Home Explore M1-Allens Made Maths Theory + Exercise [II]

# M1-Allens Made Maths Theory + Exercise [II]

## Description: M1-Allens Made Maths Theory + Exercise [II]

JEE-Mathematics Note : Given a fixed point A(a, b) and a moving point P(x, ƒ (x)) on the curve y = ƒ (x). Then A(a,b) AP will be maximum or minimum if it is normal to the curve at P. P Proof : F(x) = (x – a)2 + (ƒ (x) – b))2  F '(x) = 2(x – a) + 2(ƒ (x) – b) . ƒ '(x) (x, f(x))  ƒ '(x) =   x a  . Also m = f(x)  b . Hence ƒ '(x) . m = –1.  f(x )b  AP x a AP Illustration 11 : Find the co-ordinates of the point on the curve x2 = 4y, which is at least distance from the line y = x – 4. Solution : Let P(x1y1) be a point on the curve x2 = 4y at which normal is also a perpendicular to the line y = x – 4. Slope of the tangent at (x1, y1) is dy  dy  x1 P (x1, y1) 2x = 4 dx (x1,y1 ) 2 dx  x1 = 1  x1 = 2 2  x21 = 4y1  y1 = 1 Hence required point is (2, 1)  2  x12 9 2  Illustration 12 : Find the minimum value of (x1 – x2)2 +   x2  and x2  R+ where x1  0, 2 Solution : d2 = (x1 – x2)2 +  2  x 2  9 2  1 x2   The above expression is the square of the distance between the points x 2  , 9 x1, 2  1 ,  x 2 x2  which lie on the curves x2 + y2 = 2 and xy = 9 respectively. Now, the minimum value of the expression means square of the shortest distance between the two curves. Slope of the normal at P(x2, y2) on the curve xy = 9 dy 9  dx x2 x 2  y2  Q P(x2, y2) 2 (x1, y1) Slope of OP = x2y2 = 9 9 x2 O NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\04-Maxima-minima.p65 x 4  81  x2 = ± 3 2  y2 = ± 3 (x2, y2)  (3, 3) Now, shortest distance = PQ = OP – OQ = 3 2  2  2 2  d2 = 8 Do yourself - 5 : (i) Find the coordinates of point on the curve y2 =8x, which is at minimum distance from the line x + y = –2. 5. USEFUL FORMULAE OF MENSUR ATION TO REMEMBER : E ( a ) Volume of a cuboid = bh. ( b ) Surface area of a cuboid = 2 (b + bh + h). ( c ) Volume of a prism = area of the base × height. ( d ) Lateral surface area of prism = perimeter of the base × height. 85

JEE-Mathematics ( e ) Total surface area of a prism = lateral surface area + 2 area of the base (Note that lateral surfaces of a prism are all rectangles). 1 ( f ) Volume of a pyramid = area of the base x height. 3 1 ( g ) Curved surface area of a pyramid = (perimeter of the base) x slant height. 2 (Note that slant surfaces of a pyramid are triangles). 1 ( h ) Volume of a cone =  r2h. 3 ( i ) Curved surface area of a cylinder = 2  rh. ( j ) Total surface area of a cylinder = 2  rh + 2  r2. 4 ( k ) Volume of a sphere =   r3. 3 ( l ) Surface area of a sphere = 4  r2. 1 ( m ) Area of a circular sector = r2 , when  is in radians. 2 Illustration 13 : If a right circular cylinder is inscribed in a given cone. Find the dimension of the cylinder such that its volume is maximum. Solution : Let x be the radius of cylinder and y be its height V = x2y x, y can be related by using similar triangles yh h   y = (r – x) rx r r h h h  V(x) = x2 (r – x)  V(x) = (rx2 – x3) y r r x V'(x) = h (2rx – 3x2) r r V'(x) = 0  2r x = 0, 3 V''(x) = h 2r  6x r V''(0) = 2h  x = 0 is point of minima V ''  2r   2 h  x  2r is point of maxima NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\04-Maxima-minima.p65  3  3 Thus volume is maximum at x =  2r and y = h .  3  3 Illustration 14 : Among all regular square pyramids of volume 36 2 cm3. Find dimensions of the pyramid having least lateral surface area. Solution : Let the length of a side of base be x cm and y cm be the perpendicular height of the pyramid 1 V = area of base × height 3 1  V= x2y = 36 2 y 3 x/2 108 2  y = x2 86 E

JEE-Mathematics 1 and S = perimeter of base × slant height 2 = 1 (Ax) .  2 but  = x2  y2  S  2x x2  y2  x4  4x2y2 44  S x4  4 x2  108 2 2  S(x) = x4  8 . 1 0 8 2  x2  x2 Let f(x) = x4  8.108 2 x2 for minimizing f(x) f'(x) = 4x3 – 16 1082 0  f'(x) = 4  x6  66 0 x3 x3  x = 6 which a point of minima Hence x = 6 cm and y = 3 2 . Do yourself - 6 : ( i ) If ab = 2a + 3b where a > 0, b > 0, then find the minimum value of ab. ( i i ) Of all closed right circular cylinders of a given volume of 100 cubic centimetres, find the dimensions of cylinder which has minimum surface area. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\04-Maxima-minima.p65 6 . POINT OF INFLECTION : Y A point where the graph of a function has a tangent line and where the concavity changes is called a point of inflection. P• If function y = ƒ(x) is double differentiable then the point Point of d2y inflection at which dx2  0 & changes its sign is the point of inflection. X O d2y d2y Note : If at any point dx2 does not exist but sign of dx2 changes about this point then it is also called point of inflection. e.g. for y = x1/3, x = 0 is point of inflection. For a given function f(x) graphs of f'(x) and f''(x) can be drawn as shown in the adjacent figure. Here point C & E are point of inflection. y D y' C' x y'' F'' E D' E' x f(x) F f'(x) A'' C x E' f''(x) B' A O B'' B C'' E'' A' O D'' E 87

JEE-Mathematics 5 Illustration 15 : The point of inflexion for the curve y  x 3 is - (A) (1, 1) (B) (0, 0) (C) (1, 0) (D) (0, 1) d2y 10 Solution : Here  dx2 9 x1 / 3 From the given points we find that (0, 0) is the point of the curve where d2y d2y dx2 does not exist but sign of dx2 changes about this point.  (0, 0) is the required point Ans. (B) Illustration 16 : Find the inflection point of f(x) = 3x4 – 4x3. Also draw the graph of f(x) giving due importance to maxima, minima and concavity. Solution : f(x) = 3x4 – 4x3 f'(x) = 12x3 – 12x2 ––+ f'(x) = 12x2(x – 1) 01 f'(x) = 0  x = 0, 1 examining sign change of f'(x) +–+ thus x = 1 is a point of local minima 0 2/3 f''(x) = 12(3x2 – 2x) f''(x) = 12x(3x – 2) f(x) 2 2/3 1 x f''(x) = 0  x = 0, 2, 16 4/3 3 3 27 Again examining sign of f''(x) 2 thus x = 0, are the inflection points 3 Hence the graph of f(x) is Do yourself - 7 : ( i ) Find the point of inflection for the curve y = x3 – 6x2 + 12x + 5 (ii) Find the intervals for f(x) = x4 5x3  3x2 7 in which it is (a) concave upward (b) concave downward.  12 6 Miscellaneous Illustrations :  x NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\04-Maxima-minima.p65 Illustration 17 : Find all the values of a for which the function f(x) = (a2 – 3a + 2) cos  2  +(a – 1)x possesses critical points. Solution : Given f(x) = (a2 – 3a + 2) cos  x  + (a – 1)x  2   f'(x) = – a2  3a  2 sin  x   a  1 = a 1  1 a 2  x   2  2  1 2  2       sin  f'(x) 0 then 1 and x 2 Put = a = sin  2  = a 2 but 1  sin  x  1  2  or sin  x  1 2  2   a  2  1  |a – 2|  2  a – 2  2 and a – 2  – 2  a  4 and a  0  a (–, 0]  [4, ) Hence a  (–, 0]  {1}  [4, ) 88 E

JEE-Mathematics Illustration 18 : Divide 64 into two parts such that sum of the cubes of two parts is minimum. Solution : Let x and y be two positive number such that x + y = 64 ......... (i) Let u = x3 + y3 ......... (ii) Eliminate x from (ii) with the help of (i), then u = (64 – y)3 + y3  du  3 64  y2  3y2 ......... (iii) dy and d2u  6 64  y  6y  384  0 ......... (iv) dy2 For maximum or minimum of u, du 0 dy Then 3(64)(2y – 64) = 0  y = 32 From (i), x = 32 It is clear from (iv), u is minimum. Hence x = 32, y = 32. Illustration 19 : The three sides of a trapezium are equal each being 6cm long; find the area of trapezium when it is maximum. Solution : Let ABCD be the given trapezium. D6 C Let AM = BN = x cm then DM = CN = 36  x2  6 6  Area of trapezium ABCD is AxM N xB 1 36  x2  S = (6 + x + 6 + x) × 2 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\04-Maxima-minima.p65 = (6 + x) 36  x2  or S2 = (6 + x)2(36 – x2) Let y = (6 + x)2(36 – x2) dy  = (6 + x)2(–2x) + (36 – x2).2(6 + x) dx = 2(6 + x)2(6 – 2x) = 4(3 – x)(6 + x)2 d2 y and dx2 = – 12x(6 + x) dy For max. or min. of y, dx = 0 then x = 3 (x  – 6) d2 y   324  0 dx2 x3  y is maximum at x = 3 then S is also maximum at x = 3  S = (6 + 3) 36  9  27 3 cm2 E 89

JEE-Mathematics Illustration 20 : Show that the triangle of maximum area that can be inscribed in a given circle is an equilateral triangle. Solution :  1 Area = (Base) × (Height) 2 For maximum area height must be maximum. Height will be maximum if triangle is an isosceles triangle. A Let ABC is isosceles. Let AB = AC Let B = C =  then A =  – 2  COM  BOM    2 O If r be the radius of circle  OM = rcos( – 2) and MC = rsin( – 2) 1 B MC  Area of ABC   BC  AM 2  = 1  2r sin(  2)  r  r cos   2 2 = r2sin2(1 – cos2) Let S= r2 sin 2  1 sin 4  2   dS  r2 2 cos 2  2 cos 4 and d2S  r2 4 sin 2  8 sin 4 d2 d For max. or min. of S, dS  0 or 4  2  2 d or cos2  cos 4  and d2S 3r2  0   d2  / 3  6 3   is point of maxima  3  B  C  and A    2    3 33  Area of triangle is maximum if triangle is equilateral. ANSWERS FOR DO YOURSELF NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\04-Maxima-minima.p65 1 : ( i ) local max. at x = –1, local min. at x = 1 (ii) 121 1 1 2 : (i) e (ii) e2 3 : ( i ) local minima at x = 3 4 : ( i ) 15 & 45 (ii) 9 & 9 5 : (i) (2,–4) 6 : (i) 24 (ii) r   50 1 / 3 cm. & h  2  50 1 / 3 cm.       7 : (i) x = 2 ( i i ) (a) (–, 2)  (3, ) ( b ) (2, 3) 90 E

JEE-Mathematics EXERCISE - 01 CHECK YOUR GRASP SELECT THE CORRECT ALTERNATIVE (ONLY ONE CORRECT ANSWER) 1 . On the interval [0,1] the function x25 (1–x)75 takes its maximum value at the point - (A) 0 (B) 1/3 (C) 1/2 (D) 1/4 2 . The value of 'a' so that the sum of the squares of the roots of the equation x2 – (a – 2)x – a + 1 = 0 assume the least value is - (A) 2 (B) 0 (C) 3 (D) 1 3 . The slope of the tangent to the curve y = –x3 + 3x2 + 9x – 27 is maximum when x equals - (A) 1 (B) 3 (C) 1/2 (D) –1/2 4 . The real number x when added to it's reciprocal gives the minimum value of the sum at x equal to - (A) 1 (B) – 1 (C) – 2 (D) 2 5 . If the function f(x) = 2x3 – 9ax2 + 12a2x + 1 where a > 0, attains it’s maximum and minimum at p and q respectively such that p2 = q then ‘a’ equals - (A) 1 (B) 2 (C) 1/2 (D) 3 6 . If f(x) = 1 + 2x2 + 4x4 + 6x6 + .....+ 100x100 is a polynomial in a real variable x, then f(x) has - (A) neither a maximum nor a minimum (B) only one maximum (C) only one minimum (D) none 7 . For all a, b  R the function f(x) = 3x4 – 4x3 + 6x2 + ax + b has - (A) no extremum (B) exactly one extremum (C) exactly two extremum (D) three extremum M8 . Let f(x) L sin x , 0  x  12 MMN3  2x , x  1 then : (A) f(x) has local maxima at x = 1 (B) f(x) has local minima at x = 1 (C) f(x) does not have any local extrema at x = 1 (D) f(x) has a global minima at x = 1 9 . Two sides of a triangle are to have lengths ‘a’ cm & ‘b’ cm. If the triangle is to have the maximum area, then the length of the median from the vertex containing the sides ‘a’ and ‘b’ is - NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\04-Maxima-minima.p65 (A) 1 a2  b2 2a  b a2  b2 a  2b 2 (B) (C) (D) 3 2 3 1 0 . The difference between the greatest and the least value of f(x)  cos2 x sin x, x  [0, ] is - 2 33 3 3 1 (A) (B) (C) 8 (D) 2 2 8 8 1 1 . Equation of a straight line passing through (1,4) if the sum of its positive intercepts on the coordinate axis is the smallest is - (A) 2x + y – 6 = 0 (B) x + 2y – 9 = 0 (C) y + 2x + 6 = 0 (D) none 1 2 . A rectangle has one side on the positive y-axis and one side on the positive x-axis. The upper right hand vertex on the curve y  nx . The maximum area of the rectangle is - x2 (A) e–1 (B) e–1/2 (C) 1 (D) e1/2 E 91

