Question:medium

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

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When analyzing the behavior of a function, especially in terms of increasing or decreasing, it’s important to check the derivative at the boundary points of the given interval. In this case, by ensuring \( f'(x) \geq 0 \) at the endpoints \( x = 1 \) and \( x = 2 \), we can determine the least value of \( b \) that ensures the function is increasing on the entire interval. Always remember to compare the conditions at all boundary points to find the solution.

Updated On: Mar 27, 2026
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The Correct Option is C

Solution and Explanation

The function \( f(x) = x^2 + bx + 1 \) increases on the interval \( [1, 2] \) if its derivative \( f'(x) \) is non-negative across this interval. The derivative of \( f(x) \) is:

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

To ensure \( f'(x) \geq 0 \) for all \( x \in [1, 2] \), we evaluate the derivative at the interval's endpoints:

For \( x = 1 \):

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

For \( x = 2 \):

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

The most restrictive condition, which satisfies both inequalities, requires \( b \) to be at least \( -2 \). Therefore, the smallest value \( b \) can take is \( -2 \).

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