JEE-Mathematics 1 0 . Area of shaded region belongs to - [JEE 05 (screening) 3M] P(2–1,2) (A) z : |z + 1| > 2, |arg (z + 1)| < /4 (B) z : |z – 1| > 2, |arg (z – 1)| < /4 (C) z : |z + 1| < 2, |arg (z + 1)| < /2 A /4 (D) z : |z – 1| < 2, |arg (z – 1)| < /2 (–1,0) (1,0) Q(2–1,– 2) 1 1 . If one of the vertices of the square circumscribing the circle |z – 1| = 2 is 2 3i . Find the other vertices of square. [JEE 05 (Mains) 4 out of 60] 1 2 . If w = i where 0 and z 1, satisfies the condition that w wz is purely real, then the set of values of 1z z is - [JEE 06, 3M] (A) {z : |z|=1} (B) {z : z = z } (C) {z : z 1} (D) {z : |z| = 1, z 1} 1 3 . A man walks a distance of 3 units from the origin towards the north-east (N 45° E) direction. From there, he walks a distance of 4 units towards the north-west (N 45° W) direction to reach a point P. Then the position of P in the Argand plane is : [JEE 07, 3M] (D) (3 + 4i)ei/4 (A) 3ei/4 + 4i (B) (3 – 4i)ei/4 (C) (4 + 3i)ei/4 z [JEE 07, 3M] 1 4 . If |z| = 1 and z ± 1, then all the values of 1 z2 lie on : (A) a line not passing through the origin (B) |z| = 2 (C) the x-axis (D) the y-axis Comprehension (for 15 to 17) : Let A, B, C be three sets of complex numbers as defined below [JEE 2008, 4M, –1M] A z : Im z 1 B z :| z 2 i| 3 C z : Re((1 i)z) 2 1 5 . The number of elements in the set A B C is - (A) 0 (B) 1 (C) 2 (D) 1 6 . Let z be any point in A B C. Then |z + 1 – i|2 + |z – 5 – i|2 lies between - (A) 25 and 29 (B) 30 and 34 (C) 35 and 39 (D) 40 and 44 1 7 . Let z be any point in A B C and let be any point satisfying | – 2 – i| < 3. Then, |z|–|| + 3 lies between - NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\02.Complex number.p65 (A) –6 and 3 (B) –3 and 6 (C) –6 and 6 (D) –3 and 9 18. A particle P starts from the point z = 1 + 2i, where i = 1 . It moves first horizontally away from origin 0 by 5 units and then vertically away from origin by 3 units to reach a point z . From z the particle moves 11 2 units in the direction of the vector ˆi ˆj and then it moves through an angle in anticlockwise direction 2 on a circle with centre at origin, to reach a point z . The point z is given by - [JEE 2008, 3M, –1M] 22 (A) 6 + 7i (B) –7 + 6i (C) 7 + 6i (D) –6 + 7i 15 [JEE 2009, 3M, –1M] 1 9 . Let z = cos + i sin . Then the value of Im(z2m1 ) at = 2° is - m 1 1 1 1 1 (A) (B) (C) (D) sin 2 3 sin 2 2 sin 2 4 sin 2 2 0 . Let z = x + iy be a complex number where x and y are integers. Then the area of the rectangle whose vertices are the roots of the equation zz 3 zz3 350 is - [JEE 2009, 3M, –1M] (D) 80 (A) 48 (B) 32 (C) 40 E 61
JEE-Mathematics 2 1 . Match the conics in Column I with the statements/ expressions in Column II. [JEE 2009, 8M] Column I Column II (A) Circle (P) The locus of the point (h, k) for which the line (B) Parabola hx + ky = 1 touches the circle x2 + y2 = 4 (C) Ellipse (D) Hyperbola (Q) Points z in the complex plane satisfying | z + 2 | – | z – 2 |= ± 3 (R) Points of the conic have parametric representation x 3 1 t2 , y = 2t t2 t2 1 1 (S) The eccentricity of the conic lies in the interval 1 x < (T) Points z in the complex plane satisfying Re (z + 1)2 = | z | 2 + 1 2 2 . Let z1 and z2 be two distinct complex numbers and let z = (1 – t)z1 + tz2 for some real number t with 0 < t < 1. If Arg(w) denotes the principal argument of a nonzero complex number w, then (A) |z – z |+|z – z |=|z – z | (B) Arg(z – z ) = Arg(z – z ) [JEE 10, 3M] 1 2 12 1 2 z z1 z z1 0 (D) Arg(z – z1) = Arg(z2 – z1) (C) z2 z1 z2 z1 23. Let be the complex number cos 2 i sin 2 . Then the number of distinct complex numbers z satisfying 3 3 z 1 2 z 2 1 0 is equal to [JEE 10, 3M] 2 1 z 2 4 . Match the statements in Column-I with those in Column-II. [JEE 10, 8M] [Note : Here z takes values in the complex plane and Im z and Re z denote, respectively, the imaginary part and the real part of z.] Column I Column II (A) The set of points z satisfying z i z z i z 4 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\02.Complex number.p65 (p) an ellipse with eccentricity is contained in or equal to (B) The set of points z satisfying 5 (q) the set of points z satisfying Im z = 0 |z + 4| + |z – 4| = 10 is contained in or equal to (t) the set of points z satisfying |Im z| < 1 (C) If |w|= 2, then the set of points z w 1 is contained in or equal to w (D) If |w| = 1, then the set of points (s) the set of points z satisfying |Re z| 2 (t) the set of points z satisfying |z| 3 z w 1 is contained in or equal to w 25. Comprehension (3 questions together) Let a,b and c be three real numbers satisfying 1 9 7 ...(E) a b c8 2 7 0 0 0 7 3 7 (i) If the point P(a,b,c), with reference to (E), lies on the plane 2x + y + z = 1, then the value of 7a+b+c is (A) 0 (B) 12 (C) 7 (D) 6 62 E
JEE-Mathematics ( i i ) Let be a solution of x3 – 1 = 0 with Im() > 0. If a = 2 with b and c satisfying (E), 313 then the value of is equal to - a b c (A) –2 (B) 2 (C) 3 (D) –3 (iii) Let b = 6, with a and c satisfying (E). If and are the roots of the quadratic equation 1 1 n ax2 + bx + c = 0, then is - n 0 (A) 6 (B) 7 6 (D) (C) 7 [JEE 2011, 3+3+3] 26 . If z is any complex number satisfying |z – 3 – 2i| < 2, then the minimum value of |2z – 6 + 5i| is [JEE 2011, 4M] 2 7 . Let ei/ 3 , and a, b, c, x, y, z be non-zero complex numbers such that a + b + c = x a + b + c2 = y a + b2 + c = z. | x|2 | y|2 | z|2 [JEE 2011, 4M] Then the value of | a|2 | b|2 | c|2 is Column II 2 8 . Match the statements given in Column I with the values given in Column II Column I (p) 6 (A) If ˆj 3kˆ, ˆj 3 kˆ and 2 3kˆ form a triangle, a b c then the internal angle of the triangle between a and b is (B) If b (ƒ( x ) 3x)dx a2 b2 , then the value of ƒ is 2 a 6 (q) 3 (C) 2 5 6 is The value of n3 (r) 3 se c( x ) dx 7/6 (D) The maximum value of A rg 1 1 z for (s) |z| = 1, z 1 is given by (t) 2 [JEE 2011, 2+2+2+2M] NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\02.Complex number.p65 2 9 . Match the statements given in Column I with the intervals/union of intervals given in Column II Column I Column II (A) The set e 1 2iz 2 : z is a co m plex n um ber, | z| = 1, z (p) (, 1) (1,) R z 1 is (B) The domain of the function ƒ( x ) sin 1 8(3)x2 is (q) (,0) (0, ) 1 32( x 1 ) 1 tan 1 2 (C) If ƒ() tan 1 tan , then the set ƒ() : 0 is (r) [2,) 1 tan 1 (D) If ƒ(x) = x3/2(3x – 10), x 0, then ƒ(x) is increasing in (s) (, 1] [1, ) (t) (,0] [2,) [JEE 2011, 2+2+2+2M] E 63
JEE-Mathematics 3 0 . Let z be a complex number such that the imaginary part of z is nonzero and a = z2 + z + 1 is real. Then a cannot take the value - [JEE 2012, 3M, –1M] (A) –1 1 1 3 (B) 3 (C) (D) 2 4 31. Let complex numbers and 1 lie on circles (x – x0)2 + (y – y0)2 = r2 and (x – x0)2 + (y – y0)2 = 4r2 respectively. If z0 = x0 + iy0 satisfies the equation 2|z0|2 = r2 + 2, then || = [JEE(Advanced) 2013, 2M] 1 1 1 1 (A) (B) (C) 7 (D) 3 2 2 3 2 . Let be a complex cube root of unity with 1 and P = [pij] be a n × n matrix with pij = i+j. Then P2 0, when n = [JEE(Advanced) 2013, 3, (–1)] (A) 57 (B) 55 (C) 58 (D) 56 33. Let w 3 i and P = {wn : n = 1, 2, 3, .....}. Further H1 = z C : Re z 1 and H 2 z C : Re z 1 , 2 2 2 where C is the set of all complex numbers. If z1 P H1, z2 P H2 and O represents the origin, then z1Oz2 = [JEE-Advanced 2013, 4, (–1)] 2 5 (A) (B) 6 (C) 3 (D) 6 2 Paragraph for Question 34 and 35 Let S = S1 S2 S3, where S1= {z C : |z| < 4}, S2 z C : Im z 1 3i 0 and 1 3i S3 = {z C : Re z > 0}. 34. min|1 3i z| [JEE(Advanced) 2013, 3, (–1)] zS (A) 2 3 (B) 2 3 (C) 3 3 (D) 3 3 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\02.Complex number.p65 2 2 2 2 3 5 . Area of S = 20 16 [JEE(Advanced) 2013, 3, (–1)] 10 (B) 3 (C) 3 32 (A) 3 (D) 3 PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE-5 [B] 1 . (a) A (b) A 2 . (a) C, (b)D 3 . (a) B ; (b) B 4. A 7. B 9. B 10. A 8. k2 & 1 | k2|2 (k2 | |2 | |2 )(k2 1) 1 k2 k2 1 1 1 . ( 3 i) , (1 3 ) + i and (1 3 ) – i 12. D 13. D 14. D 15. B 1 6 . C 1 7 . D 1 8 . D 1 9 . D 2 0 . A 21. A (P) ; B (S, T) ; C (R) ; D (Q, S) 2 2 . A,C,D 2 3 . 1 2 4 . (A) (q,r), (B) (p), (C) (p,s,t), (D) (q,r,s,t) 2 5 . (i) D, (ii) A, (iii) B 26.5 2 7 . Bonus 28. (A) (q); (B) (p) or (p, q, r, s, t); (C) (s); (D) (t) 2 9 . (A) (s); (B) (t); (C) (r); (D) (r) 30. D 31. C 32 . B,C,D 33 . C,D 34.C 35. B 64 E
JEE-Mathematics CONTINUITY 1 . CONTINUOUS FUNCTIONS : A function for which a small change in the independent variable causes only a small change and not a sudden jump in the dependent variable are called continuous functions. Naively, we may say that a function is continuous at a fixed point if we can draw the graph of the function around that point without lifting the pen from the plane of the paper. a a a a a continuous discontinuous discontinuous discontinuous discontinuous Continuity of a function at a point : A function f(x) is said to be continuous at x = a, if Lim f(x) f(a) . Symbolically f is continuous at x = a if xa b gLim f(a h) Lim f(a h) f a , h > 0 h0 h0 i.e. (LHL = RHL ) equals value of ‘f’ at x = a. It should be noted that continuity of a function at x = a can x=a x=a be discussed only if the function is defined in the immediate neighbourhood of x = a, not necessarily at x = a. Ex. Continuity at x = 0 for the curve can not be discussed. o Illustration 1 : If f(x) sin x , x 1 then find whether f(x) is continuous or not at x = 1, where [ ] denotes 2 [x] x 1 greatest integer function. f(x) sin x , x 1 2 Solution : [x] , x 1 For continuity at x = 1, we determine, f(1), lim f(x) and lim f(x). x 1 x 1 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\02.CONTINUITY\\CONTINUITY.P65 Now, f(1) = [1] = 1 x lim f(x) = lim sin = sin =1 and lim f(x) = lim [x] =1 x 1 x 1 2 2 x 1 x 1 so f(1) = lim f(x) = lim f(x) x 1 x 1 f(x) is continuous at x = 1 8x 4x 2x 1 , x 0 Define the function at x = 0 if possible, so that f(x) Illustration 2 : Consider f(x ) x2 e x sin x x kn4, x 0 becomes continuous at x = 0. Solution : f(0+) = lim 8h 4h 2h 1 = lim 4h (2h 1) (2h 1) h0 h0 h2 h2 = lim (4h 1) (2h 1) = n4 . n2 h0 h h f(0–) = lim ex sin x x kn4 = kn4 x0 f(x) is continuous at x = 0, f(0+) = f(0–) = f(0) n4.n2 kn4 k = n2 f(0) = (n4)(n2) 28 E
JEE-Mathematics a(1 x sin x) b cos x 5 x0 x 0 x2 Illustration 3 : Let f(x) = 3 x 0 Solution : 1 cx dx 3 x 1 x2 If f is continuous at x = 0, then find out the values of a, b, c and d. Since f(x) is continuous at x = 0, so at x = 0, both left and right limits must exist and both must be equal to 3. Now a (1 x sin x) b cos x 5 (a b 5) a b x2 ... x2 2 Lim = Lim = 3 (By the expansions of sinx and cosx) x2 x 0 x 0 b If lim f(x) exists then a + b + 5 = 0 and –a – = 3 a = –1 and b = – 4 x0 2 1 cx dx3 x cx dx3 1 x2 x2 since lim exists lim =0c=0 x0 x0 Now lim 1 = lim 1 d = ed (1 dx) dx x0 (1 dx) x x 0 So ed = 3 d = n 3, Hence a = – 1, b = – 4, c = 0 and d = n 3. Do yourself -1 : (i) If ƒ( x ) cos x; x 0 find the value of k if ƒ (x) is continuous at x = 0. x k; x 0 | x 2| ; x 2 then discuss the continuity of ƒ (x) at x=–2 (ii) If ƒ( x ) tan 1 (x 2) 2 ; x 2 2. CONTINUITY OF THE FUNCTION IN AN INTERVAL : E ( a ) A function is said to be continuous in (a,b) if f is continuous at each & every point belonging to (a, b). Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\02.CONTINUITY\\CONTINUITY.P65 ( b ) A function is said to be continuous in a closed interval [a,b] if : ( i ) f is continuous in the open interval (a,b) (ii) f is right continuous at ‘a’ i.e. Lim f x f a = a finite quantity (iii) Note : b g b gxa f is left continuous at ‘b’ i.e. Lim f x f b a finite quantity xb (i) Obseve that lim ƒ(x) and lim ƒ(x) do not make sense. As a consequence of this definition, if ƒ(x) is xa xb defined only at one point, it is continuous there, i.e., if the domain of ƒ(x) is a singleton, ƒ(x) is a continuous function. Example : Consider ƒ(x) = a x x a . ƒ(x) is a singleton function defined only at x = a. Hence ƒ(x) is a continuous function. (ii) All polynomials, trigonometrical functions, exponential & logarithmic functions are continuous in their domains. (iii) If ƒ (x) & g(x) are two functions that are continuous at x = c then the function defined by : b g b g b g b g b g b g b g b gF1 x f x g x ; F2 x K f x , where K is any real number ; F3 x f x .g x are also continuous at x = c. b gf x b g b gFurther, if g(c) is not zero, then F4 x is also continuous at x = c. g x 29
JEE-Mathematics (iii) Some continuous functions : Function f(x) Interval in which f(x) is continuous (–) Constant function (–) xn, n is an integer 0 (–) – {0} x–n, n is a positive integer (–) (–) |x – a| (–) – {x : q(x) = 0} p(x) = a xn + a xn – 1 + a xn – 2 + ..... + a 01 2 n (–) (–) – {(2n + 1)/2 : n I} p(x) (–) – {n : n I} q(x) , where p(x) and q(x) are polynomial in x (0) sinx, cosx, ex Points of discontinuity tanx, secx Every Integer cotx, cosecx x , 3 , ..... nx 22 (iv) Some Discontinuous Functions : x 0 , , 2 ,..... Functions x=0 [x], {x} tanx, secx cotx, cosecx 1 11 sin , cos , , e1/x x xx x 1 , x 2 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\02.CONTINUITY\\CONTINUITY.P65 Illustration 4 : Discuss the continuity of f(x) = 2 x 3 , 2 x 0 , 0x3 x2 3 x3 15 , x 3 x 1 , x 2 Solution : We write f(x) as f(x) = 2x 3 , 2 x 0 , 0x3 x2 3 x3 15 , x 3 As we can see, f(x) is defined as a polynomial function in each of intervals (– , –2), (–2,0), (0,3) and (3,). Therefore, it is continuous in each of these four open intervals. Thus we check the continuity at x = –2,0,3. At the point x = –2 lim f(x) = lim (–x – 1) = + 2 – 1 = 1 x2 x 2 lim f(x) = lim (2x + 3) = 2. (–2) + 3 = – 1 x2 x 2 30 E
JEE-Mathematics Therefore, lim f(x) does not exist and hence f(x) is discontinuous at x = –2. x 2 At the point x = 0 lim f(x) = lim (2x + 3) = 3 x0 x0 lim f(x) = lim (x2 + 3) = 3 x0 x0 f(0) = 02 + 3 = 3 Therefore f(x) is continuous at x = 0. At the point x = 3 lim f(x) = lim (x2 + 3) = 32 + 3 = 12 x3 x3 lim f(x) = lim (x3 – 15) = 33 – 15 = 12 x3 x3 f(3) = 33 – 15 = 12 Therefore, f(x) is continuous at x = 3. We find that f(x) is continuous at all points in R except at x = – 2 Do yourself -2 : x2 ; 0 x 1 ; 1 x 2 then find the value of a & b if ƒ (x) is continuous in [0,) a 1 (i) If ƒ( x ) 2 b 2 4b x2 ; 2 x | x 3| ; 0 x 1 ; 1x Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\02.CONTINUITY\\CONTINUITY.P65 (ii) Discuss the continuity of ƒ (x) = sin x 2 in [0,3) ; x 3 log x 2 2 3 . REASONS OF DISCONTINUITY : ( a ) Limit does not exist i.e. Lim f(x) Lim f(x) xa xa ( b ) f(x) is not defined at x = a ( c ) Lim f(x) f(a) 0 1 2 34 xa Geometrically, the graph of the function will exhibit Lim f(x) f(1) a break at x = a, if the function is discontinuous at x = a. The graph as shown is discontinuous at x 1 x = 1 , 2 and 3. Lim f(x) does not exist x 2 f(x) is not defined at x = 3 E 31
JEE-Mathematics 4. TYPES OF DISCONTINUITIES : Type-1 : (Removable type of discontinuities) : - In case Lim f(x) exists but is not equal to f(a) then the xa function is said to have a removable discontinuity or discontinuity of the first kind. In this case we can redefine the b gfunction such that Lim f(x) f a & make it continuous at x = a. Removable type of discontinuity can be further xa classified as: (a) Missing point discontinuity : Where Lim f(x) exists but f(a) is not defined. xa (b) Isolated point discontinuity : Where Lim f(x) exists & f(a) also exists but; Lim f(x) f(a). xa xa x 1 , x 0 Illustration 5 : Examine the function , f(x) = 1 / 4 , x 0 . Discuss the continuity, and if discontinuous remove x2 1 , x 0 the discontinuity by redefining the function (if possible). Solution : Graph of f(x) is shown, from graph it is seen that y lim f(x) = lim f(x) = – 1 , but f(0) = 1/4 x0 x0 Thus, f(x) has removable discontinuity and f(x) could 1/4 1 x be made continuous by taking f(0) = – 1 O –1 x 1 , x 0 f(x) = 1 , x0 x2 1 , x 0 y = f(x) before redefining Do yourself -3 : 1 ; 0x2 ; 2x4 x 1 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\02.CONTINUITY\\CONTINUITY.P65 x2 3 , then discuss the types of discontinuity for the function. ; x4 5 (i) If ƒ( x ) x1/2 ; x 4 14 2 Type-2 : (Non-Removable type of discontinuities) :- In case Lim f(x) does not exist then it is not possible to make the function continuous by redefining it. Such a xa discontinuity is known as non-removable discontinuity or discontinuity of the 2nd kind. Non-removable type of discontinuity can be further classified as : ( i ) Finite type discontinuity : In such type of discontinuity left hand limit and right hand limit at a point exists but are not equal. (ii) Infinite type discontinuity : In such type of discontinuity atleast one of the limit viz. LHL and RHL is tending to infinity. 32 E
JEE-Mathematics (iii) Oscillatory type discontinuity : b ge.g. f x sin at x = 0 x b gf x sin y= sin(/x) x y 1 –1 x 1 –1 f(x) has non removable oscillatory type discontinuity at x = 0 Example : From the adjacent graph note that (i) f is continuous at x = –1 -1 0 12 (ii) f has isolated discontinuity at x = 1 (iii) f has missing point discontinuity at x = 2 (iv) f has non removable (finite type) discontinuity at the origin. Note : In case of non-removable (finite type) discontinuity the non-negative difference between the value of the RHL at x = a & LHL at x = a is called the jump of discontinuity. A function having a finite number of jumps in a given interval I is called a piece wise continuous or sectionally continuous function in this interval. e1/ x 1 ; when x 0 ; when x 0 Illustration 6 : Show that the function, f(x) = e1 / x 1 has non-removable discontinuity at x = 0. 0, Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\02.CONTINUITY\\CONTINUITY.P65 e1/ x 1 ; when x 0 ; when x 0 Solution : We have, f(x) = e1 / x 1 0, 1 1 1 e1/h lim f(x) = lim f(0 + h) = lim eh 1 = lim 1 = 1 [ e1/h ] x 0 h0 h0 1 h0 eh 1 1 e1/ h lim f(x) = lim e 1 / h 1 = 0 1 =–1 [ h 0 ; e–1/h 0] e 1 / h 1 0 1 x0 h0 lim f(x) = – 1 x0 lim f(x) lim f(x). Thus f(x) has non-removable discontinuity. x0 x0 co s 1 {co t x } x 2 Illustration 7 : f(x) = ; find jump of discontinuity, where [ ] denotes greatest integer & x [x] 1 2 { } denotes fractional part function. E 33
JEE-Mathematics c o s 1 {cot x } x 2 Solution : f(x) = x [x] 1 2 cos–1 {cot x} = cot h lim f(x) = lim 2 x x lim co s 1 lim co s 1 tanh 2 h 0 h 0 22 = lim f(x) lim [x] 1 lim h 1 1 2 x x h 0 22 jump of discontinuity = – 1 – 2 = 2 – 1 Do yourself -4 : 1 ; x 1 (i) Discuss the type of discontinuity for ƒ( x ) | x| ; 1 x 1 ; x 1 (x 1) 5 . THE INTERMEDIATE VALUE THEOREM : y f(b) Suppose f(x) is continuous on an interval I, and a and b are any two points of I. Then if y0 y is a number between f(a) and f(b), there f(a) 0 exists a number c between a and b such that f(c) = y x 0 0 a cb The function f, being continuous on [a,b] takes on every value between f(a) and f(b) Note that a function f which is continuous in [a,b] possesses the following properties : (i) If f(a) & f(b) posses opposite signs, then there exists atleast one root of the equation Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\02.CONTINUITY\\CONTINUITY.P65 f(x) = 0 in the open interval (a,b). (ii) If K is any real number between f(a) & f(b), then there exists atleast one root of the equation f(x) = K in the open interval (a,b). Note : In above cases the number of roots is always odd. Illustration 8 : Show that the function, f(x) = (x – a)2(x – b)2 + x, takes the value a b for some x0 (a, b) 2 Solution : f(x) = (x – a)2(x – b)2 + x f(a) = a f(b) = b ab & (f(a), f(b)) 2 By intermediate value theorem, there is atleast one x0 (a, b) such that f(x0) =ab . 2 Illustr ation 9 : Let f : [0, 1] onto [0, 1] be a continuous function, then prove that f(x) = x for atleast one x [0, 1] 34 E