JEE-Mathematics 1 3 . A solid rectangular brick is to be made from 1 cu feet of clay. The brick must be 3 times as long as it is wide. The width of brick for which it will have minimum surface area is a. Then a3 is -  2 1 / 3 2 8 3 (B) 9 (C) 729 (D)  (A)  9  2 1 4 . Let h be a twice continuously differentiable positive function on an open interval J. Let g(x) = n(h(x) for each x J Suppose (h'(x))2 > h''(x)h(x) for each x J. Then (A) g is increasing on J (B) g is decreasing on J (C) g is concave up on J (D) g is concave down on J 1 5 . Function f(x), g(x) are defined on [–1, 3] and f''(x) > 0, g''(x) > 0 for all x  [–1, 3], then which of the following is always true ? (A) f(x) – g(x) is concave upwards on (–1, 3) (B) f(x) g(x) is concave upwards on (–1, 3) (C) f(x) g(x) does not have a critical point on (–1, 3) (D) f(x) + g(x) is concave upwards on (–1, 3) 1 6 . If the point (1,3) serves as the point of inflection of the curve y = ax3 + bx2 then the value of ‘a’ and ‘b’ are - (A) a = 3/2 & b = – 9/2 (B) a = 3/2 & b = 9/2 (C) a = –3/2 & b = – 9/2 (D) a = – 3/2 & b = 9/2 1 7 . The set of value (s) of ‘a’ for which the function f(x )  ax3  (a  2)x2  (a  1) x  2 possess a negative point of 3 inflection - (A) (, 2)  (0,) (B) {–4/5} (C) (–2,0) (D) empty set 1 8 . Which of the following statements is true for the general cubic function ƒ (x) = ax3 + bx2 + cx + d (a  0) I. If the derivative ƒ '(x) has two distinct real roots then cubic has one local maxima and one local minima. II. If the derivative ƒ '(x) has exactly one real root then the cubic has exactly one relative extremum. III. If the derivative ƒ '(x) has no real roots, then the cubic has no relative extrema (A) only I & II (B) only II and III (C) only I and III (D) all I, II, III are correct. SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THEN ONE CORRECT ANSWERS) 1 9 . If y = an|x| + bx2 + x has its extremum values at x = –1 and x = 2, then :- (A) a = 2, b = – 1 1 1 (D) none of these (B) a = 2, b = – (C) a = –2, b = 2 2 2 0 . Let S be the set of real values of parameter  for which the function ƒ (x) = 2x3 – 3(2 + )x2 + 12x has exactly one local maxima and exactly one local minima. Then the subset of S is - (A) (5, ) (B) (–4, 4) (C) (3, 8) (D) (–, –1) NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\04-Maxima-minima.p65 21. The value of 'a' for which the function ƒ (x)   x 3  co s 1 a, 0  x 1 has a local minimum at x = 1, is -  x 1 x 2 , (A) –1 (B) 1 (C) 0 (D) 1  2 2 2 . If a continuous function ƒ (x) has a local maximum at x = a, then - (A) ƒ '(a+) may be 0 (B) ƒ '(a+) may be – (C) ƒ '(a+) may be non-zero finite real number (D) ƒ '(a–) may be – CHECK YOUR GRASP ANSWER KEY EXERCISE-1 Que. 1 23 4567 8 9 10 11 12 13 14 15 Ans. D DA ABCB A A AAABDD Que. 16 17 18 19 20 21 22 Ans. D AC B A,C,D A,D A,B,C E 92

EXERCISE - 02 JEE-Mathematics BRAIN TEASERS SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THEN ONE CORRECT ANSWERS) 1 . The set of values of p for which the points of extremum of the function, f(x) = x3 –3px2 + 3 (p2 –1)x + 1 lie in the interval (–2,4) is - (A) (–3,5) (B) (–3,3) (C) (–1,3) (D) (–1,5) 2. Min {f(t) : 0  t  x} ; 0  x  1 then g  1   g  3  g  5 If f (x)= 4x3 - x2 - 2x +1 and g(x )  3  x ; 1  x  2  4   4   4  has the value equal to : 7 9 13 5 (A) (B) (C) (D) 4 4 4 2 3 . The function ‘f’ is defined by f(x)  xp (1  x)q for all x  R, where p, q are positive integers, has a maximum value, for x equal to : pq (B) 1 p (D) 0 (A) p  q (C) p  q 4. If the point of minima of the function , f(x) =1 + a2x – x3 satisfy the inequality x2  x  2  0 , then ‘a’ must lie x2  5x  6 in the interval :  (A) 3 3, 3 3  (C) 2 3, 3 3  (C) 2 3, 3 3    (D) 3 3, 2 3  2 3, 3 3 x (B) it is an increasing function (D) the point (0,0) is the point of inflection 5 . The function f(x)   1  t4 dt is such that : 0 (A) it is defined on the interval [–1,1] (C) it is an odd function sin(x  a) 6 . The function sin(x  b) has no maxima or minima if - (A) b – a = n n  I (B) b – a = (2n+1) n  I (C) b – a = 2n  n  I (D) none of these. 7 . The coordinates of the point P on the graph of the function y = e–|x| , where area of triangle made by tangent NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\04-Maxima-minima.p65 and the coordinate axis has the greatest area, is - (A) 1, 1  (B)  1, 1  (C) (e, e e ) (D) none e   e  8. The least value of ‘a’ for which the equation 41 a has atleast one solution on the interval (0,  / 2)  sin x 1  sin x is - (A) 3 (B) 5 (C) 7 (D) 9 9 . Read of the following mathematical statements carefully : I. A differentiable function 'f' with maximum at x = c  f''(c) < 0. II. Antiderivative of a periodic function is also a periodic function. TT III. If f has a period T then for any a  R,  f(x)dx   f(x  a)dx 00 IV. If f(x) has a maxima at x = c, then 'f' is increasing in (c – h, c) and decreasing in (c, c + h) as h  0 for h>0 E 93

JEE-Mathematics Now indicate the correct alternative. (A) exactly one statement is correct (B) exactly two statements are correct (C) exactly three statements are correct (D) all the four statements are correct 1 0 . The lateral edge of a regular rectangular pyramid is ‘a’ cm long. The lateral edge makes an angle  with the plane of the base. The value of  for which the volume of the pyramid is greatest, is -  (B) sin 1 2 (C) cot1 2  (A) 4 3 (D) 3 1 1 . P and Q are two points on a circle of centre C and radius , the angle PCQ being 2 then the radius of the circle inscribed the triangle CPQ is maximum when - 3 1 (B) sin   5  1 (C) sin   5  1 (D) sin   5  1 (A) sin   2 2 4 22 1 2 . In a regular triangular prism the distance from the centre of one base to one of the vertices of the other base is  . The altitude of the prism for which the volume is greatest - (A) /2 (B)  / 3 (C) /3 (D) /4 1 3 . Let P(x) = a0 + a1x2 + a2x4 + ......... + anx2n be a polynomial in a real variable x with 0 < a0 < a1 < a2 < .... < an. The function P(x) has- [JEE 1986] (A) neither a maximum nor a minimum (B) only one maximum (C) only one minimum (D) only one maximum and only one minimum 1 4 . If g(x) = 7x2ex2  x  R , then g(x) has - (A) local maximum at x = 0 (B) local minima at x = 0 (C) local maximum at x = –1 (D) two local maxima and one local minima 15 . The coordinates of the point on the parabola y2 = 8x, which is at minimum distance from the circle x2 + (y + 6)2 = 1 are - (A) (2, –4) (B) (18, –2) (C) (2, 4) (D) none of these BRAIN TEASERS ANSWER KEY EXERCISE-2 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\04-Maxima-minima.p65 Que. 1 2 34567 8 9 10 Ans. C D C D A,B,C,D A,B,C A,B D AC Que. 11 12 13 14 15 Ans. B B C B,C,D A E 94

JEE-Mathematics EXERCISE - 03 MISCELLANEOUS TYPE QUESTIONS MATCH THE COLUMN Following questions contains statements given in two columns, which have to be matched. The statements in Column-I are labelled as A, B, C and D while the statements in Column-II are labelled as p, q, r and s. Any given statement in Column-I can have correct matching with ONE statement in Column-II. 1 . Four points A, B, C and D lie in the order on the parabola y = ax2 + bx + c and the coordinates of A, B and D are known A(–2, 3); B(–1, 1); D(2, 7). On the basis of above information match the following : Column-I Column-II (A) The value of a + b + c = (p) –1 (B) If roots of the equation ax2 + bx + c = 0 (q) 8 are  &  then 19 + 7 = (C) If the value of function (a + 2)x2 + (b  2) c (r) 3 2+ x at minima is L then L – 3 is equal to (D) If area of quadrilateral ABCD is greatest (s) 7 and co-ordinates of C are (p, q) then 2p + 4q = 2 . For the function f(x) = x4(12nx – 7) match the following : Column-I Column-II (A) If (a, b) is the point of inflection then a – b (p) 3 is equal to (q) 1 (B) If et is a point of minima then 12t is equal to (r) 4 (C) If graph is concave downward in (d, e) then (s) 8 d + 3e is equal to (D) If graph is concave upward in (p, ) then the least value of p is equal to ASSERTION & REASON These questions contains, Statement I (assertion) and Statement II (reason). NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\04-Maxima-minima.p65 (A) Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I. (B) Statement-I is true, Statement-II is true ; Statement-II is NOT a correct explanation for statement-I. (C) Statement-I is true, Statement-II is false. (D) Statement-I is false, Statement-II is true. 1. Statement-I : Consider the function, f(x)=  x , x0  2 ; f(x) has local minima at x = 0 7x  8 x  0 Because Statement-II : If f(a) < f(a – h) & f(a) < f(a + h) where 'h' is sufficiently small; then f(x) has local minima at x = a. (A) A (B) B (C) C (D) D 2. Statement-I : The largest term in the sequence an = n3 n2 , n N is the 7th term.  200 Because x2 Statement-II : The function f(x) = x3  200 attains local maxima at x = 7. (A) A (B) B (C) C (D) D E 95

JEE-Mathematics 3 . Statement-I : e > e. Because Statement-II : The function f(x) = x1/x attains global maxima at x = e. (A) A (B) B (C) C (D) D 4 . Consider an acute angled triangle ABC. Statement-I : Minimum value of secA + secB + secC is 6. Because Statement-II : If a continuous curve is concave upward then centroid of the triangle inscribed in the curve always lies above the curve. (A) A (B) B (C) C (D) D 5 . Let y = ƒ (x) be a thrice derivable function such that ƒ (a) ƒ (b) < 0, ƒ (b) ƒ (c) < 0, ƒ (c) ƒ (d) < 0 where a < b < c < d. Also the equations ƒ (x) = 0 & ƒ ''(x) = 0 have no common roots. Statement-I : The equation ƒ (x)(ƒ ''(x))2 + ƒ (x) ƒ '(x) ƒ '''(x) + (ƒ '(x))2 ƒ ''(x) = 0 has atleast 5 real roots. Because Statement-II : The equation ƒ (x) = 0 has atleast 3 real distinct roots & if ƒ (x) = 0 has k real distinct roots, then ƒ '(x) = 0 has atleast k – 1 distinct roots. (A) A (B) B (C) C (D) D COMPREHENSION BASED QUESTIONS Comprehension # 1 Suppose f(x) is a real valued polynomial function of degree 6 satisfying the following condition (a) 'f' has minimum value at x = 0 & 2 (b) 'f' has maximum value at x = 1 Lim 1 n f(x)/ x 1 0 x0 x (c) for all x, 0 1/x 1 = 2 1 0 1/x On the basis of above information, answer the following questions : 1 . Number of solutions of the equation 8f(x) – 1 = 0 is - (A) one (B) two (C) three (D) four 2 . Range of f(x) is - (A)  32 ,  (B)  4 ,  (C)  , 2  (D) none of these 15 15  15   a NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\04-Maxima-minima.p65 3 . If the area bounded by y = f(x), x-axis, x = ± 1; is , where a & b are relatively prime then the value b of tan–1 (a – b) is - (A) /4 (B) –/4 (C) /3 (D) /6 MISCELLANEOUS TYPE QUESTION ANSWER KEY EXERCISE-3  Match the Column 1 . (A) (r), (B) (p), (C) (s), (D) (q) 2 . (A) (s), (B) (r), (C) (p), (D) (q)  Assertion & Reason 1. D 2. C 3. A 4. A 5. A  Comprehension Based Questions Comprehension # 1 : 1. D 2. A 3. B 96 E

