Chapter 4.3 Maxima and Minima PDF

60 Application of Derivatives 201 4.3.1 Introduction. E3 In this chapter we shall study those points of the domain of a function where its graph ID changes its direction from upwards to downwards or from downwards to upwards. At such points the derivative of the function, if it y exists, is necessar...

60 Application of Derivatives 201 4.3.1 Introduction. E3 In this chapter we shall study those points of the domain of a function where its graph ID changes its direction from upwards to downwards or from downwards to upwards. At such points the derivative of the function, if it y exists, is necessarily zero. x U O D YG 4.3.2 Maximum and Minimum Values of a Function. By the maximum / minimum value of function f (x ) we should mean local or regional maximum/minimum and not the greatest / least value attainable by the function. It is also possible in a function that local maximum at one point is smaller than local minimum at another point. Sometimes we use the word extreme for Y f(a–h) U maxima and minima. Definition: A function f (x ) is said to have a maximum at x  a if f (a) is greatest of all values in the suitably small Increasi ng f(a ) f(a+h) decreasi ng f(b Increasin g f(b+h) f(b–h) X O a– aa+h b b+h (b– h ) h) ST neighbourhood of a where x  a is an interior point in the domain of f (x ). Analytically this means f (a)  f (a  h) and f (a)  f (a  h) where h  0. (very small quantity). Similarly, a function y  f (x ) is said to have a minimum at x  b. If f (b ) is smallest of all values in the suitably small neighbourhood of b where x  b is an interior point in the domain of f (x ). Analytically, f (b )  f (b  h) and f (b )  f (b  h) where h  0. (very small quantity). Y f(a ) f(a+h) decreasi ng f(b f(a–h) increasin g O f(b–h) ) f(b+h) a–h aa+h b–h b b– h X 202 Application of Derivatives Hence we find that, y 60  f (a)  f (a  h)  0 (i) x  a is a maximum point of f (x )   f (a)  f (a  h)  0  f (b )  f (b  h)  0 x  b is a minimum point of f (x )   f (b )  f (b  h)  0 (iii) x  c is neither a maximum point nor a minimum point,  f (c)  f (c  h) and    have opposite signs.  f (c)  f (c  h)  a c b x ID 4.3.3 Local Maxima and Local Minima. E3 (ii) (1) Local maximum : A function f (x ) is said to attain a local maximum at x  a if there exists a neighbourhood (a   , a   ) of a such that f ( x )  f (a) for all x  (a   , a   ), x  a U or f (x )  f (a)  0 for all x  (a   , a   ), x  a. In such a case f (a) is called the local maximum value of f (x ) at x  a. D YG (2) Local minimum: A function f (x ) is said to attain a local minimum at x  a if there exists a neighbourhood (a   , a   ) of a such that f ( x )  f (a) for all x  (a   , a   ), x  a or f (x )  f (a)  0 for all x  (a   , a   ), x  a The value of function at x  a i.e., f (a) is called the local minimum value of f (x ) at x  a. U The points at which a function attains either the local maximum values or local minimum values are known as the extreme points or turning points and both local maximum and local minimum values are called the extreme values of f (x ). Thus, a ST function attains an extreme value at x  a if f (a) is either a local y maximum value or a local minimum value. Consequently at an extreme point ' a' f (x )  f (a) keeps the same sign for all values of x in a deleted nbd of a. In fig. we observe that the x-coordinates of the points A, C, E are points of local maximum and the values at these points i.e., their ycoordinates are the local maximum values of f (x ). The x-coordinates A y= f(x) C E D B O x of points B and D are points of local minimum and their y-coordinates are the local minimum values of f (x ). Application of Derivatives 203 Note :  By a local maximum (or local minimum) value of a function at a point x  a we mean the greatest (or the least) value in the neighbourhood of point x  a and not the absolute maximum (or the absolute minimum). In fact a function may have any 60 number of points of local maximum (or local minimum) and even a local minimum value may be greater than a local maximum value. In fig. the minimum value at D is greater than the maximum value at A. Thus, a local maximum value may not be the greatest value and a local minimum value may not be the least value of the function in its domain. E3  The maximum and minimum points are also known as extreme points.  