Rolle's Theorem PDF

60 Application of Derivatives 233 4.5.1 Definition. E3 Let f be a real valued function defined on the closed interval [a, b] such that, (1) f(x) is continuous in the closed interval [a, b] (2) f (x ) is differentiable in the open interval ]a, b[ and ID (3) f (a)  f (b) Then there is atleast one val...

60 Application of Derivatives 233 4.5.1 Definition. E3 Let f be a real valued function defined on the closed interval [a, b] such that, (1) f(x) is continuous in the closed interval [a, b] (2) f (x ) is differentiable in the open interval ]a, b[ and ID (3) f (a)  f (b) Then there is atleast one value c of x in open interval ]a, b[ for which f (c)  0. U 4.5.2 Analytical Interpretation. y Now, Rolle's theorem is valid for a function such that D YG (1) f (x ) is continuous in the closed interval [a, b] (2) f (x ) is differentiable in open interval ]a, b[ and (3) f (a)  f (b) y=c So, generally two cases arises in such circumstances. O a b x Case I: f(x) is constant in the interval [a, b] then f (x )  0 for all x  [a, b ]. Hence, Rolle's theorem follows, and we can say, f (c)  0, where a  c  b ST U Case II: f(x) is not constant in the interval [a, b] and since f (a)  f (b). y C Increas e A O x= a positive decreas e B x=cx=b x The function should either increase or decrease when x takes values slightly greater than a. Now, let f (x ) increases for x  a Since, f (a)  f (b) , hence the function must seize to increase at some value x  c and decreasing upto x  b. 234 Application of Derivatives Clearly at x  c function has maximum value. Now let h be a small positive quantity then, from definition of maximum value of the function, f (c  h)  f (c)  0 But, if lim h 0 f (c  h)  f (c) f (c  h)  f (c)  0 and lim 0 h 0 h h.....(i) E3 lim h 0 60 f (c  h)  f (c) f (c  h)  f (c)  0 and 0 h h  So, and f (c  h)  f (c)  0 f (c  h)  f (c) f (c  h)  f (c) ,  lim h 0 h h The Rolle's theorem cannot be applicable because in such case, ID RHD at x  c  LHD at x  c. Hence, f (x ) is not differentiable at x  c, which contradicts the condition of Rolle's theorem.  Only one possible solution arises, when lim f (c  h)  f (c) f (c  h)  f (c)  lim 0 h 0 h h U h 0 Which implies that, f (c)  0 where a  c  b D YG Hence, Rolle's theorem is proved.y A x= a Increas e Decrea see O B minimu m x x = b x=c U Similarly, the case where f (x ) decreases in the interval a  x  c and then increases in the interval c  x  b, f (c)  0. But when x  c, the minimum value of f (x ) exists in the interval [a, b]. ST 4.5.3 Geometrical Interpretation. Consider the portion AB of the curve y  f (x ) , lying between x  a y and x  b , such that C1 (1) It goes continuously from A to B. C2 B (2) It has tangent at every point between A and B and A D (3) Ordinate of A = ordinate of B From figure, it is clear that f (x ) increases in the interval AC 1 , which implies that f (x )  0 in this region and decreases in the O x=a x= x=b c x Application of Derivatives 235 interval C1 B which implies f ( x )  0 in this region. Now, since there is unique tangent to be drawn on the curve lying in between A and B and since each of them has a unique slope i.e., unique value of f (x ).  Due to continuity and differentiability of the function f (x ) in the region A to B. There is a 60 point x  c where f (c)  0. Hence, f (c)  0 where a  c  b Thus Rolle's theorem is proved. Similarly the other parts of the figure given above can be explained, establishing Rolle's theorem throughout. E3 Note : On Rolle's theorem generally two types of problems are formulated.  To check the applicability of Rolle's theorem to a given function on a given interval.  