Application of Derivatives: Class 12 Maths Chapter 6 Hinglish mein

Rates, badhna aur ghatna, aur sabse achhi possible value

Mathematics Part 1 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

140 min
82 practice questions
21 MCQs

Is chapter mein aap kya seekhenge

Derivative ko rate of change ki tarah padho aur saath-saath badalti quantities ke rates ko chain rule se jodo, jismein marginal cost aur marginal revenue bhi shaamil hain
First derivative ke sign se tay karo ki function kahan increasing hai aur kahan decreasing, aur is test ko mean value theorem se justify karo
Function ke critical points nikaalo aur first derivative test se har ek ko local maximum, local minimum ya dono mein se koi nahi, aise classify karo
Jahan second derivative test lagta ho wahan use use karo, aur jab woh fail ho to first derivative test par wapas aao
Closed interval par continuous function ka absolute maximum aur minimum nikaalo, critical points aur end points par uski values compare karke
Geometry, mensuration, business aur motion se aane wale optimisation problems ko model karo aur solve karo

NCERT Mathematics Class 12, Chapter 6, pp. 147–186 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

6.1 Introduction

Derivative sirf compute karne wala ek symbol nahi hai. Yeh practical sawaalon ke jawaab deta hai. Ek puddle abhi is pal kitni tezi se fail raha hai? Kya company ka profit abhi bhi badh raha hai ya girna shuru ho gaya hai? Card ki ek sheet se bane kaun se size ke box mein sabse zyada samaan aayega? Chapter 5 mein tumne kai tarah ke functions ke derivatives nikaalna seekha. Is chapter mein tum unhe use karoge.

Hum teen dhaagon ko follow karenge (Fig. 6.1). Pehla, derivative ek rate of change naapta hai, aur chain rule ki madad se ek known rate se hum doosra rate nikaal sakte hain. Doosra, $f^{'} (x)$ ka sign batata hai ki graph kahan upar jaata hai aur kahan neeche. Teesra, jin jagahon par $f^{'} (x)$ zero hai ya exist nahi karta, sirf wahi interior jagahen hain jahan function peak ya dip tak pahunch sakta hai. Isse humein kisi quantity ki sabse badi ya sabse chhoti value nikaalne ka ek method milta hai.

3/3Peaks aur dips sirf wahin ho sakte hain jahan $f^{'} (x) = 0$ ho ya $f^{'}$ exist na kare (Section 6.4).

Fig. 6.1Teen baatein jo derivative humein graph ke baare mein batata hai.

Ek smooth curve upar jaakar peak tak pahunchta hai, neeche girkar dip tak aata hai aur phir upar jaata hai. Pehle step mein ek tangent line curve ko left mein chhooti hai, saath mein ek chhota dashed slope triangle hai jis par rate likha hai, jo f prime of x ko darshata hai. Doosre step mein badhte hisson par ek mota solid band bana hai jis par badhta likha hai, aur girte hisse par ek mota dashed band jis par girta likha hai. Teesre step mein ek upar wala triangle peak ko mark karta hai aur ek neeche wala triangle dip ko.

6.2 Quantities ka Rate of Change

Maan lo ek quantity $y$ kisi doosri quantity $x$ par ek rule $y = f (x)$ ke through depend karti hai. Derivative $\frac{d y}{d x}$ , jise $f^{'} (x)$ bhi likhte hain, batata hai ki us pal $x$ mein har ek unit change par $y$ kitne units badalta hai. Kisi ek khaas input par evaluate karne par, ${\frac{d y}{d x} ∣}_{x = x_{0}}$ ya $f^{'} (x_{0})$ naapta hai ki jis pal $x$ , $x_{0}$ se guzarta hai us pal $y$ kitni tezi se badal raha hai.

Speed jaana-pehchaana case hai. Agar $t$ seconds ke baad train ne $s$ metres distance tay ki hai, to $\frac{d s}{d t}$ uski speed hai metres per second mein. Rate ki units hamesha "upar wali quantity ki units per neeche wali quantity ki unit" hoti hain. Inhe likh lene se kai galtiyan pakdi jaati hain.

Aksar do quantities $x$ aur $y$ dono time $t$ ke saath badalti hain, maan lo $x = f (t)$ aur $y = g (t)$ . Agar humein pata hai ki har ek time ke saath kitni tezi se badalti hai, to chain rule $x$ ke respect mein $y$ ka rate deta hai:

Ek teesre variable ke through rates

\frac{d y}{d x} = \frac{d y / d t}{d x / d t}, agar \frac{d x}{d t} \neq 0

jahan

$\frac{d y}{d t}$: $y$ time ke saath kitni tezi se badalta hai
$\frac{d x}{d t}$: $x$ time ke saath kitni tezi se badalta hai

Ulta padhein to $\frac{d y}{d t} = \frac{d y}{d x} \cdot \frac{d x}{d t}$ . Related rates problems ka yahi asli auzaar hai: ek formula do quantities ko jodta hai, ek rate pata hai, aur humein doosra chahiye.

Yeh ek picture hai jo ek aise rate ko samjhaati hai jise hum baar-baar use karenge. Jab circle ka radius thoda sa $Δ r$ badhta hai, to naya area purane circle ke chaaron taraf ek patli ring hota hai (Fig. 6.2). Ring ko kaato aur seedha karo: yeh lagbhag $2 π r$ lambi aur $Δ r$ chaudi strip hai. Isliye $Δ A \approx 2 π r Δ r$ , aur $Δ r$ se divide karne par $\frac{d A}{d r} = 2 π r$ milta hai, jo bilkul circumference hai.

3/3Kholne par ring lagbhag $2 π r$ lambi aur $Δ r$ chaudi strip hai, isliye $Δ A \approx 2 π r Δ r$ .

Fig. 6.2Circle ke area ka uske radius ke saath rate of change uski circumference kyun hota hai.

Radius r wali ek shaded disc, jismein ek radius line bani hai. Doosre step mein disc ke chaaron taraf delta r width ki ek patli ring jodi jaati hai. Teesre step mein ek arrow dikhata hai ki ring kaatkar ek lambi patli strip mein khol di gayi hai jiski length lagbhag 2 pi r, yaani circumference, hai aur width delta r hai, isliye extra area delta A lagbhag 2 pi r guna delta r hai.

Example 1NCERT Example 1 · p. 148

Can ka volume uske radius ke saath

Saare cylindrical tins $10 c m$ oonche banaye jaate hain, lekin radius $r$ chuna ja sakta hai. Jab $r = 4 c m$ ho, to radius ke respect mein volume ka rate of change nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 2NCERT Example 2 · p. 148

Pighalti hui ice ball

Ice ki ek spherical ball aise pighalti hai ki uska volume $50 c m^{3} / m i n$ se kam hota hai aur shape wahi rehti hai. Jis pal radius $5 c m$ hai, us pal uska surface area kitni tezi se kam ho raha hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 3NCERT Example 3 · p. 149

Ek oil slick

Ek tanker se leak hota oil shaant paani par ek circle ki tarah failta hai jiska radius $2 m / m i n$ se badhta hai. Jab radius $25 m$ hai, to slick ka area kitni tezi se badh raha hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Signs par ek note. Agar $x$ ke badhne par $y$ badhta hai, to $\frac{d y}{d x}$ positive hai; agar $x$ ke badhne par $y$ ghatta hai, to yeh negative hai. Isliye girti hui quantity ka rate negative hota hai. Example 2 mein answer $- 20 c m^{2} / m i n$ koi galti nahi hai: minus sign ka matlab hai "decreasing".

Example 4NCERT Example 4 · p. 149

Ek projected rectangle

Ek projector wall par light ka ek rectangle banata hai. Zoom adjust karte waqt length $x$ $4 c m / s$ se badhti hai aur width $y$ $1.5 c m / s$ se ghatti hai. Jab $x = 12 c m$ aur $y = 9 c m$ ho, to nikaalo ki (a) perimeter aur (b) area kitni tezi se badalte hain.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Business mein bhi rates maayne rakhte hain. Agar $C (x)$ $x$ units banane ki total cost hai, to marginal cost $C^{'} (x)$ hai: production ke us level par output ke saath cost badhne ka rate. Yeh ek aur unit banane ki extra cost ke kareeb hota hai. Isi tarah, agar $R (x)$ $x$ units bechne se aane wala total revenue hai, to marginal revenue $R^{'} (x)$ hai.

Example 5NCERT Example 5 · p. 150

Bicycle workshop mein marginal cost

Ek workshop ka $x$ bicycles assemble karne ka daily cost, rupees mein, $C (x) = 0.004 x^{3} - 0.3 x^{2} + 40 x + 2500$ hai. Jab 20 bicycles assemble hoti hain, to marginal cost nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 6NCERT Example 6 · p. 150

Festival stall se marginal revenue

Diyas bechne wala ek stall paata hai ki $x$ boxes bechne se uska revenue, rupees mein, $R (x) = 120 x - \frac{1}{2} x^{2}$ hai (zyada bechne ke liye use daam ghatane padte hain). Jab 40 boxes bikte hain, to marginal revenue nikaalo, aur batao iska matlab kya hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Try karke dekho

Cone ko ek steady rate se bharo

Tumhe chahiye

paper ki ek sheet aur tape
ek chammach aur ek cup paani
ek ruler aur ek marker

Pehle guess karo

Har 5 seconds mein volume utna hi badhta hai. Har chammach par water level ki rise ka kya hota hai?

Woh same rehti haiWoh chhoti hoti jaati haiWoh badi hoti jaati hai

Paper ki sheet se ek cone banao aur use tape karo. Use nok neeche karke pakdo.
Ek chhote cup se usmein paani daalo, har 5 seconds mein ek chammach.
Har chammach ke baad cone par water level mark karo.
Naapo ki paas-paas ke marks kitni doori par hain.

Table 6.1Tumhari readings

Daale gaye chammach	Paani ki height (cm)	Pichhle mark se rise (cm)
1
2
3
4
5

Tumhari entries apne aap save hoti hain.

Quick check

4 mein se 1

Ek square ki side

3 c m / s

se badhti hai. Jab side

10 c m

ho, to uska area kitni tezi se badh raha hai?

30 c m^{2} / s

60 c m^{2} / s

9 c m^{2} / s

100 c m^{2} / s

Answer dekho

Answer (B) $60 c m^{2} / s$

A = x^{2}

isliye

\frac{d A}{d t} = 2 x \frac{d x}{d t} = 2 (10) (3) = 60

. Lalchaane wala

30

x^{2}

ko differentiate karne se aane wala factor 2 bhool jaata hai.