A function may have more than one maximum and minimum points. ID  A maximum value of a function f(x) in an interval [a, b] is not necessarily its greatest value in that interval. Similarly, a minimum value may not be the least value of the function. A minimum value may be greater than some maximum value for a function. U  If a continuous function has only one maximum (minimum) point, then at this point function has its greatest (least) value.  Monotonic functions do not have extreme points. 4.3.4 Conditions for Maxima and Minima of a Function. D YG (1) Necessary condition: A point x  a is an extreme point of a function f (x ) if f (a)  0, provided f (a) exists. Thus, if f (a) exists, then x  a is an extreme point  f (a)  0 or f (a)  0  x  a is not an extreme point But its converse is not true i.e., f (a)  0, x  a is not an extreme point. U For example if f (x )  x 3 , then f (0 )  0 but x  0 is not an extreme point. (2) Sufficient condition: ST (i) The value of the function f (x ) at x  a is maximum, if f (a)  0 and f (a)  0. (ii) The value of the function f (x ) at x  a is minimum if f (a)  0 and f (a)  0. Note :  If f (a)  0 , f (a)  0, f (a)  0 then x  a is not an extreme point for the function f (x ).  If f (a)  0, f (a)  0, f (a)  0 then the sign of f (iv) (a) will determine the maximum and minimum value of function i.e., f (x ) is maximum, if f (iv) (a)  0 and minimum if f (iv) (a)  0. 4.3.5 Working rule for Finding Maxima and Minima. (1) Find the differential coefficient of f (x ) with respect to x, i.e., f (x ) and equate it to zero. 204 Application of Derivatives (2) Find differential real values of x by solving the equation f (x )  0. Let its roots be a, b, c...... (3) Find the value of f (x ) and substitute the value of a1 , a2 , a3...... in it and get the sign of f (x ) for each value of x. 60 (4) If f (a)  0 then the value of f (x ) is maximum at x  a and if f (a)  0 then value of f (x ) will be minimum at x  a. Similarly by getting the signs of f (x ) at other points b, c.....we can find the points of maxima and minima. What are the minimum and maximum values of the function x 5  5 x 4  5 x 3  10 Rajasthan PET 1995] (a) 37,  9 (b) 10, 0 (c) It has 2 minimum and 1 maximum values (d) It has 2 maximum and 1 minimum values  ID y  x 5  5 x 4  5 x 3  10 dy  5 x 4  20 x 3  15 x 2  5 x 2 (x 2  4 x  3) = 5 x 2 (x  3) (x  1) dx dy  0 , gives x  0, 1, 3 dx......(i) U Solution: (a) [DCE 1999; E3 Example: 1 3 d y d 2y 2  20 x 3  60 x 2  30 x  10 x (2 x 2  6 x  3) and  10 (6 x  12 x  3) 3 dx 2 dx D YG Now, For x  0 : dy d 2y d 3y  0, 2  0, 3  0 ,  Neither minimum nor maximum dx dx dx For x  1 , For x  3 , dx 2 d 2y dx 2  10 =negative,  Maximum value y max.  9  90 =positive,  Minimum value y min.  37. The maximum value of sin x (1  cos x ) will be at  U Example: 2 d 2y (a) x  y  sin x (1  cos x )  sin x  ST Solution: (c) 2   6 (c) x   3 (d) x   1 sin 2 x 2 dy d 2y  cos x  cos 2 x and   sin x  2 sin 2 x dx dx 2 On putting  x dy  0, cos x  cos 2 x  0  cos x   cos 2 x  cos(  2 x )  x    2 x dx  d 2y   3 3 3 3 1  2   2.  ,   2  which is negative.   sin    2 sin   = 2 2 2 3 3 3      dx  x  / 3   at x  Example: 3 (b) x  [UPSEAT 1999]  3 the function is maximum. If y  a log x  bx 2  x has its extremum value at x  1 and x  2 , then (a, b)  Application of Derivatives 205 dy a  dy   a  2b  1  0  a  2b  1   2bx  1    dx x  dx  x 1 a  2b  1  4b  1  0 2 x is (a) (e)e (c) (e ) e (b) (e )1 / e 1 1  1  f (x )     f (x )     log  1  x x  x  f ( x )  0  log  3b  1 2  b 1 6 and 1999; 1 (d)   e  e ID x x 1 0 2 [DCE Karnataka CET 1999; UPSEAT 2003] 1 1 1  1  log e   e  x . Therefore, maximum value of function is e 1 / e. x x e Maximum slope of the curve y   x 3  3 x 2  9 x  27 is (a) 0 (b) 12 y  f (x )   x  3 x  9 x  27 3 Solution: (b) U Example: 5  b  4b  1 2. 