The value of `(tan x +2 tan 2x)/(tan x)` cannot lie between 1 and 5. Is it true or false ?

Answer

To determine whether the value of \((\tan x + 2 \tan 2x)/(\tan x)\) can lie between 1 and 5, we will analyze the expression step-by-step.

### Step 1: Simplify the Expression
We start with the expression:
\[ \frac{\tan x + 2 \tan 2x}{\tan x} \]
We can separate this into two parts:
\[ \frac{\tan x}{\tan x} + \frac{2 \tan 2x}{\tan x} = 1 + 2 \frac{\tan 2x}{\tan x} \]

### Step 2: Use the Double Angle Formula
Recall the double angle formula for tangent:
\[ \tan 2x = \frac{2 \tan x}{1 - \tan^2 x} \]
Substituting this into our expression gives:
\[ 2 \frac{\tan 2x}{\tan x} = 2 \cdot \frac{\frac{2 \tan x}{1 - \tan^2 x}}{\tan x} = \frac{4}{1 - \tan^2 x} \]
Thus, the expression simplifies to:
\[ 1 + \frac{4}{1 - \tan^2 x} \]

### Step 3: Analyze the Expression
Let \( p = \tan x \). The expression now becomes:
\[ f(p) = 1 + \frac{4}{1 - p^2} \]
We need to analyze the behavior of \( f(p) \) as \( p \) varies.

### Step 4: Determine the Domain
The expression \( 1 - p^2 \) must be positive for \( f(p) \) to be defined, which means:
\[ 1 - p^2 > 0 \implies p^2 < 1 \implies -1 < p < 1 \]
Thus, \( \tan x \) must lie within the interval \((-1, 1)\).

### Step 5: Find the Range of \( f(p) \)
Now we evaluate the limits of \( f(p) \) as \( p \) approaches the boundaries of the interval:
- As \( p \to -1 \):
\[ f(-1) = 1 + \frac{4}{1 - (-1)^2} = 1 + \frac{4}{0} \to -\infty \]
- As \( p \to 1 \):
\[ f(1) = 1 + \frac{4}{1 - 1^2} = 1 + \frac{4}{0} \to +\infty \]

### Step 6: Determine Values Between the Limits
Since \( f(p) \) approaches \(-\infty\) as \( p \) approaches \(-1\) and \( +\infty\) as \( p \) approaches \( 1\), we can conclude that \( f(p) \) takes all values in between these limits.

### Conclusion
Since \( f(p) \) can take values less than 1 and greater than 5, the statement that the value of \((\tan x + 2 \tan 2x)/(\tan x)\) cannot lie between 1 and 5 is **false**.