📈Application of Derivativesअवकलज के अनुप्रयोगApplication of Derivatives

1. Area of a circle of radius r.त्रिज्या r वाले वृत्त का क्षेत्रफल।Radius r wale circle ka area.
   A = \pi r^2
2. Rate of change of A with respect to r = derivative of A with respect to r.r के सापेक्ष A के परिवर्तन की दर = r के सापेक्ष A का अवकलज।r ke respect mein A ka rate = r ke respect mein A ka derivative.
   \dfrac{dA}{dr} = \dfrac{d}{dr}(\pi r^2) = 2\pi r
3. Put r = 5 only now, after differentiating.अवकलन के बाद ही अब r = 5 रखिए।Differentiate karne ke baad hi ab r = 5 daaliye.
   \left(\dfrac{dA}{dr}\right)_{r = 5} = 2\pi(5) = 10\pi

✔ The area changes at the rate of 10\pi cm² per cm of radius (the book prints the unit as cm²/s; the answers at the end of the book use cm²/cm for this type).क्षेत्रफल 10\pi cm² प्रति cm त्रिज्या की दर से बदलता है (पुस्तक में मात्रक cm²/s छपा है; पुस्तक के अंत के उत्तरों में ऐसे प्रश्नों का मात्रक cm²/cm है)।Area 10\pi cm² per cm radius ki dar se badalta hai (book mein unit cm²/s chhapa hai; book ke end ke answers mein aise sawaalon ki unit cm²/cm hai).

1. Let the edge be x; V = volume, S = surface area. x changes with time t.किनारा x मान लीजिए; V = आयतन, S = पृष्ठ क्षेत्रफल। x समय t के साथ बदलता है।Edge ko x maan lijiye; V = volume, S = surface area. x time t ke saath badalta hai.
   V = x^3,\quad S = 6x^2,\quad \dfrac{dV}{dt} = 9\ \text{cm}^3/\text{s}
2. Differentiate V = x^3 with respect to t (chain rule).V = x^3 का t के सापेक्ष अवकलन कीजिए (शृंखला नियम)।V = x^3 ko t ke respect mein differentiate kijiye (chain rule).
   9 = \dfrac{dV}{dt} = 3x^2\,\dfrac{dx}{dt} \;\Rightarrow\; \dfrac{dx}{dt} = \dfrac{3}{x^2}
3. Differentiate S = 6x^2 with respect to t and use the result above.S = 6x^2 का t के सापेक्ष अवकलन कीजिए और ऊपर का परिणाम रखिए।S = 6x^2 ko t ke respect mein differentiate kijiye aur upar wala result daaliye.
   \dfrac{dS}{dt} = 12x\,\dfrac{dx}{dt} = 12x \cdot \dfrac{3}{x^2} = \dfrac{36}{x}
4. Now put the edge length 10 cm.अब किनारे की लंबाई 10 cm रखिए।Ab edge ki length 10 cm daaliye.
   x = 10:\quad \dfrac{dS}{dt} = \dfrac{36}{10} = 3.6

✔ The surface area is increasing at 3.6 cm²/s.पृष्ठ क्षेत्रफल 3.6 cm²/s की दर से बढ़ रहा है।Surface area 3.6 cm²/s ki dar se badh raha hai.

1. A = \pi r^2; differentiate with respect to time by the chain rule.A = \pi r^2; शृंखला नियम से समय के सापेक्ष अवकलन कीजिए।A = \pi r^2; chain rule se time ke respect mein differentiate kijiye.
   \dfrac{dA}{dt} = \dfrac{d}{dr}(\pi r^2)\cdot\dfrac{dr}{dt} = 2\pi r\,\dfrac{dr}{dt}
2. The wave front moves at 4 cm/s, so the radius grows at 4 cm/s.तरंग 4 cm/s से आगे बढ़ती है, इसलिए त्रिज्या 4 cm/s से बढ़ती है।Wave 4 cm/s se aage badhti hai, isliye radius 4 cm/s se badhta hai.
   \dfrac{dr}{dt} = 4\ \text{cm/s},\quad r = 10:\ \dfrac{dA}{dt} = 2\pi(10)(4) = 80\pi

✔ The enclosed area is increasing at 80\pi cm²/s when r = 10 cm.r = 10 cm पर घिरा क्षेत्रफल 80\pi cm²/s की दर से बढ़ रहा है।r = 10 cm par ghira area 80\pi cm²/s ki dar se badh raha hai.

1. Length is decreasing, so its rate gets a minus sign.लंबाई घट रही है, इसलिए उसकी दर ऋणात्मक है।Length ghat rahi hai, isliye uski rate negative hai.
   \dfrac{dx}{dt} = -3\ \text{cm/min},\quad \dfrac{dy}{dt} = 2\ \text{cm/min}
2. (a) Perimeter.(a) परिमाप।(a) Perimeter.
   P = 2(x + y) \Rightarrow \dfrac{dP}{dt} = 2\left(\dfrac{dx}{dt} + \dfrac{dy}{dt}\right) = 2(-3 + 2) = -2
3. (b) Area — product rule, then x = 10, y = 6.(b) क्षेत्रफल — गुणनफल नियम, फिर x = 10, y = 6।(b) Area — product rule, phir x = 10, y = 6.
   A = xy \Rightarrow \dfrac{dA}{dt} = \dfrac{dx}{dt}\,y + x\,\dfrac{dy}{dt} = (-3)(6) + 10(2) = 2

✔ (a) The perimeter changes at -2 cm/min (it is decreasing). (b) The area changes at 2 cm²/min (it is increasing).(a) परिमाप -2 cm/मिनट की दर से बदलता है (घट रहा है)। (b) क्षेत्रफल 2 cm²/मिनट की दर से बदलता है (बढ़ रहा है)।(a) Perimeter -2 cm/min ki dar se badalta hai (ghat raha hai). (b) Area 2 cm²/min ki dar se badalta hai (badh raha hai).