Solution : Consider g(x) = f(x) – x JEE-Mathematics g(0) = f(0) – 0 = f(0) 0 0 f(x) 1 g(1) = f(1) – 1 0 g(0) . g(1) 0 g(x) = 0 has atleast one root in [0, 1] f(x) = x for atleast one x [0, 1] Do yourself -5 : 2ƒ(a ) 3ƒ(b) ( i ) If ƒ (x) is continuous in [a,b] such that ƒ (c) = 5 , then prove that c (a,b) 6 . SOME IMPORTANT POINTS : ( a ) If f(x) is continuous & g(x) is discontinuous at x = a then the product function (x) f(x).g(x) will not necessarily be discontinuous at x = a, e.g. Lsin x 0 b g b g Mf x x & g x x NM0 x 0 f(x) is continuous at x = 0 & g(x) is discontinuous at x = 0, but f(x).g(x) is continuous at x = 0. ( b ) If f (x) and g (x) both are discontinuous at x = a then the product function (x) f(x).g(x) is not necessarily be discontinuous at x = a , e.g. L1 x 0 b g b g Mf x g x N1 x 0 f(x) & g(x) both are discontinuous at x = 0 but the product function f.g(x) is still continuous at x = 0 ( c ) If f (x) and g (x) both are discontinuous at x = a then f(x) ± g(x) is not necessarily be discontinuous at x=a ( d ) A continuous function whose domain is closed must have a range also in closed interval. ( e ) If f is continuous at x = a & g is continuous at x = f (a) then the composite g[f(x)] is continuous at x = a. eg. x sin x b g b g b gf x x2 2 & g x x x sin x will also be are continuous at x =0, hence the composite (gof) x x2 2 continuous at x = 0 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\02.CONTINUITY\\CONTINUITY.P65 Illustration 10 : If f(x) = x 1 and g(x) 1 , then discuss the continuity of f(x), g(x) and fog (x) in R. x 1 x 2 Solution : x 1 f(x) = E x 1 f(x) is a rational function it must be continuous in its domain and f is not defined at x = 1. f is discontinuous at x = 1 1 g(x) = x 2 g(x) is also a rational function. It must be continuous in its domain and g is not defined at x = 2. g is discontinuous at x = 2 Now fog(x) will be discontinuous at x = 2 (point of discontinuity of g(x)) Consider g(x) = 1 (when g(x) = point of discontinuity of f(x)) 1 =1 x= 3 x 2 fog(x) is discontinuous at x = 2 & x = 3. 35
JEE-Mathematics Do yourself -6 : ( i ) Let ƒ (x) = [x] & g(x) = sgn(x) (where [.] denotes greatest integer function) , then discuss the continuity of ƒ( x ) ƒ (x) ± g(x), ƒ (x).g(x) & g(x) at x= 0. ƒ( x ) ( i i ) If ƒ (x) = sin|x|& g(x) = tan|x| then discuss the continuity of ƒ (x) ± g(x) ; g(x) & ƒ (x) g(x) 7 . SINGLE POINT CONTINUITY : Functions which are continuous only at one point are said to exhibit single point continuity Illustration 11: If ƒ x x if x Q Solution : x , find the points where ƒ(x) is continuous if x Q Let x = a be the point at which ƒ(x) is continuous. lim ƒ(x) lim ƒ(x) xa x a through rational through irrational a = –a a = 0 function is continuous at x = 0. Do yourself -7 : (i) If g x x if x Q 0 if x Q , then find the points where function is continuous. (ii) x2 ; xQ If ƒ(x) 1 x2 ; x Q , then find the points where function is continuous. ANSWERS FOR DO YOURSELF Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\02.CONTINUITY\\CONTINUITY.P65 1. (i) 1 ( i i ) discontinuous at x =–2 2 . (i) a=–1 & b=1 ( i i ) Discontinuous at x = 1 & continuous at x = 2 3 . ( i ) Missing point removable discontinuity at x = 1, isolated point removable discontinuity at x = 4 . 4 . ( i ) Finite type non-removable discontinuity at x=–1,1 6 . ( i ) All are discontinuous at x = 0. ( i i ) ƒ (x) g(x) & ƒ (x) ± g(x) are discontinuous at x (2n 1) ; n I 2 ƒ(x) n g(x) is discontinuous at x = 2 ; n I 7. (i) x = 0 (ii) x 1 2 36 E
EXERCISE - 01 JEE-Mathematics CHECK YOUR GRASP SELECT THE CORRECT ALTERNATIVE (ONLY ONE CORRECT ANSWER) x 2 , when x 1 1 . If f x 4 x 1 , when 1 x 3 , then correct statement is - x2 5 , when x 3 (A) lim f x lim f x (B) f(x) is continuous at x = 3 x 1 x3 (D) f(x) is continuous at x = 1 and 3 (C) f(x) is continuous at x = 1 1 , x 0 , then - 2. If f x e1 / x 1 0 , x 0 (A) lim f x 1 (B) lim f x 0 x0 x0 (C) f(x) is discontinuous at x = 0 (D) f(x) is continuous 3 . If function f(x) = 1x 31x , is continuous function, then f(0) is equal to - x (A) 2 (B) 1/4 (C) 1/6 (D) 1/3 x2 a 2 x 2a , x 2 is continuous at x = 2, then a is equal to - 4. If f x x 2 2 , x 2 (A) 0 (B) 1 (C) –1 (D) 2 log(1 2ax) log(1 bx) , x 0 , is continuous at x = 0 , then k is equal to - 5. If f(x) = x k , x0 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\02.CONTINUITY\\CONTINUITY.P65 (A) 2a + b (B) 2a – b (C) b – 2a (D) a + b 6. If f(x) [x] [x], x 2 f is continuous at x = 2 then is (where [.] denotes greatest integer) - x 2 , , (A) –1 (B) 0 (C) 1 (D) 2 1 cos 4x , x0 x2 7 . If f(x) = a , x 0 , then correct statement is - x , x0 16 x 4 (A) f(x) is discontinuous at x = 0 for any value of a (B) f(x) is continuous at x = 0 when a = 8 (C) f(x) is continuous at x = 0 when a = 0 (D) none of these E 37
JEE-Mathematics 1 8 . Function f(x) = log| x| is discontinuous at - (A) one point (B) two points (C) three points (D) infinite number of points 9 . Which of the following functions has finite number of points of discontinuity in R (where [.] denotes greatest integer) (A) tan x (B) |x| / x (C) x + [x] (D) sin [ x] 10. If f(x) = 1 tan x , x , x 0, is a continuous functions, then f(/4) is equal to - 4x 4 2 (A) –1/2 (B) 1/2 (C) 1 (D) –1 1 1 . The value of f(0), so that function, f(x) = a2 ax x2 a2 ax x2 becomes continuous for all x, is given ax ax by - (A) a a (B) – a (C) a (D) –a a 12. If f(x) x ex cos2x , x 0 is continuous at x = 0, then - x2 5 (B) [f(0)] = –2 (C) {f(0)} = –0.5 (D) [f(0)].{f(0)}= –1.5 (A) f(0) = 2 where [x] and {x} denotes greatest integer and fractional part function. 1 3 . Let f(x) = x(1 a cos x) b sin x , x 0 and f(0) = 1. The value of a and b so that f is a continuous function are - x3 (A) 5/2, 3/2 (B) 5/2, –3/2 (C) –5/2, –3/2 (D) none of these b g1 4 . ‘f’ is a continuous function on the real line. Given that x2 (f x 2)x 3 .f(x) 2 3 3 0 . Then the value of e jf 3 is - 2( 3 2) (B) 2 (1 3 ) (C) zero (D) cannot be determined Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\02.CONTINUITY\\CONTINUITY.P65 (A) 3 SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS) 1 5 . The value(s) of x for which ƒ (x) = e sin x is continuous, is (are) - 4 x2 9 (A) 3 (B) –3 (C) 5 (D) all x (–, –3] [3, ) 1 6 . Which of the following function(s) not defined at x = 0 has/have removable discontinuity at the origin ? (A) f(x) 1 1 x sin x 2cot (B) f(x) cos x (C) f(x) x sin 1 x (D) f(x) = n x 38 E