1  3 3 1 Maximum value of   x Solution: (b)  60 a  dy    4b  1  0    dx  x  2 2 and E3 Solution: (d) Example: 4 1  2 (d)   ,   6  3 1  (c)  2,   2  1  (b)  , 2  2   1 (a)  1,   2 2 (c) 16 (d) 32 D YG The slope of this curve f (x )  3 x 2  6 x  9 Let g(x )  f (x )  3 x 2  6 x  9 Differentiate with respect to x, g (x )  6 x  6 Put g ( x )  0  x  1 Now, g (x )  6  0 and hence at x  1, g(x ) (Slope) will have maximum value. U  [g(1)]max.  3  1  6  9  12. Example: 6 The function f (x )   t(e x ST (a) 0 Solution: (b, d)  t  1) (t  1) (t  2)3 (t  3)5 dt has a local minimum at x  [IIT1999] 1 (b) 1 (c) 2 (d) 3 x f (x )  t(e t  1)(t  1)(t  2)3 (t  3)5 dt ,  f (x )  x (e x  1) (x  1) (x  2)3 (x  3)5 1 For local minima, slope i.e., f (x ) should change sign from – ve to +ve f ( x )  0  x  0, 1, 2, 3 If x  0  h, where h is a very small number, then f (x )  ()()(1)(1)(1)  ve If x  0  h , f (x )  ()()()(1)(1)  ve Hence at x  0 neither maxima nor minima. If x  1  h , f (x )  ()()()(1)(1)  ve If x  1  h , f (x )  ()()()(1)(1)  ve Hence, at x  1 there is a local minima. If x  2  h , f (x )  ()(1)()()()  ve 206 Application of Derivatives If x  2  h, f (x )  ()()()()(1)  ve Hence at x  2 there is a local maxima. If x  3  h , f (x )  ()()()()()  ve If x  3  h , f (x )  ()()()()()  ve Hence at x  3 there is a local minima. If the function f (x )  2 x 3  9 ax 2  12 a 2 x  1 , where a  0 attains its maximum and minimum at p and q 60 Example: 7 respectively such that p 2  q , then a equals (a) 3 (c) 2 f (x )  2 x 3  9 ax 2  12 a 2 x  1 (d) 1 2 E3 Solution: (c) (b) 1 f (x )  6 x 2  18 ax  12 a 2 f (x )  12 x  18 a For maximum and minimum, 6 x 2  18 ax  12 a 2  0  x 2  3 ax  2a 2  0 ID x  a or x  2a at x  a maximum and at x  2a minimum  p2  q a 2  2 a  a  2 or a  0 but a  0, therefore a  2. The points of extremum of the function  (x )  e x t2 / 2 (1  t 2 ) dt are U Example: 8 1 (b) x  1 Solution: (b,d) (c) x  D YG (a) x  0  (x )  e x t2 / 2 (1  t 2 )dt   (x )  e  x 2 /2 1 2 (d) x  1 (1  x 2 ) 1 Now  (x )  0  1  x 2  0  x  1 Hence, x  1 are points of extremum of  (x ). 4.3.6 Point of Inflection. U A point of inflection is a point at which a curve is changing concave upward to concave downward or vice-versa. A curve y  f (x ) has one of its points x  c as y an inflection point, if f (c)  0 or is not defined and if f (x ) changes ST sign as x increases through x  c. The later condition may be replaced by y= f(x) f (c)  0 , when f (c) exists. f(c ) Thus, x  c is a point of inflection if f (c)  0 and f (c)  0. Properties of maxima and minima O c x (i) If f (x ) is continuous function in its domain, then at least one maxima and one minima must lie between two equal values of x. (ii) Maxima and minima occur alternately, that is, between two maxima there is one minimum and vice-versa. (iii) If f (x )  as x a or b and f (x )  0 only for one value of x (say c) between a and b, then f (c) is necessarily the minimum and the least value. Application of Derivatives 207 If f (x )  as x a or b, then f (c) is necessarily the maximum and the greatest value. 4.3.7 Greatest and Least Values of a Function in a given Interval. If a function f (x ) is defined in an interval [a, b], then greatest or least values of this function occurs either at x  a or x  b or at those values of x where f ( x )  0. 60 Remember that a maximum value of the function f (x ) in any interval [a, b] is not necessarily its greatest value in that interval. Thus greatest value of f(x) in interval [a, b] = max. [ f (a), f (b), f (c)] Where x  c is a point such that f (c)  0 The maximum and minimum values of x 3  18 x 2  96 in interval (0, 9) are (a) 160, 0 Solution: (c) (b) 60, 0 (c) 160, 128 dy  3 x 2  36 x  96  0 Let y  x 3  18 x 2  96 x  dx [RPET 1999] (d) 120, 28 ID Example: 9 E3 Least value of f (x ) interval [a, b]= min. [ f (a), f (b), f (c)]  x 2  12 x  32  0  (x  4)(x  8)  0, x  4, 8 d 2y d 2y  6 x  36 at x  4 , 2  24  36  12  0 2 dx dx U Now,  at x  4 function will be maximum and [ f (x )]max.  64  288  384  160 at x  8 d 2y  48  36  12  0 dx 2 Example: 10 D YG  at x  8 function will be minimum and [ f (x )]min.  128. The minimum value of the function 2 cos 2 x  cos 4 x in 0  x   is (a) 0 Solution: (d) (b) 1 (c) 3 2 (d) – 3 y  2 cos 2 x  cos 4 x = 2 cos 2 x (1  cos 2 x )  1 = 4 cos 2 x sin 2 x  1 Obviously, sin 2 x  0 Therefore, to be least value of y, cos 2x should be least i.e., – 1. Hence least value of y is – 4 + 1 = –3. On [1, e] the greatest value of x 2 log x U Example: 11 (b) 1 1 log e e (c) e 2 log e (d) None of these ST (a) e 2 [AMU 2002] Solution: (a) f (x )  x 2 log x  f (x )  (2 log x  1)x Now f ( x )  0  x  e 1 / 2 , 0  0  e 1 / 2  1 ,  None of these critical points lies in the interval [1, e ]  So we only compare the value of f (x ) at the end points 1 and e. We have f (1)  0, f (e )  e 2  greatest value = e 2 4.3.8 Maxima and Minima of Functions of Two Variables. If a function is defined in terms of two variables and if these variables are associated with a given relation then by eliminating one variable, we convert function in terms of one variable and then find maxima and minima by known methods. 