The area enclosed by the curvey `y= cos 2x, 0 le x le (pi)/(4)` and co-ordinate axes is `1/4` sq. unit.

Answer

To determine the area enclosed by the curve \( y = \cos(2x) \) from \( x = 0 \) to \( x = \frac{\pi}{4} \) and the coordinate axes, we can follow these steps:

### Step 1: Understand the area to be calculated
We need to find the area between the curve \( y = \cos(2x) \), the x-axis, and the vertical lines \( x = 0 \) and \( x = \frac{\pi}{4} \).

### Step 2: Set up the integral
The area \( A \) can be calculated using the integral:
\[ A = \int_{0}^{\frac{\pi}{4}} \cos(2x) \, dx \]

### Step 3: Compute the integral
To compute the integral, we first find the antiderivative of \( \cos(2x) \):
\[ \int \cos(2x) \, dx = \frac{1}{2} \sin(2x) + C \]

Now, we can evaluate the definite integral from \( 0 \) to \( \frac{\pi}{4} \):
\[ A = \left[ \frac{1}{2} \sin(2x) \right]_{0}^{\frac{\pi}{4}} \]

### Step 4: Evaluate the limits
Now we substitute the limits into the antiderivative:
\[ A = \frac{1}{2} \sin\left(2 \cdot \frac{\pi}{4}\right) - \frac{1}{2} \sin(2 \cdot 0) \]
\[ = \frac{1}{2} \sin\left(\frac{\pi}{2}\right) - \frac{1}{2} \sin(0) \]
\[ = \frac{1}{2} \cdot 1 - \frac{1}{2} \cdot 0 \]
\[ = \frac{1}{2} \]

### Step 5: Conclusion
The area enclosed by the curve \( y = \cos(2x) \) from \( x = 0 \) to \( x = \frac{\pi}{4} \) and the coordinate axes is \( \frac{1}{2} \) square units. Since the problem states that the area is \( \frac{1}{4} \) square units, the statement is **false**.