Maxima & Minima (Local/Absolute) PYQs
Practice Class 12 CBSE Board Previous Year Questions (2014-2026)
Q1
2025 Board
00:00
Find the absolute maximum value of $f(x) = x^3 - 3x + 2$ in the interval $[0, 2]$. 1 Mark
$f'(x) = 3x^2 - 3 = 3(x^2 - 1) = 3(x-1)(x+1)$.
Critical point in $[0, 2]$ is $x = 1$.
Values at endpoints and critical points:
$f(0) = 0 - 0 + 2 = 2$.
$f(1) = 1 - 3 + 2 = 0$.
$f(2) = 8 - 6 + 2 = 4$.
Wait, the user says answer is 5. Let me re-calculate or check the question again.
If $f(x) = x^3 - 3x + 2$, $f(2) = 4$. If $f(x) = x^3 - 3x + 4$, $f(2)=6$.
Maybe $f(x) = x^3 - 3x + 3$? $f(2) = 8-6+3=5$.
I'll use the user's provided answer if it's a known board question variant, but 4 is the math for $x^3-3x+2$ at $x=2$.
4 (Check variant for 5)
Q2
2024
00:00
Find local maxima and local minima of the function $f(x) = \sin 4x + \cos 4x$ for $0 < x < \frac{\pi}{2}$. 4 Marks
$f'(x) = 4\cos 4x - 4\sin 4x$.
Set $f'(x) = 0 \Rightarrow \tan 4x = 1 \Rightarrow 4x = \pi/4, 5\pi/4$.
$x = \pi/16, 5\pi/16$.
$f''(x) = -16\sin 4x - 16\cos 4x = -16(\sin 4x + \cos 4x)$.
At $x = \pi/16, f''(x) < 0 \Rightarrow$ Local Maxima.
At $x = 5\pi/16, f''(x) > 0 \Rightarrow$ Local Minima.
Max at π/16, Min at 5π/16
Q3
2022 (Term 2)
00:00
Find the local maximum and local minimum values of $f(x) = \sin x (1 + \cos x)$ in $(0, \pi/2)$. 3 Marks
$f'(x) = \cos x(1+\cos x) + \sin x(-\sin x) = \cos x + \cos^2 x - \sin^2 x$.
$= \cos x + \cos 2x = 2\cos(3x/2)\cos(x/2)$.
Set $f'(x) = 0$. Since $x \in (0, \pi/2)$, $x/2 \in (0, \pi/4)$ (never zero).
$3x/2 = \pi/2 \Rightarrow x = \pi/3$.
$f''(\pi/3) = -\sin(\pi/3) - 2\sin(2\pi/3) < 0 \Rightarrow$ Local Maxima.
Max at π/3
Q4
2019
00:00
Find the absolute maximum and absolute minimum values of $f(x) = x^3 - 6x^2 + 9x + 1$ on $[-1, 2]$. 4 Marks
$f'(x) = 3x^2 - 12x + 9 = 3(x-1)(x-3)$.
Critical point in $[-1, 2]$ is $x = 1$.
$f(-1) = -1 - 6 - 9 + 1 = -15$.
$f(1) = 1 - 6 + 9 + 1 = 5$.
$f(2) = 8 - 24 + 18 + 1 = 3$.
Absolute Max = 5, Absolute Min = -15.
Abs Max: 5, Abs Min: -15
Q5
2023
00:00
Find all local maxima/minima of $f(x) = 2\sin x - x$ for $-\pi/2 \le x \le \pi/2$. 3 Marks
$f'(x) = 2\cos x - 1$. Set to zero: $\cos x = 1/2 \Rightarrow x = \pm \pi/3$.
$f''(x) = -2\sin x$.
At $x = \pi/3, f'' < 0 \Rightarrow$ Local Maxima.
At $x = -\pi/3, f'' > 0 \Rightarrow$ Local Minima.
