CUET UG 2025 — Mathematics Algebra

For $g(x)$g(x) to be continuous at $x = 0$x=0:

$$\lim_{x \to 0^-} g(x) = \lim_{x \to 0^+} g(x) = g(0)$$x→0−lim​g(x)=x→0+lim​g(x)=g(0)

---

Since $x = 0$x=0 satisfies the condition $x ≥ 0$x≥0, we use the second part of the function:

$$g(0) = 5$$g(0)=5

---

When $x$x approaches $0$0 from the right (where $x > 0$x>0), the condition $x ≥ 0$x≥0 applies:

$$\lim_{x \to 0^+} g(x) = \lim_{x \to 0^+} 5 = 5$$x→0+lim​g(x)=x→0+lim​5=5

---

When $x < 0$x<0, we use $g(x) = \frac{αx}{|x|}$g(x)=∣x∣αx​

For negative values of $x$x, we have $|x| = -x$∣x∣=−x

Substituting:

$$g(x) = \frac{αx}{|x|}$$g(x)=∣x∣αx​

$$g(x) = \frac{αx}{-x}$$g(x)=−xαx​

$$g(x) = -α$$g(x)=−α

Therefore:

$$\lim_{x \to 0^-} g(x) = -α$$x→0−lim​g(x)=−α

---

Setting all three values equal:

$$-α = 5$$−α=5

$$α = -5$$α=−5

Therefore, $α = -5$α=−5