CUET UG 2025 — Mathematics Calculus

We have a function: $f(x) = x^2 + ax + 5$f(x)=x2+ax+5

We need to find the value of $k$k where the minimum value of $a$a is $-\frac{k}{2}$−2k​, such that the function is increasing on the interval $[1, 2]$[1,2].

---

A function is increasing when its derivative is non-negative throughout the interval.

For $f(x) = x^2 + ax + 5$f(x)=x2+ax+5, the derivative is:

$$f'(x) = 2x + a$$f′(x)=2x+a

---

For the function to be increasing on $[1, 2]$[1,2]:

$f'(x) \geq 0$f′(x)≥0 for all $x \in [1, 2]$x∈[1,2]

$$2x + a \geq 0$$2x+a≥0

$$a \geq -2x$$a≥−2x

---

The condition $a \geq -2x$a≥−2x must hold for every value of $x$x in $[1, 2]$[1,2].

Since $-2x$−2x is a decreasing function (it gets more negative as $x$x increases), we need to find where it's largest.

At $x = 1$x=1: $-2(1) = -2$−2(1)=−2

At $x = 2$x=2: $-2(2) = -4$−2(2)=−4

The largest value of $-2x$−2x on $[1, 2]$[1,2] is $-2$−2 (at $x = 1$x=1).

Therefore: $a \geq -2$a≥−2

The minimum value of $a$a is $-2$−2.

---

The question states the minimum value of $a$a is $-\frac{k}{2}$−2k​.

So:

$$-\frac{k}{2} = -2$$−2k​=−2

$$\frac{k}{2} = 2$$2k​=2

$$k = 4$$k=4

---

Therefore, $k = 4$k=4