Practice · 18 questions

Practice 6.1: answers ke saath

Short answerEasyNCERT Exercise 6.1, Q1
Ek equilateral triangle ki side $a$ hai. $a$ mein per unit change par uska area kitni tezi se badalta hai jab (a) $a = 4 c m$ , (b) $a = 10 c m$ ?
Solution dikhao
Answer(a) $2 \sqrt{3} c m^{2} / c m$ (b) $5 \sqrt{3} c m^{2} / c m$
$A = \frac{\sqrt{3}}{4} a^{2}$ , isliye $\frac{d A}{d a} = \frac{\sqrt{3}}{2} a$ .
(a) $a = 4$ par: $2 \sqrt{3} c m^{2} / c m \approx 3.46 c m^{2} / c m$ . (b) $a = 10$ par: $5 \sqrt{3} c m^{2} / c m \approx 8.66 c m^{2} / c m$ .
Short answerMediumNCERT Exercise 6.1, Q2
Ice ka ek cube barabar pighalta hai, jisse uska volume $6 c m^{3} / s$ se ghatta hai. Jab har edge $4 c m$ ho, to uska total surface area kitni tezi se ghat raha hai?
Solution dikhao
Answer $6 c m^{2} / s$ se ghat raha hai.
$\frac{d V}{d t} = 3 x^{2} \frac{d x}{d t} = - 6$ , isliye $x = 4$ par: $\frac{d x}{d t} = \frac{- 6}{48} = - \frac{1}{8} c m / s$ .
$\frac{d S}{d t} = 12 x \frac{d x}{d t} = 12 (4) (- \frac{1}{8}) = - 6 c m^{2} / s$ . Surface area $6 c m^{2} / s$ se ghatta hai.
Short answerEasyNCERT Exercise 6.1, Q3
Baarish ke baad ek circular puddle aise failta hai ki uska radius $1.5 c m / s$ se badhta hai. Jab radius $12 c m$ ho, to nikaalo ki uska area kitni tezi se badhta hai.
Solution dikhao
Answer $36 π c m^{2} / s$
$\frac{d A}{d t} = 2 π (12) (1.5) = 36 π \approx 113.1 c m^{2} / s$ .
Short answerEasyNCERT Exercise 6.1, Q4
Garam karne par metal cube ka har edge $0.2 c m / s$ se failta hai. Jab edge $6 c m$ ho, to uska volume kis rate se badhta hai, nikaalo.
cm³/s
Solution dikhao
Answer $21.6 c m^{3} / s$
$\frac{d V}{d t} = 3 x^{2} \frac{d x}{d t} = 3 (36) (0.2) = 21.6 c m^{3} / s$ .
Short answerEasyNCERT Exercise 6.1, Q5
Ink ki ek boond blotting paper par ek circle ki tarah failti hai jiska radius $0.3 c m / s$ se badhta hai. Jab radius $2.5 c m$ ho, to ink wala area kitni tezi se badh raha hai?
Solution dikhao
Answer $1.5 π \approx 4.71 c m^{2} / s$
$\frac{d A}{d t} = 2 π r \frac{d r}{d t} = 2 π (2.5) (0.3) = 1.5 π \approx 4.71 c m^{2} / s$ .
Short answerEasyNCERT Exercise 6.1, Q6
Edge par aur powder daalne se ek circular rangoli ka radius $2.5 c m / s$ se badhta hai. Uski circumference badhne ka rate nikaalo. Kya answer radius par depend karta hai?
Solution dikhao
Answer $5 π c m / s$ , radius chahe jo ho.
$\frac{d C}{d t} = 2 π \frac{d r}{d t} = 5 π \approx 15.7 c m / s$ .
Yeh $r$ par depend nahi karta, kyunki $\frac{d C}{d r} = 2 π$ constant hai.
Short answerMediumNCERT Exercise 6.1, Q7
Ek rectangle ki length $x$ $3 c m / m i n$ se badhti hai jabki uski width $y$ $2 c m / m i n$ se ghatti hai. Jab $x = 15 c m$ aur $y = 10 c m$ ho, to (a) perimeter, (b) area ka rate of change nikaalo.
Solution dikhao
Answer(a) $2 c m / m i n$ (b) $0 c m^{2} / m i n$
(a) $\frac{d P}{d t} = 2 (\frac{d x}{d t} + \frac{d y}{d t}) = 2 (3 - 2) = 2 c m / m i n$ .
(b) $\frac{d A}{d t} = y \frac{d x}{d t} + x \frac{d y}{d t} = 10 (3) + 15 (- 2) = 0$ . Is pal area kuch der ke liye nahi badal raha: length se hone wala faayda width se hone wale nuksaan ko bilkul cancel kar deta hai.
Short answerMediumNCERT Exercise 6.1, Q8
Ek spherical balloon se $400 c m^{3} / s$ se hawa leak hoti hai aur woh spherical hi rehta hai. Jab radius $10 c m$ ho, to uska radius kitni tezi se sikud raha hai?
Solution dikhao
Answer $\frac{1}{π} \approx 0.32 c m / s$ se sikud raha hai.
$- 400 = 4 π (100) \frac{d r}{d t}$ , isliye $\frac{d r}{d t} = - \frac{1}{π} \approx - 0.318 c m / s$ .
Short answerEasyNCERT Exercise 6.1, Q9
Jis pal radius $7 c m$ hai, us pal sphere ka surface area per centimetre radius kitni tezi se badhta hai?
Solution dikhao
Answer $56 π c m^{2} / c m$
$r = 7$ par $\frac{d S}{d r} = 8 π r = 56 π \approx 176 c m^{2} / c m$ .
Short answerMediumNCERT Exercise 6.1, Q10
Ek $13 m$ ki ladder wall se tiki hai. Uska top wall par $0.5 m / s$ se neeche fisalta hai. Jab top zameen se $5 m$ upar ho, to uska foot wall se kitni tezi se door ja raha hai?
Solution dikhao
AnswerWall se door $\frac{5}{24} m / s \approx 0.21 m / s$ .
Jab $y = 5$ , to $x = \sqrt{169 - 25} = 12$ .
$2 x \frac{d x}{d t} + 2 y \frac{d y}{d t} = 0$ , jahan $\frac{d y}{d t} = - 0.5$ .
$\frac{d x}{d t} = - \frac{y}{x} \frac{d y}{d t} = - \frac{5}{12} (- 0.5) = \frac{5}{24} \approx 0.21 m / s$ .
Short answerMediumNCERT Exercise 6.1, Q11
Ek bead curve $y = \frac{x^{3}}{3} - 2 x$ par fisalta hai. Curve par kahan uski height $y$ uski horizontal position $x$ ke double rate se badalti hai?
Solution dikhao
Answer $(2, - \frac{4}{3})$ aur $(- 2, \frac{4}{3})$
$\frac{d y}{d t} = (x^{2} - 2) \frac{d x}{d t}$ . Ise $2 \frac{d x}{d t}$ ke barabar rakhne par ( $\frac{d x}{d t} \neq 0$ ke saath) $x^{2} - 2 = 2$ milta hai, isliye $x = \pm 2$ .
$x = 2$ par: $y = \frac{8}{3} - 4 = - \frac{4}{3}$ . $x = - 2$ par: $y = - \frac{8}{3} + 4 = \frac{4}{3}$ .
Short answerEasyNCERT Exercise 6.1, Q12
Ek soap bubble spherical rehta hai jabki uska radius $0.2 c m / s$ se badhta hai. Jab radius $3 c m$ ho, to uska volume kis rate se badh raha hai?
Solution dikhao
Answer $7.2 π \approx 22.6 c m^{3} / s$
$\frac{d V}{d t} = 4 π (9) (0.2) = 7.2 π \approx 22.6 c m^{3} / s$ .
Short answerMediumNCERT Exercise 6.1, Q13
Ek spherical ball ka diameter $3 x - 2$ hai ( $x > \frac{2}{3}$ ke liye). $\frac{d V}{d x}$ , yaani $x$ mein per unit change par uske volume ka change, $x$ ke terms mein likho.
Solution dikhao
Answer $\frac{3 π}{2} (3 x - 2)^{2}$
$V = \frac{π}{6} (3 x - 2)^{3}$ .
$\frac{d V}{d x} = \frac{π}{6} \cdot 3 (3 x - 2)^{2} \cdot 3 = \frac{3 π}{2} (3 x - 2)^{2}$ .
Short answerMediumNCERT Exercise 6.1, Q14
Ek chute se anaaj $50 c m^{3} / s$ se girta hai aur ek conical dher banata hai jiski height hamesha base ke radius ki aadhi hoti hai. Jab dher $5 c m$ ooncha ho, to height kitni tezi se badh rahi hai?
Solution dikhao
Answer $\frac{1}{2 π} \approx 0.16 c m / s$
$r = 2 h$ ke saath: $V = \frac{1}{3} π (2 h)^{2} h = \frac{4}{3} π h^{3}$ .
$\frac{d V}{d t} = 4 π h^{2} \frac{d h}{d t}$ , isliye $50 = 4 π (25) \frac{d h}{d t}$ aur $\frac{d h}{d t} = \frac{1}{2 π} \approx 0.159 c m / s$ .
Short answerEasyNCERT Exercise 6.1, Q15
$x$ sau posters print karne ka daily cost, rupees mein, $C (x) = 0.002 x^{3} - 0.05 x^{2} + 25 x + 1800$ hai. Jab $x = 30$ ho, to marginal cost nikaalo.
₹
Solution dikhao
Answer₹27.40 per sau posters
$C^{'} (x) = 0.006 x^{2} - 0.1 x + 25$ .
$C^{'} (30) = 0.006 (900) - 3 + 25 = 5.4 - 3 + 25 = 27.4$ .
Short answerEasyNCERT Exercise 6.1, Q16
Ek product ki $x$ units bechne se revenue, rupees mein, $R (x) = 9 x^{2} + 40 x + 12$ hai. Jab $x = 10$ ho, to marginal revenue nikaalo.
₹
Solution dikhao
Answer₹220
$R^{'} (x) = 18 x + 40$ , isliye $R^{'} (10) = 220$ .
Quick answerEasyNCERT Exercise 6.1, Q17
Sphere ke liye, jis pal radius $3 c m$ hai, us pal $\frac{d V}{d r}$ barabar hai
$12 π$ $27 π$ $36 π$ $108 π$
Answer dekho
Answer (C) $36 π$
$4 π r^{2} = 4 π (9) = 36 π c m^{3} / c m$ . $108 π$ to $4 π r^{3}$ hai, jismein differentiate karna bhool gaye.
Quick answerEasyNCERT Exercise 6.1, Q18
$x$ units bechne se revenue, rupees mein, $R (x) = 4 x^{2} + 20 x + 7$ hai. Jab $x = 12$ ho, to marginal revenue hai
₹96₹116₹136₹823
Answer dekho
Answer (B) ₹116
$R^{'} (x) = 8 x + 20$ , isliye $R^{'} (12) = 116$ . ₹823 to $R (12)$ hai, yaani khud revenue, uska rate nahi.

6.3 Increasing aur Decreasing Functions

Garmi ke ek din mein Jaipur ke temperature ko socho. Subah se yeh badhta hai, dopahar mein peak par hota hai, aur phir shaam bhar girta hai. Subah ke ghanton mein baad wale time par temperature hamesha zyada hota hai; shaam ko baad wale time par hamesha kam. Yahi do behaviours, increasing aur decreasing, hum ab define karenge aur phir derivatives se test karenge.

