The value of x for which the tangents to the curves `y= x cos x, y = (sin x)//x` are parallel to the axis of x are roots of (respectively)

Options

A. `sin x=x, tan x=x`
B. `cot x= x, sec x=x`
C. `cot x= x, tan x=x`
D. `tan x=x, cot x=x`

Answer

To find the values of \( x \) for which the tangents to the curves \( y = x \cos x \) and \( y = \frac{\sin x}{x} \) are parallel to the x-axis, we need to determine where the derivative (slope) of each function is equal to zero.

### Step 1: Find the derivative of \( y = x \cos x \)

Using the product rule for differentiation, we have:
\[ \frac{dy}{dx} = \frac{d}{dx}(x) \cdot \cos x + x \cdot \frac{d}{dx}(\cos x) \]
\[ = 1 \cdot \cos x + x \cdot (-\sin x) \]
\[ = \cos x - x \sin x \]

### Step 2: Set the derivative equal to zero

To find where the tangent is parallel to the x-axis, we set the derivative equal to zero:
\[ \cos x - x \sin x = 0 \]
Rearranging gives:
\[ x \sin x = \cos x \]
Dividing both sides by \( \sin x \) (assuming \( \sin x \neq 0 \)):
\[ x = \cot x \]

### Step 3: Find the derivative of \( y = \frac{\sin x}{x} \)

Using the quotient rule for differentiation, we have:
\[ \frac{dy}{dx} = \frac{x \cdot \cos x - \sin x \cdot 1}{x^2} \]
\[ = \frac{x \cos x - \sin x}{x^2} \]

### Step 4: Set the derivative equal to zero

To find where the tangent is parallel to the x-axis, we set the numerator equal to zero:
\[ x \cos x - \sin x = 0 \]
Rearranging gives:
\[ x \cos x = \sin x \]
Dividing both sides by \( \cos x \) (assuming \( \cos x \neq 0 \)):
\[ x = \tan x \]

### Conclusion

The values of \( x \) for which the tangents to the curves are parallel to the x-axis are given by the equations:
1. \( x = \cot x \) for the curve \( y = x \cos x \)
2. \( x = \tan x \) for the curve \( y = \frac{\sin x}{x} \)

### Final Answer

The roots of the equations \( x = \cot x \) and \( x = \tan x \) respectively.

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