Max at π/3, Min at -π/3
Q6
2021
00:00
Find local max/min of $f(x) = 3x^4 + 4x^3 - 12x^2 + 12$. 3 Marks
$f'(x) = 12x^3 + 12x^2 - 24x = 12x(x^2 + x - 2) = 12x(x+2)(x-1)$.
Critical points: $0, -2, 1$.
Using second derivative or sign chart: Min at $x=-2, 1$ and Max at $x=0$.
Max at x=0, Min at x=-2, 1
Q7
2018
00:00
Show that $f(x) = \sin x - \cos x$ for $0 < x < 2\pi$ has local minima at $x = 5\pi/4$. 3 Marks
$f'(x) = \cos x + \sin x$. At $x=5\pi/4$, $\cos(5\pi/4) + \sin(5\pi/4) = -1/\sqrt{2} - 1/\sqrt{2} \neq 0$. Wait, check critical points.
$\cos x + \sin x = 0 \Rightarrow \tan x = -1 \Rightarrow x = 3\pi/4, 7\pi/4$.
If $f(x) = \sin x + \cos x$, then $f' = \cos x - \sin x = 0 \Rightarrow x = \pi/4, 5\pi/4$.
$f'' = -\sin x - \cos x$. At $x=5\pi/4$, $f'' = -(-1/\sqrt{2}) - (-1/\sqrt{2}) = \sqrt{2} > 0 \Rightarrow$ Minima.
The user's question might be $f(x) = \sin x + \cos x$. I'll use the correct math.
Proved for f(x) = sin x + cos x
Q8
2017
00:00
Find absolute max/min of $f(x) = 2x^3 - 15x^2 + 36x + 1$ on $[1, 5]$. 4 Marks
$f'(x) = 6x^2 - 30x + 36 = 6(x-2)(x-3)$.
Critical points: $2, 3$. Endpoints: $1, 5$.
$f(1) = 24$. $f(2) = 29$. $f(3) = 28$. $f(5) = 56$.
Abs Max: 56 at x=5, Abs Min: 24 at x=1.
Abs Max: 56, Abs Min: 24
Q9
2015
00:00
Find local max/min of $f(x) = \sin^2 x - x$ for $-\pi/2 \le x \le \pi/2$. 3 Marks
$f'(x) = 2\sin x \cos x - 1 = \sin 2x - 1$.
Set $\sin 2x = 1 \Rightarrow 2x = \pi/2 \Rightarrow x = \pi/4$.
$f''(x) = 2\cos 2x$. At $x=\pi/4$, $f'' = 0$. Use first derivative test.
Since $\sin 2x \le 1$, $f'(x) \le 0$ always. It's a point of inflection or strictly decreasing.
No local extremum
Q10
2020
00:00
Find the local maximum and local minimum values of $f(x) = x^3 - 6x^2 + 9x + 15$. 3 Marks
$f'(x) = 3x^2 - 12x + 9 = 3(x-1)(x-3)$.
Critical points: $1, 3$.
$f''(x) = 6x - 12$.
At $x=1, f'' = -6 < 0 \Rightarrow$ Max value $= 1-6+9+15 = 19$.
At $x=3, f'' = 6 > 0 \Rightarrow$ Min value $= 27-54+27+15 = 15$.
Local Max: 19, Local Min: 15
Q11
2016
00:00
Find the absolute maximum and absolute minimum values of $f(x) = 12x^{4/3} - 6x^{1/3}$ on $[-1, 1]$. 4 Marks
$f'(x) = 12(4/3)x^{1/3} - 6(1/3)x^{-2/3} = 16x^{1/3} - 2/x^{2/3}$.
Set $f'(x) = 0 \Rightarrow 16x - 2 = 0 \Rightarrow x = 1/8$.
$f(-1) = 12(-1)^{4/3} - 6(-1)^{1/3} = 12(1) - 6(-1) = 18$.
$f(1/8) = 12(1/2)^4 - 6(1/2) = 12/16 - 3 = 0.75 - 3 = -2.25$.
$f(1) = 12(1) - 6(1) = 6$.
Abs Max: 18, Abs Min: -2.25.
Abs Max: 18, Abs Min: -2.25