JEE-Mathematics EXERCISE - 04 [A] CONCEPTUAL SUBJECTIVE EXERCISE 1 . Find the points of local maxima/minima of following functions (a) f(x) = 2x3 – 21x2 + 36x – 20 (b) f(x) = –(x – 1)3(x + 1)2 (c) f(x) = xnx (d) y  3x2  4x  4 x2  x 1  3 (b3  b2  b 1) , 0  x 1 x  . 2. Let f(x) = Find all possible real values of b such that f(x) has the smallest (b2 3b 2)  2x 3 , 1x3  value at x = 1. [JEE 1993] 1 14 3 . f(x)dx  A cubic f(x) vanishes at x = –2 & has relative minimum/maximum at x = –1 and x = 1/3. If 3 , then 1 find the cubic f(x). 4 . Find the absolute maxima/minima value of following functions (a) x2 ; x  2, 9 (b) f(x) = 3x4 – 8x3 + 12x2 – 48x + 25 ; x 0, 3 f(x) = 4 x  2  2 (c) f(x) = sin x  1 cos 2x ; x  0,  2 2  5 . Let x and y be two real variables such that x > 0 and xy = 1. Find the minimum value of x + y. [JEE 1981] 6 . A rectangular sheet of poster has its area 18 m². The margin at the top & bottom are 75 cms and at the sides 50 cms. What are the dimensions of the poster if the area of the printed space is maximum ? ax  b 7 . If y = (x 1) (x  4) has a turning value at (2, 1) find a & b and show that the turning value is a maximum. 8 . The flower bed is to be in the shape of a circular sector of radius r & central angle . If the area is fixed & perimeter is minimum, find . NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\04-Maxima-minima.p65 9 . What are the dimensions of the rectangular plot of the greatest area which can be laid out within a triangle of base 36 ft. & altitude 12 ft ? Assume that one side of the rectangle lies on the base of the triangle. 1 0 . For a given curved surface of a right circular cone when the volume is maximum, prove that the semi vertical 1 angle is sin1 . 3 6 1 1 . Of all the line tangent to the graph of the curve y = x2  3 , find the equations of the tangent lines of minimum and maximum slope. 1 2 . Suppose f(x) is a function satisfying the following conditions - (i) f(0) = 2, f(1) = 1 5 (ii) If f(x) has a minimum value at x = and 2 (iii) for all x f'(x) = 2ax 2ax 1 2ax  b  1 b b 1 1 2ax  b 2ax  2b  1 2ax  b Where a, b are some constants. Determine the constants a, b, & the function f(x). [JEE 1998] E 97

JEE-Mathematics x 1 3 . Consider the function, F(x) = (t2  t)dt, x R . Find 1 (a) The x and y intercept of F if they exist. (b) Derivatives F '(x) and F''(x). (c) The interval on which F is an increasing and the intervals on which F is decreasing. (d) Relative maximum and minimum points. (e) Any inflection point. 1 4 . The function f(x) defined for all real numbers x has the following properties f(0) = 0, f(2) = 2 and f'(x) = k(2x – x2)e–x for some constant k > 0. Find (a) the intervals on which f is increasing and decreasing and any local maximum or minimum values. (b) the intervals on which the graph f is concave down and concave up (c) the function f(x) and plot its graph. 1 5 . The circle x2 + y2 = 1 cuts the x-axis at P & Q. Another circle with centre at Q and variable radius intersects the first circle at R above the x-axis & the line segment PQ at S. Find the maximum area of the triangle QSR. x 1 6 . Investigate for maxima & minima for the function, f(x) = [2(t  1)(t  2)3  3(t  1)2 (t  2)2 ] dt . 1 1 7 . The graph of the derivative ƒ ' of a continuous function ƒ is y x shown with ƒ (0) = 0. If 4 1 2 34 5 6 7 8 9 ƒ'(x) 3 (i) ƒ is monotonic increasing in the interval 2 [a, b)  (c, d)  (e, f] and decreasing in (p, q) (r, s). 1 (ii) ƒ has a local minima at x = x1 and x = x2. 0 (iii) ƒ is concave up in (, m)  (n, t] –1 –2 (iv) ƒ has inflection point at x = k (v) number of critical points of y = ƒ (x) is 'w' NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\04-Maxima-minima.p65 Find the value of (a + b + c + d + e) + (p + q + r + s) + ( + m + n) + (x1 + x2) + (k + w). 1 8 . The graph of the derivative ƒ ' of a continuous function ƒ is shown with ƒ (0) = 0 y x (i) On what intervals is ƒ increasing or decreasing ? 1 2 34 5 67 8 9 (ii) At what values of x does ƒ have a local maximum ƒ'(x) 3 2 E or minimum ? 1 (iii) On what intervals is ƒ concave upward or downward ? (iv) State the x-coordinate(s) of the point(s) of inflection. 0 –1 –2 (v) Assuming that ƒ (0) = 0, sketch a graph of ƒ . 98

JEE-Mathematics 1 9 . A window of perimeter P(including the base of the arch) is in the form of a rectangle surmounded by a semi circle. The semi-circular portion is fitted with coloured glass while the rectangular part is fitted with clear glass. The clear glass transmits three times as much light per square meter as the coloured glass does. What is the ratio for the sides of the rectangle so that the window transmits the maximum light ? [JEE 1991] 2 0 . Find the points on the curve ax2+2bxy+ay2 = c ; c>b>a>0, whose distance from the origin is minimum. [JEE 1998] CONCEPTUAL SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(A) 1 . (a) local max. at x = 1, local min. at x = 6 1 1 (b) local max. at x =  , local min. at x = –1 (c) local min. at x = , No local max. 5 e (d) y = 4 at x = 0, y = 8 at x = –2 max min 3 2 . b (–2, –1) [1, ) 3 . f(x) = x3 + x2 – x + 2 4 . (a) max. = 8, min. = –10; (b) max. = 25, min. = –39; (c) max. = 3/4 at x = /6, min. = 1/2 at x = 0 & /2 5. 2 6 . width 2 3 m, length 3 3 m 7 . a = 1, b = 0 8 .  = 2 radians 9 . 6' x 18' 1 1 . 3x + 4y – 9 = 0 ; 3x – 4y + 9 = 0 12. a = 1;b  5 44 1 3 . (a) (–1, 0), (0, 5/6); (b) F'(x) = (x2 – x), F''(x) = 2x – 1, (c) increasing (–, 0)  (1, ), decreasing (0, 1); (d) (0, 5/6); (1, 2/3); (e) x = 1/2 1 4 . (a) increasing in (0, 2) and decreasing in (–, 0)  (2, ), local min. value = 0 and local max. value = 2 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\04-Maxima-minima.p65 (b) concave up for (–, 2 – 2 )  (2 + 2 , ) and concave down in (2 – 2 , 2 + 2 ) 1 y = (c) f(x) 2 e2x .x2 4 2x 15. 3 3 1 6 . max. at x = 1, f(1) = 0, min. at x = 7/5; f(7/5) = –108/3125 17. 74 1 8 . (i) I in (1, 6)  (8, 9) and D in (0, 1)  (6, 8) ; (ii) L. Min. at x = 1 and x = 8; L. Max. x = 6 (iii) CU in (0, 2)  (3, 5)  (7, 9) and CD in (2, 3) (5, 7); (iv) x = 2, 3, 5, 7 0 12 (v) Graph is 3 4 5 67 89 19. 6 +  : 6 20.  2 c b  , 2 c b   &  c b ,  2  c b       a  a  a  2a  E 99

JEE-Mathematics BRAIN STORMING SUBJECTIVE EXERCISE EXERCISE - 04 [B]  xnx when x  0 1 . Consider the function f(x)=  0 for x  0 (a) Find whether f is continuous at x = 0 or not. (b) Find the minima and maxima if they exist. (c) Does f'(0) exist ? Find Lim f'(x). x0 (d) Find the inflection points of the graph of y = f(x). 2 . Consider the function y = f(x) = n(1 + sin x) with –2  x  2. Find (a) the zeroes of f(x) (b) inflection points if any on the graph (c) local maxima and minima of f(x) (d) asymptotes of the graph /2 (e) sketch the graph of f(x) and compute the value of the definite integral  f(x)dx.  / 2 3 . Given two points A ( 2 , 0) & B (0 , 4) and a line y = x. Find the co-ordinates of a point M on this line so that the perimeter of the  AMB is least. 4. Find the set of values of m for the cubic x3 3 x2  5  log1/ 4 (m ) has 3 distinct solutions.  2 2 5 . If the sum of the lengths of the hypotenuse and another side of a right angled triangle is given, show that the area of the triangle is a maximum when the angle between these sides is /3. 6 . Prove that among all triangles with a given perimeter, the equilateral triangle has the maximum area. 7 . The value of ‘a’ for which f (x) = x3 + 3 (a  7) x2 + 3 (a2  9) x  1 have a positive point of maximum lies in the interval (a , a2)(a3, a ). Find the value of a+ 11a + 70a . 1 4 2 3 4 8 . Use calculus to prove the inequality, sin x  2x/ in 0  x  /2. You may use the inequality to prove that, cos x  1 – x2/ in 0  x  /2 9 . Find the maximum perimeter of a triangle on a given base 'a' and having the given vertical angle . 1 0 . What is the radius of the smallest circular disk large enough to cover every acute isosceles triangle of a given NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\04-Maxima-minima.p65 perimeter L ? 1 1 . A swimmer S is in the sea at a distance d km from the closest point A on a straight shore. The house of the swimmer is on the shore at a distance L km from A. He can swim at a speed of u km/hr and walk at a speed of v km/hr (v > u). At what point on the shore should he land so that he reaches his house in the shortest possible time ? [JEE 1983] 12. Let f(x) = sin3x + sin2x, –  <x<  . Find the intervals in which  should lie in order that f(x) has exactly 22 one minimum and exactly one maximum. [JEE 1985] 1 3 . Let A(p2, –p), B(q2, q), C(r2, –r) be the vertices of the triangle ABC. A parallelogram AFDE is drawn with vertices D, E and F on the line segments BC, CA and AB respectively. Using calculus, show that maximum 1 [JEE 1986] area of such a parallelogram is (p + q)(q + r)(p – r) 4 100 E

JEE-Mathematics 1 4 . Find the point on the curve 4x2 + a2y2 = 4a2, 4 < a2 < 8 that is farthest from the point (0, –2). [JEE 1987] 15. Determine the points of maxima and minima of the function ƒ (x) = 1 nx  bx  x2 , x > 0 where b  0 8 is a constant. [JEE 1996] 1 6 . Let S be a square of unit area. Consider any quadrilateral which has one vertex on each side of S. If a, b, c and d denote the length of the sides of the quadrilateral, prove that 2  a2 + b2 + c2 + d2  4. [JEE 1997] 1 7 . Find the co-ordinates of all the points P on the ellipse (x2/a2)+(y2/b2) = 1 for which the area of the triangle PON is maximum, where O denotes the origin and N the foot of the perpendicular from O to the tangent at P. [JEE 1999] NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\04-Maxima-minima.p65 BRAIN STORMING SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(B) 1 . (a) f is continuous at x = 0; (b) e–2; (c) does not exist, does not exist; (d) pt. of inflection x = 1 2. (a) x = –2, –, 0, , 2, (b) no inflection point, (c) maxima at x =  3 and – and no minima, 2 2 (d) x = 3 and x = –  , (e) –n2 3 . (0 , 0) 4. m   1 , 1  2 2 32 16  7. 320 9. P= a 1  cosec   10. L/4 ud max 2  11. v2  u2 12.     3 , 0    0, 3 14. (0, 2) 1 5 . Min. at x  1 (b  b2  1 ) , max. at x  1 (b  b2  1 )  2   2  44 17.  a2 , b2 a2  b2 a2  b2 E 101