Ek interval se do inputs $x_{1} < x_{2}$ lo aur outputs compare karo (Fig. 6.3). Graph ke badhte hisse par baad wala input bada output deta hai. Girte hisse par woh chhota output deta hai. Agar dono inputs peak ke alag-alag taraf hon, to koi bhi jeet sakta hai, isliye jis interval mein peak ho us par function na increasing hai na decreasing.

3/3Function peak ke left mein increasing hai aur right mein decreasing.

Fig. 6.3Do inputs

x_{1} < x_{2}

ke outputs compare karo.NCERT Fig. 6.1 · p. 152

Ek smooth curve left se badhkar beech mein ek peak tak jaata hai aur phir right ki taraf girta hai. Pehle step mein badhte hisse par do points mark hain jahan x1, x2 se chhota hai, dono axes tak dashed lines hain, aur label f of x1 less than f of x2 hai. Doosre step mein pair girte hisse par hai aur label f of x1 greater than f of x2 padhta hai. Teesre step mein badhte hisse ke neeche ka region shaded hai jis par increasing likha hai, aur girte hisse ke neeche ka region shaded hai jis par decreasing likha hai, aur peak par ek dashed line dono ko alag karti hai.

Definition

Interval par increasing aur decreasing

Maan lo $I$ real function $f$ ke domain mein ek interval hai, aur $x_{1}, x_{2}$ $I$ ke koi bhi do points hain jahan $x_{1} < x_{2}$ . Tab $f$ $I$ par increasing hai agar hamesha $f (x_{1}) \leq f (x_{2})$ , strictly increasing hai agar hamesha $f (x_{1}) < f (x_{2})$ , decreasing hai agar hamesha $f (x_{1}) \geq f (x_{2})$ , strictly decreasing hai agar hamesha $f (x_{1}) > f (x_{2})$ , aur constant hai agar $I$ ke har $x$ ke liye $f (x) = c$ .

Example $f (x) = 2^{x}$ $𝐑$ par strictly increasing hai; $f (x) = 5$ constant hai, isliye yeh non-strict sense mein increasing bhi hai aur decreasing bhi, lekin strictly dono mein se koi nahi.

Jo function poore interval par increasing ho, ya poore interval par decreasing ho, use us interval par monotonic kehte hain. Hum ek point par behaviour ki baat bhi karte hain: $f$ $x_{0}$ par increasing hai agar $x_{0}$ ko contain karne wala koi open interval ho jis par $f$ increasing hai, aur decreasing ke liye bhi isi tarah. Yeh ek local idea hai. Function kisi point par increasing ho sakta hai aur phir bhi apne domain mein kahin aur decreasing ho sakta hai.

Example 7NCERT Example 7 · p. 153

Seedha definition use karna

Sirf definition use karke dikhao ki $f (x) = 11 - 3 x$ $𝐑$ par strictly decreasing hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Har pair of points check karna seedhi lines ke liye theek hai, lekin zyaadatar functions ke liye namumkin. Derivative isse kahin tez test deta hai. Jis point par graph upar jaata hai, wahan tangent upar ki taraf slope karta hai, isliye $f^{'} (x) > 0$ ; jahan graph girta hai, wahan tangent neeche ki taraf slope karta hai aur $f^{'} (x) < 0$ (Fig. 6.4). Neeche ka theorem is picture ko proof mein badal deta hai.

4/4Har interval par $f^{'}$ ka sign note karo: $+$ , $-$ , $+$ .

Fig. 6.4Tangent ka slope,

f^{'} (x)

, wahan positive hai jahan graph upar jaata hai aur wahan negative jahan graph girta hai.

Ek cubic ka graph x equals minus 1 par ek hump tak badhta hai, x equals 1 par ek dip tak girta hai aur phir badhta hai. Doosre step mein dono badhte hisson par solid tangent lines par f prime greater than 0 likha hai, aur hump aur dip par flat tangents par f prime equals 0 likha hai. Teesre step mein girte beech wale hisse par dashed tangents par f prime less than 0 likha hai. Chauthe step mein graph ke neeche ek strip minus 1 aur 1 par plus, minus aur plus sections mein bant jaati hai, jo f prime ka sign hai.

Theorem 1Increasing aur decreasing functions ke liye derivative test

Maan lo $f$ closed interval $[a, b]$ par continuous hai aur open interval $(a, b)$ ke har point par uska derivative hai. (a) Agar $(a, b)$ ke har $x$ ke liye $f^{'} (x) > 0$ , to $f$ $[a, b]$ par strictly increasing hai; agar poore interval mein $f^{'} (x) \geq 0$ , to $f$ increasing hai. (b) Agar $(a, b)$ ke har $x$ ke liye $f^{'} (x) < 0$ , to $f$ $[a, b]$ par strictly decreasing hai; agar poore interval mein $f^{'} (x) \leq 0$ , to $f$ decreasing hai. (c) Agar $(a, b)$ ke har $x$ ke liye $f^{'} (x) = 0$ , to $f$ $[a, b]$ par constant hai.

Hum (a) prove karte hain; (b) aur (c) isi pattern par chalte hain. $[a, b]$ mein koi bhi do points $x_{1} < x_{2}$ lo. Function $[x_{1}, x_{2}]$ par continuous hai aur $(x_{1}, x_{2})$ par differentiable, isliye Chapter 5 ka mean value theorem unke beech strictly ek point $c$ deta hai jiske liye

f (x_{2}) - f (x_{1}) = f^{'} (c) (x_{2} - x_{1})

Yahan $x_{2} - x_{1} > 0$ aur $f^{'} (c) > 0$ , isliye right-hand side positive hai aur $f (x_{2}) > f (x_{1})$ . Kyunki $x_{1} < x_{2}$ koi bhi liye gaye the, $f$ strictly increasing hai. Agar humein sirf $f^{'} (c) \geq 0$ pata ho, to wahi line $f (x_{2}) \geq f (x_{1})$ deti hai. Geometry mein dekhein to dono points ko jodne wali chord $c$ par tangent ke parallel hai (Fig. 6.5), aur woh tangent upar ki taraf slope karta hai.

2/2Unke beech kisi $c$ par tangent chord ke parallel hai, isliye chord ka slope $f^{'} (c)$ ke barabar hai.

Fig. 6.5Mean value theorem pictures mein:

x_{1}

aur

x_{2}

ke beech koi tangent chord ke parallel hota hai.

Ek increasing curve jis par x1 aur x2 ke upar points A aur B hain, jo ek seedhi chord se jude hain. Doosre step mein x1 aur x2 ke beech c ke upar wale point par ek dashed tangent line banayi gayi hai; yeh chord ke parallel hai, isliye iska slope f prime of c chord ke slope ke barabar hai.

Ek kaam ka extension. Kuch mushkil points ke baad bhi conclusion bacha rehta hai. Agar $f$ ek interval par continuous hai aur end points ko chhodkar har point par $f^{'} (x) > 0$ hai, to $f$ poore interval par, end points samet, strictly increasing hai. Yahi baat tab bhi sach hai jab $f^{'}$ sirf kuch isolated points par zero ho: $f (x) = x^{3}$ ka $f^{'} (0) = 0$ hai aur phir bhi yeh $𝐑$ par strictly increasing hai.

Example 8NCERT Example 8 · p. 154

Ek cubic jo kabhi nahi mudta

Dikhao ki $f (x) = x^{3} + 6 x^{2} + 15 x - 4$ $𝐑$ par strictly increasing hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 9NCERT Example 9 · p. 154

Tez beat wala cosine

Maan lo $f (x) = \cos 3 x$ . Dikhao ki $f$ (a) $(0, \frac{π}{3})$ par strictly decreasing hai, (b) $(\frac{π}{3}, \frac{2 π}{3})$ par strictly increasing hai, aur (c) $(0, \frac{2 π}{3})$ par na increasing hai na decreasing.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Function kahan badhta ya ghatta hai, yeh nikaalne ke liye hum un points ko jahan $f^{'} (x) = 0$ hai (ya $f^{'}$ exist nahi karta) baad ke khambon ki tarah use karte hain. Yeh real line ko intervals mein kaat dete hain. Har interval ke andar, agar $f^{'}$ continuous hai to uska sign nahi badal sakta, isliye har interval mein ek point test karna kaafi hai.

Example 10NCERT Example 10 · p. 155

Neeche khulne wala parabola

Woh intervals nikaalo jin par $f (x) = 3 + 10 x - x^{2}$ (a) increasing hai, (b) decreasing hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 11NCERT Example 11 · p. 155

Do mod wala cubic

Woh intervals nikaalo jin par $f (x) = x^{3} - 6 x^{2} - 15 x + 2$ increasing hai aur woh jin par yeh decreasing hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Table 6.2Har interval par

f^{'} (x) = 3 (x + 1) (x - 5)

ke factors ke signs

Interval	$x + 1$	$x - 5$	$f^{'} (x)$	$f$
$(- \infty, - 1)$	$-$	$-$	$+$	badhta hai
$(- 1, 5)$	$+$	$-$	$-$	girta hai
$(5, \infty)$	$+$	$+$	$+$	badhta hai

Example 12NCERT Example 12 · p. 156

Ek restricted domain

Woh intervals nikaalo jin par $f (x) = \cos^{2} x$ , $x \in [0, π]$ , increasing aur decreasing hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 13NCERT Example 13 · p. 157

Sine aur cosine ka sum

$[0, 2 π]$ mein woh intervals nikaalo jin par $f (x) = \sqrt{3} \sin x + \cos x$ increasing ya decreasing hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

InteractiveSlope explorer

Ek interactive graph jismein is chapter ke chhah functions mein se chun sakte ho, jaise ek cubic, do quartics, x squared minus 4 ka absolute value, root 3 sine x plus cos x, aur x to the power five thirds minus 5 x to the power two thirds. Slope and sign mode mein ek slider ya drag curve par ek point chalata hai; ek tangent line banti hai, badhte waqt solid, girte waqt dashed, aur ek status line f of x, f prime of x batati hai aur yeh bhi ki function increasing hai, decreasing hai, ya critical point par hai. Check boxes f prime ke sign ke liye plus aur minus strip dikhate hain, f prime ka graph banate hain, aur critical points mark karte hain. Closed interval mode mein do sliders ends a aur b set karte hain; ek button end points aur andar ke critical points ko unki values ke saath ek table mein list karta hai, aur graph par greatest aur least values par ring bana deta hai.

Fig. 6.6Graph par ek point chalao, uske tangent se

f^{'} (x)

padho,

f^{'}

ki sign strip dikhao, ya ek interval

[a, b]

set karke greatest aur least values nikaalo.

6.4 Maxima aur Minima

Kai asli decisions mein kisi cheez ki best value chahiye hoti hai. Ek dhaba owner thali ka woh price chahta hai jisse din mein sabse zyada paisa aaye. Ek delivery app curved road par customer ke sabse paas wala point chahta hai. Ek engineer woh open tank chahta hai jismein steel ki fixed amount se sabse zyada samaye. Har ek wahi mathematical kaam hai: function ki sabse badi ya sabse chhoti value nikaalo, aur yeh bhi ki woh kahan aati hai.

