## Vineet Loomba is teaching live on Unacademy Plus

IIT-JEE VIDEO COURSE FOR SURE SHOT SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) IIT-JEE MATHS MADE EASY PREPARED BY: ER. VINEET LOOMBA IITian | IIT-JEE MENtOR Search vineet loomba unacademy" on GOOGLE

REVISION COURSE FOR JEE MAIN AND ADVANCED (IIT-JEE) ABOUT ME B.Tech. From IIT Roorkee IIT-JEE Mentor Since 2010 Doubts/Feedback in Comment Section * Comment other topics you want to revise. Follow me @ https://unacademy.com/user/vineetloomba to get updates or search me on Google * Share among your peers as SHARING is CARING!!

APPLICATION OF DERIVATIVES MAXIMA-MINIMA IIT-IEE MATHS MADE EASY PREPARED BY: ER. VINEET LOOMBA IITiAN | IIT-JEE MENTOR FoR SURE SHoT SUcCESS IN JEE MAIN AND ADVANCED (IIT-JEE)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Second Method (second derivative test) Calculate -0 and solve for x. Suppose one root of dy <x d" dx2 = 0 is at x = a then find dx 2 If ve for x-a, then maximum at x-a dx If - +ve for x -a, then minimum at x a MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) NTH DeRIVATIVE TEST = 0 at x = a, then find dy', 0 at x = a, then neither maximum nor minimum at x = a. f 3 f0 at x-a, then find If > 0 i.e. positive at x a, then y is minimum at x a and 1 < 0 i.e. negative at x = a, then y is maximum at x = a and so on. MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) sinx+3 cosx is maximum when (1) x = 60 (3) x 30 Ex (2) x 45 (4) x - 0 dy Sol. (3) Let y= sinx + V3 cosx --sinxV3 COSX dy For maximum, or minimum =0 cosx=V3 sinx d2y -tanx- x 30 Since <0 for x 30 3 .. (3) is the correct answer. MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) 40 Ex The minimum value of the function is 3x4 +8x3-18x2 +60 (2) 3/2 (4) None of these (3) 40/53 Sol.(l) Let y -- 18x*+ 60) dy 1 dx 40 (12x3 24x2- 36x) d2y 1 dx 40 dy and- (36x2+ 48x-36) Now-=0 x3+ 2x2-3x = 0 or x(x-1) (x + 3) = 0 or x=0,1,-3 At x 0, 36 <0 y is maximum at x 0 the given function i.e.-is minimum at x-0 40 2 60 3 .. minimum value of the function- MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Two parts of 10 such that the sum of the twice of first with the square of second is minimum, are- Ex (2) 5, 5 (3) 4, 6 Sol. (1) Let two parts bex and (10-x). If y=2x+(10-x)2 dy hen2 - 2 (10 - x) - 2x - 18 Now =0=> x-9 dx Also then - 2> 0. Hence when x 9 value of y is minimum. So required two parts of 10 are 9 and 1. MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Example Find the absolute maximum and minimum values of a function fgiven by f(x) = 2x3-15x2 + 36x + 1 on the interval [1,5]. Solution We have f(x) = 2x3-15x2 + 36x + 1 f'(x) =6f-30x + 36 = 6(x-3) (x-2) or Note that f,(x) = 0 gives x = 2 and x = 3. We shall now evaluate the value of f at these points and at the end points of the interval [1, 5], i.e., at x = 1, x = 2, x = 3 and at x = 5. So /(1)=2(1)-15(12) +36 (1)+1=24 f (2) 2(23)- 15(22) +36(2) + 1 29 f(3) 2(33) 15(3) +36(3)+1 28 f (5) 2(53) 15(52)36 (5)1 56 Thus, we conclude that absolute maximum value of fon [1, 5] is 56, occurring at x=5, and absolute minimum value of/on [1, 5] is 24 which occurs at x= 1. MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) ABOUT ME B.Tech. From IIT Roorkee IIT-JEE Mentor Since 2010 Doubts/Feedback in Comment Section * Comment other topics you want to revise. Follow me @ https://unacademy.com/user/vineetloomba to get updates. * Share among your peers as SHARING is CARING!!