JEE-Mathematics 1 7 . Function whose jump (non-negative difference of LHL & RHL) of discontinuity is greater than or equal to one, is/are - (e1 / x 1) ; x 0 x1 / 3 1 ; x 1 x 0 ; 1 x 1 (e1 / x 1) x1 / 2 1 2 nx (A) ƒ (x) (B) g(x) = (1 cos x) ; (x 1) x sin 1 2x ; x 0, 1 2 (C) u(x) = tan 1 3 x (D) v(x) = log3 (x 2) ; x 2 | sin x| ; x0 (x2 5) ; x 2 x log1 / 2 18. If 1 ƒ 2 is discontinuous at x = ƒ (x) x2 17x 66 , then x 2 (A) 2 7 24 (D) 6,11 (B) 3 (C) 11 0; x Z 1 9 . Let ƒ (x) = [x] & g(x) x2 ; x R Z , then (where [.] denotes greatest integer function) - (A) Lim g(x) exists, but g(x) is not continuous at x= 1. x 1 (B) Lim f(x) does not exist and ƒ (x) is not continuous at x=1. x 1 (C) gof is continuous for all x. (D) fog is continuous for all x. Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\02.CONTINUITY\\CONTINUITY.P65 CHECK YOUR GRASP ANSWER KEY EXERCISE-1 Que. 1 2 3 4 5 6 7 8 9 10 Ans. C C C A A A B C B A Que. 11 12 13 14 15 16 18 18 19 Ans. B D C B A,B B,C,D A,C,D A,B,C A,B,C E 39
JEE-Mathematics BRAIN TEASERS EXERCISE - 02 SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS) x if x 0 1 . Consider the piecewise defined function f(x) 0 if 0 x 4 choose the answer which best describes the continuity of this function - x 4 if x 4 (A) the function is unbounded and therefore cannot be continuous (B) the function is right continuous at x = 0 (C) the function has a removable discontinuity at 0 and 4, but is continuous on the rest of the real line (D) the function is continuous on the entire real line 2 . f(x) is continuous at x=0, then which of the following are always true ? (A) Lim f(x) 0 (B) f(x) is non continuous at x=1 x0 (C) g(x) = x2f(x) is continuous at x = 0 (D) Lim (f(x) f(0)) 0 x 0 b g3 . Indicate all correct alternatives if, f x x 1 , then on the interval [0,] 2 b g1 b g(B) tan (f(x)) & 1 are both discontinuous fx (A) tan (f (x)) & are both continuous fx (C) tan (f (x))& f –1(x) are both continuous b g1 (D) tan (f(x)) is continuous but f x is not 4 . If f(x) = sgn(cos2x – 2 sinx + 3), where sgn ( ) is the signum function, then f(x) - (A) is continuous over its domain (B) has a missing point discontinuity (C) has isolated point discontinuity (D) has irremovable discontinuity. 5. 2 cos x sin 2x ; ecos x 1 f(x) ( 2x)2 g(x) 8x 4 h(x) = f(x) for x</2 = g(x) for x>/2 then which of the followings does not holds ? (A) h is continuous at x = /2 (B) h has an irremovable discontinuity at x=/2 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\02.CONTINUITY\\CONTINUITY.P65 (C) h has a removable discontinuity at x = /2 (D) f 2 g 2 6 . The number of points where f(x) = [sinx + cosx] (where [ ] denotes the greatest integer function), x (0, 2) is not continuous is - (A) 3 (B) 4 (C) 5 (D) 6 ( x |1x| 1 x 7. On the interval I= [–2, 2], the function f(x) = 1)e (x 0) 0 (x 0) then which one of the following hold good ? (A) is continuous for all values of x I (B) is continuous for x I –(0) (C) assumes all intermediate values from f(–2) & f(2) (D) has a maximum value equal to 3/e 8. cos x 1 ; where [x] is the greatest integer function of x, then f(x) is continuous at - If f(x) = cos x 2 (A) x = 0 (B) x = 1 (C) x = 2 (D) none of these 40 E
JEE-Mathematics 9. 3 co t 1 2x3 3 for x 0 x2 for where { } & [ ] denotes the fractional part and the integral part Given f(x) x 0 x2 cos e1 / x functions respectively, then which of the following statement does not hold good - (A) f (0– ) = 0 (B) f(0+)=3 (C) f(0)=0 continuity of f at x = 0 (D) irremovable discontinuity of f at x = 0 10. Let ‘f’ be a continuous function on R. If f (1 / 4n ) (sin en )en2 n2 then f(0) is - n2 1 (A) not unique (B) 1 (C) data sufficient to find f(0) (D) data insufficient to find f(0) 1 1 . Given f(x) = b ([x]2 + [x]) + 1 for x 1 = sin ((x a)) for x < – 1 where [x] denotes the integral part of x, then for what values of a, b the function is continuous at x = – 1 ? (A) a 2n (3 / 2); b R ; n I (B) a 4n 2 ; b R ; n I (C) a 4n (3 / 2) ; b R ; n I (D) a 4n 1 ; b R ; n I x[x]2log(1+x) 2 for –1 < x < 0 1 2 . Consider f(x) = e jln ex2 2 {x} where [*] & {*} are the greatest integer function & tan x for 0 < x < 1 fractional part function respectively, then - (B) f(0) = 2 f is continuous at x = 0 (A) f(0) = ln2 f is continuous at x = 0 (D) f has an irremovable discontinuity at x = 0 (C) f(0) = e2 f is continuous at x = 0 13. Let f(x) a sin2n x for x 0 and n then - Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\02.CONTINUITY\\CONTINUITY.P65 b cos2m x 1 for x 0 and m (A) f(0–) f(0+) (B) f(0+) f(0) (C) f(0–) f(0) (D) f is continuous at x = 0 b g1 4 . f x Lim xn sin xn for x 0, x 1 f(1)=0 Consider xn sin xn then - n (A) f is continuous at x = 1 (B) f has a finite discontinuity at x = 1 (C) f has an infinite or oscillatory discontinuity at x = 1 (D) f has a removable type of discontinuity at x=1 BRAIN TEASERS ANSWER KEY EXERCISE-2 Que. 1 2 3 45 6 7 8 9 10 Ans. D C,D C,D C A,C,D C B,C,D B,C B,D B,C Que. 11 12 13 14 Ans. A,C B D A E 41
JEE-Mathematics EXERCISE - 03 MISCELLANEOUS TYPE QUESTIONS TRUE / FALSE 1 1 . is discontinuous at infinite points. ([ ] denotes greatest integer function) x [x] 2 . sin|x| + |sinx| is not continuous for all x. 3 . If f is continuous and g is discontinuous at x = a, then f(x).g(x) is discontinuous at x = a. 4 . There exists a continuous onto function f : [0, 1] [0, 10], but there exists no continuous onto function g : [0, 1] (0, 10) tan( / 4 x) 5 . If f(x) = for x , then the value which can be given to f(x) at x = so that the function cos 2 x 44 becomes continuous every where in (0, /2) is 1/4. 6 . The function f, defined by f(x) = 1 is continuous for real x. 1 2tan x 7. f(x) = lim 1 is continuous at x = 1. n 1 n sin2 x 8. If f(x) is continuous in [0, 1] and f(x) = 1 for all rational numbers in [0, 1] then f 1 = 1. 2 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 OR MORE statement(s) in Column-II. 1 . Column-I Column-II sin{x}; x 1 (p) 1 (A) If f(x) = cos x a ; x 1 where {.} denotes the fractional part function, such that f(x) is continuous at x = 1. If |k| = a 2 sin (4 ) 4 then k is Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\02.CONTINUITY\\CONTINUITY.P65 (1 cos(sin x)) (q) 0 (B) If the function f(x) = x2 is continuous at x = 0, then f(0) is x , xQ (r) –1 (C) f(x) = 1 x , x Q , then the values 1 of x at which f(x) is continuous (s) 2 (D) If f(x) = x + {–x} + [x], where [x] and {x} represents integral and fractional part E of x, then the values of x at which f(x) is discontinuous 42