208 Application of Derivatives Example: 12 x and y be two variables such that x  0 and xy  1. Then the minimum value of x  y is [Kurukshetra CEE 1988; MP PET 2002] (a) 2 xy  1  y  zx 1 x (c) 4 (d) 0 and let z  x  y dz 1 1 1 dz d 2z 2  0  1  2  0  x  1,  1 and 1 2    dx x dx x x dx 2 x 3 60 Solution: (a) (b) 3  d 2z  2    dx 2   1  2  ve ,  x  1 is point of minima.   x 1 The sum of two non-zero numbers is 4. The minimum value of the sum of their reciprocals is (a) Solution: (c) 3 4 (b) 6 5 (c) 1 Let x  y  4 or y  4  x 4 4 1 1 x y or f (x )     x y xy xy x (4  x ) 4 4. (4  2 x ) , f (x )  4x  x2 (4 x  x 2 )2 U f (x )  (d) None of these ID Example: 13 E3 x  1, y  1 ,  minimum value = x  y  2. Put f ( x )  0  4  2 x  0  x  2 and y  2 Example: 14 D YG 1 1 1 1  min.       1. x y 2 2 The real number which most exceeds its cube is 1 (a) 2 Solution: (b) (b) 1 3 [MP PET 2000] (c) 1 2 (d) None of these Let number = x, then cube = x 3 Now f (x ) = x  x 3 (Maximum)  f (x )  1  3 x 2 Put f ( x )  0  1  3 x 2  0  x   1 U 3 Because f (x )  6 x  ve. when x   1. 3 ST 4.3.9 Geometrical Results related to Maxima and Minima. The following results can easily be established. (1) The area of rectangle with given perimeter is greatest when it is a square. (2) The perimeter of a rectangle with given area is least when it is a square. (3) The greatest rectangle inscribed in a given circle is a square. (4) The greatest triangle inscribed in given circle is equilateral. (5) The semi vertical angle of a cone with given slant height and maximum volume is tan 1 2 (6) The height of a cylinder of maximum volume inscribed in a sphere of radius a is a 2a / 3. Important Tips  Equilateral triangle: Area = ( 3 / 4 )x 2 , where x is its side. Application of Derivatives 209  Square: Area = a 2 , perimeter = 4 a , where a is its side. Rectangle: Area = ab, perimeter = 2(a  b) , where a, b are its sides.  Trapezium: Area =  Circle: Area = a 2 , perimeter = 2a , where a is its radius.  Sphere: Volume = 1 (a  b )h , where a, b are lengths of parallel sides and h be the distance between them. 2 4 3 a , surface area = 4a 2 , where a is its radius. 3 60  1 r 2h, curved surface = rl , where r is the radius of its base, h is its height and 3  Right circular cone: Volume = l is its slant height. Solution: (c) The adjacent sides of a rectangle with given perimeter as 100 cm and enclosing maximum area are[MP PET 1 (a) 10 cm and 40 cm (b) 20 cm and 30 cm (c) 25 cm and 25 cm (d) 15 cm and 35 cm 2 x  2 y  100  x  y  50.....(i) ID Example: 15 E3  Cylinder: Volume = r 2h , whole surface = 2r(r  h) , where r is the radius of the base and h is its height. Let area of rectangle is A,  A  xy  y  dA A  50  2 x  50  A  50 x  x 2  dx x U From (i), x  A x dA 0 dx  50  2 x  0  x  25 and y  25 D YG for maximum area  adjacent sides are 25 cm and 25 cm. Example: 16 The radius of the cylinder of maximum volume, which can be inscribed a sphere of radius R is [AMU 1999] (a) Solution: (b) 2 R 3 2 R 3 (b) (c) 3 R 4 3 R 4 (d) 2 h If r be the radius and h the height, the from the figure, r 2     R 2  h 2  4 (R 2  r 2 ) 2 U Now, V  r 2h  2r 2 R 2  r 2 dV 1  4r R 2  r 2  2r 2. dr 2 ST  D 2r 3 R r  2 R 2  3r 2  r  2 2 L R 2  r2 dV 0 dr For max. or min.,  4 r R 2  r 2  (2r) C O R   2(R 2  r 2 )  r 2 r A 2 d 2V  ve. Hence V is max. when r  R  dr 2 3 M B 2 R. 3 Example: 17 The ratio of height of a cone having maximum volume which can be inscribed in a sphere with the diameter of sphere is [MNR 1985] (a) 2 3 (b) 1 3 (c) 3 4 (d) 1 4 210 Application of Derivatives Solution: (a) Let OM  x Then height of cone i.e., h  x  a (where a is radius of sphere) Radius of base of cone = Now, dV  1  (a 2  x 2 )(x  a)   (a  x )(a  3 x ) dx 3 3 dV a  0  x  a, dx 3 But x  a, So, x  A 60 Therefore, volume V  a2  x 2 a O a 3 a a The volume is maximum at x  Height of a cone h  a  a 3 E3 x B a 4  a 3 3 C ST U D YG U ID 4 a 2 Therefore ratio of height and diameter = 3 . 2a 3 M 1. 1 4 x on [– 1, 1] is 4  x  x2 (b) 1 3 [MP PET 2000] (c) 1 6 (d) 1 5 Maximum value of x (1  x )2 when 0  x  2 , is (a) 2 3. Basic Level The maximum value of f (x )  (a) 2. Maxima and Minima (b) 4 27 [MP PET 1997] (c) 5 The maximum value of 2 x 3  24 x  107 in the interval [– 3, 3] is (d) 0 Application of Derivatives 211 4. (a) 75 (b) 89 (c) 125 The maximum value of the function f (x )  3 sin x  4 cos x is (d) 139 (a) 3 (d) 7 (b) 4 (c) 5 5. If the function f (x )  x  62 x  ax  9 is maximum at x  1 , then the value of a is 6. (a) 120 (b) – 120 The maximum value of f ( )  a sin   b cos  is 2 (c) 52 (d) 128 [MP 2000] (a) (b) – 126 (a) x  sin x (b) x  cos x 11. 