1. Differentiate the cost with respect to output x.लागत का उत्पादन x के सापेक्ष अवकलन कीजिए।Cost ko output x ke respect mein differentiate kijiye.
   MC = \dfrac{dC}{dx} = 0.005(3x^2) - 0.02(2x) + 30
2. Put x = 3.x = 3 रखिए।x = 3 daaliye.
   x = 3:\ MC = 0.015(9) - 0.04(3) + 30 = 0.135 - 0.12 + 30 = 30.015

✔ Marginal cost ≈ ₹ 30.02 (exactly ₹ 30.015).सीमांत लागत ≈ ₹ 30.02 (ठीक ₹ 30.015)।Marginal cost ≈ ₹ 30.02 (exact ₹ 30.015).

1. Differentiate revenue with respect to x.आय का x के सापेक्ष अवकलन कीजिए।Revenue ko x ke respect mein differentiate kijiye.
   MR = \dfrac{dR}{dx} = 6x + 36
2. Put x = 5.x = 5 रखिए।x = 5 daaliye.
   x = 5:\ MR = 6(5) + 36 = 66

✔ Marginal revenue = ₹ 66.सीमांत आय = ₹ 66।Marginal revenue = ₹ 66.

1. Take any two real numbers x_1 < x_2; multiplying by the positive number 7 and subtracting 3 keep the order.कोई भी दो वास्तविक संख्याएँ x_1 < x_2 लीजिए; धनात्मक संख्या 7 से गुणा करने और 3 घटाने पर क्रम नहीं बदलता।Koi bhi do real numbers x_1 < x_2 lijiye; positive number 7 se guna karne aur 3 ghataane par order nahi badalta.
   x_1 < x_2 \Rightarrow 7x_1 < 7x_2 \Rightarrow 7x_1 - 3 < 7x_2 - 3 \Rightarrow f(x_1) < f(x_2)

✔ By Definition 1, f is strictly increasing (so also increasing) on \mathbb{R}.परिभाषा 1 से f, \mathbb{R} पर निरंतर वर्धमान (अतः वर्धमान भी) है।Definition 1 se f, \mathbb{R} par strictly increasing (isliye increasing bhi) hai.

1. Differentiate and complete the square.अवकलन कीजिए और पूर्ण वर्ग बनाइए।Differentiate kijiye aur perfect square banaiye.
   f'(x) = 3x^2 - 6x + 4 = 3(x^2 - 2x + 1) + 1 = 3(x - 1)^2 + 1
2. A square is never negative, so f'(x) > 0 in every interval of \mathbb{R}.वर्ग कभी ऋणात्मक नहीं होता, अतः \mathbb{R} के हर अंतराल में f'(x) > 0।Square kabhi negative nahi hota, isliye \mathbb{R} ke har interval mein f'(x) > 0.
   3(x - 1)^2 \ge 0 \Rightarrow f'(x) \ge 1 > 0

✔ By Theorem 1, f is increasing on \mathbb{R}.प्रमेय 1 से f, \mathbb{R} पर वर्धमान है।Theorem 1 se f, \mathbb{R} par increasing hai.

1. Derivative of \cos x.\cos x का अवकलज।\cos x ka derivative.
   f'(x) = -\sin x
2. (a) So f is decreasing in (0, \pi).(a) अतः (0, \pi) में f ह्रासमान है।(a) To (0, \pi) mein f decreasing hai.
   x \in (0, \pi):\ \sin x > 0 \Rightarrow f'(x) < 0
3. (b) So f is increasing in (\pi, 2\pi).(b) अतः (\pi, 2\pi) में f वर्धमान है।(b) To (\pi, 2\pi) mein f increasing hai.
   x \in (\pi, 2\pi):\ \sin x < 0 \Rightarrow f'(x) > 0

✔ (c) By (a) and (b), f goes down and then up in (0, 2\pi), so it is neither increasing nor decreasing there.(c) (a) और (b) से, (0, 2\pi) में f पहले घटता है फिर बढ़ता है, अतः वहाँ वह न वर्धमान है न ह्रासमान।(c) (a) aur (b) se, (0, 2\pi) mein f pehle ghatta hai phir badhta hai, isliye wahan wo na increasing hai na decreasing.

1. The point x = 2 splits the real line into (-\infty, 2) and (2, \infty).बिंदु x = 2 वास्तविक रेखा को (-\infty, 2) और (2, \infty) में बाँटता है।Point x = 2 real line ko (-\infty, 2) aur (2, \infty) mein baantta hai.
   f'(x) = 2x - 4 = 0 \Rightarrow x = 2
2. Test the sign of 2x - 4 on each side.दोनों ओर 2x - 4 का चिह्न जाँचिए।Dono taraf 2x - 4 ka sign check kijiye.
   x < 2:\ f'(x) < 0;\qquad x > 2:\ f'(x) > 0

✔ (a) f is increasing in (2, \infty); (b) decreasing in (-\infty, 2).(a) f, (2, \infty) में वर्धमान है; (b) (-\infty, 2) में ह्रासमान।(a) f, (2, \infty) mein increasing hai; (b) (-\infty, 2) mein decreasing.

1. Differentiate and factorise.अवकलन करके गुणनखंड कीजिए।Differentiate karke factorise kijiye.
   f'(x) = 12x^2 - 12x - 72 = 12(x^2 - x - 6) = 12(x - 3)(x + 2)
2. These split \mathbb{R} into (-\infty, -2), (-2, 3), (3, \infty).ये \mathbb{R} को (-\infty, -2), (-2, 3), (3, \infty) में बाँटते हैं।Ye \mathbb{R} ko (-\infty, -2), (-2, 3), (3, \infty) mein baant dete hain.
   f'(x) = 0 \Rightarrow x = -2,\ 3
3. Signs of (x - 3) and (x + 2) in each interval.हर अंतराल में (x - 3) और (x + 2) के चिह्न।Har interval mein (x - 3) aur (x + 2) ke signs.
   (-\infty, -2):\ (-)(-) > 0;\quad (-2, 3):\ (-)(+) < 0;\quad (3, \infty):\ (+)(+) > 0

✔ (a) Increasing in (-\infty, -2) and (3, \infty); (b) decreasing in (-2, 3). However, f is neither increasing nor decreasing in \mathbb{R}.(a) (-\infty, -2) और (3, \infty) में वर्धमान; (b) (-2, 3) में ह्रासमान। परंतु \mathbb{R} में f न वर्धमान है न ह्रासमान।(a) (-\infty, -2) aur (3, \infty) mein increasing; (b) (-2, 3) mein decreasing. Lekin poore \mathbb{R} mein f na increasing hai na decreasing.