To verify Rolle's theorem for a given function in a given interval. ID In both types of problems we first check whether f(x) satisfies the condition of Rolle's theorem or not. The following results are very helpful in doing so. U (i) A polynomial function is everywhere continuous and differentiable. D YG (ii) The exponential function, sine and cosine functions are everywhere continuous and differentiable. (iii) Logarithmic functions is continuos and differentiable in its domain. (iv)  3 5 tan x is not continuous and differentiable at x   ,  ,  ,....... 2 2 2 (v) |x| is not differentiable at x  0. If f (x ) tends to   as x K , then f (x ) is not differentiable at x  K. (vi) For example, if f (x )  (2 x  1) , then f (x )  U 1/2 ST So, f (x ) is not differentiable at x  Example: 1  1 1 is such that as x    f (x )  2x  1 2 1. 2 The function f (x )  x (x  3)e 1 / 2 x satisfies all the condition of Rolle's theorem in [– 3, 0]. The value of c is (a) 0 Solution: (c) (b) 1 (c) – 2 (d) – 3 To determine 'c' in Rolle's theorem, f (c)  0  1  1   1 Here f (x )  (x 2  3 x )e (1 / 2) x.    (2 x  3)e (1 / 2) x = e (1 / 2) x  (x 2  3 x )  2 x  3  =  e ( x / 2) x 2  x  6 2  2   2  f (c)  0  c 2  c  6  0  c  3,  2. But c  3 [3, 0] , Hence c = –2.  236 Application of Derivatives Example: 2  2 3 1 0 If the function f (x )  x 3  6 x 2  ax  b satisfies Rolle's theorem in the interval [1, 3] and f    3   then [MP PET 2002] (b) a  6 Solution: (a) (c) a  6 f (x )  x 3  6 x 2  ax  b  f (x )  3 x 2  12 x  a 2    1  1  1    12  2  a0   0  3 2   f (c)  0  f  2      3 3 3       (d) a  1 60 (a) a  11 D YG U ID E3   1 4  1    a  0  12  1  4 3  24  4 3  a  0  a  11.   12  2   3 4     3 3 3    Rolle's Theorem Basic Level 1. Rolle's theorem is true for the function f (x )  x 2  4 in the interval (a) [– 2, 0] ST  f (x)dx 2 is equal 1 (c) 1 (d) 2       Consider the function f (x )  e 2 x sin 2 x over the interval  0, . A real number c   0,  , as guaranteed by Roll's 2 2     theorem, such that f (c)  0 is (a) 5. x2  3x satisfies all the conditions of Rolle's theorem x 1 (b) [–3, 0] (c) [1.5, 3] (d) For no interval If f (x ) satisfies the conditions of Rolle's theorem in [1, 2] and f (x ) is continuous in [1, 2] then to [DCE 2002] (a) 3 4. (d) [0, 2] For which interval, the function (a) [0, 3] 3.  1 (c) 0,   2 U 2. (b) [– 2, 2]  8 (b) 0 (b)  6 (c)  4 (d)  3 If the function f (x )  ax 3  bx 2  11 x  6 satisfies the conditions of Rolle's theorem for the interval [1, 3] and  1    0 , then the values of a and b are respectively f  2  3   (a) 1, – 6 (b) – 2, 1 (c) –1, 1 2 (d) – 1, 6 Application of Derivatives 237 6. Rolle's theorem is not applicable to the function f (x ) | x | defined on [– 1, 1] because [AISSE 1986; MP PET 1994, 95] (a) f is not continuous on [– 1, 1] 1, 1) (c) f (1)  f (1) (b) f is not differentiable on (– (d) f (1)  f (1)  0    x ln x , x  0   Let f (x )    Rolle's theorem is applicable to f for x  [0, 1] , if    , x  0  0  9. (d) 1 2 The value of a for which the equation x 3  3 x  a  0 has two distinct roots in [0, 1] is given by (a) –1 (b) 1 (c) 3 (d) None of these Let a, be two distinct roots of a polynomial f (x ). Then there exists at least one root lying between a and b of the polynomial (a) f (x ) 10. (c) 0 If E3 8. (b) –1 [IIT-JEE Screening 2004] 60 (a) –2 (c) f (x ) (b) f (x ) (d) None of these a0 a a a  1  2 .....  n 1  an  0. Then the function f (x )  a0 x n  a1 x n 1  a2 x n  2 .....  an has in (0, 1) n 1 n n 1 2 (a) At least one zero (b) At most one zero (c) Only 3 zeros ID 7. ST U D YG U *** (d) Only 2 zeros 60 Application of Derivatives 237 2 3 4 5 6 b d b a a b 7 8 9 10 d d b a ST U D YG U ID 1 E3 Assignment (Basic and Advance Level)