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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

U
Unacademy user
sir at 12:30 min , how did n pie. cos n pie - sin n pie became (-1)to the power n × n pie
sir ,if in equation their is both algebric and trigometric function given than who to procced for monotonic increasing or decreasing?
  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)


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