CUET Mathematics 2024 - If a function f(x)=x^2+b x+1 is increasing in the interval [1,2], then the least value of b is : | PYQs + Solutions | AfterBoards
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CUET Mathematics 2024

Calculus
>
Application of Derivatives

Easy

If a function f(x)=x2+bx+1f(x)=x^{2}+b x+1 is increasing in the interval [1,2][1,2], then the least value of bb is :

✅ Correct Option: 3
For a function to be increasing, its derivative must be positive.
f′(x)=2x+bf'(x) = 2x + b
For f(x)f(x) to be increasing on [1,2][1,2], we need: \newline f′(x)≥0f'(x) \geq 0 for all x∈[1,2]x \in [1,2]
This means: \newline 2x+b≥02x + b \geq 0 for all x∈[1,2]x \in [1,2]
Since f′(x)=2x+bf'(x) = 2x + b is an increasing function of xx, the minimum value of f′(x)f'(x) on [1,2][1,2] occurs at x=1x = 1.
At x=1x = 1, we need:
f′(1)≥0f'(1) \geq 0
2(1)+b≥02(1) + b \geq 0
2+b≥02 + b \geq 0
b≥−2b \geq -2
Since we want the least value of bb, and bb must be greater than or equal to −2-2, the least value is b=−2b = -2.
To verify: When b=−2b = -2, f′(x)=2x−2f'(x) = 2x - 2 \newline At x=1x = 1: f′(1)=0f'(1) = 0 (just starts increasing) \newline At x=2x = 2: f′(2)=2>0f'(2) = 2 > 0 (still increasing)
Therefore, the least value of bb is −2-2.

Related questions:

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For the function f(x)=2x3−9x2+12x−5f(x) = 2x^3 - 9x^2 + 12x - 5, x∈[0,3]x \in [0, 3], match List-I with List-II :
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