1. x \in [0, \frac{\pi}{2}] means 3x \in [0, \frac{3\pi}{2}].x \in [0, \frac{\pi}{2}] का अर्थ है 3x \in [0, \frac{3\pi}{2}]।x \in [0, \frac{\pi}{2}] ka matlab 3x \in [0, \frac{3\pi}{2}].
   f'(x) = 3\cos 3x = 0 \Rightarrow 3x = \dfrac{\pi}{2},\ \dfrac{3\pi}{2} \Rightarrow x = \dfrac{\pi}{6},\ \dfrac{\pi}{2}
2. Cosine is positive in the first quadrant.प्रथम चतुर्थांश में cos धनात्मक है।Pehle quadrant mein cos positive hai.
   0 \le x < \tfrac{\pi}{6} \Rightarrow 0 \le 3x < \tfrac{\pi}{2} \Rightarrow f'(x) > 0
3. Cosine is negative between \frac{\pi}{2} and \frac{3\pi}{2}.\frac{\pi}{2} और \frac{3\pi}{2} के बीच cos ऋणात्मक है।\frac{\pi}{2} aur \frac{3\pi}{2} ke beech cos negative hai.
   \tfrac{\pi}{6} < x < \tfrac{\pi}{2} \Rightarrow \tfrac{\pi}{2} < 3x < \tfrac{3\pi}{2} \Rightarrow f'(x) < 0

✔ f is continuous at 0 and \frac{\pi}{6}, so by Theorem 1: (a) increasing on \left[0, \dfrac{\pi}{6}\right]; (b) decreasing on \left[\dfrac{\pi}{6}, \dfrac{\pi}{2}\right].f, 0 और \frac{\pi}{6} पर संतत है, अतः प्रमेय 1 से: (a) \left[0, \dfrac{\pi}{6}\right] पर वर्धमान; (b) \left[\dfrac{\pi}{6}, \dfrac{\pi}{2}\right] पर ह्रासमान।f, 0 aur \frac{\pi}{6} par continuous hai, to Theorem 1 se: (a) \left[0, \dfrac{\pi}{6}\right] par increasing; (b) \left[\dfrac{\pi}{6}, \dfrac{\pi}{2}\right] par decreasing.

1. Only these two values lie in [0, 2\pi].[0, 2\pi] में केवल ये दो मान हैं।[0, 2\pi] mein sirf ye do values hain.
   f'(x) = \cos x - \sin x = 0 \Rightarrow \sin x = \cos x \Rightarrow x = \dfrac{\pi}{4},\ \dfrac{5\pi}{4}
2. Check a test point in each piece, e.g. x = 0: f'(0) = 1 > 0; x = \pi: f'(\pi) = -1 < 0.हर भाग में एक परीक्षण बिंदु जाँचिए, जैसे x = 0: f'(0) = 1 > 0; x = \pi: f'(\pi) = -1 < 0।Har hisse mein ek test point check kijiye, jaise x = 0: f'(0) = 1 > 0; x = \pi: f'(\pi) = -1 < 0.
   \left[0, \tfrac{\pi}{4}\right):\ f' > 0;\quad \left(\tfrac{\pi}{4}, \tfrac{5\pi}{4}\right):\ f' < 0;\quad \left(\tfrac{5\pi}{4}, 2\pi\right]:\ f' > 0

✔ Increasing in \left[0, \dfrac{\pi}{4}\right) and \left(\dfrac{5\pi}{4}, 2\pi\right]; decreasing in \left(\dfrac{\pi}{4}, \dfrac{5\pi}{4}\right).\left[0, \dfrac{\pi}{4}\right) और \left(\dfrac{5\pi}{4}, 2\pi\right] में वर्धमान; \left(\dfrac{\pi}{4}, \dfrac{5\pi}{4}\right) में ह्रासमान।\left[0, \dfrac{\pi}{4}\right) aur \left(\dfrac{5\pi}{4}, 2\pi\right] mein increasing; \left(\dfrac{\pi}{4}, \dfrac{5\pi}{4}\right) mein decreasing.

1. A square is never negative and it is 0 at x = 0.वर्ग कभी ऋणात्मक नहीं होता और x = 0 पर 0 है।Square kabhi negative nahi hota aur x = 0 par 0 hai.
   f(x) = x^2 \ge 0\ \ \forall\, x \in \mathbb{R},\quad f(0) = 0
2. The graph keeps rising on both sides, so there is no largest value.ग्राफ़ दोनों ओर ऊपर उठता ही जाता है, अतः कोई सबसे बड़ा मान नहीं।Graph dono taraf upar uthta hi jaata hai, isliye koi sabse badi value nahi.
   x^2 \to \infty \text{ as } |x| \to \infty

✔ Minimum value 0 at x = 0; no maximum value in \mathbb{R}. (Note: on [-2, 1] the maximum value would be (-2)^2 = 4 at x = -2.)x = 0 पर निम्नतम मान 0; \mathbb{R} में कोई उच्चतम मान नहीं। (टिप्पणी: [-2, 1] पर उच्चतम मान x = -2 पर (-2)^2 = 4 होगा।)x = 0 par minimum value 0; \mathbb{R} mein koi maximum value nahi. (Note: [-2, 1] par maximum value x = -2 par (-2)^2 = 4 hogi.)

1. Distance from 0 is never negative.0 से दूरी कभी ऋणात्मक नहीं होती।0 se doori kabhi negative nahi hoti.
   |x| \ge 0\ \ \forall\, x,\quad |0| = 0

✔ Minimum value 0 at x = 0; no maximum value in \mathbb{R}. On [-2, 1] the maximum would be |-2| = 2. Note: |x| is not differentiable at 0, yet it has its minimum there — the graph shows it.x = 0 पर निम्नतम मान 0; \mathbb{R} में कोई उच्चतम मान नहीं। [-2, 1] पर उच्चतम मान |-2| = 2 होगा। ध्यान दीजिए: |x|, 0 पर अवकलनीय नहीं है, फिर भी वहीं इसका निम्नतम मान है — ग्राफ़ यह दिखाता है।x = 0 par minimum value 0; \mathbb{R} mein koi maximum nahi. [-2, 1] par maximum |-2| = 2 hogi. Dhyaan dijiye: |x|, 0 par differentiable nahi, phir bhi wahin iski minimum value hai — graph ye dikhata hai.

