The maximum slope of the curve y = -x^3 + 3x^2 + 9x - 27 is: Option 1: 0 Option 2: 12 Option 3: 16 Option 4: 32

Answer: 12. Solution: Slope = y' = -3x^2 + 6x + 9. To maximize: d/dx(y') = -6x + 6 = 0, so x = 1. Second derivative test: -6 < 0, confirming maximum. Maximum slope = -3 + 6 + 9 = 12.

Calculus

Application of Derivatives

Medium

core

The maximum slope of the curve $y = -x^3 + 3x^2 + 9x - 27$ is:

✅ Correct Option: 2

Slope $= y' = -3x^2 + 6x + 9$ . To maximize: $\frac{d}{dx}(y') = -6x + 6 = 0$ , so $x = 1$ . Second derivative test: $-6 < 0$ , confirming maximum. Maximum slope $= -3 + 6 + 9 = 12$ .

2025: 14 May Shift 2

Match List-I with List-II

The function $f(x) = 2x^3 - 15x^2 + 36x + 5$ for $x \in [2,5]$ has

List-I	List-II
(A) absolute maximum value	(I) 5
(B) absolute minimum value	(II) 60
(C) point of absolute maxima	(III) 3
(D) point of absolute minima	(IV) 32

Choose the correct answer from the options given below:

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The maximum slope of the curve y=−x3+3x2+9x−27y = -x^3 + 3x^2 + 9x - 27y=−x3+3x2+9x−27 is:

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

The maximum slope of the curve $y = -x^3 + 3x^2 + 9x - 27$ is: