## 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 JEE MAIN AND ADVANCED 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) Strictly increasing function A finction/to)is said to be a strictly increasing fiunction on (a,b)if x, < x, f(x) <f(x) for all x" x, e (a, b) Thus,f(x) is strictly increasing on (a, b) ifthe values off(x) increase with the increase in the values of x. Graphically, f(a) is increasing on (a, b) ifthe graphy (x) moves up as x moves to the right. The graph in Fig. 1 is the graph ofa strictly increasing function on (a.b). flx?) 01ax1 x2 b MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Strictly decreasing function A functionf(x) is said to be a strictly decreasing function on (a, b) if Xi < X2 f(x;) >f(x2) for all xi, X2 (a, b) Thus, f(x) is strictly decreasing on (a, b) if the values of f (x) de- crease with the increase in the values ofx. Graphically it means that f(x) is a decreasing function on (a, b) if its graph moves down as x moves to the right. The graph in Fig. 2 is O2 the graph ofa strictly decreasing function. MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Monotonically increasing (MI) / Monotonically Decreasing (MD) (i) fis strictly increasing in (a, b) if f'(x) > 0 for each xe (a, b) (ii) f is strictly decreasing in (a, b) if f'(x) < 0 for each x e (a, b) (a) A function fx) is said to be increasing in (a, b) if f(x)20 (b) f(x) is said to be decreasing in (a, b) if f(x) S 0 MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Example Show that the function f given by f(x) =x3-3x2 + 4x, xe R is strictly increasing on R. Solution Note that f'(x) = 3x2 _ 6x + 4 t)y 3(x2- 2x1) 1 -30x 1>0, in every interval of R Therefore, the function fis strictly increasing on R MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Illustration The function (logx increases on the intervwal log.r (D) none of these SOLUTION: (C) Clearly,f(x) is defined for>0 Now, log r -1 (log x) f(r)- MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Ex. Find the intervals in which fo) =-x2-2x + 15 is increasing or decreasing Sol We have, fx)-x2-2x 15 For fx) to be increasing, we must have [ _2 < 0 and ab > 0, a < 0 b<01 Thus, f(x) is increasing on the interval For f(x) to be decreasing, we must have ,-1). f(x) < 0 2(x + 1 ) < 0 [ _2 < 0 and ab < 0, a < 0 b>0] So, f(x) is decreasing on (-1, ) MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Illustration If the function f(x) 2r-kx 5 is increasing on [, 2], then k lies in the interval SOLUTION (A) f' (x) is an increasing function on [1,2] We have, f (x) 4r-k Since f(x) is an increasing function on [1,2], therefore f (1) is the least value off" (x) on [1,2] Therefore, for f' (x) to be greater than zero for all x e [1,2] its least value i.e.f (1) must also be greater than zero. Now, f(>0 4-k> 0 f (x)>0 for 1x s2 Now, f" (x) = 4 for all x e [1,2] f"x)> 0 for all x E [1,2] MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Illustration For x> 1, y -log x satisfies the inequality (A) x- (B) x-1>y (C) y>x- (D) SOLUTION : (ABD) log x < x-1 for x >1. Butx2 r for x> Let f(x) = log x-(x-1 ). Then "f'(x)-1 I -x Clearly, f'(x)<0 for x>1. -1>x -1 and log x<x-1 x <x-1 log x < x2-1. f(x) is decreasing function for x > 1 f(x) <f( 1 ) for x > 1 log x-(x-1)<0for x > 1 < log x. Similarly, it can be proved that Hence, (a), (b) and (d) are true. => MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)