Statement-1:`int_(0)^(1)(cos x)/(1+x^(2))dxgt(pi)/(4)cos1`
Statement-2: If f(x) and g(x) are continuous on [a,b], then
`int_(a)^(b) f(x) g(x)dx=f(c )int_(a)^(b)g(x)` for some `c in (a,b)`.

Options

A. Statement-1 is true, Statement-2 is True,Statement-2 is a correct explanation for Statement-1.
B. Statement-1 is True, Statement-2 is not a correct explanation for Statement-1.
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False, Statement-2 is True.

Explanation

To solve the problem, we need to analyze both statements and verify their correctness step by step.

### Step 1: Analyze Statement 2
Statement 2 states that if \( f(x) \) and \( g(x) \) are continuous on \([a,b]\), then:
\[ \int_{a}^{b} f(x) g(x) \, dx = f(c) \int_{a}^{b} g(x) \, dx \]
for some \( c \in (a,b) \).

This statement is a version of the Mean Value Theorem for integrals. Since \( f(x) \) and \( g(x) \) are continuous, this statement is indeed true.

**Hint:** Recall that the Mean Value Theorem for integrals guarantees the existence of a point \( c \) in the interval where the average value of the function is equal to the function value at that point.

### Step 2: Analyze Statement 1
Statement 1 states:
\[ \int_{0}^{1} \frac{\cos x}{1+x^2} \, dx > \frac{\pi}{4} \cos 1 \]

To verify this, we will use the Mean Value Theorem for integrals as established in Statement 2.

1. Let \( f(x) = \cos x \) and \( g(x) = \frac{1}{1+x^2} \).
2. According to the Mean Value Theorem:
\[ \int_{0}^{1} \cos x \cdot \frac{1}{1+x^2} \, dx = \cos(c) \int_{0}^{1} \frac{1}{1+x^2} \, dx \]
for some \( c \in (0, 1) \).

3. Now, we need to compute \( \int_{0}^{1} \frac{1}{1+x^2} \, dx \):
\[ \int_{0}^{1} \frac{1}{1+x^2} \, dx = \tan^{-1}(x) \Big|_{0}^{1} = \tan^{-1}(1) - \tan^{-1}(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4} \]

4. Substituting this back, we have:
\[ \int_{0}^{1} \frac{\cos x}{1+x^2} \, dx = \cos(c) \cdot \frac{\pi}{4} \]

5. Since \( c \in (0, 1) \), and \( \cos(x) \) is a decreasing function in this interval, we know:
\[ \cos(c) > \cos(1) \]

6. Therefore:
\[ \int_{0}^{1} \frac{\cos x}{1+x^2} \, dx = \cos(c) \cdot \frac{\pi}{4} > \cos(1) \cdot \frac{\pi}{4} \]

This implies:
\[ \int_{0}^{1} \frac{\cos x}{1+x^2} \, dx > \frac{\pi}{4} \cos(1) \]

### Conclusion
Both statements are true:
- Statement 1 is true because we have shown that the integral is greater than \( \frac{\pi}{4} \cos(1) \).
- Statement 2 is true as it is a valid theorem.

Thus, the correct option is that both statements are true, and Statement 2 is the correct explanation for Statement 1.