JEE-Mathematics EXERCISE - 05 [A] JEE-[MAIN] : PREVIOUS YEAR QUESTIONS 1 . If the function f(x) = 2x3 – 9ax2 + 12a2x + 1, where a > 0, attains its maximum and minimum at p and q respectively such that p2 = q, then a equals- [AIEEE-2003] (1) 1/2 (2) 3 (3) 1 (4) 2 2 . The real number x when added to its inverse gives the minimum value of the sum at x equal to- [AIEEE-2003] (1) –2 (2) 2 (3) 1 (4) –1 3 . If u = a2 cos2   b2 sin2   a2 sin2   b2 cos2  then the difference between the maximum and minimum values of u2 is given by- [AIEEE-2004] (1) 2(a2 + b2) (2) 2 a2  b2 (3) (a + b)2 (4) (a – b)2 x2 [AIEEE-2006] 4 . The function f(x) =  has a local minimum at- 2x (1) x = –2 (2) x = 0 (3) x = 1 (4) x = 2 5 . A triangular park is enclosed on two sides by a fence and on the third side by a straight river bank. The two sides having fence are of same length x. The maximum area enclosed by the park is- [AIEEE-2006] x3 (2) 1 x2 (3) x2 (4) 3 x2 (1) 2 2 8 6 . If p and q are positive real numbers such that p2 + q2 = 1, then the maximum value of (p + q) is- [ A I E E E - 2 0 0 7 ] (1) 2 1 1 (4) 2 (2) (3) 2 2 7 . Suppose the cubic x3 – px + q has three real roots where p > 0 and q > 0. Then which of the following holds ? [AIEEE-2008] pp pp (1) The cubic has minima at and maxima at – (2) The cubic has minima at – and maxima at 33 33 pp pp (3) The cubic has minima at both – & (4) The cubic has maxima at both &– 33 33 8 . Given P(x) = x4 + ax3 + bx2 + cx + d such that x=0 is the only real root of P'(x) = 0. If P(–1) < P(1), then in the interval [–1, 1] :- (1) P(– 1) is the minimum but P(1) is not the maximum of P. [AIEEE-2009] NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\04-Maxima-minima.p65 (2) Neither P(– 1) is the minimum nor P(1) is the maximum of P (3) P(– 1) is the minimum and P(1) is the maximum of P (4) P(– 1) is not minimum but P(1) is the maximum of P [AIEEE-2009] 9 . The shortest distance between the line y – x = 1 and the curve x = y2 is :- 32 3 32 23 (1) (2) (3) (4) 5 4 8 8 10. Let f : R  R be defined by f(x)  k  2x, if x  1 [AIEEE-2010] 2x  3, if x  1 If f has a local minimum at x = – 1, then a possible value of k is : (1) 1 (2) 0 1 (4) –1 (3) – 2 102 E

x   0, 5  x t sin t dt . Then f has :- JEE-Mathematics  2  f(x)  [AIEEE-2011] 0 1 1 . For , define (1) local minimum at  and local maximum at 2 [AIEEE-2011] (2) local maximum at  and local minimum at 2 (3) local maximum at  and 2 (4) local minimum at  and 2 1 2 . The shortest distance between line y – x = 1 and curve x = y2 is :- 8 4 3 32 (1) (2) 3 (3) (4) 32 4 8 tan x , x 0  13. Let f be a function defined by f(x)   x  1 , x  0 Statement - 1: x = 0 is point of minima of f. Statement - 2: f'(0) = 0. [AIEEE-2011] (1) Statement-1 is false, statement-2 is true. (2) Statement-1 is true, statement-2 is true; Statement-2 is correct explanation for statement-1. (3) Statement-1 is true, statement-2 is true; Statement-2 is not a correct explanation for statement-1. (4) Statement-1 is true, statement-2 is false. 1 4 . A spherical balloon is filled with 4500 cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of 72 cubic meters per minute, then the rate (in meters per minute) at which the radius of the balloon decreases 49 minutes after the leakage began is : [AIEEE-2012] (1) 9/2 (2) 9/7 (3) 7/9 (4) 2/9 1 5 . Let a, b R be such that the function f given by f(x) = ln |x| + bx2 + ax, x 0 has extreme values at x = – 1 and x = 2. Statement–1 : f has local maximum at x = –1 and at x = 2. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\04-Maxima-minima.p65 Statement–2 : a  1 and b  1 . [AIEEE-2012] 24 (1) Statement–1 is true, Statement–2 is false. (2) Statement–1 is false, Statement–2 is true. (3) Statement–1 is true, Statement–2 is true ; Statement–2 is a correct explanation for Statement–1. (4) Statement–1 is true, Statement–2 is true ; Statement–2 is not a correct explanation for Statement–1. 1 6 . The real number k for which the equation 2x3 + 3x + k = 0 has two distinct real roots in [0, 1] [JEE-MAIN 2013] (1) lies between 1 and 2. (2) lies between 2 and 3. (3) lies between –1 and 0 (4) does not exist PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE-5 [A] 1. 4 2. 3 3. 4 4. 4 5. 2 6. 4 7. 1 8. 4 9. 3 10. 4 11. 2 12. 4 13. 3 14. 4 15. 3 16. 4 E 103

JEE-Mathematics EXERCISE - 05 [B] JEE-[ADVANCED] : PREVIOUS YEAR QUESTIONS  x for 0  x  2 1. Let f (x) =  for x  0 . Then at x = 0, ‘f’ has - [JEE 2000 Screening, 1M out of 35] 1 (A) a local maximum (B) no local maximum (C) a local minimum (D) no extremum. 2 . Let f(x) = (1+b2)x2+2bx+1 and let m (b) the minimum value of f(x). As b varies, the range of m(b) is - [JEE 01 Screening, 1M, out of 35] (A) [0, 1] (B)  0, 1  (C) 1 ,1 (D) 0, 1  2   2 3 . A straight line L with negative slope passes through the point (8,2) and cuts the positive coordinates axes at points P and Q. Find the absolute minimum value of OP + OQ, as L varies, where O is the origin. [JEE 02 Mains, 5M out of 60] 4 . The minimum value of f(x) = x2 + 2bx + 2c2 is more than the maximum value of g(x) = –x2 – 2cx + b2, x being real, for - [JEE 03, (Scr.) 3M out of 84] (A) c  b 2 (B) 0  c  b 2 (C) b 2  c  0 (D) no values of b and c 5. For every    0,   , the value of x2  x  tan2  is greater than or equal to -  2  ,x  0 x2  x [JEE 03, (Scr.) 3M out of 84] (A) 2 (B) 2tan 5 (D) sec (C) 2 6 . For a circle x2+y2 = r2, find the value of ‘r’ for which the area enclosed by the tangents drawn from the point P (6, 8) to the circle and the chord of contact is maximum. [JEE 03, Mains 2M out of 60] 7 . If p(x) be a polynomial of degree 3 satisfying p(–1) = 10, p(1) = –6 and p(x) has maximum at x = –1and p'(x) has minima at x = 1. Find the distance between the local maximum and local minimum of the curve. [JEE 05, Mains 4M out of 60] 8 . f(x) is cubic polynomial which has local maximum at x = –1. If f(2) = 18, f(1) = –1 and f'(x) has local minima at x = 0 then - [JEE 06, (5M, –1M) out of 184] (A) the distance between (–1, 2) and (a, f(a)), where x = a is the point of local minima is 2 5 (B) f(x) is increasing for x  (1 , 2 5 ] NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\04-Maxima-minima.p65 (C) f(x) has local minima at x = 1 (D) the value of f(0) = 5 Rx , 0  x  1 |S z9 . f(x)  2  ex1 , 1  x  2 and g(x)  f(t) dt , x [0, 3] then g(x) hasx T|x  e , 2  x  3 0 (A) local maxima at x = 1 + ln2 and local minima at x = e [JEE 06, (3M, –1M) out of 184] (B) local maxima at x = 1 and local minima at x = 2 (C) no local maxima (D) no local minima (2  x)3, 3  x  1 10. The total number of local maxima and local minima of the function ƒ(x) =  is :-  x2/3 1  x  2 (A) 0 (B) 1 (C) 2 (D) 3 104 [JEE 08, (3M, –1M)] E

1 1 . Match the column : JEE-Mathematics Column I [JEE 08, 6M] Column II (A) The minimum value of x2 2x 4 is (p) 0 (q) 1 x 2 (r) 2 (B) Let A and B be 3 × 3 matrices of real numbers, where A is (s) 3 symmetric, B is skew-symmetric, and (A+B)(A–B) = (A–B)(A+B). If (AB)t = (–1)k AB, where (AB)t is the transpose of the matrix AB, then the possible value of k are (C) Let a = log3 log3 2. An integer k satisfying 1  2(k3a )  2 , must be less than (D) If sin = cos, then the possible values of 1        are   2  Paragraph for Question 12 to 14 Consider the function ƒ : (–, )  (–, ) defined by x2  ax  1 ƒ (x) = x2  ax  1 , 0 < a < 2. 1 2 . Which of the following is true ? [JEE 2008, (4M, –1M)] (A) (2 + a)2 ƒ \"(1) + (2 – a)2 ƒ \"(–1) = 0 (B) (2 – a)2 ƒ \"(1) – (2 + a)2 ƒ \"(–1) = 0 (C) ƒ '(1) ƒ '(–1) = (2 – a)2 (D) ƒ '(1) ƒ '(–1) = –(2 + a)2 1 3 . Which of the following is true ? [JEE 2008, (4M, –1M)] (A) ƒ(x) is decreasing on (–1, 1) and has a local minimum at x = 1 (B) ƒ (x) is increasing on (–1, 1) and has a local maximum at x = 1 (C) ƒ (x) is increasing on (–1, 1) but has neither a local maximum nor a local minimum at x = 1 (D) ƒ(x) is decreasing on (–1, 1) but has neither a local maximum nor a local minimum at x = 1 1 4 . Let ex ƒ '(t) [JEE 2008, (4M, –1M)] NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\04-Maxima-minima.p65 g(x) = 0 1  t2 dt (B) g'(x) is negative on (–, 0) and positive on (0, ) (D) g'(x) does not change sign on (–, ) Which of the following is true ? (A) g'(x) is positive on (–, 0) and negative on (0, ) (C) g'(x) charges sign on both (–, 0) and (0, )  1 5 . The maximum value of the function f(x) = 2x3 – 15 x2 + 36 x – 48 on the set A = x| x2  20  9 x is [JEE 2009, 4M, –1M]  px 16. Let p(x) be a polynomial of degree 4 having extremum at x=1, 2 and lim 1  =2, then the value x2  x0 of p(2) is [JEE 2009, 4M, –1M] 17. Let f, g and h be real-valued functions defined on the interval [0, 1] by f(x) = ex2  ex2 , g(x) = xex2  ex2 and h(x) = x2ex2  ex2 . If a, b and c denote, respectively, the absolute maximum of f, g and h on [0, 1], E then [JEE 10, 3M, –1M] (A) a = b and c  b (B) a = c and a  b (C) a  b and c  b (D) a = b = c 105

JEE-Mathematics 1 8 . Let f be a function defined on R (the set of all real numbers) such that f'(x) = 2010 (x–2009) (x–2010)2 (x–2011)3 (x – 2012)4, for all x  R. If g is a function defined on R with values in the interval (0,) such that ƒ (x) = n (g(x)), for all x  R, then the number of points in R at which g has a local maximum is - [JEE 10, 3M] 1 9 . Let ƒ : IR  IR be defined as ƒ(x) = |x| + |x2 – 1|. The total number of points at which ƒ attains either a local maximum or a local minimum is [JEE 2012, 4M] 2 0 . Let p(x) be a real polynomial of least degree which has a local maximum at x = 1 and a local minimum at x = 3. If p(1) = 6 and p(3) = 2, then p'(0) is [JEE 2012, 4M] 2 1 . A rectangular sheet of fixed perimeter with sides having their lengths in the ratio of 8 : 15 is converted into an open rectangular box by folding after removing squares of equal area from all four corners. If the total area of removed squares is 100, the resulting box has maximum volume. Then the lengths of the sides of the rectangular sheet are [JEE 2013, 4M,–1M] (A) 24 (B) 32 (C) 45 (C) 60 22 . The function ƒ(x) = 2|x| + |x + 2| – ||x + 2| – 2|x|| has a local minimum or a local maximum at x = [JEE 2013, 3M,–1M] (A) –2 2 (C) 2 2 (B) 3 (D) 3 PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE-5[B] NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\04-Maxima-minima.p65 1. A 2. D 3. 18 4. A 5. B 6. 5 7. Distance between (–1, 10) and (3, –22) is 4 65 units 8 . B, C 9. A 1 0 . C 1 1 . (A) – (r), (B) – (q, s), (C) – (r, s), (D) – (p, r) 12. A 13. A 1 4 . B 1 5 . 7 1 6 . 0 1 7 . D 1 8 . 1 19. 5 2 0 . 9 21. A,C 22. A,B E 106

JEE-Mathematics METHODS OF DIFFERENTIATION The process of calculating derivative is called differentiation. 1 . DERIVATIVE OF f(x) FROM THE FIRST PRINCIPLE : y f(x  x)  f(x) dy Obtaining the derivative using the definition Lim  Lim  f '(x)  is called calculating x0 x x0 x dx derivative using first principle or ab initio or delta method. Illustration 1 : Differentiate each of following functions by first principle : (i) f(x) = tanx (ii) f(x) = esinx Solution : (i) f'(x) = lim tan(x  h)  tan x = lim tan(x  h  x)1  tan x tan(x  h) h h h0 h0 = lim tanh . (1 + tan2x) = sec2x. Ans. h0 h (ii) f'(x) = lim esin(xh)  esin x = lim esin x esin(xh )sin x  1  sin(x  h)  sin x sin(x  h)  sin x  h  h0 h h0 = esin x lim sin(x  h)  sin x = esinxcosx Ans. h0 h Do yourself -1 : ( i ) Differentiate each of following functions by first principle: (a) f(x) = nx 1 (b) f(x) = x 2. DERIVATIVE OF STANDARD FUNCTIONS : E Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\04.MOD\\MOD.p65 f(x) f'(x) (ii) f(x) f'(x) (iv) ex (i) x n nxn–1 (vi) nx ex (iii) a x axna, a > 0 (viii) sinx 1/x (v) logax (1/x) logae, a > 0, a  1 (x) tanx cosx (vii) cosx – sinx (xii) cosecx sec2x (ix) secx secx tanx constant – cosecx . cotx (xi) cotx – cosec2x (xiv) 0 cos–1 x (xiii) sin–1 x 1 (xvi) 1 , 1  x  1 , 1  x 1 sec–1 x 1  x2 (xviii) 1  x2 cot–1 x 1 ,| x|1 | x | x2 1 tan–1 x 1 x R (xv) cosec–1 x , 1 (xvii) 1  x2 , x R 1 1  x2 , | x|1 | x | x2 1 77