Pehle hum "sabse bada" aur "sabse chhota" ko precise banate hain, phir derivatives se unhe dhoondhte hain. Bonus yeh hai ki graph kahan mudta hai yeh pata hone se use sketch karna bahut aasaan ho jaata hai.

Definition

Interval par maximum, minimum aur extreme value

Maan lo $f$ ek interval $I$ par defined hai. Agar $I$ mein koi point $c$ hai jiske liye $I$ ke har $x$ par $f (c) \geq f (x)$ , to $f (c)$ $I$ par $f$ ki maximum value hai aur $c$ maximum value ka point hai. Agar $I$ ke har $x$ ke liye $f (c) \leq f (x)$ , to $f (c)$ minimum value hai aur $c$ minimum value ka point. Dono mein se koi bhi extreme value hai, aur tab $c$ ek extreme point hai.

Graph aksar in values ko ek nazar mein dikha deta hai, un points par bhi jahan function ka derivative nahi hota. Agle teen examples koi bhi calculus use karne se pehle graphs ko dekhte hain.

Example 14NCERT Example 14 · p. 161

Poori line par neeche khulne wala parabola

Kya $f (x) = 5 - x^{2}$ , $x \in 𝐑$ , ki koi greatest value hai? Koi least value? Agar domain ko kaatkar $[- 1, 3]$ kar diya jaaye to kya badalta hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 15NCERT Example 15 · p. 161

V-shape wala graph

Tay karo ki $f (x) = ∣ 2 x - 3 ∣$ , $x \in 𝐑$ , ki koi greatest aur least value hai ya nahi, aur jo exist karti hain unhe nikaalo. $[0, 4]$ par kya hota hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 16NCERT Example 16 · p. 162

Open interval apne extremes se bach sakta hai

Kya $f (x) = 2 x + 1$ open interval $(- 1, 3)$ par koi greatest ya least value tak pahunchta hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 16 do facts ki taraf ishaara karta hai jinhe hum bina proof ke maan lete hain. Closed interval par ek monotonic function apna maximum aur minimum end points par leta hai. Zyada general baat: closed interval par continuous function us interval mein kahin na kahin hamesha ek greatest aur ek least value tak pahunchta hai. Closed intervals par hum Section 6.4.1 mein wapas aayenge.

Ab kai mod wale ek graph ko dekho (Fig. 6.7). Har hump ke top par function aas-paas ke saare points se bada hai, lekin zaroori nahi ki baaki har jagah se bada ho. Har dip ke bottom par yeh aas-paas ke saare points se chhota hai. Aise points ko local maximum aur local minimum ke points kehte hain (relative maximum aur minimum bhi), aur wahan ki values local maximum aur minimum values hain.

5/5Greatest value end $b$ par hai, kisi hump par nahi; least value neeche wale dip par hai. Na left hump na right dip overall extreme hai.

Fig. 6.7Local maxima aur minima ek point ko sirf uske padosiyon se compare karte hain.NCERT Fig. 6.11 · p. 162

x equals a se x equals b tak ek lehardaar curve. Doosre step mein upar wale triangles dono humps ko mark karte hain, har ek par local max likha hai. Teesre step mein neeche wale triangles dono dips ko mark karte hain, jin par local min likha hai. Chauthe step mein c minus h se c plus h tak ek shaded window oonche hump ko gherti hai, aur note likha hai yahan highest: window ke andar woh point apne saare padosiyon se aage hai, phir bhi aur right mein end point b usse bhi ooncha hai. Aakhri step mein humps aur dips phir se mark hote hain, sabse bada point, right end b par, gole mein hai aur us par greatest likha hai, aur neeche wala dip dashed gole mein hai jis par least likha hai.

Definition

Local maximum aur local minimum

$f$ ke domain ka ek interior point $c$ lo. $c$ ko point of local maximum kaho jab $h > 0$ ke saath koi open interval $(c - h, c + h)$ ho jismein aisa koi $x$ nahi jahan $f (x)$ , $f (c)$ se zyada ho; number $f (c)$ ek local maximum value hai. $c$ ko point of local minimum kaho jab aise kisi interval mein aisa koi $x$ na ho jahan $f (x)$ , $f (c)$ se kam ho; tab $f (c)$ ek local minimum value hai.

$c$ par local maximum ke paas derivative kya karta hai? Bilkul left mein graph chadhkar $c$ tak aata hai, isliye wahan $f^{'} (x) > 0$ ; bilkul right mein woh neeche girta hai, isliye $f^{'} (x) < 0$ . Agar $f^{'} (c)$ exist karta hai, to woh na positive ho sakta hai na negative, isliye use zero hona hi padega. Local minimum par bhi yahi reasoning lagti hai, bas signs ulte. Ek corner, jahan $f^{'} (c)$ exist nahi karta, bhi peak ya dip ho sakta hai, jaise Example 15 mein.

Theorem 2Local extremes kahan ho sakte hain

Maan lo $f$ ek open interval $I$ par defined hai aur $I$ ke ek point $c$ par uska local maximum ya local minimum hai. Tab derivative $f^{'} (c)$ zero hai, ya exist nahi karta.

Converse galat hai. $f (x) = (x - 2)^{3} + 1$ lo. Tab $f^{'} (x) = 3 (x - 2)^{2}$ , isliye $f^{'} (2) = 0$ , lekin 2 ke dono taraf $f^{'}$ positive hai. Graph $x = 2$ par ek pal ke liye flat hota hai aur phir chadhta rehta hai, isliye 2 na local maximum hai na local minimum.

Definition

Critical point

$f$ ka critical point uske domain ka woh point $c$ hai jahan derivative zero ho ya exist na kare.

Example $∣ x^{2} - 4 ∣$ ke critical points $- 2$ , $0$ aur $2$ hain: $f^{'} (0) = 0$ , aur $\pm 2$ par corners hain.

To Theorem 2 kehta hai: interior points par local extremes sirf critical points par ho sakte hain. Critical points candidates hain, guarantee nahi (Fig. 6.8). Kaun se candidates peaks hain, kaun se dips aur kaun se dono nahi, yeh tay karne ke liye hum dekhte hain ki har ek se $x$ guzarte waqt $f^{'}$ ka sign kaise behave karta hai.

4/4Inmein se har ek critical point hai; sirf $c$ ke aas-paas $f^{'}$ ka sign inhe alag pehchaanta hai.

Fig. 6.8Teen tarah ke critical point.

Teen panels aamne-saamne. Pehle mein ek smooth hump hai jiske top par flat dashed tangent hai, label f prime of c equals 0, smooth top, max. Doosre mein neeche ek V-shape corner hai, label f prime of c undefined, corner, min. Teesre mein ek curve hai jo horizontal tangent ke saath flat hota hai aur phir badhta rehta hai, label f prime of c equals 0, flat step, neither. Steps baari-baari har panel ko highlight karte hain aur phir teenon ko saath.

Theorem 3First derivative test

Maan lo $f$ ek open interval $I$ par defined hai, $c$ $I$ mein ek critical point hai, aur $f$ $c$ par continuous hai. $c$ ke dono taraf ek chhote hisse par $f^{'}$ ka sign dekho. (i) Left mein positive aur right mein negative ( $+$ se $-$ ): $c$ point of local maximum hai. (ii) Left mein negative aur right mein positive ( $-$ se $+$ ): $c$ point of local minimum hai. (iii) Dono taraf same sign: $c$ dono mein se koi nahi, aur ise point of inflection kehte hain.

Fig. 6.9 test ko kaam karte dikhata hai. $f$ ke graph ke neeche $f^{'}$ ka graph banao. Jahan bhi $f^{'}$ axis ko neeche jaate hue cross karta hai, wahan $f$ ka peak hai; jahan woh upar jaate hue cross karta hai, wahan $f$ ka dip hai. Agar $f^{'}$ sirf axis ko chhoo kar usi taraf laut jaaye, to $f$ bas thodi der rukta hai.

3/3 $x = - 1$ par $f^{'}$ $+ \to -$ badalta hai: local maximum. $x = 3$ par yeh $- \to +$ badalta hai: local minimum.

Fig. 6.9Critical point ke aas-paas

f^{'}

ka sign uska type tay karta hai.NCERT Fig. 6.14 · p. 165

Upar wala panel y equals f of x dikhata hai, ek cubic jo badhta hai, x equals minus 1 par mudta hai, girta hai, x equals 3 par mudta hai aur badhta hai. Doosre step mein neeche wala panel y equals f prime of x dikhata hai, ek upar khulne wala parabola jo minus 1 se pehle positive (shaded, plus sign), minus 1 aur 3 ke beech negative (shaded, minus sign), aur 3 ke baad positive hai. Teesre step mein dashed lines minus 1 aur 3 par f prime ke zeros ko f ke mod se jodti hain: minus 1 par plus se minus sign change par max likha hai, aur 3 par minus se plus par min likha hai.

Example 17NCERT Example 17 · p. 165

First derivative test use karna

$f (x) = 2 x^{3} - 3 x^{2} - 12 x + 1$ ke saare points of local maximum aur local minimum, aur local maximum aur minimum values nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 18NCERT Example 18 · p. 166

Ek critical point jo extreme nahi hai

$f (x) = 3 x^{4} - 4 x^{3} + 1$ ke saare points of local maximum aur minimum nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Continuous functions ke liye first derivative test hamesha kaam karta hai, lekin dono taraf signs check karna dheema ho sakta hai. Jab $f$ twice differentiable ho, to aksar ek tez tarika hota hai: dekho ki slope khud kaise badal raha hai.

Cup jaise dip ke bottom par tangent slope negative se, zero se hote hue, positive ho jaata hai: slope badh raha hai, isliye $f^{''} (c) > 0$ . Cap jaise peak ke top par slope positive se zero se hote hue negative ho jaata hai, isliye $f^{''} (c) < 0$ (Fig. 6.10).

3/3 $f^{'} (c) = 0$ ke saath: $f^{''} (c) > 0$ ka matlab local minimum, $f^{''} (c) < 0$ ka matlab local maximum.

Fig. 6.10Second derivative batata hai ki slope badh raha hai (cup) ya ghat raha hai (cap).

Do panels. Left mein ek cup jaise curve par teen tangents hain: left side par minus mark wala dashed tangent, bottom par 0 mark wala flat tangent, aur right side par plus mark wala solid tangent; uske neeche slope badhe aur f double prime of c greater than 0 likha hai. Right mein ek cap jaise curve par left se right plus, 0 aur minus mark wale tangents hain, saath mein slope ghate aur f double prime of c less than 0 likha hai. Steps pehle cup, phir cap, phir dono ko highlight karte hain.

