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

Answer: (C)

-2

Answer 2

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 $b = -2$.