Lagrange's Mean Value Theorem PDF

60 238 Application of Derivatives If a function f (x ) , (1) Is continuous in the closed interval [a, b] and (2) Is differentiable in the open interval (a, b) E3 4.6.1 Definition. 4.6.2 Analytical Interpretation. D YG   (x ) is also continuous in [a,b] f (b)  f (a) x b a U First form: Consider t...

60 238 Application of Derivatives If a function f (x ) , (1) Is continuous in the closed interval [a, b] and (2) Is differentiable in the open interval (a, b) E3 4.6.1 Definition. 4.6.2 Analytical Interpretation. D YG   (x ) is also continuous in [a,b] f (b)  f (a) x b a U First form: Consider the function,  (x )  f (x )  Since, f (x ) is continuous in [a, b] f (b)  f (a) b a ID Then there is atleast one value c  (a, b ) , such that; f (c)  since, f (x ) exists in (a, b) hence  (x ) also exists in (a, b) and  (x )  f (x )  f (b)  f (a) b a.....(i) Clearly,  (x ) satisfies all the condition of Rolle's theorem  There is atleast one value of c of x between a and b such that  (c)  0 substituting x  c in (i) we get, f (b)  f (a) which proves the theorem. b a U f (c)  ST Second form: If we write b  a  h then a  c  b, c  a  h where 0    1 Thus, the mean value theorem can be stated as follows: If (i) f (x ) is continuous in closed interval [a, a+h] (ii) f (x ) exists in the open interval (a, a  h) then there exists at least one number  (0    1) Such that f (a  h)  f (a)  h f (a  h). 4.6.3 Geometrical Interpretation. Let f (x ) be a function defined on [a, b] and let APB be the curve represented by y  f (x ). Then co-ordinates of A and B are (a, f (a)) and (b, f (b)) respectively. Suppose the chord AB makes an angle  with the axis of x. Then from the triangle ARB, we have Application of Derivatives 239 tan   BR f (b)  f (a)  tan   b a AR By Lagrange's Mean value theorem, we have, f (c)  f (b)  f (a)  tan   f (c) b a 60 y B[b, f(b)]  [a, f(a)] A P[c, M O N x E3 R  f(c)] In the mean-value theorem 15 6 (a) 1  (b) 1  15 (c) 1  From mean value theorem f (c)  21 6 (d) 1  21 f (b)  f (a) b a D YG Solution: (c) f (b)  f (a) 1  f (c) , if a  0 , b  and f (x )  x (x  1)(x  2), the value of c is [MP PET 200 2 b a U Example: 1 ID  slope of the chord AB = slope of the tangent at (c, f (c)) a  0, f (a)  0  b  1 3 , f (b )  2 8 f (x )  (x  1)(x  2)  x (x  2)  x (x  1) , f (c)  (c  1)(c  2)  c(c  2)  c(c  1) = c 2  3 c  2  c 2  2c  c 2  c , f (c)  3c 2  6 c  2 According to mean value theorem U 3  0 3 5 8 f (b)  f (a) 2   3c 2  6 c   0  f (c)   3c  6c  2    4 4 b a 1  0 2 6  36  15 6  21 21  1. 23 6 6 ST c Example: 2 From mean value theorem f (b)  f (a)  (b  a) f (x1 ), a  x1  b if f ( x )  (a) Solution: (a) Example: 3 ab (b) 2 ab ab (c) 1 then x 1 x ab 2 (d) ba ba 1 1  1 1 b a 1   x 1  ab. f (x 1 )  2 ,  2  b a ab x1 x1 The abscissae of the points of the curve y  x 3 in the interval [–2, 2], where the slope of the tangent can be obtained by mean value theorem for the interval [– 2, 2] are (a)  Solution: (a) 2 3 (b)  3 2 Given that equation of curve y  x 3  f (x ) (c)  3 (d) 0 240 Application of Derivatives So f (2)  8 and f (2)  8 8  (8 ) 2 f (2)  f (2) .  3x 2; x   4 2  (2) 3 ST U D YG U ID E3 60 Now f (x )  3 x 2  f (x )  60 Application of Derivatives 241 Basic Level For the function x  (a) 1 (b) 3 (c) 2 [MP PET 1997] (d) None of these For the function f (x )  e x , a  0, b  1 , the value of c in mean value theorem will be (a) log x (b) log (e  1) (c) 0 [MP PET 1999] (d) a  x1  b 1 , x  [1, 3] , the value of c for the mean value theorem is x D YG 3. (c) a  x1  b (b) a  x1  b (a) a  x1  b 2. f (b)  f (a) , then b a ID If from mean value theorem, f (x1 )  U 1. E3 Lagrange's Mean Value Theorem [DCE 2002] (d) 1 Advance Level 4. If the function f (x )  x 3  6 ax 2  5 x satisfies the conditions of Lagrange's mean value theorem for the interval [1, U 2] and the tangent to the curve y  f (x ) at x  7 is parallel to the chord that joins the points of intersection of 4 the curve with the ordinates x  1 and x  2. Then the value of a is (b) 35 / 48 ST (a) 35 / 16 5. If f (x )  cos x , 0  x  (a) 6.  6  2 (c) 7 / 16 (d) 5 / 16 , then the real number 'c' of the mean value theorem is (b)  4 2 (c) sin 1     [MP PET 1994]  (d) cos 1     Let f (x ) satisfy all the conditions of mean value theorem in [0, 2]. If f (0 )  0 and | f ( x )|  1 for all x in [0, 2], 2 then (a) f (x )  2 (b) | f (x )|  1 (c) f ( x )  2 x (d) f ( x )  3 for at least one x in [0, 2] 242 Application of Derivatives 7. The function f (x )  (x  3)2 satisfies all the conditions of mean value theorem in [3, 4]. A point on y  (x  3)2 , where the tangent is parallel to the chord joining (3, 0) and (4, 1) is 7 1 (a)  ,  2 2 (c) (1, 4 ) Let f (x ) and g(x ) are defined and differentiable for x  x 0 and f (x 0 )  g(x 0 ), f (x )  g(x ) for x  x 0 , then (a) f (x )  g(x ) for some x  x 0 (b) f (x )  g(x ) for some x  x 0 E3 (c) f (x )  g(x ) for all x  x 0 (d) None of these Let f be differentiable for all x. If f (1)  2 and f ( x )  2 for all x  [1, 6] then (b) f (6 )  8 (a) f (6 )  8 10. (c) f (6 )  5 ID 9. (d) (4, 1) 60 8. 7 1  (b)  ,  2 4 (d) f (6 )  5 The value of c in Lagrange's theorem for the function |x| in the interval [– 1, 1] is (a) 0 (b) 1/2 (c) –1/2 U interval ST U D YG *** (d) Non-existent in the 2 3 4 5 6 c b b b c b 7 8 9 10 b c b d ST U D YG U ID 1 E3 Assignment (Basic and Advance Level) 60 242 Application of Derivatives