1. Whatever point you call "closest to 0", half of it is even closer — so no point gives the minimum.जिस बिंदु को आप "0 के सबसे निकट" कहें, उसका आधा और भी निकट है — अतः कोई बिंदु निम्नतम मान नहीं देता।Jis point ko aap "0 ke sabse paas" kahein, uska aadha aur bhi paas hai — isliye koi point minimum nahi deta.
   x_0 \in (0, 1) \Rightarrow \dfrac{x_0}{2} < x_0 \text{ and } \dfrac{x_0}{2} \in (0, 1)
2. Similarly no point is "closest to 1", so no maximum.इसी प्रकार कोई बिंदु "1 के सबसे निकट" नहीं, अतः उच्चतम मान नहीं।Isi tarah koi point "1 ke sabse paas" nahi, isliye maximum nahi.
   x_1 \in (0, 1) \Rightarrow \dfrac{x_1 + 1}{2} > x_1 \text{ and } \dfrac{x_1 + 1}{2} \in (0, 1)

✔ f has neither a maximum nor a minimum value in (0, 1). On [0, 1] it would have minimum 0 (at x = 0) and maximum 1 (at x = 1).(0, 1) में f का न उच्चतम मान है न निम्नतम। [0, 1] पर निम्नतम मान 0 (x = 0 पर) और उच्चतम मान 1 (x = 1 पर) होगा।(0, 1) mein f ki na maximum value hai na minimum. [0, 1] par minimum 0 (x = 0 par) aur maximum 1 (x = 1 par) hogi.

1. These are the only critical points.केवल यही क्रांतिक बिंदु हैं।Sirf yahi critical points hain.
   f'(x) = 3x^2 - 3 = 3(x - 1)(x + 1) = 0 \Rightarrow x = 1,\ -1
2. Sign changes from - to + at x = 1 → local minima; f(1) = 1 - 3 + 3 = 1.x = 1 पर चिह्न - से + → स्थानीय निम्नतम; f(1) = 1 - 3 + 3 = 1।x = 1 par sign - se + → local minima; f(1) = 1 - 3 + 3 = 1.
   \text{near } 1:\ f'(0.9) < 0,\ f'(1.1) > 0
3. Sign changes from + to - at x = -1 → local maxima; f(-1) = -1 + 3 + 3 = 5.x = -1 पर चिह्न + से - → स्थानीय उच्चतम; f(-1) = -1 + 3 + 3 = 5।x = -1 par sign + se - → local maxima; f(-1) = -1 + 3 + 3 = 5.
   \text{near } -1:\ f'(-1.1) > 0,\ f'(-0.9) < 0

✔ x = -1: local maxima, local maximum value 5. x = 1: local minima, local minimum value 1.x = -1: स्थानीय उच्चतम, स्थानीय उच्चतम मान 5। x = 1: स्थानीय निम्नतम, स्थानीय निम्नतम मान 1।x = -1: local maxima, local maximum value 5. x = 1: local minima, local minimum value 1.

1. Only critical point: x = 1.केवल एक क्रांतिक बिंदु: x = 1।Sirf ek critical point: x = 1.
   f'(x) = 6x^2 - 12x + 6 = 6(x - 1)^2 = 0 \Rightarrow x = 1
2. The sign does not change through x = 1.x = 1 से होकर चिह्न नहीं बदलता।x = 1 se guzarte hue sign nahi badalta.
   f'(x) = 6(x - 1)^2 \ge 0;\ \ f'(x) > 0 \text{ just left and just right of } 1

✔ x = 1 is neither a point of local maxima nor of local minima — it is a point of inflection. Since f' never changes sign on \mathbb{R}, the graph has no turning points.x = 1 न स्थानीय उच्चतम का बिंदु है न स्थानीय निम्नतम का — यह नति परिवर्तन बिंदु है। क्योंकि \mathbb{R} पर f' कभी चिह्न नहीं बदलता, ग्राफ़ में कोई वर्तन बिंदु नहीं।x = 1 na local maxima hai na local minima — ye point of inflection hai. Kyunki \mathbb{R} par f' kabhi sign nahi badalta, graph mein koi turning point nahi.

1. So the second derivative test cannot be used; 0 is a critical point. Use the first derivative test.अतः द्वितीय अवकलज परीक्षण लागू नहीं; 0 एक क्रांतिक बिंदु है। प्रथम अवकलज परीक्षण लगाइए।Isliye second derivative test nahi lagega; 0 critical point hai. First derivative test lagaiye.
   f \text{ is not differentiable at } x = 0
2. Sign of f' changes from - to + at 0.0 पर f' का चिह्न - से + बदलता है।0 par f' ka sign - se + badalta hai.
   x < 0:\ f(x) = 3 - x,\ f'(x) = -1 < 0;\qquad x > 0:\ f(x) = 3 + x,\ f'(x) = 1 > 0

✔ x = 0 is a point of local minima and the local minimum value is f(0) = 3.x = 0 स्थानीय निम्नतम का बिंदु है और स्थानीय निम्नतम मान f(0) = 3 है।x = 0 local minima ka point hai aur local minimum value f(0) = 3 hai.

1. Critical points.क्रांतिक बिंदु।Critical points.
   f'(x) = 12x^3 + 12x^2 - 24x = 12x(x - 1)(x + 2) = 0 \Rightarrow x = 0,\ 1,\ -2
2. Second derivative.द्वितीय अवकलज।Second derivative.
   f''(x) = 36x^2 + 24x - 24 = 12(3x^2 + 2x - 2)
3. Apply the second derivative test at each critical point.हर क्रांतिक बिंदु पर द्वितीय अवकलज परीक्षण लगाइए।Har critical point par second derivative test lagaiye.
   f''(0) = -24 < 0,\quad f''(1) = 36 > 0,\quad f''(-2) = 72 > 0
4. Find the values of f at these points.इन बिंदुओं पर f के मान निकालिए।In points par f ki values nikaaliye.
   f(0) = 12,\quad f(1) = 3 + 4 - 12 + 12 = 7,\quad f(-2) = 48 - 32 - 48 + 12 = -20

✔ Local maximum value f(0) = 12 at x = 0; local minimum values f(1) = 7 at x = 1 and f(-2) = -20 at x = -2.x = 0 पर स्थानीय उच्चतम मान f(0) = 12; x = 1 पर स्थानीय निम्नतम मान f(1) = 7 और x = -2 पर f(-2) = -20।x = 0 par local maximum value f(0) = 12; x = 1 par local minimum value f(1) = 7 aur x = -2 par f(-2) = -20.