JEE-Mathematics 3 . FUNDA MENTAL THEOREMS : If f and g are derivable functions of x, then, d df dg d df (a) (f  g)   (b) (cf)  c , where c is any constant dx dx dx dx dx (c) d dg df known as “PRODUCT RULE” (fg)  f  g dx dx dx d f g  df   f  dg   dx   dx  (d) dx  g   where g  0 known as “QUOTIENT RULE” g2 dy dy du (e) If y = f(u) & u = g (x) then  . known as “CHAIN RULE” dx du dx dy du Note : In general if y = f(u) then  f '(u). . dx dx Illustration 2 : If y = ex tan x + xlogex, find dy dx . Solution : y = ex.tan x + x · logex On differentiating we get, dy = ex · tan x + ex · sec2x + 1 · log x + x · 1 dx x Hence, dy = ex(tanx + sec2 x) + (logx + 1) Ans. dx Illustration 3 : log x + ex sin2x + log5x, find dy If y = x dx . Solution : On differentiating we get, 1 dy d log x ·x  log x . 1 1 dx  dx  x  + d (ex sin 2x)  d (log5 x) = x + ex sin2x + 2ex . cos2x + x loge 5  dx dx x2 Hence, dy  1  log x + ex(sin2x + 2cos2x) + 1 Ans. dx  x2  x loge 5  1  y  x2  dy Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\04.MOD\\MOD.p65  tan  x2   dx Illustration 4 : If x = exp   , then equals - (A) x [1 + tan (log x) + sec2 x] (B) 2x [1 + tan (log x)] + sec2 x (C) 2x [1 + tan (log x)] + sec x (D) 2x + x[1 + tan(logx)]2 Taking log on both sides, we get Solution :  y  x2  tan (log x) = (y – x2) / x2 log x = tan–1  x2    y = x2 + x2 tan (log x) On differentiating, we get dy  = 2x + 2x tan (log x) + x sec2 (log x)  2x [1 + tan (log x)] + x sec2 (log x) dx = 2x + x[1 + tan(logx)]2 Ans. (D) 78 E

JEE-Mathematics dy Illustr ation 5 : If y = loge (tan1 1  x2 ) , find dx . Solution : y = loge (tan1 1  x2 ) On differentiating we get, 1 11 = . . .2x tan 1 1  x2 1  ( 1  x2 )2 2 1  x2 xx       = = Ans. 2 tan 1 1  x2 2  x2  1  x2 tan 1 1  x2 1  1  x2 1  x2 Do yourself -2 : (b) y = e5x tan(x2 + 2) dy (i) Find dx if - (a) y = (x + 1) (x + 2) (x + 3) 4 . LOGARITHMIC DIFFERENTIATION : To find the derivative of a function : (a) which is the product or quotient of a number of functions or (b) of the form [f(x)] g (x) where f & g are both derivable. It is convenient to take the logarithm of the function first & then differentiate. Illustration 6 : If y = (sin x)n x, find dy dx Solution : n y = n x. n (sin x) On differentiating we get, 1 dy 1 n (sinx) + n x. cos x  dy = (sinx)n x  n (sin x )  cot x  n x  Ans.  sin x dx  x  Ans. y dx x Illustration 7 : x1 / 2 (1  2 x)2 / 3 dy If y = (2  3x)3 / 4 (3  4x)4 / 5 find Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\04.MOD\\MOD.p65 dx Solution : n y = 1 n x + 2 n (1 – 2x) – 3 n (2 – 3x) – 4 n (3 – 4x) 2 3 4 5 On differentiating we get, 1 dy 1 4 9 16  y dx  2 x – 3(1  2x)  4(2  3x)  5(3  4 x) dy = 1  4  9  16  dx y  2x 3(1  2 x) 4(2  3x) 5(3  4 x) Do yourself -3 : (ii) Find dy if y  e x .e x2 .e x3 .e x4 dy dx ( i ) Find if y = xx 79 dx E

JEE-Mathematics 5 . DIFFERENTIATION OF IMPLICIT FUNCTIONS : (x, y)  0 (a) To find dy /dx of implicit functions, we differentiate each term w.r.t. x regarding y as a function of x & then collect terms with dy/dx together on one side. (b) In the case of implicit functions, generally, both x & y are present in answers of dy/dx. Illustration 8 : If xy + yx = 2, then find dy . dx Solution : Let u = xy and v = yx u + v = 2  du dv  0 Now u = xy  n u = y nx dx dx and v = yx and n v = x n y  1 du y + nx dy and 1 dv = n y + x dy  v dx y dx u dx x dx  du = x y  y  nx dy  and dv  y x   n y  x dy  dx  x dx  dx  y dx   y x ny  x y . y   x   xy  y  n x dy  + y x  n y  x dy  = 0  dy   Ans.  x dx   y dx  dx  x y n x  y x . x   y  Illustration 9 : If y = sin x , prove that dy 1  y cos x  y sin x cos x dx  1  2y  cos x  sin x . 1  sin x 1 1  cos x..... Solution : sin x (1  y ) sin x Given function is y = cos x = 1  y  cos x 1 1y or y + y2 + y cos x = (1 + y) sin x Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\04.MOD\\MOD.p65 Differentiate both sides with respect to x, dy dy dy = (1 + y) cosx + dy sin x  2y  cos x  y sin x dx dx dx dx dy dx (1 + 2y + cosx – sinx) = (1 + y) cosx + ysinx dy 1  y  cos x  y sin x Ans. or  E dx 1  2y  cos x  sin x Do yourself -4 : dy ( i ) Find dx , if x + y = sin(x – y) ( i i ) If x2 + xey + y = 0, find y', also find the value of y' at point (0,0). 80

JEE-Mathematics 6 . PAR A METRIC DIFFERENTIATION : If y  f() & x  g() where  is a parameter, then dy dy / d .  dx dx / d dy  Illustration 10 : If y = a cos t and x = a(t – sint) find the value of at t = dx 2 Solution : dy a sin t  dy Ans.  dx t   1 dx a(1  cos t) 2 Illustration 11 : Prove that the function represented parametrically by the equations. x 1t ; 32 t3 y  2t2 t satisfies the relationship : x(y’)3 = 1 + y’ dy (where y’ = dx ) Solution : Here x = 1t  1 1 t3 t3  t2 Differentiating w.r. to t dx 3 2   dt t4 t3 32 y  2t2 t Differentiating w.r. to t dy 3 2   dt t3 t2 dy  dy / dt  t  y ' dx dx / dt Since x = 1t 1y' or x(y')3 = 1 + y' Ans. t3  x (y ')3 Do yourself -5 : (i) Find dy t if y = cos4t & x = sin4t . at dx 4 ( i i ) Find the slope of the tangent at a point P(t) on the curve x = at2 , y=2at. Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\04.MOD\\MOD.p65 7 . DERIVATIVE OF A FUNCTION W.R.T. ANOTHER FUNCTION : dy dy / dx f '(x) Let y= f (x) ; z = g (x) then   dz dz / dx g '(x) 8 . DERIVATIVE OF A FUNCTION AND ITS IN VERSE FUNCTION : If g is inverse of f, then (a) g{f(x)} = x (b) f{g(x)} = x g'{f(x)}f'(x)=1 f '{g(x)}g'(x) = 1 Illustration 12 : Differentiate loge (tan x) with respect to sin–1(ex). d(loge tan x) d (loge tan x ) cot x.sec2 x = ex 1  e2x d(sin 1 (e x )) dx ex.1 / 1  e2x sin x cos x Solution : = = Ans. d sin 1 (e x ) dx E 81

JEE-Mathematics 1 Illustration 13 : If g is inverse of f and f'(x) = 1  xn , then g'(x) equals :- (A) 1 + xn (B) 1 + [f(x)]n (C) 1 + [g(x)]n (D) none of these Ans. (C) Solution : Since g is the inverse of f. Therefore f(g(x)) = x for all x d  f(g(x))  1 for all x dx 1  f'(g(x)) g'(x) = 1      g'(x) = f '(g(x)) = 1 + (g(x))n Do yourself -6 : ( i ) Differentiate xnx with respect to nx. ( i i ) If g is inverse of ƒ and ƒ (x) = 2x + sinx; then g’(x) equals: 31 1   (D) 2  cos(g(x)) (A) x2 1  x2 (B) 2 + sin–1x (C) 2 + cos g(x) 9 . HIGHER ORDER DERIVATIVES : Let a function y = ƒ (x) be defined on an interval (a, b). If ƒ (x) is differentiable function, then its derivative ƒ '(x) [or (dy/dx) or y'] is called the first derivative of y w.r.t. x. If ƒ '(x) is again differentiable function on (a, b), then its derivative ƒ \"(x) [or d2y/dx2 or y\"] is called second derivative of y w.r.t. x. Similarly, the 3rd order derivative of y d3y d  d2y  w.r.to x, if it exists, is defined by dx3  dx  dx2  and denoted by ƒ '''(x) or y''' and so on. Note : If x = f() and y = g() where '' is a parameter then dy  dy / d & d2 y  d  dy  dx dx dx / d dx2 d  dx  d dn y d  dn1 y  dx    In general dxn d  dx n 1  d Illustration 14 : If f(x) = x3 + x2 f'(1) + xf''(2) + f'''(3) for all x  R. Then find f(x) independent of f'(1), f''(2) and Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\04.MOD\\MOD.p65 f'''(3). Solution : Here, f(x) = x3 + x2 f'(1) + xf''(2) + f'''(3) put f'(1) = a, f''(2) = b, f'''(3) = c f'(1) = 3 + 2a + b .......(i)  f(x) = x3 + ax2 + bx + c .......(ii)  f'(x) = 3x2 + 2ax + b or  f''(x) = 6x + 2a or f''(2) = 12 + 2a .......(iii)  f'''(x) = 6 or f'''(3) = 6 .......(iv) from (i) and (iv), c = 6 from (i), (ii) and (iii) we have, a = –5, b = 2  f(x) = x3 – 5x2 + 2x + 6 Ans. 82 E

JEE-Mathematics d2y Illustration 15 : If x = a (t + sin t) and y = a(1 – cos t), find dx2 . Solution : Here x = a (t + sin t) and y = a (1–cos t) Differentiating both sides w.r.t. t, we get : dx dy = a (sin t) = a(1 + cos t) and dt dt tt 2 sin .cos  dy = a sin t  22   t dx t tan  2  a 1  cos t 2 cos2 2 Again differentiating both sides, we get, d2y 1 1 sec 2  t  dx2 2  co  2  = sec2  t   1  dt = sec 2  t / 2   = 1   2  2 dx a 1 s t  2a  t   2  2 cos2 Hence, d2y = 1  sec4  t  Ans. dx2 4a  2  Illustration 16 : y = f(x) and x = g(y) are inverse functions of each other then express g'(y) and g''(y) in terms of derivative of f(x). dy dx Solution :  f '(x) and  g '(y) dy dx  g '(y)  1 ...........(i) f '(x) Again differentiating w.r.t. to y g ''(y )  d 1   d  f 1 )  . dx =  f ''(x) .  1  dy  f '(x ) dx  '( x  dy (f '(x))2  '( x )  f f ''(x)  g ''(y)   ...........(ii) (f '(x))3 d2y Which can also be remembered as d2x dx2 Ans. =– dy2 3 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\04.MOD\\MOD.p65  d y   d x  Do yourself : 7 ( i ) If y = xex2 then find y''. ( i i ) Find y\" at x = /4, if y = x tan x. (i i i ) Prove that the function y= ex sin x satisfies the relationship y'' – 2y' + 2y = 0. 1 0 . DIFFERENTIATION OF DETERMINANTS : f(x) g(x) h(x) If F(x)  l(x) m(x) n(x) , where f, g, h. l, m, n, u, v, w are differentiable functions of x then u(x) v(x) w(x) f'(x) g'(x) h'(x) f(x) g(x) h(x) f(x) g(x) h(x) F '(x)  l(x) m (x) n(x) + l '(x) m '(x) n '(x) + l(x) m(x) n(x) u(x) v(x) w(x) u(x) v(x) w(x) u '(x) v '(x) w '(x) E 83

