CUET UG 2024 — Mathematics Calculus

For a function to be increasing, its derivative must be positive.

$$f'(x) = 2x + b$$f′(x)=2x+b

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For $f(x)$f(x) to be increasing on $[1,2]$[1,2], we need:
$f'(x) \geq 0$f′(x)≥0 for all $x \in [1,2]$x∈[1,2]

This means:
$2x + b \geq 0$2x+b≥0 for all $x \in [1,2]$x∈[1,2]

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Since $f'(x) = 2x + b$f′(x)=2x+b is an increasing function of $x$x, the minimum value of $f'(x)$f′(x) on $[1,2]$[1,2] occurs at $x = 1$x=1.

At $x = 1$x=1, we need:

$$f'(1) \geq 0$$f′(1)≥0

$$2(1) + b \geq 0$$2(1)+b≥0

$$2 + b \geq 0$$2+b≥0

$$b \geq -2$$b≥−2

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Since we want the least value of $b$b, and $b$b must be greater than or equal to $-2$−2, the least value is $b = -2$b=−2.

To verify: When $b = -2$b=−2, $f'(x) = 2x - 2$f′(x)=2x−2
At $x = 1$x=1: $f'(1) = 0$f′(1)=0 (just starts increasing)
At $x = 2$x=2: $f'(2) = 2 > 0$f′(2)=2>0 (still increasing)

Therefore, the least value of $b$b is $-2$−2.