Let `f (x)= [{:(2x+3,x gt 1),(alpha ^(2) x+1,x le1):}` If range of `f (x)=R` (set of real numbers) then number orf integral value(s), which `alpha` any take :

Options

A. 2
B. 3
C. 4
D. 5

Answer

To solve the problem, we need to analyze the function \( f(x) \) defined as follows:

\[ f(x) = \begin{cases} 2x + 3 & \text{if } x > 1 \\ \alpha^2 x + 1 & \text{if } x \leq 1 \end{cases} \]

We are tasked with finding the integral values of \( \alpha \) such that the range of \( f(x) \) covers all real numbers \( \mathbb{R} \).

### Step 1: Analyze the function for \( x > 1 \)

For \( x > 1 \), the function is given by:

\[ f(x) = 2x + 3 \]

As \( x \) approaches infinity, \( f(x) \) also approaches infinity. At \( x = 1 \):

\[ f(1) = 2(1) + 3 = 5 \]

Thus, for \( x > 1 \), the function \( f(x) \) will take values from \( 5 \) to \( \infty \).

### Step 2: Analyze the function for \( x \leq 1 \)

For \( x \leq 1 \), the function is given by:

\[ f(x) = \alpha^2 x + 1 \]

At \( x = 1 \):

\[ f(1) = \alpha^2(1) + 1 = \alpha^2 + 1 \]

### Step 3: Ensure continuity at \( x = 1 \)

For the function to cover all real numbers, the value of \( f(1) \) from both pieces must be consistent. Therefore, we set:

\[ \alpha^2 + 1 \geq 5 \]

This simplifies to:

\[ \alpha^2 \geq 4 \]

### Step 4: Solve the inequality

The inequality \( \alpha^2 \geq 4 \) gives us:

\[ \alpha \geq 2 \quad \text{or} \quad \alpha \leq -2 \]

### Step 5: Determine integral values of \( \alpha \)

The integral values for \( \alpha \) that satisfy \( \alpha \geq 2 \) are:

\[ 2, 3, 4, \ldots \quad (\text{and so on, up to } +\infty) \]

The integral values for \( \alpha \) that satisfy \( \alpha \leq -2 \) are:

\[ -2, -3, -4, \ldots \quad (\text{and so on, down to } -\infty) \]

### Step 6: Count the integral values

Since both conditions provide an infinite number of integral values, we conclude that there are infinitely many integral values that \( \alpha \) can take.

### Final Answer

The number of integral values that \( \alpha \) can take is infinite.

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