The number of points of intersection of `2y =1 ` and `y =cos x` in `- pi //2 le x lt pi//2` is

Options

A. 1
B. 2
C. 3
D. 4

Answer

To find the number of points of intersection between the lines \(2y = 1\) and \(y = \cos x\) in the interval \(-\frac{\pi}{2} \leq x < \frac{\pi}{2}\), we can follow these steps:

### Step 1: Rewrite the equations
First, we rewrite the equation \(2y = 1\) in terms of \(y\):
\[ y = \frac{1}{2} \]

### Step 2: Set up the intersection condition
Next, we set the two equations equal to each other to find the points of intersection:
\[ \cos x = \frac{1}{2} \]

### Step 3: Solve for \(x\)
Now we solve the equation \(\cos x = \frac{1}{2}\). The cosine function equals \(\frac{1}{2}\) at specific angles. Within the interval \(-\frac{\pi}{2} \leq x < \frac{\pi}{2}\), we need to find the angles where this condition holds true.

The general solutions for \(\cos x = \frac{1}{2}\) are:
\[ x = \pm \frac{\pi}{3} + 2k\pi \quad (k \in \mathbb{Z}) \]

However, we only consider the solutions within the interval \(-\frac{\pi}{2} \leq x < \frac{\pi}{2}\). The relevant solution in this interval is:
\[ x = \frac{\pi}{3} \]

### Step 4: Count the solutions
Next, we check if there are any other solutions in the given interval. The cosine function is continuous and decreases from \(1\) to \(-1\) as \(x\) moves from \(0\) to \(\frac{\pi}{2}\) and increases from \(-1\) to \(1\) as \(x\) moves from \(-\frac{\pi}{2}\) to \(0\).

Since \(\cos x = \frac{1}{2}\) has only one solution in the interval \(-\frac{\pi}{2} \leq x < \frac{\pi}{2}\), we conclude that there is only one point of intersection.

### Final Answer
Thus, the number of points of intersection of the lines \(2y = 1\) and \(y = \cos x\) in the interval \(-\frac{\pi}{2} \leq x < \frac{\pi}{2}\) is:
\[ \text{Number of points of intersection} = 1 \]

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