Question

If $f(x) = x^2 + bx + 1$ is increasing in the interval $[1, 2]$, then the least value of $b$ is:

• A. 5
• B. 0
• C. -2
• D. -4

Answer 1

The Correct Option is C

The function \( f(x) = x^2 + bx + 1 \) is increasing if \( f'(x) \geq 0 \) for all \( x \in [1, 2] \). Differentiating \( f(x) \):

\[ f'(x) = 2x + b. \]

For \( f'(x) \geq 0 \) in \([1, 2]\), check the boundary points:

At \( x = 1 \):

\[ 2(1) + b \geq 0 \implies b \geq -2. \]

At \( x = 2 \):

\[ 2(2) + b \geq 0 \implies b \geq -4. \]

Thus, the least \( b \) satisfying both conditions is \( b = -2 \).