177 11 (b)  (b) 16 If A  B   2 1 2 (b) 23 11 [EAMCET 2003] 9 (d) 4 (c) 20 [MNR 1995] (d) 24 [AMU 1999] (c) 1 (d) 4 3 [Rajasthaan PET 2001] (c)  c ab (d)  2c ab U c2 6 (b) ab c3 ab [Rajasthan PET 2001] c3 (c) 2ab (d) c3 2 ab The function f (x )  2 x  15 x  36 x  4 is maximum at 2 ST 3 (b) x  4 [Karnataka CET 2001] (c) x  0 (d) x  3 x The function f (x )  x , (x  R) attains a maximum value at x  (b) 3 (c) 1 e (d) 1 The function y  a(1  cos x ) is maximum when x  (b)  2 [Kerala (Engg.) 2002] (c)  2 (d)  6 250   The minimum value of  x 2   is x   (a) 75 19. 3 4 (b) 2c ab 2 4 (a)  18. 9 (c) 8 (d) If a x  b y  c , then maximum value of xy is 2 4 (a) 2 17. 23 11 , the maximum value of cos Acos B is (a) x  2 16. (c)  [UPSEAT 1999] (d) x  tan x If xy  c 2 , then minimum value of ax  by is (a) 15.  3 UPSEAT (d) None of these The maximum value of xy subject to x  y  8 , is (a) c ab 14. 177 11 The minimum value of 2 x 2  x  1 is 1 3 (a) (b) 2 4 (a) 13. (c) x  The minimum value of the expression 7  20 x  11 x 2 is (a) 8 12. (c) – 120 E3 (a) – 128 x is maximum at 1  x tan x 1999; a2  b 2 (d) ab The minimum value of the function y  2 x 3  21 x 2  36 x  20 is (a) 10. (c) 2 ID 9. a b 2 U 8. a (b) D YG 7. a b PET 60 4 In the graph of the function (b) 50 [Haryana CEE 2002] (c) 25 (d) 55 3 sin x  cos x the maximum distance of a point from x-axis is 212 Application of Derivatives (a) 4 [AMU 2000] (a) A minimum but no maximum minimum (b) A (c) Neither maximum nor minimum minimum (d) Both The point for the curve y  xe x (b) x  0 is minimum (c) x  1 is maximum (b) 9, 4 (c) 3, 12 The necessary condition to be maximum or minimum for the function is (a) f ( x )  0 and it is sufficient (b) (c) f ( x )  0 but it is not sufficient (d) (c) 6, – 9 ID (b) – 16, 39 2 (b) – 1 (b) 2 f ( x )  0 and f (x )  ve (d) None of these (d) 4 (d) 6 The maximum value of the function x  x  x  4 is D YG 2 (c) Does not have a maximum value (b) 4 (d) None of these The function x  5 x  5 x  10 has a maximum when x = 5 (a) 3 29. f ( x )  0 and it is sufficient (c) – 2 (c) 4 3 (a) 127 28. (d) None of these 1 5 The minimum value of | x |  | x  |  | x  3 |  | x  | is 2 2 (a) 0 27. 4 3 (b) 2 (c) 1 (d) 0 If x  2 y  4 , the minimum value of xy is (a) – 2 (b) 2 [UPSEAT 2003] (c) 0 (d) – 3 1 is at 1  x2 The minimum value of x 2  (d) x  3 31. (a) x  0 (b) x  1 (c) x  4 The maximum and minimum value of the function | sin 4 x  3 | are (a) 1, 2 (d) – 1, 1 U 30. (b) 4, 2 ST 32. (c) 2, 4 2 (b) 177 (c) 77 (d) None of these Let f (x )  (x  p)  (x  q)  (x  r). Then f (x ) has a minimum at x   , where  is equal to 2 (a) 34. [UPSEAT 2003] The maximum value of function x  12 x  36 x  17 in the interval [1, 10] is 3 (a) 17 33. and If f (x )  2 x  3 x  12 x  5 and x  [2, 4 ] , then the maximum value of function is at the following value of x[MP PET 1987, 3 (a) 2 26. maximum no The maximum and minimum value of the function 3 x 4  8 x 3  12 x 2  48 x  25 in the interval [1, 3] (a) 16, – 39 25. but (d) x  0 is maximum U 24. maximum 36 is factorized into two factors in such a way that sum of factors is minimum, then the factors are (a) 2, 18 23. 3 The function f (x )  x  sin x has (a) x  1 is minimum 22. (d) 60 21. (c) 1 E3 20. (b) 2 2 pqr 3 2 (b) 3 pqr (c) 3 1 1 1   p q r (d) None of these The function x 2 log x in the interval (1, e) has (a) A point of maximum (b) A point of minimum (c) Points of maximum as well as of minimum (d) Neither a point of maximum nor minimum 35. The two parts of 100 for which the sum of double of first and square of second part is minimum, are 36. Of the given perimeter, the triangle having maximum area is (a) 50, 50 (b) 99, 1 (c) 98, 2 (d) None of these Application of Derivatives 213 (a) Isosceles triangle 38. (b) Right angled triangle (c) Equilateral (d) None of these The function x  5 x  5 x  1 is 4 3 (a) Maximum at x  3 and minimum at x  1 (b) Minimum at x  1 (c) Neither maximum nor minimum at x  0 (d) Maximum at x  0 If u  a 2 cos 2   b 2 sin 2   a 2 sin 2   b 2 cos 2  then the difference between the maximum and minimum values of u 2 is given by 60 37. 