A function `y = f(x)` satisfies the differential `(dy)/(dx)*sin x -y cos x+sin^2 x/x^2=0` such that `y -> 0` as `x -> oo`. Then:

Options

A. (a) `lim_(xto0)f(x)=1`
B. (b) `int_(0)^(pi//2)f(x)dx` is lesser than `(pi)/(2)`
C. (c) `int_(0)^(pi//2)f(x)dx` is greater than unity
D. (d) `f(x)` is an odd function

Answer

To solve the given differential equation

\[ \frac{dy}{dx} \sin x - y \cos x + \frac{\sin^2 x}{x^2} = 0, \]

we will follow these steps:

### Step 1: Rearranging the Equation
We can rearrange the equation as follows:

\[ \frac{dy}{dx} \sin x = y \cos x - \frac{\sin^2 x}{x^2}. \]

### Step 2: Dividing by \(\sin x\)
Next, we divide both sides by \(\sin x\):

\[ \frac{dy}{dx} = \frac{y \cos x}{\sin x} - \frac{\sin x}{x^2}. \]

### Step 3: Recognizing the Quotient Rule
We can rewrite the left-hand side using the quotient rule. Notice that:

\[ \frac{d}{dx} \left( \frac{y}{\sin x} \right) = \frac{\sin x \frac{dy}{dx} - y \cos x}{\sin^2 x}. \]

Thus, we can express the equation as:

\[ \frac{d}{dx} \left( \frac{y}{\sin x} \right) = -\frac{1}{x^2}. \]

### Step 4: Integrating Both Sides
Now, we integrate both sides with respect to \(x\):

\[ \int \frac{d}{dx} \left( \frac{y}{\sin x} \right) dx = \int -\frac{1}{x^2} dx. \]

This gives us:

\[ \frac{y}{\sin x} = \frac{1}{x} + C, \]

where \(C\) is the constant of integration.

### Step 5: Solving for \(y\)
Multiplying both sides by \(\sin x\), we find:

\[ y = \sin x \left( \frac{1}{x} + C \right) = \frac{\sin x}{x} + C \sin x. \]

### Step 6: Applying the Limit Condition
We know that \(y \to 0\) as \(x \to \infty\). The term \(\frac{\sin x}{x}\) approaches \(0\) as \(x\) approaches infinity. For \(y\) to approach \(0\), the term \(C \sin x\) must also approach \(0\). Since \(\sin x\) oscillates between \(-1\) and \(1\), we conclude that \(C\) must be \(0\).

Thus, we have:

\[ y = \frac{\sin x}{x}. \]

### Step 7: Evaluating the Options
Now, we evaluate the options given in the problem:

1. **Limit as \(x \to 0\)**:
\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \quad \text{(True, option A is correct)} \]

2. **Integral from \(0\) to \(\frac{\pi}{2}\)**:
We know that \(\frac{\sin x}{x} < 1\) for \(x \in (0, \frac{\pi}{2})\). Thus:
\[ \int_0^{\frac{\pi}{2}} \frac{\sin x}{x} dx < \int_0^{\frac{\pi}{2}} dx = \frac{\pi}{2} \quad \text{(True, option B is correct)} \]

3. **Integral greater than 1**:
Since \(\frac{\sin x}{x}\) approaches \(1\) as \(x\) approaches \(0\) and is less than \(1\) for \(x > 0\), we can conclude:
\[ \int_0^{\frac{\pi}{2}} \frac{\sin x}{x} dx > 1 \quad \text{(True, option C is correct)} \]

4. **Odd function**:
The function \(\frac{\sin x}{x}\) is an even function, not an odd function. Thus, option D is incorrect.

### Conclusion
The correct options are A, B, and C.