1. Same function as Example 18, now with the second derivative.उदाहरण 18 वाला ही फलन, अब द्वितीय अवकलज के साथ।Example 18 wala hi function, ab second derivative ke saath.
   f'(x) = 6(x - 1)^2,\quad f''(x) = 12(x - 1)
2. The second derivative test fails here.यहाँ द्वितीय अवकलज परीक्षण असफल है।Yahan second derivative test fail hai.
   f'(1) = 0,\quad f''(1) = 0

✔ Go back to the first derivative test (Example 18): x = 1 is neither a point of local maxima nor of local minima; it is a point of inflection.प्रथम अवकलज परीक्षण पर लौटिए (उदाहरण 18): x = 1 न स्थानीय उच्चतम का बिंदु है न स्थानीय निम्नतम का; यह नति परिवर्तन बिंदु है।First derivative test par wapas jaaiye (Example 18): x = 1 na local maxima hai na local minima; ye point of inflection hai.

1. Let one number be x; the other is 15 - x.एक संख्या x मानिए; दूसरी 15 - x है।Ek number x maaniye; doosra 15 - x hai.
   S(x) = x^2 + (15 - x)^2 = 2x^2 - 30x + 225
2. By the second derivative test, x = \frac{15}{2} gives a local minimum of S.द्वितीय अवकलज परीक्षण से x = \frac{15}{2} पर S का स्थानीय निम्नतम है।Second derivative test se x = \frac{15}{2} par S ka local minimum hai.
   S'(x) = 4x - 30 = 0 \Rightarrow x = \dfrac{15}{2},\qquad S''(x) = 4 > 0

✔ The numbers are \dfrac{15}{2} and 15 - \dfrac{15}{2} = \dfrac{15}{2}. (In general, for sum k they are \dfrac{k}{2} and \dfrac{k}{2}.)संख्याएँ \dfrac{15}{2} और 15 - \dfrac{15}{2} = \dfrac{15}{2} हैं। (व्यापक रूप में योग k के लिए \dfrac{k}{2} और \dfrac{k}{2}।)Numbers \dfrac{15}{2} aur 15 - \dfrac{15}{2} = \dfrac{15}{2} hain. (General: sum k ke liye \dfrac{k}{2} aur \dfrac{k}{2}.)

1. Take any point (h, k) on the parabola; since k = h^2, write D in terms of k only.परवलय पर कोई बिंदु (h, k) लीजिए; k = h^2 से D को केवल k में लिखिए।Parabola par koi point (h, k) lijiye; k = h^2 se D ko sirf k mein likhiye.
   D = \sqrt{h^2 + (k - c)^2},\quad k = h^2 \Rightarrow D(k) = \sqrt{k + (k - c)^2}
2. Set the derivative to zero.अवकलज को शून्य रखिए।Derivative ko zero rakhiye.
   D'(k) = \dfrac{1 + 2(k - c)}{2\sqrt{k + (k - c)^2}} = 0 \Rightarrow k = \dfrac{2c - 1}{2}
3. First derivative test: - to +, so D is minimum there (the first derivative test is easier here than the second).प्रथम अवकलज परीक्षण: - से +, अतः वहाँ D न्यूनतम है (यहाँ प्रथम अवकलज परीक्षण द्वितीय से आसान है)।First derivative test: - se +, isliye wahan D minimum hai (yahan first test second se aasaan hai).
   k < \tfrac{2c - 1}{2} \Rightarrow D'(k) < 0;\qquad k > \tfrac{2c - 1}{2} \Rightarrow D'(k) > 0
4. Substitute k = \frac{2c - 1}{2}.k = \frac{2c - 1}{2} रखिए।k = \frac{2c - 1}{2} daaliye.
   D = \sqrt{\dfrac{2c - 1}{2} + \left(\dfrac{2c - 1}{2} - c\right)^2} = \sqrt{\dfrac{2c - 1}{2} + \dfrac{1}{4}} = \dfrac{\sqrt{4c - 1}}{2}

✔ The shortest distance is \dfrac{\sqrt{4c - 1}}{2}.न्यूनतम दूरी \dfrac{\sqrt{4c - 1}}{2} है।Sabse kam doori \dfrac{\sqrt{4c - 1}}{2} hai.

1. Let AR = x m, so RB = (20 - x) m; use Pythagoras in the two right triangles.मान लीजिए AR = x m, तो RB = (20 - x) m; दोनों समकोण त्रिभुजों में पाइथागोरस लगाइए।Maan lijiye AR = x m, to RB = (20 - x) m; dono right triangles mein Pythagoras lagaiye.
   RP^2 = x^2 + 16^2,\quad RQ^2 = (20 - x)^2 + 22^2
2. Add them.दोनों जोड़िए।Dono jodiye.
   S(x) = x^2 + 256 + (20 - x)^2 + 484 = 2x^2 - 40x + 1140
3. Second derivative test → minimum.द्वितीय अवकलज परीक्षण → न्यूनतम।Second derivative test → minimum.
   S'(x) = 4x - 40 = 0 \Rightarrow x = 10,\qquad S''(x) = 4 > 0

✔ AR = 10 m (R is the midpoint of AB).AR = 10 m (R, AB का मध्य-बिंदु है)।AR = 10 m (R, AB ka midpoint hai).

