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Examples on Monotonicity Part-1 (in Hindi)
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Vineet Loomba is teaching live on Unacademy Plus

Vineet Loomba
IITian | No. 1 Educator in IIT-JEE (Maths) | 550+ free Lessons | 8+ Years Experience |vineetloomba.com | Youtuber | Read Reviews!!

U
Unacademy user
sir is it sufficient for Sbi po 2019.....please tell me.
Kapil Kathpal
2 months ago
yes..more than sufficient
Akhilesh Pandey
2 months ago
thanku sir .....u r doing a great job. again thanku so much.
sir what about sin x and cos x graph
Vineet Loomba
a month ago
Wht abt tht ?
nice explanation sir
Vineet Loomba
3 months ago
Thanks .. there are 29 more courses on my profile
If A Function Belongs To - 1 to plus infinity then function decrease. How?
Vineet Loomba
4 months ago
your statement is incomplete... range of function doesnot tell us decreasing nature.. provide me the complete ques
sir why are tangents are said as derivative of the function in a graph can u plzz tell me that in brief about that plzz
When this course will be completed??
  1. 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


  2. 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!!


  3. 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)


  4. 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)


  5. 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)


  6. 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)


  7. 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)


  8. 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)


  9. 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)


  10. 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)


  11. 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)