Range ofthe function `f(x)=cos(Ksin x)` is `[-1,1]`, then the least positive integral value of K will be

Options

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

Answer

To find the least positive integral value of \( K \) such that the range of the function \( f(x) = \cos(K \sin x) \) is \([-1, 1]\), we can follow these steps:

### Step 1: Understand the Range of the Cosine Function
The cosine function, \( \cos(y) \), has a range of \([-1, 1]\) for any real number \( y \). Therefore, for \( f(x) = \cos(K \sin x) \) to have a range of \([-1, 1]\), the argument \( K \sin x \) must cover all values that \( \cos(y) \) can take.

### Step 2: Determine the Range of \( K \sin x \)
The sine function, \( \sin x \), has a range of \([-1, 1]\). Therefore, \( K \sin x \) will range from:
\[ K \cdot (-1) \text{ to } K \cdot 1 \]
This means:
\[ -K \text{ to } K \]

### Step 3: Set the Range of \( K \sin x \)
For \( \cos(K \sin x) \) to achieve its full range of \([-1, 1]\), the argument \( K \sin x \) must cover the interval \([- \pi, \pi]\) because \( \cos(y) \) achieves all values from \(-1\) to \(1\) when \( y \) varies from \(-\pi\) to \(\pi\).

### Step 4: Establish the Inequality
From the previous step, we need:
\[ -K \leq -\pi \quad \text{and} \quad K \geq \pi \]
This can be rearranged to:
\[ K \geq \pi \]

### Step 5: Find the Least Positive Integral Value of \( K \)
Since \( \pi \) is approximately \( 3.14 \), the least positive integral value of \( K \) that satisfies \( K \geq \pi \) is:
\[ K = 4 \]

### Conclusion
Thus, the least positive integral value of \( K \) is \( 4 \).