Class 12 Maths · Chapter 6
Short answer:
Is chapter me hum derivatives ko real-life aur geometry problems solve karne ke liye use karte hain — rate of change (related rates), increasing/decreasing functions, tangent-normal equations, approximations, aur maxima-minima (first derivative test, second derivative test, absolute max/min, optimization word problems). Neeche 20 exercise-style questions hain jo har sub-topic cover karte hain — rate of change ke 3 problems, monotonicity ke 3, tangent/normal ke 2, approximation ke 3, local max-min ke 3, absolute max-min ke 2, aur optimization word problems ke 4.
Application of Derivatives chapter derivative ko ek tool ki tarah use karta hai — rate measure karne ke liye, curve ka behaviour samajhne ke liye, aur best possible value (maximum ya minimum) dhoondne ke liye. Board exam me ye chapter high-weightage hota hai kyunki isme conceptual clarity aur calculation dono test hote hain. Maxima-minima wale questions especially favourite hain examiners ke, kyunki inme first derivative test aur second derivative test dono use ho sakte hain — dono method aana chahiye.
Chapter 6 Summary — 5 Minute Revision
1. Rate of Change of Quantities
Agar y ek quantity hai jo x pe depend karti hai, to dy/dx us rate ko represent karta hai jis rate se y, x ke respect me change ho rahi hai. Jab dono quantities time t ke function hon, to related rates problems me chain rule use karke inko connect karte hain.
- Agar y = f(x), to y ke change ki rate = dy/dt = f'(x) · dx/dt (chain rule)
- Related rates problems me pehle ek equation banao jo dono variables ko connect kare, phir dono sides ka time ke respect me derivative lo
2. Increasing and Decreasing Functions
Interval (a, b) me function f:
- Increasing hai agar us interval me har x1 < x2 ke liye f(x1) ≤ f(x2), yaani f'(x) ≥ 0
- Decreasing hai agar f'(x) ≤ 0
- Strictly increasing/decreasing ke liye strict inequality f'(x) > 0 ya f'(x) < 0 chahiye
Method: f'(x) nikalo, usko zero karke critical points dhoondo, phir number line pe intervals bana ke f'(x) ka sign check karo.
3. Tangent and Normal
Curve y = f(x) pe point (x1, y1) par:
- Tangent ka slope = f'(x1) = dy/dx at (x1, y1)
- Tangent equation: y − y1 = f'(x1)(x − x1)
- Normal, tangent ke perpendicular hoti hai, isliye normal ka slope = −1/f'(x1)
- Normal equation: y − y1 = −1/f'(x1) · (x − x1)
4. Approximations
Chhote change Δx ke liye, function ki value me approximate change:
- Δy ≈ f'(x)·Δx, yaani f(x + Δx) ≈ f(x) + f'(x)·Δx
- Ye tab useful hai jab hume sqrt, cube root, ya kisi function ki value ek known point ke paas approximate karni ho
5. Maxima and Minima
Critical points: jahan f'(x) = 0 ya f'(x) exist nahi karta.
First Derivative Test
| Condition around critical point c | Result |
|---|---|
| f'(x) > 0 before c, f'(x) < 0 after c | Local maximum at x = c |
| f'(x) < 0 before c, f'(x) > 0 after c | Local minimum at x = c |
| f'(x) ka sign change nahi hota | Neither max nor min (point of inflection) |
↔ Table ko side me swipe karein
Second Derivative Test
Second derivative test se maxima-minima jaldi mil jaata hai — f''(x) < 0 pe maxima, f''(x) > 0 pe minima. Agar f''(x) = 0 aaye to test fail ho jaata hai, phir first derivative test use karna padta hai.
