## 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) Ex. Find the local maxima or local minima, if any, of the function f(x) = sinx + cosx. 0 < x <-using the first derivative test So. We have, dy y f(x) sinx + cosx = cosx-sinx For a local maximum or a local minimum, we have dy 4 Now, we will see whether x =-is a point of local maximum or a point of local minimum or none of these In the left nbd of x =-, we have x <--> cosx > sinx cosx-sinx > 0 -> 0 In the right nbd of x- we Thus, So, f(x) attains a local maximum at x 2 4 have x > --> cosx < sinx cosx-sinx < 0 < 0 4 changes its sign from positive to negative as x increases through dx 4 + 4 MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Find the points at which the function f given by fx-(x-2f(x + 1 has i) local maxima Ex. (ii) local minima i) points of inflexion Sol.We have, 2 Now, f,(x)=0=> x=2,-1. Since (x 2(x +1) is always positive. So, sign of f(x) depends upon the sign of (x -2) (7x - 2). The changes in signs of f(x) as x increases through and 2 are shown in fig 2 Clearly, f(x) changes its sign from positive to negative as increases through 2/7 so, xis a point of local maximum. We observe that f(x) changes its sign from negative to positive as x increases through2 So, x 2 is a point of local minimum. There is no change in the sign of f(x) as increases through -1 So, x =-1 is a point of inflexion. MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)

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) 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!!