STATEMENT-1 : If two sides of a triangel are given , then its area will be maximum if the angle between the given side be `(pi)/(2)` .
STATEMENT-2 : The function `f(x)=(1)/(1+x^(2))` is increases in the interval `(0,oo)`
STATEMENT-3: The lenght of the subnormal of the curve `y^(3)=12ax` at y=1 is 4a

Options

A. TFT
B. TTT
C. FFF
D. FFT

Answer

To solve the given statements, we will analyze each statement one by one, providing a step-by-step solution.

### Statement 1:
**Statement:** If two sides of a triangle are given, then its area will be maximum if the angle between the given sides is \( \frac{\pi}{2} \).

**Solution:**
1. Let the two sides of the triangle be \( b \) and \( c \), and let the angle between them be \( \theta \).
2. The area \( A \) of the triangle can be expressed as:
\[ A = \frac{1}{2} b c \sin \theta \]
3. To maximize the area, we differentiate \( A \) with respect to \( \theta \):
\[ \frac{dA}{d\theta} = \frac{1}{2} b c \cos \theta \]
4. Set the derivative equal to zero to find critical points:
\[ \frac{1}{2} b c \cos \theta = 0 \]
This implies:
\[ \cos \theta = 0 \quad \Rightarrow \quad \theta = \frac{\pi}{2} \]
5. To confirm that this is a maximum, we check the second derivative:
\[ \frac{d^2A}{d\theta^2} = -\frac{1}{2} b c \sin \theta \]
Evaluating at \( \theta = \frac{\pi}{2} \):
\[ \frac{d^2A}{d\theta^2} = -\frac{1}{2} b c \cdot 1 = -\frac{1}{2} b c < 0 \]
Since the second derivative is negative, the area is maximized at \( \theta = \frac{\pi}{2} \).

**Conclusion:** Statement 1 is **True**.

### Statement 2:
**Statement:** The function \( f(x) = \frac{1}{1 + x^2} \) is increasing in the interval \( (0, \infty) \).

**Solution:**
1. To determine if \( f(x) \) is increasing, we find the derivative:
\[ f'(x) = \frac{d}{dx} \left( \frac{1}{1 + x^2} \right) = -\frac{2x}{(1 + x^2)^2} \]
2. The sign of \( f'(x) \) determines whether the function is increasing or decreasing. Since \( -\frac{2x}{(1 + x^2)^2} \) is negative for \( x > 0 \), \( f(x) \) is **decreasing** in the interval \( (0, \infty) \).

**Conclusion:** Statement 2 is **False**.

### Statement 3:
**Statement:** The length of the subnormal of the curve \( y^3 = 12ax \) at \( y = 1 \) is \( 4a \).

**Solution:**
1. First, we differentiate the curve implicitly:
\[ 3y^2 \frac{dy}{dx} = 12a \quad \Rightarrow \quad \frac{dy}{dx} = \frac{12a}{3y^2} = \frac{4a}{y^2} \]
2. The length of the subnormal is given by:
\[ \text{Length of subnormal} = |y| \cdot \left| \frac{dy}{dx} \right| \]
3. Substitute \( y = 1 \):
\[ \frac{dy}{dx} \bigg|_{y=1} = \frac{4a}{1^2} = 4a \]
4. Thus, the length of the subnormal is:
\[ |1| \cdot |4a| = 4a \]

**Conclusion:** Statement 3 is **True**.

### Final Summary:
- Statement 1: True
- Statement 2: False
- Statement 3: True