Theorem 4Second derivative test

Maan lo $f$ ek interval $I$ par defined hai, $c \in I$ , $f^{'} (c) = 0$ , aur second derivative $f^{''} (c)$ exist karta hai. (i) Negative value, $f^{''} (c) < 0$ , $c$ ko point of local maximum banati hai, jiski local maximum value $f (c)$ hai. (ii) Positive value, $f^{''} (c) > 0$ , $c$ ko point of local minimum banati hai, jiski local minimum value $f (c)$ hai. (iii) Zero value, $f^{''} (c) = 0$ , kuch tay nahi karti; first derivative test par wapas jao.

Example 19NCERT Example 19 · p. 166

Do tarah ke critical points

$f (x) = ∣ x^{2} - 4 ∣$ , $x \in 𝐑$ , ke local maxima aur minima nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 20NCERT Example 20 · p. 167

Second derivative test use karna

$f (x) = x^{4} - 4 x^{3} + 4 x^{2} + 1$ ki local maximum aur local minimum values nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 21NCERT Example 21 · p. 167

Jab second test chup rahe

$f (x) = 3 x^{4} - 4 x^{3} + 1$ (Example 18) ke liye har critical point par second derivative test lagao. Tum kya conclude kar sakte ho?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Optimisation problems solve karna

Jo word problems "sabse bada", "sabse kam", "sabse paas" ya "sabse sasta" poochte hain, woh ek hi routine follow karte hain. Jis quantity ko optimise karna hai uska naam rakho. Problem ki conditions use karke use ek variable ka function likho, aur note karo ki woh variable kaun si values le sakta hai. Phir critical points nikaalo, tay karo ki kaun sa required extreme deta hai, aur problem ke apne shabdon aur units mein answer do.

Example 22NCERT Example 22 · p. 167

Ek asmaan batwara

Do positive numbers ka sum 12 hai. Unhe aise nikaalo ki pehle ka square plus doosre ke square ka double jitna ho sake utna chhota ho.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Ise $x + y = k$ hone par $x^{2} + y^{2}$ ko minimise karne se compare karo: wahi steps $x = y = \frac{k}{2}$ dete hain, yaani barabar batwara. Doosre square par weight 2 best batwaare ko chhote doosre number ki taraf dhakel deta hai.

Example 23NCERT Example 23 · p. 168

Parabola par sabse paas ka point

Parabola $y^{2} = 4 x$ ka woh point nikaalo jo point $(5, 0)$ ke sabse paas hai, aur least distance bhi.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 24NCERT Example 24 · p. 169

Water tank kahan lagayein

Ek seedhi road ke paas teen ghar hain. Ek junction se road ke saath naapne par, road ke unke sabse paas wale points 2 km, 7 km aur 15 km par hain, aur ghar road se 1 km, 3 km aur 2 km door hain. Road par ek tank aise lagana hai ki teenon gharon tak uski straight-line distances ke squares ka sum least ho. Tank kahan lagna chahiye?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 25NCERT Example 25 · p. 169

Semicircle mein sabse bada rectangle

Ek rectangle ki ek side radius $5 c m$ wale semicircle ke diameter par hai aur uske baaki do corners arc par hain. Sabse bada possible area nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

2/3Half-width $x = \frac{5}{\sqrt{2}} \approx 3.54$ : area 25, sabse bada.

Fig. 6.11Rectangle ko chauda karne se uski chhat neeche aati hai: area beech mein sabse bada hota hai.NCERT Fig. 6.17 · p. 170

Radius 5 ka ek semicircle apne diameter par rakha hai, centre O. Pehle step mein ek ooncha patla rectangle diameter par khada hai aur uske top corners arc par hain; half-width x equals 1.5 aur area 14.3 hai. Doosre step mein ek balanced rectangle ki half-width 3.54 hai, jo uski height ke barabar hai, aur area 25.0 hai. Teesre step mein ek chauda chapta rectangle hai jiski half-width 4.7 aur area 16.0 hai. Har step mein O se ek top corner tak ek dashed radius jaata hai.

Example 26NCERT Example 26 · p. 170

Triangle mein rectangle

Base $B$ aur height $H$ wali ek triangular sheet se ek rectangle kaata jaata hai, jiski ek side base ke saath hai. Prove karo ki aisa sabse bada rectangle triangle ke area ka aadha hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Try karke dekho

Best box mod kar banao

Tumhe chahiye

card ya squared paper
kainchi aur tape
ek ruler

Pehle guess karo

Tumhare hisaab se kaun sa cut sabse bada volume dega?

x = 1

x = 3

x = 5

$20 c m$ by $20 c m$ ka card ka square kaato (ya squared paper use karo). Har corner se side $x$ ka square kaato, flaps ko upar modo aur tape lagakar ek open box banao. $x = 1, 2, 3, 4, 5 c m$ ke saath boxes banao (ya unke volumes compute karo) aur table bharo.

Table 6.3Box ke volumes

Cut $x$ (cm)	Base ki side $20 - 2 x$ (cm)	Volume (cm³)
1
2
3
4
5

Tumhari entries apne aap save hoti hain.

Quick check

3 mein se 1

Do positive numbers ka product 64 hai. Unke sum ki sabse kam possible value kya hai?

Answer dekho

Answer 16

Sum

s (x) = x + \frac{64}{x}

x = 8

par

s^{'} (x) = 1 - \frac{64}{x^{2}} = 0

s^{''} (8) > 0

. Least sum

8 + 8 = 16

6.4.1 Closed Interval mein Function ki Maximum aur Minimum Values

Local maximum sirf apne padosiyon se jeet-ta hai. Kai sawaal overall winner chahte hain: interval par kahin bhi $f$ ki greatest value, jise uska absolute maximum kehte hain (global maximum, ya greatest value bhi), aur least value, uska absolute minimum (global minimum, least value). Fig. 6.7 mein greatest value right end par hai, dono humps se upar, aur least value dono dips mein se neeche wale par hai.

Yeh values exist karti hain ya nahi, yeh end points par depend kar sakta hai. $h (x) = x^{2}$ lo. Open interval $(- 1, 2)$ par iski least value $h (0) = 0$ hai, lekin koi greatest value nahi: jaise-jaise $x$ 2 ke paas aata hai, $h (x)$ dheere-dheere 4 ki taraf badhta hai aur wahan kabhi nahi pahunchta. Closed interval $[- 1, 2]$ par greatest value $h (2) = 4$ ek end point par mil jaati hai. Agle do theorems theek closed intervals par hi lagte hain.

Theorem 5Closed interval par extreme values

Agar $f$ closed interval $I = [a, b]$ par continuous hai, to $I$ mein aise points $c$ aur $d$ hain ki $I$ ke har $x$ ke liye $f (d) \leq f (x) \leq f (c)$ . Shabdon mein: $f$ ek greatest value (at $c$ ) aur ek least value (at $d$ ) tak pahunchta hai, ho sakta hai ek se zyada points par.

Theorem 6Interior extremes par tangent flat hota hai

Maan lo $f$ ek closed interval $I$ par differentiable hai. Agar $I$ par $f$ ki greatest value $I$ ke strictly andar kisi point $c$ par aati hai, to $f^{'} (c) = 0$ . Agar least value kisi interior point par aaye, to bhi yahi sach hai.

Dono ko saath rakhein to yeh kehte hain ki greatest aur least values exist karti hain, aur har ek ya to end point par hai ya andar kisi critical point par. Isliye humein sirf candidates ki ek chhoti list check karni hai (Fig. 6.12).

4/4Steps 3 aur 4: saare candidates par $f$ compare karo. Yahan greatest value $f (b)$ ek end par hai, aur least $f (c_{2})$ ek dip par.

Fig. 6.12Sirf end points aur critical points hi greatest ya least value rakh sakte hain.NCERT Fig. 6.19 · p. 172

Ek curve sirf x equals a aur x equals b ke brackets ke beech bana hai. Doosre step mein do critical points c1 aur c2 squares aur flat tangents se mark hain: c1 ek hump hai aur c2 ek dip. Teesre step mein end points a aur b circles se mark hain. Aakhri step mein dashed level lines chaaron candidates se right mein ek list tak jaati hain, jo unhe upar se neeche f of b, f of c1, f of a aur f of c2 ke order mein rakhti hai. Greatest, end point par f of b, solid circle se ghera hai, aur least, dip par f of c2, dashed circle se.

Step by step

[a, b] par absolute maximum aur minimum ka working rule

Step 1 / 4

Critical points nikaalo

$(a, b)$ mein har woh $x$ dhoondho jahan $f^{'} (x) = 0$ ho ya jahan $f$ differentiable na ho.

1 / 4

Example 27NCERT Example 27 · p. 173

Closed interval par ek cubic

$[- 2, 4]$ par $f (x) = x^{3} - 3 x^{2} - 9 x + 4$ ki greatest aur least values nikaalo, aur yeh bhi ki woh kahan aati hain.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 28NCERT Example 28 · p. 173

Bina derivative wala critical point

$[- 1, 5]$ par $f (x) = x^{5 / 3} - 5 x^{2 / 3}$ apni greatest aur least values kahan leta hai, aur woh kya hain?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 29NCERT Example 29 · p. 174

Ek drone aur ek camera

Ek survey drone path $y = x^{2}$ par udta hai (distances km mein). Ek pahaadi par camera point $(16, \frac{1}{2})$ par hai. Drone aur camera ke beech least distance nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

[0, 6]

par ek continuous

f

ke sirf critical points 2 aur 5 hain, aur

f (0) = 4

f (2) = 9

f (5) = 1

f (6) = 12

. Uska absolute maximum kya hai?

x = 2

par12,

x = 6

par4,

x = 0

parKoi nahi hai

Answer dekho

Answer (B) 12, $x = 6$ par

Chaaron candidates compare karo: sabse bada

f (6) = 12

hai, ek end point par.

x = 2

par local maximum 9 greatest value nahi hai.

Miscellaneous Examples

Yeh examples chapter ke ideas ko milate hain: formula se rates, similar triangles ke saath related rates, chaar intervals wali sign table, aur teen optimisation problems.

Example 30NCERT Example 30 · p. 177

Do stops ke beech ek e-rickshaw

Ek e-rickshaw $t = 0$ par stop P se rest se chalta hai aur agle stop Q par ruk jaata hai. $t$ seconds ke baad tay ki gayi distance $x = 0.3 t^{2} - 0.005 t^{3}$ metres hai. Yeh Q kab pahunchta hai, stops kitni door hain, aur iski top speed kya hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 31NCERT Example 31 · p. 178

V-shape wala trough bharna

Ek water trough $4 m$ lamba hai. Iske ends isosceles triangles hain, nok neeche, top par $60 c m$ chaude aur $40 c m$ gehre. Paani $0.02 m^{3} / m i n$ se pump kiya jaata hai. Jab paani $20 c m$ gehra ho, to level kitni tezi se badh raha hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

2/2Paani ka triangle end ke similar hai, isliye har depth par $w = 1.5 h$ .