5 [AIEEE 2004] (a) (a  b)2 (c) (a  b)2 (b) 2 a2  b 2 (d) 2(a 2  b 2 ) 39. The minimum value of 2 x  3 y , when xy  6 , is 40. The real number x when added to its inverse gives the minimum value of the sum at x equal to (b) 9 (c) 8 2000; AIEEE 2003] (a) – 2 1 is maximum at x (b) x  1 f (x )  (1  x ) e is minimum at 2 x (b) x  1 (a) x  1 43. [Rajasthan PET (d) – 1 [Rajasthan PET 1991] (a) x  1 42. (c) 1 (c) x  2 (d) x  2 (c) x = 0 (d) x  2 ID x (b) 2 The maximum value of the function x  12 x  45 x is 3 (a) 0 2 U 41. (d) 6 E3 (a) 12 [MP PET 2003] (b) 50 (c) 54 [Rajasthan PET 1994] (d) 70 44. D YG Advance Level | x | , 0 | x |  2 Let f (x )   , then at x  0 f has x 0 1 , (a) A local maximum 46. If f (x )  (b) Is not attained even though f is bounded (c) Is equal to 1 (d) Is equal to –1 The number of values of x where the function f (x )  cos x  cos( 2 x ) attains its maximum is (b) 1 ST (b) 25 1 2 75 (d) Infinite takes its maximum value at the point (c) 1 3 [IIT 1995] (d) 1 4 x x has a stationary point at (a) x  e (b) x   te x 49. (c) 2 On the interval [0, 1] the function x (1  x ) (a) 0 48. (d) No extremum (a) Does not exist because f is unbounded (a) 0 47. (c) A local minimum x 1 , for every real number x, then the minimum value of f x2 1 2 U 45. (b) No local maximum [IIT Screening 2000] A minimum value of t 2 1 e (c) x  1 (d) x  e (c) 3 (d) 0 dt is 0 (a) 1 50. (b) 2 The sum of two numbers is fixed. Then its multiplication is maximum, when (a) Each number is half of the sum respectively of the sum (b) Each number is 1 2 and 3 3 214 Application of Derivatives (c) Each number is 51. 1 3 and respectively of the sum 4 4 (d) None of these The value of a so that the sum of the squares of the roots of the equation x 2  (a  2)x  a  1  0 assume the least value, is (a) 2 2 (c) 10, 8 is (b) e 2 2 (c) 12 e 2 One maximum point of sin p x cos q x is The minimum value of e log 2 2 [Rorkee 1998] (d) 0 (b) (5, 15) (c) (13, 7) (d) None of these (b) exp (2  3) [AMU 1999] (c) exp(4) (d) 1 log x in the interval [2, ) x (b) Is zero [Roorkee 1999] (c) Is 1 e (d) Does not exist b , a, b, x  0 takes on the least value at x equal to x (b) a (c) b [AMU 2000] (d) b /a ST (b) 3 : 2 (c) 4 : 3 (d) 1 : 1 The denominator of a fraction number is greater than 16 of the square of numerator, then least value of the number is [Rajasthan PET 2000] 1 (b) 8 1 (c) 12 1 (d) 16 If for a function f (x ) , f (a)  0, f (a)  0, f (a)  0 , then at x  a, f (x ) is (a) Minimum 64. (c) 1 (d) 12, 8 The area of a rectangle of given perimeter is maximum, when ratio of its length and breadth is 1 (a) 4 63. 1 e U The function f (x )  ax  (a) 2 : 1 62. is D YG The minimum value of (a) b 61. (c) 8, 12 [Rajasthan PET 1997] The minimum value of exp( 2  3 cos x  sin x ) is (a) Is 60. (d) x  tan 1 (q / p) Divide 20 into two parts such that the product of one part and the cube of the other is maximum. The two parts are [DCE 1999] (a) exp(2) 59. (2 x 2  2 x 1) sin 2 x (b) (a) (10, 10) 58. ID (b) 16, 4 (a) e 57. (d) 4 e 2 20 is divided into two parts so that product of cube of one quantity and square of the other quantity is maximum. The parts are (a) 10, 10 56. (c) x  tan 1 ( p / q) (b) x  tan 1 (q / p) U 55. (d) 13, 5 [Rajasthan PET 1997; AMU 2000] (a) x  tan 1 (p / q) 60 (b) 9, 9 The maximum value of x 4 e  x (a) e 54. (d) 0 If from a wire of length 36 metre a rectangle of greatest area is made, then its two adjacent sides in metre are[MP PET (a) 6, 12 53. (c) 3 E3 52. (b) 1 (b) Maximum [MP PET 1994] (c) Not an extreme points (d) Extreme point The least value of the sum of any positive real number and its reciprocal is (a) 1 (b) 2 (c) 3 [MP PET 1994] (d) 4 Application of Derivatives 215 65. If x is real, then greatest and least values of x2  x 1 are x2  x 1 [Rajasthan PET 1999; AMU 1999; UPSEAT 2002] (a) 3,  (b) 3, (c) – 3,  (c) Both (a) and (b) The function x 1  x , (x  0) has (b) A local minima If x  y  16 and x  y 2 2 is minimum, the value of x and y are (a) 3, 13 (b) 4, 12 (c) 6, 10 (d) 8, 8 The area of a rectangle will be maximum for the given perimeter. When rectangle is a (b) Trapezium (c) Square log x Local maximum value of the function is x (d) None of these ID (a) Parallelogram 70. (d) None of these (d) Neither (a) nor (b) 60 (b) Square (c) Neither a local maxima nor a local minima 69. (d) None of these 2 (a) A local maxima 68. 1 3 A wire of constant length is given. In which shape it should be bent to surround maximum area (a) Circle 67. 