1. Drop perpendiculars DP and CQ on the base AB; \triangle APD \sim \triangle BQC and use Pythagoras.आधार AB पर लंब DP और CQ डालिए; \triangle APD \sim \triangle BQC और पाइथागोरस लगाइए।Base AB par perpendicular DP aur CQ daaliye; \triangle APD \sim \triangle BQC aur Pythagoras lagaiye.
   AP = QB = x,\quad DP = QC = \sqrt{100 - x^2}
2. Area = ½ (sum of parallel sides) × height; the base is 2x + 10.क्षेत्रफल = ½ (समांतर भुजाओं का योग) × ऊँचाई; आधार 2x + 10 है।Area = ½ (parallel sides ka sum) × height; base 2x + 10 hai.
   A(x) = \tfrac{1}{2}(2x + 10 + 10)\sqrt{100 - x^2} = (x + 10)\sqrt{100 - x^2}
3. Product rule, then a common denominator.गुणनफल नियम, फिर एक समान हर।Product rule, phir common denominator.
   A'(x) = \sqrt{100 - x^2} - \dfrac{x(x + 10)}{\sqrt{100 - x^2}} = \dfrac{-2x^2 - 10x + 100}{\sqrt{100 - x^2}}
4. A distance cannot be negative.दूरी ऋणात्मक नहीं हो सकती।Doori negative nahi ho sakti.
   2x^2 + 10x - 100 = 0 \Rightarrow x = 5 \ \ (x = -10 \text{ rejected})
5. Second derivative test → maximum at x = 5.द्वितीय अवकलज परीक्षण → x = 5 पर उच्चतम।Second derivative test → x = 5 par maximum.
   A''(x) = \dfrac{2x^3 - 300x - 1000}{(100 - x^2)^{3/2}},\quad A''(5) = \dfrac{-2250}{75\sqrt{75}} = \dfrac{-30}{\sqrt{75}} < 0
6. Put x = 5.x = 5 रखिए।x = 5 daaliye.
   A(5) = 15\sqrt{75} = 75\sqrt{3}

✔ The maximum area is 75\sqrt{3} cm².अधिकतम क्षेत्रफल 75\sqrt{3} cm² है।Maximum area 75\sqrt{3} cm² hai.

1. Cone: radius OC = r, height OA = h. Cylinder: radius OE = x, height QE; \triangle QEC \sim \triangle AOC.शंकु: त्रिज्या OC = r, ऊँचाई OA = h। बेलन: त्रिज्या OE = x, ऊँचाई QE; \triangle QEC \sim \triangle AOC।Cone: radius OC = r, height OA = h. Cylinder: radius OE = x, height QE; \triangle QEC \sim \triangle AOC.
   \dfrac{QE}{OA} = \dfrac{EC}{OC} \Rightarrow \dfrac{QE}{h} = \dfrac{r - x}{r} \Rightarrow QE = \dfrac{h(r - x)}{r}
2. Curved surface area of the cylinder = 2\pi \times radius \times height.बेलन का वक्र पृष्ठ = 2\pi \times त्रिज्या \times ऊँचाई।Cylinder ka curved surface = 2\pi \times radius \times height.
   S(x) = 2\pi x \cdot \dfrac{h(r - x)}{r} = \dfrac{2\pi h}{r}(rx - x^2)
3. Second derivative is negative for all x → maximum.द्वितीय अवकलज सभी x के लिए ऋणात्मक → उच्चतम।Second derivative sabhi x ke liye negative → maximum.
   S'(x) = \dfrac{2\pi h}{r}(r - 2x) = 0 \Rightarrow x = \dfrac{r}{2},\qquad S''(x) = -\dfrac{4\pi h}{r} < 0

✔ The curved surface is greatest when the cylinder's radius is \dfrac{r}{2} — half the radius of the cone.वक्र पृष्ठ तब अधिकतम है जब बेलन की त्रिज्या \dfrac{r}{2} हो — शंकु की त्रिज्या की आधी।Curved surface tab maximum hai jab cylinder ka radius \dfrac{r}{2} ho — cone ke radius ka aadha.

1. Critical points (both inside [1, 5]).क्रांतिक बिंदु (दोनों [1, 5] के अंदर)।Critical points (dono [1, 5] ke andar).
   f'(x) = 6x^2 - 30x + 36 = 6(x - 3)(x - 2) = 0 \Rightarrow x = 2,\ 3
2. Values at the critical points and at the end points 1 and 5.क्रांतिक बिंदुओं और अंत्य बिंदुओं 1 और 5 पर मान।Critical points aur end points 1 aur 5 par values.
   f(1) = 24,\quad f(2) = 29,\quad f(3) = 28,\quad f(5) = 56

✔ Absolute maximum value 56 at x = 5; absolute minimum value 24 at x = 1.निरपेक्ष उच्चतम मान 56, x = 5 पर; निरपेक्ष निम्नतम मान 24, x = 1 पर।Absolute maximum value 56, x = 5 par; absolute minimum value 24, x = 1 par.

1. Differentiate and take a common denominator.अवकलन करके एक समान हर लीजिए।Differentiate karke common denominator lijiye.
   f'(x) = 16x^{1/3} - \dfrac{2}{x^{2/3}} = \dfrac{2(8x - 1)}{x^{2/3}}
2. So the critical points are x = 0 and x = \frac{1}{8}.अतः क्रांतिक बिंदु x = 0 और x = \frac{1}{8} हैं।Isliye critical points x = 0 aur x = \frac{1}{8} hain.
   f'(x) = 0 \Rightarrow x = \dfrac{1}{8};\qquad f'(0) \text{ not defined}
3. Values at the critical points and the end points ((-1)^{4/3} = 1, (-1)^{1/3} = -1).क्रांतिक बिंदुओं और अंत्य बिंदुओं पर मान ((-1)^{4/3} = 1, (-1)^{1/3} = -1)।Critical points aur end points par values ((-1)^{4/3} = 1, (-1)^{1/3} = -1).
   f(-1) = 12 + 6 = 18,\quad f(0) = 0,\quad f\left(\tfrac{1}{8}\right) = 12\cdot\tfrac{1}{16} - 6\cdot\tfrac{1}{2} = -\tfrac{9}{4},\quad f(1) = 6

✔ Absolute maximum value 18 at x = -1; absolute minimum value -\dfrac{9}{4} at x = \dfrac{1}{8}.निरपेक्ष उच्चतम मान 18, x = -1 पर; निरपेक्ष निम्नतम मान -\dfrac{9}{4}, x = \dfrac{1}{8} पर।Absolute maximum value 18, x = -1 par; absolute minimum value -\dfrac{9}{4}, x = \dfrac{1}{8} par.

1. The helicopter is at (x, x^2 + 7).हेलिकॉप्टर (x, x^2 + 7) पर है।Helicopter (x, x^2 + 7) par hai.
   \text{distance} = \sqrt{(x - 3)^2 + (x^2 + 7 - 7)^2} = \sqrt{(x - 3)^2 + x^4}
2. Minimise the square of the distance (same point, simpler algebra).दूरी के वर्ग को न्यूनतम कीजिए (वही बिंदु, आसान बीजगणित)।Doori ke square ko minimum kijiye (wahi point, aasaan algebra).
   f(x) = (x - 3)^2 + x^4,\quad f'(x) = 2(x - 3) + 4x^3 = 2(x - 1)(2x^2 + 2x + 3)
3. Its discriminant is 4 - 24 < 0.इसका विविक्तकर 4 - 24 < 0 है।Iska discriminant 4 - 24 < 0 hai.
   f'(x) = 0 \Rightarrow x = 1 \quad (2x^2 + 2x + 3 = 0 \text{ has no real roots})
4. Only one critical point; comparing with another value shows \sqrt{5} is the minimum.केवल एक क्रांतिक बिंदु; किसी दूसरे मान से तुलना करने पर \sqrt{5} न्यूनतम सिद्ध होता है।Sirf ek critical point; kisi aur value se compare karne par \sqrt{5} minimum nikalta hai.
   f(1) = (1 - 3)^2 + 1^4 = 5,\quad \text{distance} = \sqrt{5};\qquad \sqrt{f(0)} = 3 > \sqrt{5}

✔ The nearest distance between the soldier and the helicopter is \sqrt{5} units.सैनिक और हेलिकॉप्टर के बीच निकटतम दूरी \sqrt{5} इकाई है।Soldier aur helicopter ke beech sabse kam doori \sqrt{5} units hai.

1. Velocity is the rate of change of distance.वेग दूरी के परिवर्तन की दर है।Velocity distance ka rate of change hai.
   v = \dfrac{dx}{dt} = 4t - t^2 = t(4 - t)
2. The car is at rest at P (t = 0) and at Q, so at Q, t = 4.कार P (t = 0) और Q पर विराम में है, अतः Q पर t = 4।Car P (t = 0) aur Q par ruki hai, isliye Q par t = 4.
   v = 0 \Rightarrow t = 0 \text{ or } t = 4
3. Distance covered in 4 seconds.4 सेकंड में तय दूरी।4 second mein tay doori.
   x\big|_{t = 4} = 4^2\left(2 - \dfrac{4}{3}\right) = 16 \cdot \dfrac{2}{3} = \dfrac{32}{3}

✔ The car reaches Q after 4 seconds; PQ = \dfrac{32}{3} m.कार 4 सेकंड बाद Q पर पहुँचती है; PQ = \dfrac{32}{3} m।Car 4 second baad Q par pahunchti hai; PQ = \dfrac{32}{3} m.

1. r = radius of the water surface, h = depth, \alpha = semi-vertical angle.r = पानी की सतह की त्रिज्या, h = गहराई, \alpha = अर्ध-शीर्ष कोण।r = paani ki surface ka radius, h = gehraai, \alpha = semi-vertical angle.
   \tan\alpha = \dfrac{r}{h} = 0.5 \Rightarrow r = \dfrac{h}{2}
2. Volume of water in terms of h only.पानी का आयतन केवल h में।Paani ka volume sirf h mein.
   V = \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3}\pi\left(\dfrac{h}{2}\right)^2 h = \dfrac{\pi h^3}{12}
3. Chain rule; \frac{dV}{dt} = 5 m³/h, h = 4 m.शृंखला नियम; \frac{dV}{dt} = 5 m³/h, h = 4 m।Chain rule; \frac{dV}{dt} = 5 m³/h, h = 4 m.
   \dfrac{dV}{dt} = \dfrac{\pi}{4}h^2\,\dfrac{dh}{dt} \Rightarrow 5 = \dfrac{\pi}{4}(4)^2\,\dfrac{dh}{dt}
4. Solve for \frac{dh}{dt}.\frac{dh}{dt} के लिए हल कीजिए।\frac{dh}{dt} ke liye solve kijiye.
   \dfrac{dh}{dt} = \dfrac{5}{4\pi} = \dfrac{35}{88}\ \text{m/h} \quad \left(\pi = \dfrac{22}{7}\right)

✔ The water level is rising at \dfrac{35}{88} m/h.पानी का तल \dfrac{35}{88} m/h की दर से ऊपर उठ रहा है।Paani ka level \dfrac{35}{88} m/h ki dar se upar uth raha hai.

1. AB = lamp post, MN = man, MS = s = shadow, AM = l; \triangle MSN \sim \triangle ASB.AB = खंभा, MN = आदमी, MS = s = छाया, AM = l; \triangle MSN \sim \triangle ASB।AB = khambha, MN = aadmi, MS = s = parchhaai, AM = l; \triangle MSN \sim \triangle ASB.
   \dfrac{MS}{AS} = \dfrac{MN}{AB} \Rightarrow \dfrac{s}{AS} = \dfrac{2}{6} \Rightarrow AS = 3s
2. Differentiate with respect to time.समय के सापेक्ष अवकलन कीजिए।Time ke respect mein differentiate kijiye.
   l = AM = 3s - s = 2s \Rightarrow \dfrac{dl}{dt} = 2\,\dfrac{ds}{dt}
3. The man walks at 5 km/h.आदमी 5 km/h से चलता है।Aadmi 5 km/h se chalta hai.
   \dfrac{dl}{dt} = 5 \Rightarrow \dfrac{ds}{dt} = \dfrac{5}{2}

✔ The shadow lengthens at \dfrac{5}{2} km/h.छाया \dfrac{5}{2} km/h की दर से बढ़ती है।Parchhaai \dfrac{5}{2} km/h ki dar se badhti hai.

1. Differentiate, take \frac{6}{5} common and factorise (x = 1 is a root of x^3 - 2x^2 - 5x + 6).अवकलन करके \frac{6}{5} उभयनिष्ठ लीजिए और गुणनखंड कीजिए (x = 1, x^3 - 2x^2 - 5x + 6 का मूल है)।Differentiate karke \frac{6}{5} common lijiye aur factorise kijiye (x = 1, x^3 - 2x^2 - 5x + 6 ka root hai).
   f'(x) = \dfrac{6}{5}x^3 - \dfrac{12}{5}x^2 - 6x + \dfrac{36}{5} = \dfrac{6}{5}(x - 1)(x + 2)(x - 3)
2. Test points: x = -3 gives (-4)(-1)(-6) < 0; x = 0 gives (-1)(2)(-3) = 6 > 0; x = 2 gives (1)(4)(-1) < 0; x = 4 gives > 0.परीक्षण बिंदु: x = -3 से (-4)(-1)(-6) < 0; x = 0 से (-1)(2)(-3) = 6 > 0; x = 2 से (1)(4)(-1) < 0; x = 4 से > 0।Test points: x = -3 se (-4)(-1)(-6) < 0; x = 0 se (-1)(2)(-3) = 6 > 0; x = 2 se (1)(4)(-1) < 0; x = 4 se > 0.
   (-\infty, -2): -;\quad (-2, 1): +;\quad (1, 3): -;\quad (3, \infty): +

✔ (a) Increasing in (-2, 1) and (3, \infty); (b) decreasing in (-\infty, -2) and (1, 3).(a) (-2, 1) और (3, \infty) में वर्धमान; (b) (-\infty, -2) और (1, 3) में ह्रासमान।(a) (-2, 1) aur (3, \infty) mein increasing; (b) (-\infty, -2) aur (1, 3) mein decreasing.

1. Chain rule; (\sin x + \cos x)^2 = 1 + \sin 2x.शृंखला नियम; (\sin x + \cos x)^2 = 1 + \sin 2x।Chain rule; (\sin x + \cos x)^2 = 1 + \sin 2x.
   f'(x) = \dfrac{\cos x - \sin x}{1 + (\sin x + \cos x)^2} = \dfrac{\cos x - \sin x}{2 + \sin 2x}
2. The denominator is positive and the numerator \cos x - \sin x > 0 there.हर धनात्मक है और वहाँ अंश \cos x - \sin x > 0 है।Denominator positive hai aur wahan numerator \cos x - \sin x > 0 hai.
   2 + \sin 2x > 0;\qquad 0 < x < \tfrac{\pi}{4} \Rightarrow \tan x < 1 \Rightarrow \cos x > \sin x

✔ f'(x) > 0 in \left(0, \dfrac{\pi}{4}\right), so f is increasing there.\left(0, \dfrac{\pi}{4}\right) में f'(x) > 0, अतः वहाँ f वर्धमान है।\left(0, \dfrac{\pi}{4}\right) mein f'(x) > 0, isliye wahan f increasing hai.

1. Chain rule.शृंखला नियम।Chain rule.
   A = \pi r^2 \Rightarrow \dfrac{dA}{dt} = 2\pi r\,\dfrac{dr}{dt}
2. Put r = 3.2 cm and \frac{dr}{dt} = 0.05 cm/s.r = 3.2 cm और \frac{dr}{dt} = 0.05 cm/s रखिए।r = 3.2 cm aur \frac{dr}{dt} = 0.05 cm/s daaliye.
   = 2\pi(3.2)(0.05) = 0.320\pi

✔ The area is increasing at 0.320\pi cm²/s.क्षेत्रफल 0.320\pi cm²/s की दर से बढ़ रहा है।Area 0.320\pi cm²/s ki dar se badh raha hai.

1. Side of removed square = x m = height; length 8 - 2x, breadth 3 - 2x.काटे गए वर्ग की भुजा = x m = ऊँचाई; लंबाई 8 - 2x, चौड़ाई 3 - 2x।Kaate gaye square ki side = x m = height; length 8 - 2x, breadth 3 - 2x.
   V(x) = x(3 - 2x)(8 - 2x) = 4x^3 - 22x^2 + 24x
2. Critical points.क्रांतिक बिंदु।Critical points.
   V'(x) = 12x^2 - 44x + 24 = 4(x - 3)(3x - 2) = 0 \Rightarrow x = 3,\ \tfrac{2}{3}
3. So x = \frac{2}{3} (we need 0 < x < 1.5).अतः x = \frac{2}{3} (हमें 0 < x < 1.5 चाहिए)।Isliye x = \frac{2}{3} (humein 0 < x < 1.5 chahiye).
   x \ne 3 \quad (\text{breadth } 3 - 2x \text{ would be negative})
4. Second derivative test → maximum.द्वितीय अवकलज परीक्षण → उच्चतम।Second derivative test → maximum.
   V''(x) = 24x - 44,\quad V''\left(\tfrac{2}{3}\right) = 16 - 44 = -28 < 0
5. Largest volume.अधिकतम आयतन।Maximum volume.
   V\left(\tfrac{2}{3}\right) = 4\left(\tfrac{2}{3}\right)^3 - 22\left(\tfrac{2}{3}\right)^2 + 24\left(\tfrac{2}{3}\right) = \dfrac{200}{27}

✔ Cut squares of side \dfrac{2}{3} m; the largest volume is \dfrac{200}{27} m³.\dfrac{2}{3} m भुजा के वर्ग काटिए; सबसे बड़ा आयतन \dfrac{200}{27} m³ है।\dfrac{2}{3} m side ke squares kaatiye; sabse bada volume \dfrac{200}{27} m³ hai.

1. Selling price and cost price of x items.x वस्तुओं का विक्रय मूल्य और लागत मूल्य।x items ka selling price aur cost price.
   S(x) = \left(5 - \dfrac{x}{100}\right)x = 5x - \dfrac{x^2}{100},\qquad C(x) = \dfrac{x}{5} + 500
2. Profit = selling price − cost price.लाभ = विक्रय मूल्य − लागत मूल्य।Profit = selling price − cost price.
   P(x) = S(x) - C(x) = \dfrac{24}{5}x - \dfrac{x^2}{100} - 500
3. Second derivative test → maximum.द्वितीय अवकलज परीक्षण → उच्चतम।Second derivative test → maximum.
   P'(x) = \dfrac{24}{5} - \dfrac{x}{50} = 0 \Rightarrow x = 240,\qquad P''(x) = -\dfrac{1}{50} < 0

✔ He should sell 240 items to earn maximum profit.अधिकतम लाभ के लिए उसे 240 वस्तुएँ बेचनी चाहिए।Maximum profit ke liye use 240 items bechne chahiye.