Fig. 6.13Paani ka cross-section trough ke end ki ek chhoti copy hai.NCERT Fig. 6.20 · p. 178

Trough ka end view: neeche nok wala ek triangle jo top par 60 cm chauda aur 40 cm gehra hai, jis par trough end likha hai. Doosre step mein paani neeche wala hissa depth h tak bharta hai, surface width w mark hai, aur saath mein ratio w over h equals 60 over 40 likha hai, kyunki paani ka triangle poore end ke similar hai.

Example 32NCERT Example 32 · p. 178

Patang ki dor dheeli karna

Ek patang $60 m$ ki steady height par udti hai aur use udaane wale bachche se $4 m / s$ se horizontally door jaati hai. Jab $100 m$ dor bahar ho, to dor kitni tezi se chhodni padegi? (Dor ko seedha maano aur bachche ka haath zameen ke level par.)

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 33NCERT Example 33 · p. 179

Teen roots se chaar intervals

Woh intervals nikaalo jin par $f (x) = \frac{x^{4}}{4} - x^{3} - 2 x^{2} + 12 x + 1$ increasing aur decreasing hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 34NCERT Example 34 · p. 180

Ek inverse-tangent function

Dikhao ki $f (x) = \tan^{- 1} x - \frac{x}{1 + x^{2}}$ $𝐑$ par increasing hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 35NCERT Example 35 · p. 181

Garam ki gayi square plate

Ek square metal plate garam ki jaati hai aur har side $0.02 c m / s$ se failti hai. Jis pal side $8 c m$ hai, us pal area kitni tezi se badh raha hai? Agle aadhe second mein mile extra area ka estimate lagao.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 36NCERT Example 36 · p. 181

Card se best box

$24 c m$ by $15 c m$ ke card ke corners se barabar squares kaatkar aur sides upar modkar ek open box banaya jaata hai (Fig. 6.14). Sabse bada possible volume nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

3/3Flaps upar modo: box $x$ ooncha hai aur base $(24 - 2 x)$ by $(15 - 2 x)$ hai.

Fig. 6.14Flat card se open box tak: cut

x

height ban jaata hai.NCERT Fig. 6.23 · p. 182

Ek rectangular card jis par 24 by 15 sheet likha hai, aur har corner par side x ka chhota square mark hai. Doosre step mein chaaron corner squares khisak kar fade ho jaate hain, aur base ke saath dashed fold lines wala ek cross-shape net bachta hai. Teesre step mein right mein ek open box perspective mein dikhta hai, jismein height x, length 24 minus 2x aur width 15 minus 2x likhe hain.

Example 37NCERT Example 37 · p. 182

Sabse zyada profit ke liye cakes ka daam

Ek bakery paati hai ki woh din mein $x$ cakes ₹ $(400 - \frac{x}{2})$ each par bech sakti hai. $x$ cakes banane ka uska daily cost ₹ $(100 x + 5000)$ hai. Greatest profit ke liye use din mein kitne cakes banane aur bechne chahiye, aur kis daam par?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Ice ki ek ball pighalte waqt spherical rehti hai. Agar uska volume constant rate se ghatta hai, to uska radius ghatta hai

constant rate se hichhota hote-hote aur tezi sechhota hote-hote aur dheere

Answer dekho

Answer (B) chhota hote-hote aur tezi se

\frac{d r}{d t} = \frac{d V / d t}{4 π r^{2}}

;

\frac{d V}{d t}

fixed ho to chhota

r

∣ \frac{d r}{d t} ∣

ko bada kar deta hai. Constant rate ke liye

r^{2}

ka constant hona zaroori hota.

Recap

Summary aur key points

$y = f (x)$ ke liye $\frac{d y}{d x}$ naapta hai ki $x$ mein per unit change par $y$ kitni tezi se badalta hai; $f^{'} (x_{0})$ input $x_{0}$ par woh rate hai (Section 6.2).
Agar $x$ aur $y$ dono $t$ par depend karte hain, to $\frac{d y}{d t} = \frac{d y}{d x} \cdot \frac{d x}{d t}$ ; negative rate ka matlab hai quantity ghat rahi hai (Section 6.2).
Marginal cost $C^{'} (x)$ hai aur marginal revenue $R^{'} (x)$ hai (Example 5).
$f$ $I$ par increasing hai agar $x_{1} < x_{2}$ se $f (x_{1}) \leq f (x_{2})$ mile, strictly increasing agar $f (x_{1}) < f (x_{2})$ ; decreasing bhi isi tarah (Section 6.3).
Agar $f$ $[a, b]$ par continuous hai aur $(a, b)$ par $f^{'} > 0$ , to $f$ $[a, b]$ par strictly increasing hai; agar $f^{'} < 0$ , to strictly decreasing; agar $f^{'} = 0$ , to constant (Theorem 1).
$f$ kahan badhta ya girta hai yeh nikaalne ke liye domain ko $f^{'}$ ke zeros (aur jahan woh exist nahi karta) par baanto aur har tukde par sign test karo (Example 11).
Critical point woh $c$ hai jahan $f^{'} (c) = 0$ ho ya $f$ $c$ par differentiable na ho; interior local extremes sirf wahin hote hain (Theorem 2).
First derivative test: $c$ par $f^{'}$ $+ \to -$ badle: local maximum; $- \to +$ : local minimum; koi change nahi: dono nahi, point of inflection (Theorem 3).
Second derivative test: $f^{'} (c) = 0$ ke saath $f^{''} (c) < 0$ local maximum deta hai, $f^{''} (c) > 0$ local minimum; $f^{''} (c) = 0$ koi conclusion nahi deta (Theorem 4).
Closed interval par continuous function ka ek absolute maximum aur ek absolute minimum hota hai (Theorem 5).
$[a, b]$ par working rule: andar ke critical points aur end points ki list banao, har ek par $f$ evaluate karo, aur sabse bada aur sabse chhota chuno (Section 6.4.1).
Optimisation mein constraint use karke target ko ek variable ka function likho, uske critical points nikaalo, extreme ko justify karo aur context mein answer do (Optimisation problems solve karna).

Flash cards · 9

Jaldi se yaad karo

1 / 9

Chapter 6 ki Miscellaneous Exercise

Practice · 16 questions

Practice questions answers ke saath (NCERT “Miscellaneous exercise”)