Absolute Maxima and Minima (closed interval [a, b])
- f'(x) = 0 karke critical points dhoondo jo [a, b] ke andar hain
- f ki value in critical points pe nikalo
- f(a) aur f(b) — dono endpoints pe bhi value nikalo
- Sabse badi value = absolute maximum, sabse choti = absolute minimum
6. Optimization Word Problems
Real-life problems (maximum area, maximum volume, minimum cost) me steps:
- Ek variable (jaise x) define karo aur baaki quantities usme express karo (constraint equation use karke)
- Quantity jo maximize/minimize karni hai usko x ke function ki tarah likho
- Derivative zero karke critical point nikalo
- Second derivative test ya first derivative test se confirm karo ki max hai ya min
Poore Class 12 Maths ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q20)
Q1. Radius of a circle is increasing at the rate of 3 cm/s. Find the rate at which the area is increasing when the radius is 10 cm.
Area A = πr². Chain rule se dA/dt = 2πr · dr/dt.
dr/dt = 3 cm/s, r = 10 cm
dA/dt = 2π(10)(3) = 60π cm²/s
Q2. A 5 m long ladder is leaning against a wall. The bottom is pulled away from the wall at the rate of 2 cm/s. Find the rate at which the top of the ladder slides down when the bottom is 4 m from the wall.
Let x = distance of bottom from wall, y = height of top on wall. x² + y² = 25 (ladder length constant).
2x·dx/dt + 2y·dy/dt = 0
Given x = 4, so y² = 25 − 16 = 9, y = 3
dx/dt = 2 cm/s
dy/dt = −(x/y)(dx/dt) = −(4/3)(2) = −8/3 cm/s
Negative sign shows the top is sliding down at 8/3 cm/s.
Q3. A spherical balloon's radius is increasing at 5 cm/s. Find the rate of increase of its volume when the radius is 6 cm.
Volume V = (4/3)πr³.
dV/dt = 4πr² · dr/dt
dV/dt = 4π(36)(5) = 720π cm³/s
Q4. Find the intervals in which f(x) = 2x³ − 15x² + 36x + 1 is increasing and decreasing.
f'(x) = 6x² − 30x + 36 = 6(x² − 5x + 6) = 6(x − 2)(x − 3).
f'(x) = 0 at x = 2, x = 3
x < 2: f'(x) > 0 → increasing
2 < x < 3: f'(x) < 0 → decreasing
x > 3: f'(x) > 0 → increasing
So f is increasing on (−∞, 2) ∪ (3, ∞) and decreasing on (2, 3).
Q5. Find the intervals in which f(x) = x³ − 12x² + 36x + 17 is increasing or decreasing.
f'(x) = 3x² − 24x + 36 = 3(x² − 8x + 12) = 3(x − 2)(x − 6).
Critical points x = 2, x = 6
x < 2: f'(x) > 0 → increasing
2 < x < 6: f'(x) < 0 → decreasing
x > 6: f'(x) > 0 → increasing
Increasing on (−∞, 2) ∪ (6, ∞), decreasing on (2, 6).
Q6. Show that the function f(x) = x³ − 6x² + 12x − 1 is increasing on R.
f'(x) = 3x² − 12x + 12 = 3(x² − 4x + 4) = 3(x − 2)².
(x − 2)² ≥ 0 for all real x, so f'(x) ≥ 0 for all x
f'(x) = 0 sirf ek isolated point x = 2 par hota hai, baaki har jagah f'(x) > 0. Isliye f(x) R par increasing hai.
Q7. Find the equations of the tangent and normal to the curve y = x³ at the point (1, 1).
dy/dx = 3x². At x = 1, slope of tangent = 3.
Tangent: y − 1 = 3(x − 1) ⇒ y = 3x − 2
Slope of normal = −1/3
Normal: y − 1 = −(1/3)(x − 1) ⇒ x + 3y = 4
Q8. Find the equations of the tangent and normal to the curve y = x² + 4x + 1 at the point where it crosses the y-axis.