1 3 E3 66. 1 2 [MNR 1984; Rajasthan PET 1997, 2002; DCE 2002; Karnataka CET 2000; UPSEAT 2001; MP PET 2002] (a) e (b) 0, – 4 (c) 4,0 D YG 2 (c) f (x ) has maximum at x  1 (b) 3 2 2 U Let f (x )  [IIT Screening 2003] (c) | c | | b | 2 (d) | c | | b | 2  (c) 9 (d) 8 ST [Haryana CEE 2002] (a) Maxima when n  2,  4,  6..... (b) Maxima n  1,  3,  5,...... (c) Minima when n  0, 2, 4,...... (d) Minima when n  1, 3, 5,...... The function f (x )  2 x  3 x  12 x  4 has 3 2 [DCE 2002] (b) One maximum 4 3 (b) and one (d) Two minima 1 , then its maximum value is If f (x )  2 4 x  2x  1 (a) 79. (d) 1 cos t dt , x  0 then f (x ) has t (a) No maxima and minima minimum (c) Two maxima 78. (c) 2 2 (b) 1 x 0 77. (d) f (x ) has no maxima or minima The minimum value of [(5  x )(2  x )] /[1  x ] for non-negative real x is (a) 12 76. f (x ) has maximum at x  6 (b) If f (x )  x  2bx  2c and g(x )   x  2cx  b such that min f (x )  max g(x ) , then the relation between b and c is 2 (a) No real value of b and c(b) 0  c  b 2 75. (d) None of these If sum of two numbers is 3, then maximum value of the product of first and the square of second is (a) 4 74. (d) 2e If f (x )  2 x  21 x  36  30 , then which one of the following is correct 3 (a) f (x ) has minimum at x  1 73. 1 e Local maximum and local minimum values of the function (x  1)(x  2)2 are (a) – 4, 0 72. (c) U 71. (b) 1 2 3 [Rajasthan PET 2002] (c) 1 (d) 3 4 If PQ and PR are the two sides of a triangle, then the angle between them which gives maximum area of the triangle is 216 Application of Derivatives [Kerala (Engg.) 2002] 80.   (a)  (b) (c) 4 3 If ab  2a  3b, a  0, b  0 then the minimum value of ab is (a) 12 (b) 24 (c)  (d) 2 [Orissa JEE 2002] 1 4 (d) None of these The perimeter of a sector is p. The area of the sector is maximum when its radius is [Karnataka CET 2002] 1 p p (a) p (b) (c) (d) 4 2 p 82. The maximum area of the rectangle that can be inscribed in a circle of radius r is 83. (b) r 2 (c) 2  3 x  12 x  1 ,  1  x  2 If f (x )   , then  , 2x3  37  x rr 2 (d) 2r 2 4 [IIT 1993] f (x ) is maximum at x  2 (d) (a) f (x ) is increasing [1, 2] (b) f (x ) is continuous in [– 1, 3] (c) 2 p 1 [EAMCET 1994] E3 (a) r 2 60 81. If f (x )  (x  a) (x  b) 85. x a (a) x  a is a point of minimum (b) is a point of maximum (c) x  a is not a point of maximum or minimum (d) None of these N characters of information are held on magnetic tape, in batches of x characters each, the batch processing ID 84. when n and p are positive integers, then U 2n time is   x 2 seconds,  and  are constants. The optical value of x for fast processing is 86.   (b) (b) f ( x )  2 The maximum and minimum values of y  dy d 2y  0 and 0 dx dx 2 U (c)  6 (b)  4 (d) f ( x )  1 8 (b) ax 2  2bx  c  y( Ax 2  2 Bx  C) is a perfect square (d) None of these (c) The greatest value of the function f (x )  (a) 91. 1 4 minimum at x  0 for (a) All a and all b (b) All b if a  0 (c) All b  0 (d) All a  0 An isosceles triangle of vertical angle 2 is inscribed in a circle of radius a. Then area of the triangle is maximum when  (a) 90. (c) f ( x )  A differentiable function f (x ) has a relative minimum at x  0 , then the function y  f (x )  ax  b has a relative ST 89.   (d) ax 2  2bx  c are those for which Ax 2  2 Bx  C (a) ax 2  2bx  c  y( Ax 2  2 Bx  C) is equal to zero 88.   (c) If f (x )  sin6 x  cos 6 x , then (a) f ( x )  1 87.   D YG (a) 1 2 (b)  3 (d)  2 sin 2 x   on the interval 0,  is    2 sin x   4  2 (d)  2 (c) 1 The longest distance of the point (a, 0) from the curve 2 x 2  y 2  2 x  0 , is given by (a) 1  2a  a2 (b) 1  2a  2a2 (c) 1  2a  a2 (d) 1  2a  2a2 Application of Derivatives 217 92.  x The function f (x )  {2(t  1)(t  2)3  3(t  1)2 (t  2)2 dt attains its maximum at x  1 (a) 1  14. Then f (x ) is 3 1 f (x )dx  96. If A  0, B  0 and A  B  (a) x  2 and has relative minimum/maximum at x  1 and x  (b) x 3  x 2  x  1 (a) x 3  x 2  x 1  3 (c) x 3  x 2  x  2 1 3 such that (d) x 3  x 2  x  2 , then the maximum value of tan A tan B is (b) 3 60 f (x ) vanishes at (d) None of these E3 A cubic 1 (c) (1  2n / 2 )2 (b) 2 29   (d)   ,  7   (c) (, 7) 1  1   The minimum value of  1   1   is sin n    cos n    (a) 1 95. (d) 4 2  29  (b) a  (,  3)   3,   7  (a) a  (3, )  (,  3) 94. (c) 3 If the function f (x )  x  3(a  7)x  3(a  9)x  1 has a positive point of maximum, then 2 ID 93. (b) 2 3 1 3 (c) 3 (d) 3 