JEE-Mathematics x x2 x3 Illustration 17 : If f(x) = 1 2x 3x2 , find f'(x). 0 2 6x Solution : x x2 x3 Here, f(x) = 1 2x 3x2 0 2 6x On differentiating, we get,  f'(x) = d d d x3 x x2 x3 + x x2 x3 (x) (x2 ) 1 2x 3x2 dx d 1 d 2x d 3x2  dx dx 3x2 + d 0 d 2 d 6x 1 2x dx dx dx 0 2 6x 0 2 6x dx dx dx 1 2x 3x2 x x2 x3 x x2 x3 or f'(x) = 1 2x 3x2  0 2 6x  1 2x 3x2 0 2 6x 0 2 6x 0 0 6 As we know if any two rows or columns are equal, then value of determinant is zero. =0+0+ x x2 x3  f'(x) = 6 (2x2 – x2) Therefore, 1 2x 3x2 Ans. 00 6 f'(x) = 6x2 Do yourself : 8 ( i ) If ƒ(x)  ex x2 , then find ƒ '(1). ( i i ) 2x x2 x3 If ƒ(x )  x2  2x 1 3x  1 then find ƒ ' (1). nx sin x 1  3x2 2x 5x 1 1 . L ' HOˆ P ITAL ' S R ULE : 0 (a) This rule is applicable for the indeterminate forms of the type 0 ,  . If the function f(x) and g(x) are differentiable in certain neighbourhood of the point 'a', except, may be, at the point 'a' itself and g'(x)  0, and if lim f(x)  lim g(x )  0 or lim f(x )  lim g(x)   , Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\04.MOD\\MOD.p65 xa xa xa xa f(x) f '(x) then lim  lim xa g(x ) xa g '(x) provided the limit f '(x) exists (L' Hôpital's rule). The point 'a' may be either finite or improper lim xa g '(x ) (+  or –). 0 (b) Indeterminate forms of the type 0.  or  –  are reduced to forms of the type or by algebraic 0 transformations. (c) Indeterminate forms of the type 1, 0 or 00 are reduced to forms of the type 0 ×  by taking logarithms or by the transformation [f(x)](x) = e(x).nf(x). 84 E

JEE-Mathematics Illustration 18 : Evaluate lim x sin x x 0 Solution : lim x sin x = lim e  esin x loge x lim loge x x0 cos ecx x 0 x 0 1/x (applying L'Hôpital's rule) lim = e x0 cos ecx cot x sin2 x  s in x 2  x   = e = e = e  e  1lim  x0 x cos x lim     12 0 0 Ans. x0  x   cos x  Illustration 19 : Solve lim logsin x sin 2x. x 0 Solution : Here lim logsin x sin 2x x 0 log sin 2x    = lim   form x0 log sin x 1 2 cos2x = lim sin 2x {applying L'Hôpital's rule} x0 1  cos x sin x  2x  lim  sin 2x cos 2x = = cos 2 x Ans. lim  1 Ans. x0  x  x0 cos x  sin x  cos x  en 1/ n Illustration 20 : Evaluate lim  n     .  en 1/ n Solution : Here, A = lim  (0 form) n     1  en  n log e  log      log A = lim log  = lim   form n n   n n log e  0 {applying L'Hôpital's rule} = lim n 1 e1 or lim  en 1/n logA = 1A= n     =e Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\04.MOD\\MOD.p65 Do yourself : 9 ( i ) Using L' Hoˆpital 's rule find (a) lim tan x  x (b) lim ex  x 1 x 0 xx 0 3 x2 (ii) Using L' Hoˆpital 's rule verify that : (a) lim sin x  tan x = 1 (b) lim n(1  x)  1 x3  x0 x x 0 2 INTERESTING FACT : In 1694 John Bernoulli agreed to accept a retainer of 300 pounds per year from his former student L'Hôpital to solve problems for him and keep him up to date on calculus. One of the problems was the so-called 0/0 problem, which Bernoulli solved as agreed. When L'Hôpital published his notes on calculus in book form in 1696, the 0/0 rule appeared as a theorem. L'Hôpital acknowledged his debt to Bernoulli and, to avoid claiming authorship of the book's entire contents, had the book published anonymously. Bernoulli nevertheless accused L'Hôpital of plagiarism, an accusation inadvertently supported after L'Hôpital's death in 1704 by the publisher's promotion of the book as L'Hôpital's. By 1721, Bernoulli, a man so jealous he once threw his son Daniel out of the house for accepting a mathematics prize from the French Academy of Sciences, claimed to have been the author of the entire work. As puzzling and fickle as ever, history accepted Bernoulli's claim (until recently), but still named the rule after L'Hôpital. E 85

JEE-Mathematics 1 2 . ANALYSIS AND GR APHS OF SOME IN VERSE TRIGONOMETRIC FUNCTIONS : 2 tan 1 x | x | 1  (a) y  f(x)  sin 1  2x     2 tan 1 x x 1  1  x2  x  1  1 y (   2 ta n x) /2 Important points : (i) Domain is x  R & range is   ,   2 2  D (ii) f is continuous for all x but not differentiable 1 at x = 1, –1 x 2 for | x | 1 –1 x 1  x2 D x (iii) dy  existent for | x | 1  non E dx    /2  2  x 2 for | x | 1 1 (iv) Increasing in ( –1 , 1) & Decreasing in (, 1)  (1, ) (b) Consider y  f(x )  cos1 1  x2   2 tan 1 x if x  0  1  x2   if x  0 2 tan 1 x Important points : f(x) (i) Domain is x  R & range is [0, ) (ii) Continuous for all x but not differentiable at x = 0  2 for x 0  /2 1 for x 0 –1 0 1  x 2 for x 0 dy   non existent (iii) dx   1 2 2  x (iv) Increasing in (0, ) & Decreasing in (,0) 2 tan 1 x | x | 1  x  1 (c) y  f(x)  tan 1 2x    2 tan 1 x x 1 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\04.MOD\\MOD.p65 1  x2   2 tan 1 x ) ( Important points : f(x)  /2 (i) Domain is R – {1, – 1} & range is    ,   2 2  -1 0 1 –/2 (ii) It is neither continuous nor differentiable at x = 1, –1 1 2 | x|1  (iii) dy   x2 dx  non existent | x |  1 (iv) Increasing  x in its domain (v) It is bounded for all x 86

JEE-Mathematics (  3 sin 1 x) if 1  x   1 y  2 2 ( d ) y  f(x)  sin 1 (3x  4 x3 )  3 sin1 x  if  1  x  1   3 sin 1 x 22 Important points : 1 if  x  1 2 (i) Domain is x [1, 1] & range is   ,  D 2 2  (ii) Continuous everywhere in its domain 3 1 I 2 2 11  1 3 x x , 22 x (iii) Not derivable at 22 –1 0 1 3 if x  ( 1 , 1 ) I 22 D dy  1  x2 11 dx 3 (iv)   if x  (1,  )  ( ,1)  22   1  x2  2 (v) Increasing in   1 , 1 and Decreasing in 1,  1    1 ,1  2 2  2   2 3 cos1 x  2 if 1  x   1  2 ( e ) y  f(x)  cos1 (4 x3  3x )  2  3 cos1 x  if 1  x  1 3 cos1 x  22 if 1  x  1 2 Important points : y (i) Domain is x [1, 1] & range is [0, ]  (ii) Continuous everywhere in its domain (iii) Not derivable at 1 1 DI D x , /2 22 3 if x   1 , 1   2 2  Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\04.MOD\\MOD.p65 dy   1  x2 (iv) dx   3  1   1  I   2   2   1  x2 if x 1,   ,1 –1 – 3 –1 O 13 2 2 22 (v) Increasing in   1 , 1 &  2 2  Decreasing in 1,  1   1 ,1 2   2 GENERAL NOTE : Concavity is decided by the sign of 2nd derivative as : d2y d2y dx2  0  Concave upwards ; 0  Concave downwards dx2 E 87

JEE-Mathematics Illustration 21 : d  s i n 2  cot 1 1  x  = dx     1  x   1 (B) 0 1 (D) –1 (A)  (C) 2 2 Solution : Let  1x x = cos 2    0,  y = sin2  cot1 1  x  . Put  2   1  cos 2   y = sin2 cot–1  1  cos 2  = sin2 cot–1 (cot ) 1  cos2 1  x 1 x  y = sin2  = = = 2 2 22 dy 1 Ans (A)   . dx 2  2x  Illustration 22 : If f(x) = sin–1  1  x2  then find (i) f'(2) 1 (iii) f'(1) x = tan, (ii) f'  2  y = sin–1(sin2) Solution : where        22   2   2      2 tan1 x x 1  2  y = 2   2    f(x) =  2 tan 1 x 1  x  1  22 x  1 (  2)   2       2 ta n 1 x ) ( 2  2 x 1 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\04.MOD\\MOD.p65   x2 1  f'(x) = 2 1  x  1   1  x2  2 x  1   1  x2 (i) 2 (ii) 1  8 (iii) f'(1+) = – 1 and f'(1–) = +1   f'(1) does not exist Ans. f'(2) =  f'  2  5 E 5 Do yourself : 10 ( i ) If y = cos–1(4x3 – 3x) Then find (a)  3 (b) ƒ ' (0), (c)  3 ƒ '  2  , ƒ ' 2  . 88

JEE-Mathematics Miscellaneous Illustrations : dy 1  y2 Illustration 23 : If 1  x2  1  y2  a(x  y) , then prove that  - dx 1  x2 Solution : Put x = sin   = sin–1(x) y = sin   = sin–1(y)  cos + cos = a(sin – sin)  2 cos     cos       2a cos      sin       2   2   2   2   cot       a  2       2 cot1 (a)  sin–1x – sin–1y = 2cot–1(a) differentiating w.r.t. to x. 1 1 dy  0 1  x2 1  y2 dx dy 1  y2   hence proved Ans. dx 1  x2 Ans. Illustration 24 : Find second order derivative of y = sinx with respect to z = ex. Ans. dy dy / dx cos x Solution :  dz dz / dx ex d2 y d  cos x  dx = ex sin x  cos xex . 1  dz2  dx  ex  . dz ex 2 ex d2 y sin x  cos x  dz2   e2x Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\04.MOD\\MOD.p65 Illustration 25 : If y = (tan–1x)2 then prove that (1 + x2)2 d2y +2x (1 + x2) dy =2 dx2 dx Solution : y = (tan–1x)2 Differentiating w.r.t. x dy 2 tan 1 x  dx 1  x2   1  x2 dy  2 tan 1 (x) dx Again differentiating w.r.t. x d2y dy 2 1  x2 2 d2 y  2 x(1  x2 ) dy  2 1  x2 dx2 dx       1  x2 dx2  2x   dx E 89

JEE-Mathematics Illu str ation 26 : Obtain differential coefficient of tan–1 1  x2  1 with respect to cos–1 1  1  x2 x 2 1  x2 Solution : Assume u = tan–1 1  x2  1 , v = cos–1 1  1  x2 x 2 1  x2 The function needs simplification before differentiation Let x = tan  u = t an–1  sec   1  = t an–1  1  cos   = t a n –1  ta n     tan    sin    2  = 2 v = cos–1 1  sec  = cos–1 1  cos  = cos–1  cos   =  u = v 2 sec  2  2  2 du  = 1. Ans. dv ANSWERS FOR DO YOURSELF 1 1 1 : (i) (a) (b)  x2 x 2 : (i) ( a ) 3x2 + 12x + 11 ( b ) 5e5x tan (x2 + 2) + 2xe5x sec2(x2 + 2) Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\04.MOD\\MOD.p65 3 : ( i ) xx (nx + 1) (ii) y(1 + 2x + 3x2 + 4x3) cos(x  y) 1 (ii) y '    2x  ey  , –1 4 : (i ) cos(x  y)  1  xey 1    5. (i) –1 1 (ii) 6 : ( i ) 2(xnx)(nx) 7 : ( i ) y'' = 4y + 2xy' t 8 : ( i ) e( sin 1 + cos 1) – 1 (ii) D (ii)  + 4 (ii) 9 1 1 ( c ) –6 9. (i) (a) 3 (b) 10 : (i) (a) – 6 2 (b) 3 90 E

EXERCISE - 01 JEE-Mathematics CHECK YOUR GRASP SELECT THE CORRECT ALTERNATIVE (ONLY ONE CORRECT ANSWER) 1. If y  sec x  tan x then dy equals - sec x  tan x dx (A) 2 sec x (sec x – tan x) (B) –2sec x (sec x – tan x)2 (C) 2 sec x (sec x + tan x)2 (D) –2 sec x (sec x + tan x)2 1 x2  x4 dy 2 . If y = 1  x  x2 and dx = ax + b, then values of a & b are - (A) a = 2 , b = 1 (B) a = –2 , b = 1 (C) a = 2 , b = –1 (D) a = –2 , b = –1 3 . Which of the following could be the sketch graph of y = d xnx ? y dx y yy 1 (A) (B) x' x (C) x' 0 1/e x (D) x' 0e x x' x 01 0 4. Let y' x +3 ln(x – 2) & g(x) = x + y5' ln(x - 1), then the set y' satisfying the inequality y' g'(x) is - f(x) = of x f'(x) < (A)  2, 7 (B) 1, 2    7 ,  (C) (2, ) (D)  7 ,   2   2  2 111  m  n  mn  m  n  mn Differential coefficient of  x m n  . x n   . x m  5. w.r.t. x is - (A) 1 (B) 0 (C) –1 (D) xlmn 1 11 dy emnp 6. If y   then at x = is equal to - 1  xnm  xpm 1  xmn  xpn 1  xmp  xnp dx (A) e mnp (B) emn / p (C) enp / m (D) none of these  x2  y2  dy 7 . If cos–1  x2  y2  = log a then dx = Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\04.MOD\\MOD.p65 (A) x y y x y (B)  (C) (D) y x x  If f(x) = 100 x  n n101n ; then f 101 f '101 n 1 8 . = (A) 5050 1 (C) 10010 1 (B) (D) 5050 10010 HGF KJI9.  If  f(x)  | x| |sin x| , then f ' 4 is - FGH  IKJ 1/ 2 FGH 2 4 22 KIJ GFH  JIK 1/ 2 HFG 2 log 4 2 2 IJK 4 2 log  4 2  (A)  (B)   FHG JKI GHF I2 log   2 2 FHG  KJI 1/ 2 2 log  2 2 HGF  KJI 1/ 2 JK2 4  4 2 4  4 (C)  (D) E 91