2 . Column-I JEE-Mathematics Column-II (A) If f(x) = 1/(1–x), then the points at which 1 (p) 2 the function fofof(x) is discontinuous (q) 0 11 (B) f(u) = u2 u 2 , where u= x 1 . The values of x at which 'f' is discontinuous (C) f(x) = u2, where u x 1, x 0 (r) 2 = x 1, x 0 The number of values of x at which (s) 1 'f' is discontinuous (D) The number of value of x at which the 2x5 8x2 11 function f(x) = x4 4x3 8x2 8x 4 is discontinuous ASSERTION & REASON These questions contains, 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 : f(x) = sinx + [x] is discontinuous at x = 0 Because Statement-II : If g(x) is continuous & h(x) is discontinuous at x = a, then g(x) + h(x) will necessarily be discontinuous at x = a (A) A (B) B (C) C (D) D 2 sin(a cos1 x) if x (0,1) Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\02.CONTINUITY\\CONTINUITY.P65 2 . Consider ƒ (x) = 3 if x 0 ax b if x 0 Statement-I : If b = 2 then ƒ (x) is continuous in (–, 1) 3 and a = 3 Because Statement-II : If a function is defined on an interval I and limit exist at every point of interval I then function is continuous in I. (A) A (B) B (C) C (D) D cos x ex2 / 2 x3 3. Let ƒ( x ) , x 0 then 0 , x 0 Statement-I : ƒ (x) is continuous at x = 0. Because Statement-II : lim cos x ex2 / 2 1 . xx0 4 12 (A) A (B) B (C) C (D) D E 43
JEE-Mathematics x3 1 4 . Statement-I : The equation sin x 3 2 has atleast one solution in [–2, 2] 43 Because Statement-II : If f:[a, b] R be a function & let 'c' be a number such that f(a) < c < f(b), then there is atleast one number n (a, b) such that f(n) = c. (A) A (B) B (C) C (D) D 5. Statement-I : Range of ƒ (x) = e2x e 2 x x2 x4 is not R. x e2x e 2 x Because Statement-II : Range of a continuous even function can not be R. (A) A (B) B (C) C (D) D (D) D Ax B x 1 6 . Let ƒ (x) = 2x2 3Ax B x (1, 1] 4 x 1 31 Statement-I : ƒ (x) is continuous at all x if A = 4 , B = – 4 . Because Statement-II : Polynomial function is always continuous. (A) A (B) B (C) C COMPREHENSION BASED QUESTIONS Comprehension # 1 If x x2 x2n and x > 1 Sn (x) x 1 (x 1)(x2 1) ......... (x 1)(x2 1).....(x2n 1) lim S n (x) n Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\02.CONTINUITY\\CONTINUITY.P65 ax b 1 , x x 0 g(x) 1 , x 0 h : R R h(x) = x9 – 6x8 – 2x7 + 12x6 + x4 – 7x3 + 6x2 + x – 7 On the basis of above information, answer the following questions : 1 . If g(x) is continuous at x = 0 then a + b is equal to - (A) 0 (B) 1 (C) 2 (D) 3 2 . If g(x) is continuous at x = 0 then g'(0) is equal to - (A) h(6) (C) a – 2b (D) does not exist (B) 2 3 . Identify the incorrect option - (A) h(x) is surjective (B) domain of g(x) is [–1/2, ) (D) = 1 (C) h(x) is bounded 44 E
JEE-Mathematics Comprehension # 2 Road B N WE A man leaves his home early in the morning to have a walk. He arrives at a junction of road A & road B as shown in S figure. He takes the following steps in later journey : (a) 1 km in north direction (b) changes direction & moves in north-east direction for 2 2 kms. Road A (c) changes direction & moves southwards for distance of 2 km. Home (d) finally he changes the direction & moves in south-east direction to reach road A again. Visible/Invisible path :- The path traced by the man in the direction parallel to road A & road B is called invisible path, the remaining path traced is visible. Visible points :- The points about which the man changes direction are called visible points except the point from where he changes direction last time Now if road A & road B are taken as x-axis & y-axis then visible path & visible point represents the graph of y = f(x). On the basis of above information, answer the following questions : 1 . The value of x at which the function is discontinuous - (A) 2 (B) 0 (C) 1 (D) 3 2 . The value of x at which fof(x) is discontinuous - (A) 0 (B) 1 (C) 2 (D) 3 3 . If f(x) is periodic with period 3, then f(19) is - (A) 2 (B) 3 (C) 19 (D) none of these Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\02.CONTINUITY\\CONTINUITY.P65 MISCELLANEOUS TYPE QUESTION ANSWER KEY EXERCISE -3 True / False 5. F 6. F 7. F 8. T 1. T 2. F 3. F 4. T Match the Column 1. (A) (p, r); (B) (s); (C) (s); (D) (p, q, r) 2. (A) (q, s); (B) (p, r, s); (C) (q); (D) (q) Assertion & Reason 1. A 2. C 3. A 4. C 5. A 6. B Comprehension Based Questions C omp r eh ens i on # 1 : 1 . D 2 . B 3 . C C omp r eh ens i on # 2 : 1 . A 2 . B,C 3. A E 45
JEE-Mathematics CONCEPTUAL SUBJECTIVE EXERCISE EXERCISE - 4 [A] x2, when x 0 5 x 4, when 0 1. If f(x) x 1 , discuss the continuity of f(x) in R. 4 x 2 3x, when 1 x 2 3x 4, when x 2 2 sin x for x 2 x . If f is continuous on , then find the values of a & b. 22 2. Let f x a sin x b for for x 2 cos x sin a 1 x sin x for x 0 x 3 . Determine the values of a,b & c for which the function f x c for x 0 1/2 x1 / 2 x bx2 bx 3 / 2 for x 0 is continuous at x = 0 21 / x 1 4 . Determine the kind of discontinuity of the function y 21 / x 1 at the point x = 0 b g5 . fx , x3 Suppose that f x x3 3x2 4x 12 and h x x 3 then K x 3 (a) find all zeros of ‘f’ (b) find the value of K that makes ‘h’ continuous at x =3 (c) using the value of K found in (b) determine whether ‘h’ is an even function. b g6 . Draw the graph of the function f x x x x2 , 1 x 1 & discuss the continuity or discontinuity of f in the interval 1 x 1 . Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\02.CONTINUITY\\CONTINUITY.P65 b g b g7 . If f x sin 3x A sin 2x B sin x x 0 is continuous at x = 0, then find A & B. Also find f(0). x5 8 . (a) Let f(x + y) = f(x) + f(y) for all x, y & if the function f(x) is continuous at x = 0, then show that f(x) is continuous at all x. (b) If f(x . y) = f(x) . f(y) for all x, y and f(x) is continuous at x = 1. Prove that f(x) is continuous for all x except at x = 0. Given f(1) 0. 9 . Examine the continuity at x= 0 of the sum function of the infinite series : b gb g b gb gx x x .......... x 1 x 1 2x 1 2x 1 3x 1 1 0 . Show that : (a) a polynomial of an odd degree has at least one real root (b) a polynomial of an even degree has at least two real roots if it attains at least one value opposite in sign to the coefficient of its highest-degree term. CONCEPTUAL SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(A) 1 . continuous every where except at x = 0 2 . a = –1 b = 1 E 3 . a = –3/2, b 0 , c = 1/2 4 . non-removable - finite type 5 . (a) –2, 2, 3 (b) K= 5 (c) even 6 . f is continuous in –1 x 1 7 . A = –4, B = 5, f( 0) = 1 9 . discontinuous at x = 0 46