Total number of parallel tangents of f f1 (x )  x 2  x  1 and x 3  x 2  2 x  1 is equal to 98. (a) 2 (b) 3 (c) 4 (d) None of these Function f (x ) | px  q | r | x | (p  0, q  0, r  0) attains its minimum value only at one point, if (b) q  r D YG (a) p  q 99. U 97. (c) r  p (d) p  q  r The height of right circular cylinder of maximum volume inscribed in a sphere of radius a is a 2a (a) (b) 3 a (c) (d) 2 3 a 3 3 100. A line is drawn through a fixed point (a, b), (a  0, b  0) to meet the positive direction of the coordinate axes in P,Q respectively. The minimum value of OP  OQ is a b (a) (a) (c) ( a  b )3 (d) None of these C4 a2 b2 , the maximum value of r is   2 2 r sin  cos 2  U 101. For the curve (b) ( a  b )2 c2 ab (b) ab c2 (c) c2 ab (d) c 2 (a  b) ST 102. The coordinates of a point situated on the curve 4 x 2  a2 y 2  4 a2 (4  a2  8) , which are at maximum distance from the point (0, – 2) is (a) (a, 0) (b) (2a,  4 ) 103. For what value of k, the function: f (x )  kx 2  (a) 9 2 (c) (0, 2) (d) None of these 9 2k 2  81 x  12 , is maximum at x  4 2 (b) 9 (c) 9 2 (d) 9 104. If    , then correct statement is (a)   sin     sin  (b)   sin     sin  (c) sin      sin    (d) None of these 105. The difference between two numbers is a if their product is minimum, then number are a a a 2 a a 4 a , , , (a) (b) a, 2a (c) (d) 2 2 3 3 3 3 218 Application of Derivatives 106. If ,  be real numbers such that x 3  x 2  x  6  0 has its roots real and positive then the minimum value of  is (a) 3  3 36 (b) 11 (c) 0 (d) None of these 107. Let the tangent to the graph of y  f (x ) at the point x  a be parallel to the x-axis, let f (a  h)  0 and f (a  h)  0 , where h is a very small positive number. Then the ordinate of the point is (b) A minimum (c) Both a maximum and a minimum (d) Neither a maximum nor a minimum 108. If a  b  0 , the minimum value of a sec   b tan  is (a) b  a a2  b 2 (b) a2  b 2 (c) f (x )  sin 1   x 2 , 0  x  1 ; 2 x, x  1 f (x ) can have a minimum at x  1 if the value of  is (a) 1 (b) – 1 (d) 2 a 2  b 2 E3 109. Let the function f (x ) be defined as below: 60 (a) A maximum (c) 0 (d) None of these 110. Let f (x )  ax  bx  cx  1 have extreme at x  ,  such that   0 and f ( ). f ( )  0. Then the equation f ( x )  0 3 ID 2 has (a) Three equal real roots (c) One positive root if f ( )  0 and f ( )  0 Let f (x )  1  2 x 2  2 2 x 4 ......  210 x 20 ; Then f (x ) has U 111. (b) Three distinct real roots (d) One negative root if f ( )  0 and f ( )  0 (a) More than one minimum (b) 112. Let the function f (x ) be defined as follows: D YG f (x )  x 3  x 2  10 x ,  1  x  0 Exactly one minimum cos x , 0  x   2 ; 1  sin x ,  2 (c) At least one maximum(d)  x  Then f (x ) has (a) A local minimum at x   2 (c) An absolute minimum at x  1 x   2 (b) A local maximum at x  (d) An absolute maximum at U 113. Two part of 64 such that the sum of their cubes is minimum will be (a) 44, 20 (b) 16, 48 (c) 32, 32 114. If x be real then the minimum value of f (x )  3 ST (a) 2 115. The minimum value of e (2 x (a) e x 1 3 ( x 1) (b) 6 2 is (c)  2 x 1) sin 2 x (d) 50, 14 2 3 (d) 7 9 is [Roorkee 1998] 1 (b) e (c) 1 (d) 0 116. The semi-vertical angle of a right circular cone of given slant height and maximum volume is (a) tan 1 2 117. (b) tan 1 2 (c) tan 1 1 / 2 (d) tan 1 1 / 2 If 0  a  x , then the minimum value of log a x  log a a is (a) 2 (b) – 2 118. Which point of the parabola y  x (a) (– 1, 1) 2 [IIT 1984] (c) 2a (d) Does not exist is nearest to the point (3, 0) (b) (1, 1) 119. The point of inflexion for the curve y  x (c) (2, 4) 5/2 is (d) (– 2, 4) [Rajasthan PET 1989, 1992] Application of Derivatives 219 (a) (1, 1) (b) (0, 0) (c) (1, 0) (d) (0, 1) ST U D YG U ID E3 60 *** 60 218 Application of Derivatives 2 3 4 5 6 7 8 9 10 11 c a d c a d a b c c b 21 22 23 24 25 26 27 28 29 30 a d c a d d c c a a 41 42 43 44 45 46 47 48 49 b a c a d b d b d 61 62 63 64 65 66 67 68 69 d b c b b a a d 81 82 83 84 85 86 87 88 d d d c c a,c b,c b 101 102 103 104 105 106 107 a c b b a a a 13 14 15 16 17 18 19 20 a b c a c a a b c 31 32 33 34 35 36 37 38 39 40 b b a d b c c a a c 50 51 52 53 54 55 56 57 58 59 60 a b b d a d c b d d d 70 71 72 73 74 75 76 77 78 79 80 U c c b c a d c a,d b a d b 89 90 91 92 93 94 95 96 97 98 99 100 a c d a b c c b d d c b 108 109 110 111 112 113 114 115 116 117 118 119 c d b,c b b c a c b d b b D YG U ST 12 ID 1 E3 Assignment (Basic & Advance Level)