CUET UG 2025 — Mathematics Calculus

For a function to be continuous at $x = 1$x=1, the left-hand limit must equal the right-hand limit at that point.

---

When approaching $x = 1$x=1 from the left (where $x < 1$x<1), the function uses $f(x) = 2x + 1$f(x)=2x+1.

The left-hand limit at $x = 1$x=1:

$$\lim_{x \to 1^-} f(x) = 2(1) + 1$$x→1−lim​f(x)=2(1)+1

$$= 3$$=3

---

When approaching $x = 1$x=1 from the right (where $x > 1$x>1), the function uses $f(x) = ax - 1$f(x)=ax−1.

The right-hand limit at $x = 1$x=1:

$$\lim_{x \to 1^+} f(x) = a(1) - 1$$x→1+lim​f(x)=a(1)−1

$$= a - 1$$=a−1

---

For continuity at $x = 1$x=1:

$$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x)$$x→1−lim​f(x)=x→1+lim​f(x)

$$3 = a - 1$$3=a−1

$$a = 4$$a=4

Therefore, $a = 4$a=4.