EXERCISE - 4 [B] JEE-Mathematics BRAIN STORMING SUBJECTIVE EXERCISE b g FHG JIK HFG KIJn x x 1. Given f x tan sec ; r,n N 2r 2r 1 r 1 ln f x tan x f x tan x n . sin tan x 2n 2n 2 ; x /4 g x Lnim x n 2n 1 f x tan K ; x / 4 GF JIwhere [ ] denotes the greatest integer function and the domain of g(x) is 0, . Find the value of k, if possible, H K2 b gso that g(x) is continuous at x / 4 . Also state the points of discontinuity of g(x) in 0, / 4 , if any. 2. Let f(x)= 1 x3 , x0 g(x) ( x 1)1 / 3 , x0 Discuss the continuity of g(f(x)). 3. ; 1)1 / 2 , x 0 4. x 2 1, ( x 5. x 0 6. L 4x 5 [x] for x 1 7. b g MDiscuss the continuity of ‘f’ in [0,2] where f x ; where [x] is the greatest integer not E N[cos x] for x 1 greater than x. Also draw the graph b g b gln 2 x x2n sin x Discuss the continuity of the function f x Lim 1 x2n at x = 1 n 1 a x xa xn a for x 0 where a > 0. Consider the function g(x)= axx2 for x 0 2x ax xn2 xn a 1 x2 Find the value of ‘a’ & ‘g(0)’ so that the function g(x) is continuous at x = 0. sin1 1 {x}2 .sin1 1 {x} Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\02.CONTINUITY\\CONTINUITY.P65 2 {x} {x}3 2 for x 0 where {x} is the fractional part of x. Let f(x) = for x 0 2 Consider another function g(x); such that f(x) for x 0 g(x) = 2 2 f(x) for x < 0 Discuss the continuity of the functions f(x) & g(x) at x = 0. f(x) asin x atan x for x 0 tan x sin x = n(1 x x2 ) n(1 x x2 ) for x < 0, if ‘f’ is continuous at x = 0, find ‘a’ sec x cos x now if g(x)= n 2 x .cot(x – a) for x a, a 0, a > 0. If ‘g’ is continuous at x = a then show that g(e–1) = –e a 47
JEE-Mathematics 8 . Let [x] denote the greatest integer function & f(x) be defined in a neighbourhood of 2 by [x1] , x2 exp x 2 n4 4 16 f(x) 4x 16 1 cos(x 2) , x 2 A (x 2) tan(x 2) Find the value of A & f(2) in order that f(x) may be continuous at x = 2. 9 . If g : [a, b] onto [a, b] is continuous show that there is some c [a, b] such that g(c) = c. b g e j e j10. x2 x2 x2 x2 b g b gandyx yn x . Discuss the continuity of Let yn x 1 x2 1 x2 2 ......... 1 x2 n1 Lim n b g b gy n x n 1,2,3.....n and y (x) at x = 0 BRAIN STORMING SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(B) Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\02.CONTINUITY\\CONTINUITY.P65 b g M1 . k 0 ; g x LM b gln tan x if 0 x 4 . Hence g(x) is continuous everywhere. M0 if x MN 4 2 2 . gof is discontinuous at x = 0, 1 and –1 1 3. the function ‘f’ is continuous everywhere in [0,2] except for x = 0, ,1 & 2 2 4 . discontinuous at x = 1 1 n22 5 . a = , g(0)= 28 6. f(0+) = ; f(0–) = ‘f’ is dicontinuous at x = 0 ; g(0+) = g(0–) = g(0) = ‘g’ is continuous at x = 0 24 2 2 7 . a = e–1 8 . A = 1 ; f(2) = 1/2 1 0 . y (x) is continuous at x = 0 for all n and y (x) is discontinuous at x = 0 E n 48
EXERCISE - 05 [A] JEE-Mathematics JEE-[MAIN] : PREVIOUS YEAR QUESTIONS 1. x x Q If f(x) = x , then f is continuous at- x Q [AIEEE 2002] (1) Only at zero (2) only at 0, 1 (3) all real numbers (4) all rational numbers x |1x| 1 x 2. If f(x) = e , x 0 then f(x) is- [AIEEE 2003] 0 , x 0 (1) discontinuous everywhere (2) continuous as well as differentiable for all x (3) continuous for all x but not differentiable at x=0 (4) neither differentiable nor continuous at x = 0 3. Let f(x) = 1 tan x x 0, , If f(x) is continuous in 0, , then f is- [AIEEE 2004] ,x 4 , 2 2 4 4x (1) 1 (2) 1/2 (3) –1/2 (4) –1 12 4 . The function f : R/{0} R given by f(x) = x – e2x 1 can be made continuous at x = 0 by defining f(0) as- [AIEEE 2007] (1) 2 (2) –1 (3) 0 (4) 1 sin(p 1)x sin x , x0 x q , x0 5 . The values of p and q for which the function f(x) = is continuous for all x in R, x x2 x , x0 3 x 2 are:- [AIEEE 2011] 31 13 13 51 (1) p = – , q = (2) p = , q = (3) p = , q = – (4) p = , q = 22 22 22 22 6. Define F(x) as the product of two real functions f1(x) = x, x IR, and f2 (x) sin 1 , if x 0 as follows: x 0, if x 0 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\02.CONTINUITY\\CONTINUITY.P65 F (x) f1 ( x ).f2 (x ) if x 0 0, if x 0 [AIEEE 2011] Statement-1 : F(x) is continuous on IR. Statement-2 : f1(x) and f2(x) are continuous on IR. (1) Statemen-1 is false, statement-2 is true. (2) Statemen-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 7 . Consider the function, f(x) = |x – 2| + |x – 5|, x R. Statement–1 : f'(4) = 0. Statement–2 : f is continuous in [2, 5], differentiable in (2, 5) and f(2) = f(5). [AIEEE 2012] (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. PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE-5 [A] Que. 1 2345 6 7 Ans 1 3341 4 4 E 49
JEE-Mathematics JEE-[ADVANCED] : PREVIOUS YEAR QUESTIONS EXERCISE - 05 [B] e1 /( x 1) 2 , x 1 1. Discuss the continuity of the function ƒ( x ) e1 /( x 1 ) 2 at x = 1. 1, x 1 [REE 2001 (Mains), 3] 2. For every integer n, let a and b be real numbers. Let function ƒ : IR IR be given by n n ƒ( x ) an sin x, for x 2n,2n 1 for bn cos x, x 2n 1,2n , for all integers n. If ƒ is continuous, then which of the following holds(s) for all n ? [JEE 2012, 4] (D) a – b = –1 (A) a – b = 0 (B) a – b = 1 (C) a – b = 1 n–1 n–1 nn n n+1 n–1 n PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE-5 [B] Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\02.CONTINUITY\\CONTINUITY.P65 1 . Discontinuous at x = 1 ; f(1+) = 1 and f(1–) = –1 2 . B,D E 50
JEE-Mathematics DEFINITE INTEGR ATION b A definite integral is denoted by f(x)dx which represent the algebraic area bounded by the curve y = f(x), the a ordinates x = a, x = b and the x axis. 1 . THE FUNDA MENTAL THEOREM OF CALCULUS : The Fundamental Theorem of Calculus is appropriately named because it establishes a connection between the two branches of calculus : differential calculus and integral calculus. Differential calculus arose from the tangent problem, whereas integral calculus arose from a seemingly unrelated problem, the area problem. Newton's teacher at Cambridge, Isaac Barrow (1630-1677), discovered that these two problems are actually closely related. In fact, he realized that differentiation and integration Y are inverse processes. The Fundamental Theorem of Calculus gives the precise inverse relationship between the derivative y = f(t) and the integral. It was Newton and Leibnitz who exploited this relationship and used it to develop calculus into a area = g(x) systematic mathematical method. In particular, they saw that the Fundamental Theorem enabled them to compute areas and integrals very easily without having to compute them as Oa x b t limits of sums. The Fundamental Theorem of Calculus, Part 1 If f is continuous on [a, b], then the function g defined by x axb g(x) f(t)dt a is continuous on [a, b] and differentiable on (a, b), and g'(x) = f(x). NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\02-Definite integration.p65 The Fundamental Theorem of Calculus, Part 2 If f is continuous on [a, b], then b f(x)dx F(b) F(a) a where F is any antiderivative of f, that is, a function such that F '= f. 3. b E Note : If f(x)dx 0 then the equation f(x) = 0 has atleast one root lying in (a, b) provided f is a continuous a function in (a,b). PROPERTIES OF DEFINITE INTEGRAL : bb ( a ) f(x)dx f(t) dt provided f is same aa ba (b ) f(x)dx f(x) dx ab b cb ( c ) f(x)dx f(x) dx f(x) dx , where c may lie inside or outside the interval [a,b ]. This property is to be a ac used when f is piecewise continuous in (a, b). 31