JEE-Mathematics 1 0 . If y = xxxxxx ......... then dy - a b a b a b dx a b a b (A) ab  2ay (B) ab  2by (C) ab  2by (D) ab  2ay 11. If y  xx2 then dy = dx (A) 2 n x.x x2 (B) (2 n x  1).x x2 (C) (n x  1).x x2 1 (D) x x2 1.n (ex2 ) y xy 1 2 . If xm . yn = (x + y)m+n, then dy is - (C) (D) xy dx x (A) xy x (B) y dy 1 3 . If x (1  y )  y (1  x)  0 , then d x equals - 1 1 11 (A) (1  x)2 (B)  (1  x)2   (1  x )2 (C) (1  x) (D) none of these 1 4 . If x2 ey + 2xyex + 13 = 0, then dy equals - dx 2xeyx  2y(x  1) 2xexy  2y(x  1)  2xexy  2y(x  1)  (B) x(xe yx  2) x(xeyx  2) (A) x(xeyx  2) (C) (D) none of these 15. If x  e y e y ...........to  , x > 0 then dy is - dx x 1 x 1 x 1 (A) 1  x (B) x (C) x (D) x 1 1 dy = 16. If x =  –  and y=+ , then  dx x y x y (A) y (B) (C) y (D) x x 17. The derivative of sin 1  x w.r.t. cos 1  1  x 2  , (x > 0) is -   1  x 2   1  x2   (A) 1 (B) 2 1 1 (C) 2 (D)  2 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\04.MOD\\MOD.p65 x10  1 8 . Let g is the inverse function of f & f '(x)  . If g (2) = a then g'(2) is equal to - 1  x2 5 1  a2 a10 1  a10 (A) 210 (B) a10 (C) 1  a2 (D) a2 19. Let f(x) = sinx ; g(x) = x2 & h(x)=loge x & F(x) = h[g(f(x))] then d2F is equal to - dx2 (A) 2 cosec3x (B) 2 cot (x2)–4x2 cosec2 (x2) (C) 2x cot x2 (D) –2 cosec2x x 1 2 0 . If ƒ (x) = x2  1 , g(x) = x2  1 and h(x) = 2x – 3, then ƒ '(h'(g'(x)) = (A) 0 1 2 x (B) x2  1 (C) (D) x2  1 5 92 E

JEE-Mathematics 2 1 . If ƒ & g are the functions whose graphs are as shown, let u(x) = ƒ (g(x)); w(x) = g(g(x)), y then the value of u'(1) + w'(1) is - 5 ƒ (2,4) (A) 1 (B) 3   4 2 2 (6,3) (C)  5 (D) does not exist 4 3 2g 1 0 1 2 3 4 5 6x 2 2 . f'(x) = g(x) and g'(x) = - f(x) for all real x and f(5) = 2 = f'(5) then f2 (10) + g2 (10) is - (A) 2 (B) 4 (C) 8 (D) none of these 2 3 . If f(x) = xn, then the value of f(1)  f '(1)  f ''(1)  f '''(1)  .......  (1)n f '''''''''.......(n times) (1) - 1! 2! 3! n! (A) 2n – 1 (B) 0 (C) 1 (D) 2n 2 4 . A function y = f(x) has second order derivative f\"(x) = 6(x – 1). If its graph passes through the point (2, 1) and at that point the tangent to the graph is y = 3x – 5, then the function is - (A) (x + 1)3 (B) (x + 1)2 (C) (x – 1)2 (D) (x – 1)3 25. If ƒ (x) = x + x2  x3 xn , then ƒ (0) + ƒ '(0) + ƒ ''(0) + ........ + ƒ''''...... n times(0) is equal to - 1! 2!  ........ (n 1)! n(n 1) (n2 1)  n(n  1) 2 n(n  1)(2n  1) (A) (B) (C)  2  (D) 6 2 2 cos x x 1 26. Let f (x) = 2 sin x x2 2x f '(x) . Then Limit  x0 x tan x x 1 (A) 2 (B) –2 (C) –1 (D) 1 2 7 . If f is differentiable in (0, 6) & f'(4) =5 then Lim f(4)  f(x2 ) = x2 2  x (A) 5 (B) 5/4 (C) 10 (D) 20 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\04.MOD\\MOD.p65 2 8 . If f(4) = g(4) = 2 ; f’ (4) = 9 ; g’ (4) = 6 then Limit f(x)  gx is equal to - x4 x  2 (A) 3 2 3 (C) 0 (D) none of these (B) 2 SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS) 2 9 . The slope(s) of common tangent(s) to the curves y = e–x & y = e–x sinx can be - (A) e  / 2 (B) e   (D) 1 (C) 2 3 0 . If y + n(1 + x) = 0, which of the following is true ? (A) ey = xy' + 1 1 (C) y' + ey = 0 (D) y' = ey (B) y' =  (x  1) E 93

JEE-Mathematics 3 1 . If y = 23x , then y' equals - (A) 3x n3 n2 (B) y(log2y) n3 n2 (C) 23x . 3x n6 (D) 23x . 3x n3 n2 d2y (D) None of these 3 2 . If y = 3t2 & x = 2t then dx2 equals- ƒ '(1) (A) 3t (B) 3 3 (D) (ƒ(1))2 (C) 2 3 3 . If g is inverse of ƒ and ƒ (x) = x2 + 3x – 3 (x > 0) then g'(1) equals- 1 (B) –1 1 (A) 2 g(1)  3 (C) 5 CHECK YOUR GRASP ANSWER KEY EXERCISE-1 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\04.MOD\\MOD.p65 Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Ans. B C C D B D C B A D D C B A C Que. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 A ns . A C B D C B C B D A B D A A , B A,B,C Que. 31 32 33 Ans. B,D C A,C 94 E

EXERCISE - 02 JEE-Mathematics BRAIN TEASERS SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS) 1 . If y = fofof (x) and f (0) = 0, f '(0) = 2, then find y'(0) - (A) 6 (B) 7 (C) 8 (D) 9 2. If y2 = p(x) is a polynomial of degree 3, then 2 d  y 3 d2y  is equal to - dx  dx2  (A) p'''(x) . p'(x) (B) p''(x) . p'''(x) (C) p(x) . p'''(x) (D) none of these 3 . If y is a function of x then d2 y  y dy  0 . If x is a function of y then the equation becomes - dx2 dx d2x dx d2x  dx 3 d2 x  dx 2 d2x  dx 2 x 0 (B) dy2  y  dy   0 (C) dy2  y  dy   0 (D) dy2  x  dy   0 (A) dy2 dy dy 4 . If y = tanx tan 2x tan 3x then dx is equal to- (A) 3 sec2 3x tan x tan 2x + sec2x tan 2x tan 3x + 2 sec2 2x tan 3x tan x (B) 2y (cosec 2x + 2 cosec 4x + 3 cosec 6x) (C) 3 sec2 3x – 2 sec2 2x – sec2 x (D) sec2x + 2 sec2 2x + 3 sec2 3x 5 . If y  e x  e x then dy equals - dx e x  e x (B) e x  e x (C) 1 y2  4 1 y2  4 (A) 2x 2x (D) 2x 2x dy 6 . Let y  x  x  x  ...... then - dx 1 x 1 y (A) 2y  1 (B) x  2y (C) (D) 2x  y 1  4x 2 x (1  2 y ) (D) 2 y (2x  1) 7 . If 2x + 2y = 2 x+y then dy has the value equal to - dx Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\04.MOD\\MOD.p65 2y 1 (C) 1–2y (A)  2x (B) 1  2x 8 . The functions u = ex sin x ; v = ex cos x satisfy the equation - (A) du dv  u2  v2 d2u d2 v (D) none of these v u (B) dx2  2v (C) dx2  2u dx dx 9 . Two functions f & g have first & second derivatives at x = 0 & satisfy the relations, f(0)  2 , f'(0) = 2 g'(0) = 4g (0), g''(0) = 5 f''(0) = 6 f(0) = 3 then - g(0) fx 15 (B) if k(x) = f(x) . g(x) sinx then k'(0) = 2 (A) if h (x) = g x then h'(0) = 4 (D) none of these (C) Limit g'x 1  95 x 0 f 'x 2 E

JEE-Mathematics 1 0 . If y2 + b2 = 2xy, then dy equals - dx 1 y xy  b2 xy  b2 (A) xy  b2 (B) y  x (D) y (C) y  x2 y  x  c , then dy 11. If yx  is equal to - dx 2x x (C) y  y2  x2 c2 (A) c2 (B) y  y2  x2 x (D) 2y (D) non existent   1 2 . Lim xxx  xx is - (C) equal to –1 x 0  (A) equal to 0 (B) equal to 1 1 3 . Select the correct statements - 2x2  3 for x  1 (A) The function f defined by f(x) = 3x  2 for x  1 is neither differentiable nor continuous at x = 1. (B) The function f(x) = x2|x| is twice differentiable at x = 0 (C) If f is continuous at x = 5 and f(5) = 2 then Lim f(4x2–11) exists. x 2 (D) If Lim (f(x)+g(x)) = 2 and Lim (f(x) – g(x)) = 1 then Lim f(x). g(x) may not exist. xa xa xa 1 4 . Let   Lim xm nxn where m, n  N then - (B)  is independent of m and depends on m x 0 (A)  is independent of m and n (C)  is independent of n and depends on m (D)  is dependent on both m and n 1 5 . Lim logsin2 x cos x has the value equal to - x 0 x log cos 2 x sin2 2 (A) 1 (B) 2 (C) 4 (D) none of these BRAIN TEASERS ANSWER KEY EXERCISE-2 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\04.MOD\\MOD.p65 Que. 1 2 3 4 5 6 7 8 9 10 Ans. C C Que. 11 12 C A,B,C A,C A,C,D A,B,C,D A,B,C A,B,C B,C Ans. A,B,C C 13 14 15 B,C A C 96 E

EXERCISE - 03 JEE-Mathematics MISCELLANEOUS TYPE QUESTIONS TRUE / FALSE 1. Let u(x) and v(x) are differentiable functions such that u u '(x) = p and  u(x) ' = q then pq =1 v (x) = 7 If v '(x)  v(x)  pq 2 . If f(x) = |x – 2|, then f'(f(x)) = 1 for x > 20 c 3 . If f(0) = a, f'(0) = b, g(0) = 0 and (fog)'(0) = c, then g'(0) = b 1 4 . The differential coefficient of f(logx) w.r.t. logx where f(x) = logx is log x 5 . f'(sinx) = (f(sinx))' 6. If x = t2 + 3t – 8, y = 2t2 – 2t –5, then dy at (2, –1) is 6 dx 7 MATCH THE COLUMN Following questions contains statements given in two columns, which have to be matched. The statements in Column-I are labelled as A, B, C and D while the statements in Column-II are labelled as p, q, r and s. Any given statement in Column-I can have correct matching with ONE statement in Column-II. 1 . Column-I Column-II Graph of f(x) Graph of f'(x) (A) (p) (B) (q) Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\04.MOD\\MOD.p65 (C) (r) (D) (s) E 97

JEE-Mathematics 2 . Column-I Column-II (A) If f(x) = x3 + x + 1, then f'(x2 + 1) at (p) 1 x = 0 is (q) 0 (B) If f(x) = logx2 (log x) , then f'(ee) is equal to (r) 28 (C) For the function y = n tan    x  (s) 4  4 2  dy if dx = secx + p, then p is equal to (D) If f(x) = |x3 – x2 + x – 1| sin x, then 4f'(28f(f())) is equal to ASSERTION & REASON These questions contain Statement I (assertion) and Statement II (reason). (A) Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I. (B) Statement-I is true, Statement-II is true ; Statement-II is NOT a correct explanation for statement-I. (C) Statement-I is true, Statement-II is false. (D) Statement-I is false, Statement-II is true. 1 . Statement-I : Let f(x) is a continuous function defined from R to Q and f(5) = 3 then differential coefficient of f(x) w.r.t. x will be 0. Because Statement-II : Differentiation of constant function is always zero. (A) A (B) B (C) C (D) D 2. Statement-I : Derivative of sin 1  2 x  with respect to cos 1  1  x 2  is 1 for 0 < x < 1.   x   1  x 2  1 2 Because Statement-II : sin 1  2x   cos 1 1  x2  for –1  x  1.  1  x2   1  x2  (A) A (B) B (C) C (D) D x Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\04.MOD\\MOD.p65 3 . Consider ƒ (x) = x2 1 & g(x) = ƒ ''(x). Statement-I : Graph of g(x) is concave up for x > 1. Because dn (ƒ (x))  (1)n n !  1  1  dxn   1)n1  1)n 1 , Statement-II : 2 (x (x  n N (A) A (B) B (C) C (D) D COMPREHENSION BASED QUESTIONS Comprehension # 1 f(x  y) f(x) f(y) 1 Let = + xy,, xy  R. f(x) is differentiable and f'(0) = 1. Let g(x) be a derivable function 22 at x = 0 and follows the functional rule g  x  y  = g(x)  g(y) (k  R, k  0, 2)  k  k Let g'(0) =   0 98 E