Curve crosses y-axis at x = 0, so y = 1. Point is (0, 1).
dy/dx = 2x + 4, at x = 0, slope = 4
Tangent: y − 1 = 4(x − 0) ⇒ y = 4x + 1
Slope of normal = −1/4
Normal: y − 1 = −(1/4)x ⇒ x + 4y = 4
Q9. Using derivatives, find the approximate value of √25.3.
Let f(x) = √x, x = 25, Δx = 0.3.
f'(x) = 1/(2√x)
f(25) = 5, f'(25) = 1/10 = 0.1
Δy ≈ f'(x)·Δx = 0.1 × 0.3 = 0.03
√25.3 ≈ f(25) + Δy = 5 + 0.03 = 5.03
Q10. Using derivatives, find the approximate value of (66)^(1/3).
Let f(x) = x^(1/3), x = 64, Δx = 2 (since 64 is a perfect cube close to 66).
f'(x) = 1/(3x^(2/3))
f(64) = 4, f'(64) = 1/(3×16) = 1/48
Δy ≈ f'(x)·Δx = (1/48)(2) = 1/24 ≈ 0.0417
(66)^(1/3) ≈ 4 + 0.0417 = 4.0417
Q11. Find the approximate change in the volume of a cube of side x metres caused by increasing the side by 2% (take x = 10 cm, Δx = 0.02x).
V = x³, dV/dx = 3x².
x = 10 cm, Δx = 0.02 × 10 = 0.2 cm
ΔV ≈ 3x²·Δx = 3(100)(0.2) = 60 cm³
Note: agar Δx directly 0.02 cm diya ho (jaise 10 se 10.02 cm), to ΔV = 3(100)(0.02) = 6 cm³ — dono version practice karo.
Q12. Find the local maximum and local minimum values of f(x) = x³ − 6x² + 9x + 15 using the first derivative test.
f'(x) = 3x² − 12x + 9 = 3(x² − 4x + 3) = 3(x − 1)(x − 3).
Critical points x = 1, x = 3
x < 1: f'(x) > 0 (increasing)
1 < x < 3: f'(x) < 0 (decreasing) → x = 1 is local maximum
x > 3: f'(x) > 0 (increasing) → x = 3 is local minimum
f(1) = 1 − 6 + 9 + 15 = 19 (local max value)
f(3) = 27 − 54 + 27 + 15 = 15 (local min value)
Q13. Find the local maxima and local minima of f(x) = x³ − 3x using the second derivative test.
f'(x) = 3x² − 3 = 3(x² − 1) = 0 ⇒ x = ±1.
f''(x) = 6x
At x = 1: f''(1) = 6 > 0 → local minimum, f(1) = 1 − 3 = −2
At x = −1: f''(−1) = −6 < 0 → local maximum, f(−1) = −1 + 3 = 2
Q14. Find the local maximum and local minimum values of f(x) = 2x³ − 21x² + 36x − 20 using the second derivative test.
f'(x) = 6x² − 42x + 36 = 6(x² − 7x + 6) = 6(x − 1)(x − 6).
Critical points x = 1, x = 6
f''(x) = 12x − 42
At x = 1: f''(1) = −30 < 0 → local maximum
f(1) = 2 − 21 + 36 − 20 = −3 (local max value)
At x = 6: f''(6) = 30 > 0 → local minimum
f(6) = 432 − 756 + 216 − 20 = −128 (local min value)
Q15. Find the absolute maximum and absolute minimum values of f(x) = x³ − 3x on the interval [−2, 2].
f'(x) = 3x² − 3 = 0 ⇒ x = ±1, both lie in [−2, 2].
f(−2) = −8 + 6 = −2
f(−1) = −1 + 3 = 2
f(1) = 1 − 3 = −2
f(2) = 8 − 6 = 2
Comparing all values (−2, 2, −2, 2): absolute maximum = 2 (at x = −1 and x = 2), absolute minimum = −2 (at x = −2 and x = 1).
Q16. Find the absolute maximum and absolute minimum values of f(x) = 2x³ − 15x² + 36x + 1 on the interval [1, 5].
f'(x) = 6(x − 2)(x − 3) = 0 ⇒ x = 2, 3, both lie in [1, 5].
f(1) = 2 − 15 + 36 + 1 = 24
f(2) = 16 − 60 + 72 + 1 = 29
f(3) = 54 − 135 + 108 + 1 = 28
f(5) = 250 − 375 + 180 + 1 = 56
Comparing 24, 29, 28, 56: absolute maximum = 56 at x = 5; absolute minimum = 24 at x = 1.
Q17. Find two positive numbers x and y such that x + y = 60 and xy³ is maximum.
y = 60 − x. Let P = xy³ = x(60 − x)³.
dP/dx = (60 − x)³ + x·3(60 − x)²(−1) = (60 − x)²[(60 − x) − 3x] = (60 − x)²(60 − 4x)
dP/dx = 0 ⇒ x = 60 (rejected, gives y = 0) or x = 15
Second derivative test confirms x = 15 gives maximum P
So the numbers are x = 15 and y = 45.
Q18. A square piece of tin of side 18 cm is to be made into a box without a top by cutting a square of side x cm from each corner and folding up the flaps. Find the value of x for which the volume is maximum, and find the maximum volume.
Length of box = 18 − 2x, height = x, so Volume V = x(18 − 2x)².
dV/dx = (18 − 2x)² + x·2(18 − 2x)(−2) = (18 − 2x)[(18 − 2x) − 4x] = (18 − 2x)(18 − 6x)
dV/dx = 0 ⇒ x = 9 (rejected, box collapses) or x = 3
d²V/dx² at x = 3 is negative → maximum
Maximum volume V = 3(18 − 6)² = 3(12)² = 3 × 144 = 432 cm³
Q19. Divide 36 into two parts such that the sum of their squares is minimum.
Let the parts be x and 36 − x. S = x² + (36 − x)².
dS/dx = 2x − 2(36 − x) = 4x − 72
dS/dx = 0 ⇒ x = 18
d²S/dx² = 4 > 0 → minimum
So the two parts are 18 and 18, and minimum sum of squares = 18² + 18² = 648.
Q20. Show that the rectangle of maximum area that can be inscribed in a circle of radius r is a square.
Let the rectangle have half-sides x and y, inscribed in circle: x² + y² = r².
Area A = (2x)(2y) = 4xy = 4x√(r² − x²)
Maximize A² = 16x²(r² − x²); let u = x²
d(A²)/du = 16(r² − 2u) = 0 ⇒ u = r²/2 ⇒ x = r/√2
Then y² = r² − x² = r²/2 ⇒ y = r/√2
Since x = y, all sides of the rectangle are equal — it is a square, with side = 2x = r√2. This confirms area is maximum when the inscribed rectangle is a square.
Important Equations — Ek Nazar Me
| Concept | Formula |
|---|---|
| Rate of change | dy/dt = f'(x) · dx/dt (chain rule) |
| Increasing function | f'(x) > 0 on the interval |
| Decreasing function | f'(x) < 0 on the interval |
| Equation of tangent at (x1, y1) | y − y1 = f'(x1)(x − x1) |
| Equation of normal at (x1, y1) | y − y1 = −1/f'(x1) · (x − x1) |
| Approximation | Δy ≈ f'(x)·Δx, so f(x + Δx) ≈ f(x) + f'(x)·Δx |
| First derivative test (max) | f' changes + to − at x = c |
| First derivative test (min) | f' changes − to + at x = c |
| Second derivative test (max) | f'(c) = 0 and f''(c) < 0 |
| Second derivative test (min) | f'(c) = 0 and f''(c) > 0 |
| Absolute max/min on [a, b] | Compare f at critical points and at x = a, x = b |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Tangent aur normal ka slope relation ulta karna. Normal ka slope tangent ke slope ka negative reciprocal hota hai (−1/f'(x1)), khud f'(x1) nahi. Ye galti bahut common hai jab student jaldi me answer likhte hain.
- Closed interval ke endpoints check nahi karna. Absolute maxima/minima nikalte waqt sirf critical points pe f(x) check karna kaafi nahi hai — f(a) aur f(b), dono endpoints ki value bhi zaroor compare karo.
- Increasing/decreasing ke sign test me galti. f'(x) ka factorised form nikal ke number line pe sign check karte waqt sign flip karna bhool jaana — har critical point ke aage-peeche alag interval me sign dobara test karo, assume mat karo.
- Critical point domain se bahar hai ye verify na karna. f'(x) = 0 solve karke jo bhi roots milte hain unme se sirf woh lo jo given interval/domain ke andar hain — domain se bahar wale roots ko discard karna zaroori hai.
- Rate of change word problems me units bhool jaana. dr/dt, dA/dt jaise answers me units (cm/s, cm²/s, cm³/s) likhna zaroori hai — sirf numeric value dena incomplete answer maana jaata hai.
- Second derivative test me f''(c) = 0 aane par bhi conclusion nikal lena. Agar f''(c) = 0 aata hai to second derivative test inconclusive hai — is case me first derivative test use karna hi padta hai, galat se maxima/minima ka answer likhna mat.
Board-Style Important Questions
- 1 mark: Find the slope of the normal to the curve y = x² at the point (1, 1).
- 2 marks: Find the rate of change of the area of a circle with respect to its radius r when r = 6 cm.
- 3 marks: Find the intervals in which the function f(x) = x² − 4x + 6 is strictly increasing or strictly decreasing.
- 4 marks: Find the equation of the tangent and normal to the curve y = x³ − 3x + 2 at the point where it crosses the x-axis.
- 5 marks: Show that of all the rectangles inscribed in a given circle, the square has the maximum area.
- 5 marks: A wire of length 28 m is cut into two pieces; one is bent into a square and the other into a circle. Find how the wire should be cut so that the combined area is minimum.
Aksar Poochhe Jaane Wale Sawaal
Application of Derivatives chapter me maxima-minima kyun important hai?
Kyunki optimization word problems (max area, max volume, min cost) real-life applications hain aur board exam me isse 4-5 mark ka question aksar aata hai. First derivative aur second derivative test dono aana chahiye.
Tangent aur normal ka slope relation kya hai?
Tangent ka slope curve ke derivative f'(x1) ke barabar hota hai. Normal, tangent ke perpendicular hoti hai, isliye normal ka slope tangent ke slope ka negative reciprocal hota hai, yaani −1/f'(x1).
First derivative test aur second derivative test me kaunsa use karna chahiye?
Second derivative test usually fast hota hai jab f''(x) easily nikal sake aur zero na aaye. Agar f''(c) = 0 aa jaaye ya second derivative nikalna mushkil ho, to first derivative test (sign change check) use karo.
Absolute maxima/minima aur local maxima/minima me kya farak hai?
Local maxima/minima sirf ek chhoti neighbourhood me sabse badi/chhoti value hoti hai. Absolute maxima/minima poore given interval [a, b] me sabse badi/chhoti value hoti hai — isme endpoints bhi check karne padte hain.
Approximation wale questions me formula kaise yaad rakhein?
Δy ≈ f'(x)·Δx yaad rakho — matlab function ki value me chhota change, derivative × input ka chhota change ke barabar hota hai. Isse √, cube root jaisi values bina calculator ke approximate ki ja sakti hain.
Optimization word problems solve karne ka sahi tarika kya hai?
Pehle constraint equation banao jo variables ko connect kare, ek variable ko doosre me express karo, quantity (area/volume/cost) ko single variable ka function banao, derivative zero karke critical point nikalo, aur second derivative test se confirm karo max hai ya min.
Class 12 Maths — Saare Chapters
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