor \frac{x^2}{a} \right\rfloor = 0 \).

6. **Finding the Condition on \( a \)**:
The condition \( \left\lfloor \frac{x^2}{a} \right\rfloor = 0 \) implies:
\[ 0 \leq \frac{x^2}{a} < 1 \]
This leads to:
\[ 0 \leq x^2 < a \]
Since \( x \) ranges from \(-20\) to \(20\), the maximum value of \( x^2 \) is \( 400 \). Thus, we have:
\[ a > 400 \]

7. **Conclusion**:
Therefore, the set of values for \( a \) is:
\[ a \in (400, \infty) \]

### Final Answer:
The set of values of \( a \) is \( (400, \infty) \).