CUET UG 2025 — Mathematics Calculus

To determine if the function is increasing or decreasing, the derivative $f'(x)$f′(x) is examined. If $f'(x) > 0$f′(x)>0 for all $x$x, the function is increasing. If $f'(x) < 0$f′(x)<0 for all $x$x, the function is decreasing.

---

$$f(x) = x^3 + 3x^2 + 4x + 4$$f(x)=x3+3x2+4x+4

Using the power rule:

$$f'(x) = 3x^2 + 6x + 4$$f′(x)=3x2+6x+4

---

To determine the sign of $f'(x) = 3x^2 + 6x + 4$f′(x)=3x2+6x+4, the discriminant is calculated.

For the quadratic $ax^2 + bx + c$ax2+bx+c where $a = 3$a=3, $b = 6$b=6, $c = 4$c=4:

$$D = b^2 - 4ac$$D=b2−4ac

$$D = (6)^2 - 4(3)(4)$$D=(6)2−4(3)(4)

$$D = 36 - 48$$D=36−48

$$D = -12$$D=−12

---

Since $D < 0$D<0, the quadratic $f'(x) = 3x^2 + 6x + 4$f′(x)=3x2+6x+4 has no real roots.

Since the coefficient of $x^2$x2 is positive ($a = 3 > 0$a=3>0), the parabola opens upward.

Therefore, $f'(x) > 0$f′(x)>0 for all $x \in \mathbb{R}$x∈R.

---

Since $f'(x) > 0$f′(x)>0 for all $x \in \mathbb{R}$x∈R, the function $f(x) = x^3 + 3x^2 + 4x + 4$f(x)=x3+3x2+4x+4 is increasing on $\mathbb{R}$R.