JEE-Mathematics On the basis of above information, answer the following questions : 1 . Domain of n(f(x)) is- (A) R+ (B) R – {0} (C) R (D) R– 2 . Range of y = log3/4(f(x)) (A) (–, 1] (B) 3 ,   (C) (–, ) (D) R  4  3 . If the graphs of y = f(x) and y = g(x) intersect in coincident points the  can take values- (A) 3 (B) 1 (C) –1 (D) 4 Comprehension # 2 Limits that lead to the indeterminate forms 1, 00, 0 can sometimes be solved taking logarithm first and then using L' Hoˆpital 's rule Let L im (f ( x ))g ( x ) is in the form of 0, it can be written as e  elim g ( x )nf ( x ) L xa xa where L = n f ( x ) is  form and can be solved using L' Hoˆpital 's rule. lim  xa 1 / g(x) On the basis of above information, answer the following questions : 1. Lim x1 /(1x ) - x 1 (A) –1 (B) e–1 (C) –2 (D) e–2 2 . Lim (nx)1 / 2x  x1 / xn   n  N - x  (A) 2 (B) 0 (C) e1/2 (D) e (C) 2 (D) does not exist 3 . Lim (sin x)2sin x x 0  (A) 1 (B) 0 Comprehension # 3 Left hand derivative and right hand derivative of a function f(x) at a point x = a are defined as f' (a–) = lim f(a) f(a h) = hlim0 f(a h) f(a) and h h h 0 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\04.MOD\\MOD.p65 f'(a+) = lim f(a  h)  f(a) = lim f(a)  f(a h) = lim f(a)  f(x) respectively h 0  h 0  xa a  x h h Let f be a twice differentiable function. We also know that derivative of an even function is odd function and derivative of an odd function is even function. On the basis of above information, answer the following questions : 1 . If f is odd, which of the following is Left hand derivative of f at x = –a (A) lim f(a h) f(a) (B) hlim0 f(h a) f(a) (C) lim f(a)  f(a h) (D) lim f(a)  f(–a  h) h h h h h 0  h 0  h 0  2 . If f is even, which of the following is Right hand derivative of f' at x = a (A) lim  f '(a)  f '(a  h) (B) lim f '(a)  f '(–a  h) h h h 0 h 0  (C) lim –f '(–a)  f '(–a  h) (D) lim f '(a)  f '(–a  h) h h h 0  h 0  E 99

JEE-Mathematics 3. The statement lim f(x) f(x h) = lim f(x)f(x h) implies that for all x  R h h h 0 h 0 (A) f is odd (B) f is even (C) f is neither odd nor even (D) nothing can be concluded Comprehension # 4 An operator  is defined to operate on differentiable functions defined as follows. If ƒ (x) is a differentiable function then  ƒ x   lim ƒ3 x  h   ƒ3 x  h0 h g(x) is a differentiable function such that the slope of the tangent to the curve y = g(x) at any point (a, g(a)) is equal to 2ea (a+1) also g(0)=0. On the basis of above information, answer the following questions : 1 . gx at x=n2 is – (A) 24n2 {2n2+ 2} (B) n 4e2  n2 2 (C) 96n 4e2  n2 2 (D) 192n(4e) n22 (D) 26·34 2.  ( (x + 2)) (B) 29·35 (C) 24·35 x=0 (A) 25·39 3. lim g x x0 n cos 2x  (A) –12 (B) 12 (C) 24 (D) –24 MISCELLANEOUS TYPE QUESTION ANSWER KEY EXERCISE-3 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\04.MOD\\MOD.p65  True / False E 1. T 2. T 3. T 4. T 5. F 6. T  Match the Column 1 . (A)  (q); (B)  (s); (C) (p); (D) (r) 2 . (A)  (s); (B)  (q); (C) (q); (D) (s)  Assertion & Reason 1. A 2. C 3. A  Comprehension Based Questions Comprehension # 1 : 1. C 2. A 3 . A,C Comprehension # 2 : 1 . B 2. A 3. A Comprehension # 3 : 1 . A 2. A 3. B Comprehension # 4 : 1 . C 2. D 3. A 100

EXERCISE - 04 [A] JEE-Mathematics CONCEPTUAL SUBJECTIVE EXERCISE a  bx3 /2 dy a 1. If y = x 5 / 4 and vanishes at x = 5 then find . dx b x4  4 dy 2 . If y  x2  2 x  2 then find dx x1 2 3. If f'(x) = 2x2 1 and y = f(x2) then find dy at x = 1. dx 1 + nt 3 + 2nt dy  dy 2 4 . If x = t2 and y = t . Show that y dx = 2x  dx  + 1.  5 . d If fn(x) = e fn1 ( x ) for all n  N and f0(x) = x then show that dx fn (x )  f1 (x ).f2 (x ).........fn (x ) . 6. x2 1 x2  1  n x x2 1 prove that 2y = xy' + ny', where y' denotes the derivative of y w.r.t. x. If y   x 22 7 . Let f(x) = x + 1 2x  1 1 2x  2x  ........ Compute the value of f(100).f'(100). If y  tan 1 u x  sec1 1  1  1  dy 8. 1  u2 & 2u2 1 , u  0, 2    2 ,1 ; prove that 2 1  0 dx 9. If y = tan–1 x + sin  2 ta n 1 1 x  then find dy for x  (–1, 1).  1 x  , dx 1 1x2 1 0 . If x = 2 cost – cos2t & y = 2 sint – sin2t, find the value of (d2y / dx2) when t  ( / 2) . 11. If ax2 bx c y'  1  a a x  b b x  c c  [JEE 98] y     1 , Prove that y x     x  (x  a)(x  b)(x  c) (x  b)(x  c) (x  c) Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\04.MOD\\MOD.p65 12. If arcsin y d2y  2(x2  y2 ) x > 0. dx2 , x2  y2  e x2 y2 . Prove that (x  y)3 1 3 . Let f(x)  x2  4 x  3, x  2 and let g be the inverse of f. Find the value of g' at f(x) = 9 14. If y = xn[(ax)–1 + a–1], prove that x(x + 1) d2 y dy =y–1 dx2 + x dx 15. If x  sec   cos  ; y  secn   cosn  , then show that (x2  4)  dy 2  n2 (y2  4)  0 .  dx  1 6 . (a) Differentiate y  c o s 1 1  x2 w. r. t. tan-1x, stating clearly where function is not differentiable. (b) 1  x2 If y  sin 1 (3x  4 x3 ) find dy/dx stating clearly where the function is not derivable in ( –1,1). E 101

JEE-Mathematics 1 7 . Suppose f and g are two functions such that f, g : R  R,    f(x) = n 1  1  x2 and g(x) = n x  1  x2 then find the value of x.eg(x)  f  1  ' + g'(x) at x = 1.   x   1 8 . Determine the values of a, b and c so that Lim (a  b cos x )x  c sin x  1 x0 x5 Solve using L' Hoˆp it a l 's rule or series expansion. (Q.18 – Q.21) 1 9 . Lim x cos x  n(1  x) x0 x2 20. Lim  1  1   x2 sin2 x  x 0 21. If Lim ax  xa  1 find 'a'. xx  aa xa 22. Lim log tan2 x (tan2 2 x ) x 0 (x  a)4 (x  a)3 1 (x  a)4 (x  a)2 1 2 3 . If f(x)  (x  b)4 (x  b)3 1 then f '(x)  . (x  b)4 (x  b)2 1 Find the value of  . (x  c)4 (x  c)3 1 (x  c)4 (x  c)2 1 CONCEPTUAL SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(A) Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\04.MOD\\MOD.p65 1. 5 2. 3 3. 2 7. 100 12x 3 1 9. 10.  13. 2 1x2 2 8 17. zero 1 6 . (a) Not differentiable at x = 0 (b) Not derivable at x  1 / 2 1 8 . a = 120; b = 60; c = 180 1 20. 1 21. a = 1 22. 1 23. 3 19.  3 2 102 E

JEE-Mathematics EXERCISE - 04 [B] BRAIN STORMING SUBJECTIVE EXERCISE 1. If x = 1 and y = f(x), show that : d2f  2z3 dy  z4 d2y z dx2 dz dz2 2. Prove that if |a sin x + a sin 2x +.....+ a sin nx ||sin x| for x R, then|a + 2a + 3a +.....+ na | 1 1 2 n 1 2 3 n 3. Show that the substitution z = n  tan x  changes the equation d2y  cot x dy  4y cosec2 x  0 to  2  dx2 dx (d2y/dz2) + 4y = 0 4 . Find a polynomial function f(x) such that f(2x) = f'(x) f''(x). 5 . If Y = sX and Z = tX, where all the letters denotes the function of x and suffixes denotes the differentiation w.r.t. XYZ  X 3 s1 t1 s2 t2 x then prove that X1 Y1 Z1 X2 Y2 Z2 1  x6  1  y 6  a 3 .(x3  y3 ) , prove that dy x2 1  y6 6. If  . dx y2 1  x6 7 . If  be a repeated root of a quadratic equation f(x) = 0 & A(x), B(x), C(x) be the polynomials of degree A(x) B(x) C(x) 3, 4 & 5 respectively, then show that A () B() C () is divisible by f(x), where dash denotes the derivative. A '() B '() C '() 8. If y  tan 1 x2 1 1  tan 1 x2  1  3  tan 1 x2  1  7  tan 1 x2 1  13 +......... upto n terms. x 3x 5x  7x Find dy/dx, expressing your answer in 2 terms. g(x) , x0 9. Let g(x) be a polynomial, of degree one & f(x) be defined by f(x) =  1  x 1 / x   , x 0  2  x  Find the continuous function f(x) satisfying f'(1) = f(–1) 10. Let f(x  y)  f(x)  f(y)  a  xy for all real x and y. If f(x) is differentiable and f'(0) exists for all real permissible 22 values of 'a' and is equal to 5a  1  a2 . Prove that f(x) is positive for all real x. Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\04.MOD\\MOD.p65 12 ,x  0 11. Find the value f(0) so that the function f(x) =  1 is continuous at x = 0 & examine the differentiability x e2x of f(x) at x = 0. 12. If Lim a sin x  bx  cx2  x3 exists & is finite, find the value of a, b, c & the limit. x0 2x2 .n(1  x)  2x3  x4 BRAIN STORMING SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(B) 4x3 11  2  1  n 3  x if x0 4.   3  6 1 2  if x 0 8. 9. f ( x ) =   /x 9 1  (x  n)2 1  x2 x  1  x 2  1 1 . f(0) = 1, differentiable at x = 0, f'(0+) = –(1/3); f'(0–) = –(1/3) 1 2 . a = 6, b = 6, c = 0 ; 3/40 E 103

JEE-Mathematics JEE-[MAIN] : PREVIOUS YEAR QUESTIONS EXERCISE - 05 [A] 1 . If f(1) = 1, f'(1) = 2, then lim f(x) 1  (3) 3 [AIEEE - 2002] x1 x 1 (4) 4 (1) 2 (2) 1 2. lim log xn [x] , n  N, (where [x] denotes greatest integer less than or equal to x)- x [x] (1) Has value -1 (2) Has values 0 (3) Has value 1 [AIEEE - 2002] (4) Does not exist dy [AIEEE-2002] 3 . If y = logyx, then dx = 1 1 1 1 (1) (2) log x(1  y ) (3) x(1  log y ) (4) y  log x x  log y dy [AIEEE-2002] 4 . If x = 3cos – 2cos3 and y = 3sin – 2sin3, then dx = (1) sin (2) cos (3) tan (4) cot  n [AIEEE-2002] 5 . If y = x  1  x2 then (1 + x2)y2 + xy1 = (1) ny2 (2) n2y (3) n2y2 (4) None of these 6 . If f(x) = xn, then the values of f(1) – f '(1)  f \"(1)  f '\"(1)  ...  (1)n fn (1) is- [AIEEE-2003] 1! 2! 3! n! (1) 1 (2) 2n (3) 2n–1 (4) 0 7 . Let f(x) be a polynomial function of second degree. If f(1) = f(–1) and a, b, c are in A.P. then f'(a), f'(b) and f'(c) are in- [AIEEE-2003] (1) Arithmetic-Geometric Progression (2) Arithmetic progression (A.P.) (3) Geometric progression (G.P.) (4) Harmonic progression (H.P.) 8. If x = e y  ey  .....to , x > 0, then dy is - [AIEEE-2004] dx Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\04.MOD\\MOD.p65 x 1 1x 1x (1) 1  x (2) (3) (4) x x x dy [AIEEE-2006] 9 . If xm.yn = (x + y)m+n, then dx is- xy (2) xy x y (1) xy (3) y (4) x 1 0 . Let y be an implicit function of x defined by x2x – 2xx cot y – 1 = 0. Then y'(1) equals :- [AIEEE-2009] (1) log 2 (2) –log2 (3) –1 (4) 1 1 1 . Let f : (–1, 1)  R be a differentiable function with f(0) = – 1 and f'(0) = 1. Let g(x) = [f(2f(x) + 2)]2. Then g'(0) :- [AIEEE-2010] (1) 4 (2) –4 (3) 0 (4) –2 104 E