Number of stationary points in `[0,pi]` for the function `f (x) = sin x + tan x-2x` is:

Options

A. 0
B. 1
C. 2
D. 3

Answer

To find the number of stationary points of the function \( f(x) = \sin x + \tan x - 2x \) in the interval \([0, \pi]\), we need to follow these steps:

### Step 1: Find the derivative of the function
The first step is to differentiate the function \( f(x) \):
\[ f'(x) = \frac{d}{dx}(\sin x) + \frac{d}{dx}(\tan x) - \frac{d}{dx}(2x) \]
Using the derivatives:
- \( \frac{d}{dx}(\sin x) = \cos x \)
- \( \frac{d}{dx}(\tan x) = \sec^2 x \)
- \( \frac{d}{dx}(2x) = 2 \)

Thus, we have:
\[ f'(x) = \cos x + \sec^2 x - 2 \]

### Step 2: Set the derivative equal to zero
To find the stationary points, we set the derivative equal to zero:
\[ \cos x + \sec^2 x - 2 = 0 \]
This can be rewritten as:
\[ \cos x + \frac{1}{\cos^2 x} - 2 = 0 \]

### Step 3: Multiply through by \(\cos^2 x\) to eliminate the fraction
Multiplying the entire equation by \(\cos^2 x\) (note that \(\cos x \neq 0\) in the interval \((0, \pi)\)):
\[ \cos^3 x + 1 - 2\cos^2 x = 0 \]
Rearranging gives:
\[ \cos^3 x - 2\cos^2 x + 1 = 0 \]

### Step 4: Let \( y = \cos x \) and solve the cubic equation
Substituting \( y = \cos x \), we have:
\[ y^3 - 2y^2 + 1 = 0 \]
We can find the roots of this cubic equation.

### Step 5: Use the Rational Root Theorem or synthetic division
Testing for rational roots, we can try \( y = 1 \):
\[ 1^3 - 2(1^2) + 1 = 1 - 2 + 1 = 0 \]
So \( y = 1 \) is a root. We can factor the cubic:
\[ (y - 1)(y^2 - y - 1) = 0 \]

### Step 6: Solve the quadratic equation
Now we solve the quadratic \( y^2 - y - 1 = 0 \) using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2} \]

### Step 7: Find the values of \( x \)
Now we have three potential values for \( y \):
1. \( y = 1 \) (which gives \( \cos x = 1 \) or \( x = 0 \))
2. \( y = \frac{1 + \sqrt{5}}{2} \) (not valid since \( y \) must be in \([-1, 1]\))
3. \( y = \frac{1 - \sqrt{5}}{2} \) (also not valid since \( y < -1\))

### Conclusion
The only valid solution in the interval \([0, \pi]\) is \( x = 0 \).

Thus, the number of stationary points in the interval \([0, \pi]\) is:
\[ \boxed{1} \]