Question

The least value of \( a \) such that the function \( x^2 + ax + 1 \) is increasing on \([1,2]\) is

• A. 4
• B. 2
• C. -2
• D. 1

Hint:

For monotonicity on a closed interval, always check the minimum value of the derivative on that interval.

Answer 1

The Correct Option is C

Step 1: Condition for an increasing function.
A function is increasing on an interval if its derivative is non-negative on that interval.
So, for the function \( f(x) = x^2 + ax + 1 \), we require:
\[ f'(x) \geq 0 \quad \text{for all } x \in [1,2]. \]

Step 2: Differentiate the function.
\[ f'(x) = 2x + a. \]

Step 3: Apply the condition on the interval.
We need:
\[ 2x + a \geq 0 \quad \forall x \in [1,2]. \]

Step 4: Find the minimum value of the derivative.
Since \( 2x + a \) is a linear function increasing in \( x \), its minimum value on \([1,2]\) occurs at \( x = 1 \).
So,
\[ \min (2x+a) = 2(1) + a = 2 + a. \]

Step 5: Ensure non-negativity.
For the function to be increasing:
\[ 2 + a \geq 0. \]

Step 6: Solve for \( a \).
\[ a \geq -2. \]

Step 7: Find the least value.
The least possible value of \( a \) satisfying the condition is:
\[ a = -2. \]
Final Answer:
\[ \boxed{-2}. \]