Short answerMediumNCERT Miscellaneous Exercise 6, Q1
Dikhao ki $f (x) = \frac{\log x}{x^{2}}$ , $x > 0$ , ka maximum $x = \sqrt{e}$ par hai, aur maximum value nikaalo.
Solution dikhao
AnswerMaximum $\frac{1}{2 e}$ , $x = \sqrt{e}$ par.
$f^{'} (x) = \frac{x^{2} \cdot \frac{1}{x} - 2 x \log x}{x^{4}} = \frac{1 - 2 \log x}{x^{3}}$ .
$x > 0$ ke liye sign $1 - 2 \log x$ jaisa hai: $x < \sqrt{e}$ ke liye positive, $x > \sqrt{e}$ ke liye negative. Isliye $x = \sqrt{e}$ maximum deta hai, aur yahi akela turning point hai.
$f (\sqrt{e}) = \frac{1 / 2}{e} = \frac{1}{2 e}$ .
Short answerMediumNCERT Miscellaneous Exercise 6, Q2
Ek isosceles triangle ka base $10 c m$ par fixed hai. Iski do barabar sides $2 c m / s$ se badhti hain. Jab barabar sides $13 c m$ lambi hon, to iska area kitni tezi se badh raha hai?
Solution dikhao
Answer $\frac{65}{6} \approx 10.8 c m^{2} / s$
$A = \frac{1}{2} (10) \sqrt{a^{2} - 25} = 5 \sqrt{a^{2} - 25}$ , isliye $\frac{d A}{d t} = \frac{5 a}{\sqrt{a^{2} - 25}} \frac{d a}{d t}$ .
$a = 13$ par: $\sqrt{169 - 25} = 12$ , isliye $\frac{d A}{d t} = \frac{5 (13)}{12} (2) = \frac{65}{6} \approx 10.8 c m^{2} / s$ .
Short answerMediumNCERT Miscellaneous Exercise 6, Q3
$[0, 2 π]$ mein woh intervals nikaalo jin par $f (x) = \sin x - x \cos x$ increasing aur decreasing hai.
Solution dikhao
Answer $[0, π]$ par increasing, $[π, 2 π]$ par decreasing.
$f^{'} (x) = \cos x - (\cos x - x \sin x) = x \sin x$ .
$0 < x < π$ ke liye $x > 0$ aur $\sin x > 0$ , isliye $f^{'} > 0$ . $π < x < 2 π$ ke liye $\sin x < 0$ , isliye $f^{'} < 0$ .
Short answerMediumNCERT Miscellaneous Exercise 6, Q4
Woh intervals nikaalo jin par $f (x) = x^{2} + \frac{1}{x^{2}}$ , $x \neq 0$ , increasing aur decreasing hai.
Solution dikhao
Answer $(- 1, 0)$ aur $(1, \infty)$ par increasing; $(- \infty, - 1)$ aur $(0, 1)$ par decreasing.
$f^{'} (x) = 2 x - \frac{2}{x^{3}} = \frac{2 (x^{4} - 1)}{x^{3}} = \frac{2 (x - 1) (x + 1) (x^{2} + 1)}{x^{3}}$ .
Signs: $x < - 1$ : $\frac{+}{-} < 0$ ; $- 1 < x < 0$ : $\frac{-}{-} > 0$ ; $0 < x < 1$ : $\frac{-}{+} < 0$ ; $x > 1$ : $\frac{+}{+} > 0$ .
Short answerHardNCERT Miscellaneous Exercise 6, Q5
Ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$ mein ek isosceles triangle inscribed hai, jiska apex $(0, 3)$ par hai aur base $x$ -axis ke parallel hai. Iska greatest possible area nikaalo.
Solution dikhao
Answer $\frac{45 \sqrt{3}}{4}$ square units
Area $A (s) = 5 \sqrt{1 - s^{2}} \cdot (3 - 3 s) = 15 (1 - s) \sqrt{1 - s^{2}}$ .
$A^{'} (s) = 15 [- \sqrt{1 - s^{2}} - \frac{s (1 - s)}{\sqrt{1 - s^{2}}}] = - \frac{15 (1 - s) (1 + 2 s)}{\sqrt{1 - s^{2}}}$ , jo $s = - \frac{1}{2}$ par zero hai, pehle positive aur baad mein negative.
$A (- \frac{1}{2}) = 15 \cdot \frac{3}{2} \cdot \frac{\sqrt{3}}{2} = \frac{45 \sqrt{3}}{4} \approx 19.5$ . (General roop mein $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ ke liye answer $\frac{3 \sqrt{3}}{4} a b$ hai.)
Short answerMediumNCERT Miscellaneous Exercise 6, Q6
Ek open rectangular tank $1.5 m$ gehra hai aur usmein $12 m^{3}$ aata hai. Base ki cost ₹400 per m² aur sides ki ₹250 per m² hai. Aise sabse saste tank ki cost nikaalo.
Solution dikhao
Answer₹ $(3200 + 3000 \sqrt{2}) \approx$ ₹7442.64
Base $x$ by $y$ , jahan $1.5 x y = 12$ , isliye $x y = 8$ . Side area $= 2 (1.5) (x + y) = 3 (x + \frac{8}{x})$ .
Cost $C (x) = 400 (8) + 250 \cdot 3 (x + \frac{8}{x}) = 3200 + 750 (x + \frac{8}{x})$ . $x = 2 \sqrt{2}$ par $C^{'} (x) = 750 (1 - \frac{8}{x^{2}}) = 0$ , aur $C^{''} > 0$ .
Tab $y = 2 \sqrt{2}$ (square base) aur $C = 3200 + 750 (4 \sqrt{2}) = 3200 + 3000 \sqrt{2} \approx 7442.64$ .
Short answerHardNCERT Miscellaneous Exercise 6, Q7
Ek square aur ek equilateral triangle ke perimeters ka sum ek constant $k$ hai. Prove karo ki unka total area tab least hota hai jab triangle ki side square ki side ki $\sqrt{3}$ guna ho.
Solution dikhao
$s = \frac{k - 4 x}{3}$ aur $A (x) = x^{2} + \frac{\sqrt{3}}{4} s^{2}$ , isliye $A^{'} (x) = 2 x + \frac{\sqrt{3}}{2} s \cdot \frac{d s}{d x} = 2 x - \frac{2 \sqrt{3}}{3} s$ .
$A^{'} (x) = 0$ se $2 x = \frac{2}{\sqrt{3}} s$ milta hai, yaani $s = \sqrt{3} x$ . Saath hi $A^{''} (x) = 2 + \frac{2 \sqrt{3}}{3} \cdot \frac{4}{3} > 0$ , isliye yeh minimum hai.
Short answerHardNCERT Miscellaneous Exercise 6, Q8
Ek window ek rectangle hai jiske upar ek equilateral triangle hai. Iska total outer perimeter $12 m$ hai. Woh width nikaalo jo sabse zyada roshni andar aane de (sabse bada area).
Solution dikhao
AnswerWidth $\frac{12}{6 - \sqrt{3}} \approx 2.81 m$ , rectangle ki height $\approx 1.78 m$ .
$3 x + 2 y = 12$ se $y = \frac{12 - 3 x}{2}$ milta hai, isliye $A (x) = x y + \frac{\sqrt{3}}{4} x^{2} = 6 x - \frac{3}{2} x^{2} + \frac{\sqrt{3}}{4} x^{2}$ .
$A^{'} (x) = 6 - 3 x + \frac{\sqrt{3}}{2} x = 0$ se $x = \frac{12}{6 - \sqrt{3}} = \frac{4 (6 + \sqrt{3})}{11} \approx 2.81$ milta hai; $A^{''} = - 3 + \frac{\sqrt{3}}{2} < 0$ .
Tab $y = \frac{12 - 3 x}{2} \approx 1.78$ .
Short answerHardNCERT Miscellaneous Exercise 6, Q9
Ek seedhi ladder ko zameen se, height $a$ wali fence ke upar se, us wall tak pahunchna hai jo fence ke peeche distance $b$ par khadi hai. Dikhao ki aisi sabse chhoti ladder ki length ${(a^{2 / 3} + b^{2 / 3})}^{3 / 2}$ hai, aur $a = b = 2 m$ hone par ise nikaalo.
Solution dikhao
Answer $a = b = 2 m$ hone par $4 \sqrt{2} \approx 5.66 m$ .
$L^{'} (θ) = - \frac{a \cos θ}{\sin^{2} θ} + \frac{b \sin θ}{\cos^{2} θ} = 0$ se $\tan^{3} θ = \frac{a}{b}$ milta hai, isliye $\tan θ = \frac{a^{1 / 3}}{b^{1 / 3}}$ . $θ \to 0$ ya $\frac{π}{2}$ hone par $L$ bina limit ke badhta hai, isliye yeh akela critical point minimum hai.
Tab $\sin θ = \frac{a^{1 / 3}}{\sqrt{a^{2 / 3} + b^{2 / 3}}}$ aur $\cos θ = \frac{b^{1 / 3}}{\sqrt{a^{2 / 3} + b^{2 / 3}}}$ , isliye $L = \sqrt{a^{2 / 3} + b^{2 / 3}} (a^{2 / 3} + b^{2 / 3}) = {(a^{2 / 3} + b^{2 / 3})}^{3 / 2}$ .
$a = b = 2$ ke liye: $L = {(2 \cdot 2^{2 / 3})}^{3 / 2} = 2^{3 / 2} \cdot 2 = 4 \sqrt{2} \approx 5.66 m$ .
Short answerHardNCERT Miscellaneous Exercise 6, Q10
Woh points nikaalo jin par $f (x) = (x - 1)^{3} (x + 2)^{4}$ ka (a) local maximum, (b) local minimum, (c) point of inflection hai.
Solution dikhao
Answer(a) $x = - 2$ (b) $x = - \frac{2}{7}$ (c) $x = 1$
$f^{'} (x) = 3 (x - 1)^{2} (x + 2)^{4} + 4 (x - 1)^{3} (x + 2)^{3} = (x - 1)^{2} (x + 2)^{3} (7 x + 2)$ . Critical points: $- 2$ , $- \frac{2}{7}$ , $1$ .
Signs: $x < - 2$ : $(+) (-) (-) > 0$ ; $- 2 < x < - \frac{2}{7}$ : $(+) (+) (-) < 0$ ; $- \frac{2}{7} < x < 1$ : $(+) (+) (+) > 0$ ; $x > 1$ : $(+) (+) (+) > 0$ .
Isliye $x = - 2$ par sign $+ \to -$ hota hai; $x = - \frac{2}{7}$ par $- \to +$ ; $x = 1$ par yeh nahi badalta.
Short answerMediumNCERT Miscellaneous Exercise 6, Q11
$0 \leq x \leq π$ ke liye $f (x) = \sin^{2} x - \cos x$ ki greatest aur least values kya hain?
Solution dikhao
AnswerMaximum $\frac{5}{4}$ , $x = \frac{2 π}{3}$ par; minimum $- 1$ , $x = 0$ par.
$f^{'} (x) = 2 \sin x \cos x + \sin x = \sin x (2 \cos x + 1)$ , jo $x = 0, π$ par aur jahan $\cos x = - \frac{1}{2}$ , yaani $x = \frac{2 π}{3}$ , par zero hai.
Values: $f (0) = - 1$ , $f (\frac{2 π}{3}) = \frac{3}{4} + \frac{1}{2} = \frac{5}{4}$ , $f (π) = 0 + 1 = 1$ .
Short answerHardNCERT Miscellaneous Exercise 6, Q12
Radius $r$ wale sphere mein ek cone inscribed hai. Maan lo $x$ sphere ke centre se cone ke base tak ki distance hai, isliye cone ki height $r + x$ hai. Dikhao ki volume tab greatest hota hai jab height $\frac{4 r}{3}$ ho.
Solution dikhao
$0 \leq x < r$ ke liye $V (x) = \frac{π}{3} (r^{2} - x^{2}) (r + x)$ .
$V^{'} (x) = \frac{π}{3} [- 2 x (r + x) + (r^{2} - x^{2})] = \frac{π}{3} (r + x) (r - 3 x)$ , jo $x = \frac{r}{3}$ par zero hai, pehle positive aur baad mein negative.
Height $= r + \frac{r}{3} = \frac{4 r}{3}$ .
Short answerMediumNCERT Miscellaneous Exercise 6, Q13
Maan lo $f$ $[a, b]$ par continuous hai aur $(a, b)$ ke har $x$ ke liye $f^{'} (x) < 0$ . Prove karo ki $f$ $(a, b)$ par strictly decreasing hai.
Solution dikhao
$(a, b)$ mein $x_{1} < x_{2}$ lo. Mean value theorem se unke beech ek $c$ hai jiske liye $f (x_{2}) - f (x_{1}) = f^{'} (c) (x_{2} - x_{1})$ .
Yahan $f^{'} (c) < 0$ aur $x_{2} - x_{1} > 0$ , isliye $f (x_{2}) - f (x_{1}) < 0$ , yaani $f (x_{2}) < f (x_{1})$ . Isliye $f$ $(a, b)$ par strictly decreasing hai.
Short answerHardNCERT Miscellaneous Exercise 6, Q14
Radius $6 c m$ wali lakdi ki ball se ek cylindrical plug kharaada jaata hai. Dikhao ki radius $R$ wali ball ke liye greatest volume wale plug ki height $\frac{2 R}{\sqrt{3}}$ hai, aur $R = 6 c m$ hone par iski height aur volume nikaalo.
Solution dikhao
AnswerHeight $4 \sqrt{3} c m$ ; volume $96 \sqrt{3} π \approx 522 c m^{3}$ .
$V (h) = π (R^{2} - \frac{h^{2}}{4}) h$ , isliye $h = \frac{2 R}{\sqrt{3}}$ par $V^{'} (h) = π (R^{2} - \frac{3 h^{2}}{4}) = 0$ ; $V^{''} (h) = - \frac{3 π h}{2} < 0$ .
$R = 6$ ke liye: $h = \frac{12}{\sqrt{3}} = 4 \sqrt{3} \approx 6.93 c m$ , base radius ka square $36 - 12 = 24$ , aur $V = π (24) (4 \sqrt{3}) = 96 \sqrt{3} π \approx 522 c m^{3}$ . General roop mein $V_{\max} = \frac{4 π R^{3}}{3 \sqrt{3}}$ .
Short answerHardNCERT Miscellaneous Exercise 6, Q15
Ek cylinder ek cone ke andar khada hai jiski height $h$ aur semi-vertical angle $α$ hai, dono ka axis common hai aur cylinder ka base cone ke base par hai. Dikhao ki greatest volume wale cylinder ki height $\frac{h}{3}$ aur volume $\frac{4}{27} π h^{3} \tan^{2} α$ hai. $18 c m$ oonche aur base radius $9 c m$ wale cone ke liye ise check karo.
Solution dikhao
Height $y$ wale cylinder ka radius $(h - y) \tan α$ hai, isliye $V (y) = π \tan^{2} α (h - y)^{2} y$ .
$y = \frac{h}{3}$ par $V^{'} (y) = π \tan^{2} α (h - y) (h - 3 y) = 0$ (root $y = h$ zero volume deta hai), aur wahan $V^{'}$ $+ \to -$ badalta hai.
$V_{\max} = π \tan^{2} α {(\frac{2 h}{3})}^{2} \frac{h}{3} = \frac{4}{27} π h^{3} \tan^{2} α$ . $h = 18$ , $\tan α = \frac{1}{2}$ ke liye: height 6, radius 6, volume $216 π c m^{3}$ , aur formula $\frac{4}{27} π (5832) \cdot \frac{1}{4} = 216 π$ deta hai.
Quick answerEasyNCERT Miscellaneous Exercise 6, Q16
Radius $5 m$ wale ek cylindrical tank mein $157 m^{3} / h$ se paani bhara jaata hai. $π = 3.14$ lene par, paani ki depth badhti hai
$0.2 m / h$ $2 m / h$ $6.28 m / h$ $20 m / h$
Answer dekho
Answer (B) $2 m / h$
$\frac{d h}{d t} = \frac{d V / d t}{π r^{2}} = \frac{157}{3.14 \times 25} = 2 m / h$ .

