The function f(x) = kx^3 + 6kx^2 + 18x + 17 is increasing on R(set of real numbers) if: Option 1: k (0,3) Option 2: k (0, 3/2) Option 3: k (1/2, 3/2) Option 4: k (1/2, 5/2)

Answer: k (0, 3/2). Solution: For f(x) to be increasing on R, the derivative f'(x) ≥ 0 for all real x. f(x) = kx^3 + 6kx^2 + 18x + 17 f'(x) = 3kx^2 + 12kx + 18 This is a quadratic in x where a = 3k, b = 12k, c = 18. For a quadratic expression to be ≥ 0 for every real x, two conditions must hold: 1. Coefficient of x^2 > 0 (parabola opens upward) 2. Discriminant ≤ 0 (never goes negative) Condition 1 — Coefficient of x^2 > 0: 3k > 0 k > 0 s(i) Condition 2 — Discriminant ≤ 0: D = b^2 - 4ac ≤ 0 D = (12k)^2 - 4(3k)(18) ≤ 0 144k^2 - 216k ≤ 0 2k^2 - 3k ≤ 0 k(2k - 3) ≤ 0 The roots are k = 0 and k = 3/2. The expression is ≤ 0 between the roots: 0 ≤ k ≤ 3/2 s(ii) From (i): k > 0 From (ii): 0 ≤ k ≤ 3/2 Taking the intersection: k (0,\ 3/2)

Calculus

Application of Derivatives

Medium

applied

The function $f(x) = kx^3 + 6kx^2 + 18x + 17$ is increasing on $\mathbb{R}$ (set of real numbers) if:

$k \in (0,3)$

$k \in (0, \frac{3}{2})$

$k \in (\frac{1}{2}, \frac{3}{2})$

$k \in (\frac{1}{2}, \frac{5}{2})$

✅ Correct Option: 2

For $f(x)$ to be increasing on $\mathbb{R}$ , the derivative $f'(x) \geq 0$ for all real $x$ .

$f(x) = kx^3 + 6kx^2 + 18x + 17$

$f'(x) = 3kx^2 + 12kx + 18$

This is a quadratic in $x$ where $a = 3k$ , $b = 12k$ , $c = 18$ .

For a quadratic expression to be $\geq 0$ for every real $x$ , two conditions must hold:

Coefficient of $x^2 > 0$ (parabola opens upward)
Discriminant $\leq 0$ (never goes negative)

Condition 1 — Coefficient of $x^2 > 0$ :

$3k > 0$

$k > 0 \quad \cdots(i)$

Condition 2 — Discriminant $\leq 0$ :

$D = b^2 - 4ac \leq 0$

$D = (12k)^2 - 4(3k)(18) \leq 0$

$144k^2 - 216k \leq 0$

$2k^2 - 3k \leq 0$

$k(2k - 3) \leq 0$

The roots are $k = 0$ and $k = \dfrac{3}{2}$ . The expression is $\leq 0$ between the roots:

$0 \leq k \leq \dfrac{3}{2} \quad \cdots(ii)$

From $(i)$ : $k > 0$

From $(ii)$ : $0 \leq k \leq \dfrac{3}{2}$

Taking the intersection:

$k \in \left(0,\ \dfrac{3}{2}\right)$

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The function $f(x) = x^3 + 3x^2 + 4x + 4$ , $x \in \mathbb{R}$ (set of real numbers) :

The function f(x)=kx3+6kx2+18x+17f(x) = kx^3 + 6kx^2 + 18x + 17f(x)=kx3+6kx2+18x+17 is increasing on R\mathbb{R}R(set of real numbers) if:

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

The function $f(x) = kx^3 + 6kx^2 + 18x + 17$ is increasing on $\mathbb{R}$ (set of real numbers) if: