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Solved Examples - Monotonicity and Maxima Minima
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This lesson takes up more problems on maxima minima and monotonicity

Vineet Loomba is teaching live on Unacademy Plus

Vineet Loomba
IITian | No. 1 Educator in IIT-JEE | 6 Million Minutes Watch Time | 8+ Years Experience | Youtube: Maths Wallah | vineetloomba.com
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sir at 12:30 min , how did n pie. cos n pie - sin n pie became (-1)to the power n × n pie
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sir ,if in equation their is both algebric and trigometric function given than who to procced for monotonic increasing or decreasing?
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  1. SOLVED EXAMPLES APPLICATIONS OF DERIVATIVES JEE MAIN RANK BOOSTER COURSE PREPARED BY: ER. VINEET LOOMBA ITIAN IIT-JEE MENTOR


  2. FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) ABOUT ME B.Tech. From IIT Roorkee -$ IIT-JEE Mentor Since 2010. Youtube Channel with 10k Active Followers & Founder @ vineetloomba.com * Follow me @ https://unacademy.com/user/vineetloomba to get unacademy updates or search me on Google * Share among your peers as SHARING is CARING!!


  3. FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Other Detailed Courses Made so far on Unacademv: V Strategy for JEE Main and Advanced V Sets, Relations and Functions V Trigonometry V Applications of Derivatives Limits, Continuity. Differentiability V Indefinite Integration Definite Integration Complex Number V Logarithmic Functions V Sequences Series V Most Important Questions in IIT-JEE Permutations Combinations Binomial Theorem V Straight Lines V Applications of Integrals MathematicS V Parabola (Detailed Course) v Inverse Trigonometry V Mathematical Reasoning V Ellipse (Detailed Course) V Hyperbola V Differential Equations Circles (Detailed Course) V Probability Upcoming Courses next month: Coordinate Geometry > Differential Equations ER. VINEET LOOMBA (IIT RooRKee) MATHEMATICS FOR IIT-I


  4. 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Function f(x) = 2x3-9x2 + 12x + 29 is decreasing when (A) x < 2 Ex (B) x >2 (C) x3 Sol. x) is decreasing if f (x) <0 6x2-18x + 12 < 0 Ans.[D] SUCCESS IN JEE MAIN AND ADVANCED ER. VINEET LOOMBA (IIT RooRKEE)


  5. 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Ex. If fx) x2kx1 is increasing function in the interval 1, 2], then least value of k is- (A) 2 f(x) = 2x + k fx) is increasing function in the interval [1, 2] (B) 4 (C)-2 (D)-4 Sol. f(x)>02x + k>0 or k >-2x, when 1 s x 2 or k 2 and-4 k 2 max (-2, -4) Therefore least value of k is -2. Ans.[C SUCCESS IN JEE MAIN AND ADVANCED ER. VINEET LOOMBA (IIT RooRKEE)


  6. 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Function f(x) = x2e-x is monotonically increasing when (A) 0 < x < 2 (C) X eR [0, 2 (D)x < 0 Ex. Since f(x) is increasing (x) > 0 L. eX > 0] Ans.[A] SUCCESS IN JEE MAIN AND ADVANCED ER. VINEET LOOMBA (IIT RooRKEE)


  7. 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Ex. Function f(x) = sin x-cos x is monotonic increasing when (A) X e (0, /2) (C) X E T/4, 3T/4) (D) No where f(x) = cos x + sin x (B) x e (-r/4, /4) Sol. 2| cos x + sinx | = /2 sin( /4 + x) Now fx) is monotonic increasing when f(x) >0 sin(n/4 + x) > 0 ( sino is positive when 0 < < ) Xe (- /4, 3-4) Ans.IC] SUCCESS IN JEE MAIN AND ADVANCED ER. VINEET LOOMBA (IIT RooRKEE)


  8. 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) ify increasing function for all values of x, thern (A)-3/2 <sa1 (B) a > 1 (C) a < -3/2 Ex ax3 + 3x2 + (2a + 1)x + 1000 is strictly (D) a > 1 ora<-32 So y is increasing function 3ax2 + 6x + (2a + 1) > 0 Y X 36 4.3a (2a 1) < 0 and a>0 3 - 2a2-a < 0 and a 0 (2a3) (a - 1) >0 and a>0 - a1 or a <-3/2 and a > 0 Ans.[B] SUCCESS IN JEE MAIN AND ADVANCED ER. VINEET LOOMBA (IIT RooRKEE)


  9. 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Ex. The minimum value of the function XX (x > 0) is at - (A) x -1 (C) x = e-1 Let y = xx-log y = x log x (B) x = e (D) None of thesee (log y) = 1 + log x 2 and 2 (log y) = X= x Now for minimum value of y or log y dx 0xe1 Again for x e-1 (log y) - 0 ->1 l e0y is minimum at x Ans.[c] dx 2 dx SUCCESS IN EE MAIN AND ADVANCED ER. VINEET LOOMBA (IIT RooRKEe)


  10. 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Ex. The minimum value of 64 sec x 27 cosec x, 0<x<is (A) 91 (B) 25 (C) 125 (D) None of these Sol. Let y64 sec x27 cosec x =64 sec x tan x-27 cosec x cot x dx = 64 sec3 x + 64 sec x tan2x + 27 cosec3 x + 27 cosec x cot? x dx Now = 0 64 sec x tan x =27 cosec x cot x dx tan3 x 27/64 - tan x3/4 Also thenx So v is minimum when x = tan-1 (34) and its min. value64 5/4) 27 (5/3)125 (+0 < x < /2) Ans.[C] SUCCESS IN EE MAIN AND ADVANCED ER. VINEET LOOMBA (IIT RooRKEe)


  11. FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) ENROL for this Course (Free) Enroll 439 Recommend Lessons (Like) E Rate and Review the Course .Comments . Sharing with friends 15 1. 68 111 ratings 21 reviews Share ER. VINEET LOOMBA (IIT RooRKEE) MATHEMATICS FOR IIT-IEE