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercise 6.1, Q1	Practice 6.1 · Q1
NCERT Exercise 6.1, Q2	Practice 6.1 · Q2
NCERT Exercise 6.1, Q3	Practice 6.1 · Q3
NCERT Exercise 6.1, Q4	Practice 6.1 · Q4
NCERT Exercise 6.1, Q5	Practice 6.1 · Q5
NCERT Exercise 6.1, Q6	Practice 6.1 · Q6
NCERT Exercise 6.1, Q7	Practice 6.1 · Q7
NCERT Exercise 6.1, Q8	Practice 6.1 · Q8
NCERT Exercise 6.1, Q9	Practice 6.1 · Q9
NCERT Exercise 6.1, Q10	Practice 6.1 · Q10
NCERT Exercise 6.1, Q11	Practice 6.1 · Q11
NCERT Exercise 6.1, Q12	Practice 6.1 · Q12
NCERT Exercise 6.1, Q13	Practice 6.1 · Q13
NCERT Exercise 6.1, Q14	Practice 6.1 · Q14
NCERT Exercise 6.1, Q15	Practice 6.1 · Q15
NCERT Exercise 6.1, Q16	Practice 6.1 · Q16
NCERT Exercise 6.1, Q17	Practice 6.1 · Q17
NCERT Exercise 6.1, Q18	Practice 6.1 · Q18
NCERT Exercise 6.2, Q1	Practice 6.2 · Q1
NCERT Exercise 6.2, Q2	Practice 6.2 · Q2
NCERT Exercise 6.2, Q3	Practice 6.2 · Q3
NCERT Exercise 6.2, Q4	Practice 6.2 · Q4
NCERT Exercise 6.2, Q5	Practice 6.2 · Q5
NCERT Exercise 6.2, Q6	Practice 6.2 · Q6
NCERT Exercise 6.2, Q7	Practice 6.2 · Q7
NCERT Exercise 6.2, Q8	Practice 6.2 · Q8
NCERT Exercise 6.2, Q9	Practice 6.2 · Q9
NCERT Exercise 6.2, Q10	Practice 6.2 · Q10
NCERT Exercise 6.2, Q11	Practice 6.2 · Q11
NCERT Exercise 6.2, Q12	Practice 6.2 · Q12
NCERT Exercise 6.2, Q13	Practice 6.2 · Q13
NCERT Exercise 6.2, Q14	Practice 6.2 · Q14
NCERT Exercise 6.2, Q15	Practice 6.2 · Q15
NCERT Exercise 6.2, Q16	Practice 6.2 · Q16
NCERT Exercise 6.2, Q17	Practice 6.2 · Q17
NCERT Exercise 6.2, Q18	Practice 6.2 · Q18
NCERT Exercise 6.2, Q19	Practice 6.2 · Q19
NCERT Exercise 6.3, Q1	Practice 6.3 · Q1
NCERT Exercise 6.3, Q2	Practice 6.3 · Q2
NCERT Exercise 6.3, Q3	Practice 6.3 · Q3
NCERT Exercise 6.3, Q4	Practice 6.3 · Q4
NCERT Exercise 6.3, Q5	Practice 6.3 · Q5
NCERT Exercise 6.3, Q6	Practice 6.3 · Q6
NCERT Exercise 6.3, Q7	Practice 6.3 · Q7
NCERT Exercise 6.3, Q8	Practice 6.3 · Q8
NCERT Exercise 6.3, Q9	Practice 6.3 · Q9
NCERT Exercise 6.3, Q10	Practice 6.3 · Q10
NCERT Exercise 6.3, Q11	Practice 6.3 · Q11
NCERT Exercise 6.3, Q12	Practice 6.3 · Q12
NCERT Exercise 6.3, Q13	Practice 6.3 · Q13
NCERT Exercise 6.3, Q14	Practice 6.3 · Q14
NCERT Exercise 6.3, Q15	Practice 6.3 · Q15
NCERT Exercise 6.3, Q16	Practice 6.3 · Q16
NCERT Exercise 6.3, Q17	Practice 6.3 · Q17
NCERT Exercise 6.3, Q18	Practice 6.3 · Q18
NCERT Exercise 6.3, Q19	Practice 6.3 · Q19
NCERT Exercise 6.3, Q20	Practice 6.3 · Q20
NCERT Exercise 6.3, Q21	Practice 6.3 · Q21
NCERT Exercise 6.3, Q22	Practice 6.3 · Q22
NCERT Exercise 6.3, Q23	Practice 6.3 · Q23
NCERT Exercise 6.3, Q24	Practice 6.3 · Q24
NCERT Exercise 6.3, Q25	Practice 6.3 · Q25
NCERT Exercise 6.3, Q26	Practice 6.3 · Q26
NCERT Exercise 6.3, Q27	Practice 6.3 · Q27
NCERT Exercise 6.3, Q28	Practice 6.3 · Q28
NCERT Exercise 6.3, Q29	Practice 6.3 · Q29
NCERT Miscellaneous Exercise 6, Q1	Miscellaneous exercise · Q1
NCERT Miscellaneous Exercise 6, Q2	Miscellaneous exercise · Q2
NCERT Miscellaneous Exercise 6, Q3	Miscellaneous exercise · Q3
NCERT Miscellaneous Exercise 6, Q4	Miscellaneous exercise · Q4
NCERT Miscellaneous Exercise 6, Q5	Miscellaneous exercise · Q5
NCERT Miscellaneous Exercise 6, Q6	Miscellaneous exercise · Q6
NCERT Miscellaneous Exercise 6, Q7	Miscellaneous exercise · Q7
NCERT Miscellaneous Exercise 6, Q8	Miscellaneous exercise · Q8
NCERT Miscellaneous Exercise 6, Q9	Miscellaneous exercise · Q9
NCERT Miscellaneous Exercise 6, Q10	Miscellaneous exercise · Q10
NCERT Miscellaneous Exercise 6, Q11	Miscellaneous exercise · Q11
NCERT Miscellaneous Exercise 6, Q12	Miscellaneous exercise · Q12
NCERT Miscellaneous Exercise 6, Q13	Miscellaneous exercise · Q13
NCERT Miscellaneous Exercise 6, Q14	Miscellaneous exercise · Q14
NCERT Miscellaneous Exercise 6, Q15	Miscellaneous exercise · Q15
NCERT Miscellaneous Exercise 6, Q16	Miscellaneous exercise · Q16

Quick check: MCQs

6.2 Quantities ka Rate of Change4 questions
6.3 Increasing aur Decreasing Functions4 questions
6.4 Maxima aur Minima4 questions
Optimisation problems solve karna3 questions
6.4.1 Closed Interval mein Function ki Maximum aur Minimum Values3 questions
Miscellaneous Examples3 questions

Glossary · 13

Key terms (English + Hindi)

absolute maximumनिरपेक्ष उच्चिष्ठnirapeksh uchchishth: Diye gaye interval par kahin bhi function jo sabse badi value leta hai.Chapter mein pehli baar yahan aaya →
absolute minimumनिरपेक्ष निम्निष्ठnirapeksh nimnishth: Diye gaye interval par kahin bhi function jo sabse chhoti value leta hai.Chapter mein pehli baar yahan aaya →

critical pointक्रांतिक बिंदुkraantik bindu: Domain ka woh point $c$ jahan $f^{'} (c) = 0$ ho ya jahan $f$ differentiable na ho.

decreasing functionह्रासमान फलनhraasmaan phalan: Aisa function jismein interval mein jab bhi $x_{1} < x_{2}$ ho, $f (x_{1}) \geq f (x_{2})$ ; agar $f (x_{1}) > f (x_{2})$ to strictly decreasing.Chapter mein pehli baar yahan aaya →

extreme valueचरम मानcharam maan: Interval par function ki greatest ya least value, dono mein se koi bhi.

increasing functionवर्धमान फलनvardhamaan phalan: Aisa function jismein interval mein jab bhi $x_{1} < x_{2}$ ho, $f (x_{1}) \leq f (x_{2})$ ; agar $f (x_{1}) < f (x_{2})$ to strictly increasing.Chapter mein pehli baar yahan aaya →

local maximumस्थानीय उच्चिष्ठsthaaneeya uchchishth: Ek interior point $c$ jiske liye $c$ ke aas-paas ke kisi open interval ke har $x$ par $f (c) \geq f (x)$ ; $f (c)$ local maximum value hai.Chapter mein pehli baar yahan aaya →
local minimumस्थानीय निम्निष्ठsthaaneeya nimnishth: Ek interior point $c$ jiske liye $c$ ke aas-paas ke kisi open interval ke har $x$ par $f (c) \leq f (x)$ ; $f (c)$ local minimum value hai.Chapter mein pehli baar yahan aaya →

marginal costसीमांत लागतseemant laagat: Output badhne par total cost kitni tezi se badhta hai, $C^{'} (x)$ ; lagbhag ek aur unit banane ki extra cost.Chapter mein pehli baar yahan aaya →
marginal revenueसीमांत आयseemant aay: Zyada units bechne par total revenue kitni tezi se badhta hai, $R^{'} (x)$ ; lagbhag ek aur sale se aane wala paisa.Chapter mein pehli baar yahan aaya →
monotonic functionएकदिष्ट फलनekdisht phalan: Aisa function jo poore interval par increasing ho ya poore interval par decreasing.Chapter mein pehli baar yahan aaya →

point of inflectionनति परिवर्तन बिंदुnati parivartan bindu: Yahan, aisa critical point jahan dono taraf $f^{'}$ ka sign same rehta hai, isliye graph ek pal ke liye flat hota hai lekin na peak hai na dip.Chapter mein pehli baar yahan aaya →

rate of changeपरिवर्तन की दरparivartan ki dar: Ek quantity doosri ke mukable kitni tezi se badalti hai; $y = f (x)$ ke liye yeh $\frac{d y}{d x}$ hai, jo $y$ ki units per $x$ ki unit mein naapa jaata hai.Chapter mein pehli baar yahan aaya →

Application of Derivatives: Class 12 Maths Chapter 6 Hinglish mein

Is chapter mein aap kya seekhenge

6.1 Introduction

6.2 Quantities ka Rate of Change

Practice 6.1: answers ke saath

6.3 Increasing aur Decreasing Functions

Practice 6.2: answers ke saath

6.4 Maxima aur Minima

Optimisation problems solve karna

6.4.1 Closed Interval mein Function ki Maximum aur Minimum Values

Practice 6.3: answers ke saath

Miscellaneous Examples

Summary aur key points

Chapter 6 ki Miscellaneous Exercise

Practice questions answers ke saath (NCERT “Miscellaneous exercise”)

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye