Increasing & Decreasing Functions PYQs
Practice Class 12 CBSE Board Previous Year Questions (2014-2026)
Q1
2024
00:00
Find the intervals in which the function $f(x) = 2x^3 - 3x^2 - 36x + 7$ is strictly increasing or strictly decreasing. 3 Marks
$f'(x) = 6x^2 - 6x - 36 = 6(x^2 - x - 6) = 6(x-3)(x+2)$.
Critical points: $x = -2, 3$.
Intervals: $(-\infty, -2), (-2, 3), (3, \infty)$.
In $(-\infty, -2)$, $f'(x) > 0 \Rightarrow$ Strictly Increasing.
In $(-2, 3)$, $f'(x) < 0 \Rightarrow$ Strictly Decreasing.
In $(3, \infty)$, $f'(x) > 0 \Rightarrow$ Strictly Increasing.
Inc: (-∞, -2) ∪ (3, ∞), Dec: (-2, 3)
Q2
2021
00:00
Show that the function $f(x) = 4x^3 - 18x^2 + 27x - 7$ is always increasing on $\mathbb{R}$. 3 Marks
$f'(x) = 12x^2 - 36x + 27 = 3(4x^2 - 12x + 9)$.
$f'(x) = 3(2x - 3)^2$.
Since $(2x-3)^2 \ge 0$ for all $x \in \mathbb{R}$, $f'(x) \ge 0$.
Thus, $f(x)$ is always increasing on $\mathbb{R}$.
Proved
Q3
2019
00:00
Find the intervals in which the function $f(x) = (x+1)^3(x-3)^3$ is strictly increasing or strictly decreasing. 4 Marks
$f'(x) = 3(x+1)^2(x-3)^3 + 3(x+1)^3(x-3)^2 = 3(x+1)^2(x-3)^2 [ (x-3) + (x+1) ]$.
$f'(x) = 3(x+1)^2(x-3)^2 (2x-2) = 6(x+1)^2(x-3)^2(x-1)$.
Critical points: $x = -1, 1, 3$.
Notice $(x+1)^2$ and $(x-3)^2$ are always non-negative. Sign depends on $(x-1)$.
If $x > 1$, $f'(x) > 0$ (Increasing).
If $x < 1$, $f'(x) < 0$ (Decreasing).
Inc: (1, ∞), Dec: (-∞, 1)
Q4
2016
00:00
Prove that the function $f(x) = x^2 - x + 1$ is neither strictly increasing nor strictly decreasing on $(-1, 1)$. 2 Marks
$f'(x) = 2x - 1$.
Setting $f'(x) = 0 \Rightarrow x = 1/2$.
In $(-1, 1/2)$, $f'(x) < 0$ (Decreasing).
In $(1/2, 1)$, $f'(x) > 0$ (Increasing).
Since it is decreasing in one part and increasing in another, it is neither strictly increasing nor strictly decreasing on the whole interval $(-1, 1)$.
Proved
Q5
2023
00:00
Find the intervals in which the function $f(x) = x^4 - 8x^3 + 22x^2 - 24x + 21$ is strictly increasing. 4 Marks
$f'(x) = 4x^3 - 24x^2 + 44x - 24 = 4(x^3 - 6x^2 + 11x - 6)$.
Using hit and trial, $x=1$ is a root. $(x-1)(x^2 - 5x + 6) = (x-1)(x-2)(x-3)$.
Critical points: $x = 1, 2, 3$.
Sign of $f'(x)$:
$(-\infty, 1) \to f'(x) < 0$.
$(1, 2) \to f'(x) > 0$ (Strictly Increasing).
$(2, 3) \to f'(x) < 0$.
$(3, \infty) \to f'(x) > 0$ (Strictly Increasing).
(1, 2) ∪ (3, ∞)
Q6
2022 (Term 1 MCQ)
00:00
The function $f(x) = x^3 - 3x$ is strictly increasing in: 1 Mark
$f'(x) = 3x^2 - 3 = 3(x-1)(x+1)$.
For strictly increasing, $f'(x) > 0 \Rightarrow x \in (-\infty, -1) \cup (1, \infty)$.
(-∞, -1) ∪ (1, ∞)
Q7
2020
00:00
Find the intervals in which the function $f(x) = x^3 + x^2 - x + 2$ is strictly increasing or strictly decreasing. 3 Marks
$f'(x) = 3x^2 + 2x - 1 = (3x-1)(x+1)$.
Critical points: $x = -1, 1/3$.
Inc: $(-\infty, -1) \cup (1/3, \infty)$. Dec: $(-1, 1/3)$.
Inc: (-∞, -1) ∪ (1/3, ∞), Dec: (-1, 1/3)
Q8
2018
00:00
Find the intervals in which the function $f(x) = \frac{x^4}{4} - x^3 - 5x^2 + 24x + 12$ is strictly increasing or strictly decreasing. 4 Marks
$f'(x) = x^3 - 3x^2 - 10x + 24 = (x-2)(x-4)(x+3)$.
Critical points: $2, 4, -3$.
Intervals: $(-\infty, -3), (-3, 2), (2, 4), (4, \infty)$.
Signs: $-, +, -, +$.
Strictly Increasing: $(-3, 2) \cup (4, \infty)$. Strictly Decreasing: $(-\infty, -3) \cup (2, 4)$.
Inc: (-3, 2) ∪ (4, ∞), Dec: (-∞, -3) ∪ (2, 4)
Q9
2015
00:00
Find the intervals in which the function $f(x) = \sin x + \cos x, 0 \le x \le 2\pi$ is strictly increasing or strictly decreasing. 4 Marks
$f'(x) = \cos x - \sin x = \sqrt{2}\cos(x + \pi/4)$.
Set $f'(x) = 0 \Rightarrow x + \pi/4 = \pi/2, 3\pi/2 \Rightarrow x = \pi/4, 5\pi/4$.
Inc: $[0, \pi/4) \cup (5\pi/4, 2\pi]$. Dec: $(\pi/4, 5\pi/4)$.
Inc: [0, π/4) ∪ (5π/4, 2π], Dec: (π/4, 5π/4)
Q10
2022 (Term 2)
00:00
Find the intervals in which the function $f(x) = 2x^3 - 9x^2 + 12x + 15$ is strictly increasing or strictly decreasing. 3 Marks
$f'(x) = 6x^2 - 18x + 12 = 6(x-1)(x-2)$.
Critical points: $1, 2$.
Inc: $(-\infty, 1) \cup (2, \infty)$. Dec: $(1, 2)$.
Inc: (-∞, 1) ∪ (2, ∞), Dec: (1, 2)
Q11
2017
00:00
Show that the function $f(x) = x^3 - 3x^2 + 6x - 100$ is increasing on $\mathbb{R}$. 2 Marks
$f'(x) = 3x^2 - 6x + 6 = 3(x^2 - 2x + 2)$.
$= 3[(x-1)^2 + 1]$.
Since $(x-1)^2 + 1 > 0$ for all $x$, $f'(x) > 0$.
Thus, $f(x)$ is strictly increasing on $\mathbb{R}$.
Proved