JEE-Mathematics 3 then evaluate f(x)dx x 2 , If f(x) = Illustration 1 : 0 x 2 Solution : 3x 4, 2 x 3 0 Illustration 2 : 32 3 23 f(x)dx f(x)dx f(x)dx x2dx (3x 4)dx 00 2 02 x3 2 3x2 3 8 27 12 6 8 37 / 6 Ans. 3 0 2 4 x 2 3 2 3[x ] 5| x|, x 0 2 x If f(x)dx is equal to ([.] denotes the greatest integer function) f(x) = then 2, x 0 3 / 2 11 7 (C) –6 17 (A) – (B) – (D) 2 2 2 Solution : 3[x] – 5 x 3[x] 5 , if x > 0 y x 2 = 3[x] + 5, if x < 0 1 1 2 1 01 2 –2 –3/2 2 x –2 f(x)dx = (1)dx (2)dx (5)dx (2)dx 3 / 2 3 / 2 1 0 1 = – 1 1 3 2(1) 1( 5 ) (2) = 1 2 5 2 11 –5 Ans. (A) 2 2 2 Illustration 3 : 2 The value of (x[x2 ] [x2 ]x )dx , where [.] denotes the greatest integer function, is equal to - 1 5 3 (2 3 2 2 ) 1 (9 3 3 ) (A) 4 log 3 5 3 2 1 (2 3 2 2 ) 1 (9 3 3 ) (B) 4 3 log 2 log 3 5 2 1 (2 3 2 2 ) 1 (9 3 3 ) (C) 4 3 log 2 log 3 (D) none of these 2 23 2 Solution : We have, I = (x[x2 ] [x2 ]x )dx = (x 1)dx (x2 2x )dx (x3 3x )dx 1 12 3 x2 2 x3 2x 3 x4 3x 2 NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\02-Definite integration.p65 = 2 x1 3 log 2 2 4 log 3 3 5 3 2 1 (2 3 2 2 ) 1 (32 3 3 ) Ans. (B) = 4 3 log 2 log 3 Ans. 20 Here [.] is the greatest integer function. E Illustration 4 : Evaluate : [cot1 x]dx . 10 Solution : 20 I = [cot1 x]dx , we know cot–1 x (0, ) x R 10 3, x (, cot 3) x (cot 3, cot 2) Thus [cot–1 x] = 2, x (cot 2, cot1) 1, x (cot1, ) 0 cot 3 cot 2 cot1 20 Hence I = 3dx 2dx 1dx 0dx = 30 + cot1 + cot2 + cot3 10 cot 3 cot 2 cot1 32
JEE-Mathematics Do yourself -1 : Evaluate : 3 ( i ) | x2 x 2| dx 0 4 ( i i ) {x}dx , where {.} denotes fractional part of x. 0 /2 (iii) | sin x cos x|dx 0 2 0 x 1 2 x , where [.] denotes the greatest integer function. Evaluate ( i v ) ƒ( x ) dx If ƒ( x ) [x] 1 x 3 0 a a 0 ; if f(x) is an odd function a ( d ) f(x) a dx [f(x) f ( x )]dx 2 f(x)dx ; if f(x) is an even function 0 0 Illustration 5 : 1/2 n 1 x Evaluate 1 / 2 cos x 1 x dx Solution : f(–x) = cos(–x) n 1 x = –cos n 1 x = –f(x) 1 x 1 x f(x) is odd Hence, the value of the given integral = 0. Ans. cos x e x2 2x cos2 x / 2 /2 If f(x) = x2 sec x Illustration 6 : sin x x3 , then the value of (x2 1)(f(x) f ''(x))dx 1 2 x tan x / 2 (A) 1 (B) –1 (C) 2 (D) none of these NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\02-Definite integration.p65 Solution : cos x e x2 2x cos2 x / 2 As, f(x) = x2 sec x sin x x3 x tan x 1 2 f(–x) = – f(x) f(x) is odd f'(x) is even f''(x) is odd Thus, f(x) + f''(x) is odd function let, (x) = (x2 + 1).{f(x) + f''(x)} (–x) = (x) i.e. (x) is odd /2 Ans. (D) (x)dx 0 / 2 Do yourself -2 : (i i ) /2 n 4 sin d Evaluate : / 2 2 4 sin /2 33 ( i ) (x2 sin3 x cos x)dx / 2 E
JEE-Mathematics bb aa ( e ) f(x)dx f(a b x) dx , In particular f(x)dx f(a x) dx 00 aa Illustration 7 : If f, g, h be continuous functions on [0, a] such that f(a – x) = –f(x), g(a – x) = g(x) and a 3h(x) – 4h(a – x) = 5, then prove that f(x)g(x)h(x)dx 0 0 aa a Solution : I = f(x)g(x)h(x)dx f(a x)g(a x)h(a x)dx = f(x)g(x)h(a x)dx 00 0 7I = 3I + 4I aa = f(x)g(x) 3h(x) 4h(a x) dx = 5 f(x)g(x)dx 0 00 (since f(a – x) g(a – x) = –f(x)g(x)) I = 0 Ans. Ans. Illustration 8 : Evaluate x sin x dx ex 1 0 x sin x x sin x Solution : I = e x 1 dx 0 e x 1 dx I1 I2 0 x sin x where I1 = e x 1 dx Put x = –t dx = – dt I1 = 0 (t) sin(t)(dt) t sin t dt et t sin t dt ex x sin x dx et 1 0 et 1 0 ex 1 et 1 0 ex x sin x x sin x dx 0 ex 1 dx Hence I = I1 + I2 = 0 ex 1 I = x sin xdx x sin x dx sin xdx I 00 0 0 2I = sin xdx cos x 2 I NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\02-Definite integration.p65 0 2 dx Illu str ation 9 : Evaluate 0 (17 8 x 4 x2 )[e6(1x) 1] Solution : 2 dx Let I = 0 (17 8 x 4 x2 )[e6(1x) 1] 2 dx aa Also I = 0 (17 8 x 4 x2 )[e6(1x) 1] f(x)dx f(a x)dx 0 Adding, we get 0 2I =2 17 1 4x2 1 1 1 dx 0 8x e6(1x ) e 6 (1 x ) 1 =2 1 dx 1 2 x2 dx 0 17 8x 4x2 4 0 2x 17 / 4 34 E
JEE-Mathematics x 1 2 x 1 1 2 (x dx 1 1 21 0 1)2 21 / 4 4 2 21 log = 2 4 2 21 2 0 21 2 1 2 21 log 2 21 log log = 1 2x 2 I = 4 21 2x 2 21 0 8 21 2 21 21 2 = 1 21 2 Ans. log 4 21 2 21 /2 dx Illustration 10 : Evaluate 1 tan x 0 Solution : / 2 dx /2 cos x dx .....(i) I = = 0 1 tan x 0 cos x sin x / 2 cos( / 2 x)dx /2 sin x dx sin x cos x then I = = .....(ii) 0 cos( / 2 x) sin( / 2 x) 0 Adding (i) and (ii), we get / 2 2I = sin x /2 cos x dx 0 sin x cos x .dx + 0 sin x cos x / 2 /2 2 = 1.dx 0 0 sin x cosx .dx = x 0 / 2 0 sin x cos x Ans. 2I = 2 I = 4 Illustration 11 : 1 cot1 (1 x x2 )dx equals - 0 (A) log 2 (B) log 2 (C) – log 2 (D) none of these 2 2 1 tan 1 1 dx 1 1 x (1 x) Solution : 0 x 0 tan 1 x(1 x) dx 1 x2 NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\02-Definite integration.p65 11 dx 1 tan 1 (1 x) dx 0 [tan 1 x tan 1 (1 x )] dx tan 1 x 00 1 x tan 1 x 1 log(1 x2 ) 1 = 2 log 2 log 2 Ans. (B) 2 0 4 2 2 tan 1 x dx 2 0 / 2 a sin x b cos x Illustration 12 : dx sin x cos x 0 /2 a sin x b cos x Solution : dx sin x cos x .....(i) 0 / 2 a sin( / 2 x) b cos( / 2 x) dx / 2 a cos x b sin x dx .....(ii) 0 sin( / 2 x) cos( / 2 x) 0 sin x cos x / 2 (a b)(sin x cos x) dx /2 0 sin x cos x 2 (a b)dx (a b) / 2 (a b) / 4 Ans. 0 E 35
JEE-Mathematics 2 / 2 sin x Illustration 13 : 0 2sin x 2cos x dx equals - (A) 2 (B) (C) 4 (D) 2 2 / 2 sin x 2 / 2 sin ( / 2x ) 2 / 2 cos x 2sin x 2cos x dx dx dx Solution : 2cos x 2sin x 2 2sin x ( / 2 x ) cos ( / 2 x ) 0 0 0 2 /2 dx 0 2 4 Ans. (C) Do yourself -3 : Evaluate : 5x dx /3 dx (i) x 6x (ii) / 6 1 tan5 x 1 2a a a a ; 2 f(x) dx if f(2a x) f(x) f(x) dx f(2a (f) f(x)dx x) dx 0 0 00 0 ; if f(2a x) f(x) xdx Illustration 14 : Evaluate 0 1 cos2 x xdx ( x)dx dx I 0 1 cos2 0 1 cos2 ( x) 0 1 cos2 Solution : Let I = x = x dx / 2 dx /2 sec2 xdx 0 1 cos2 2 2 0 1 cos2 0 2 tan2 x 2I = x x = Let tan x = t so that for x 0, t 0 and for x /2, t . Hence we can write, dt 1 ta n 1 t = 2 Ans. 2 2 0 22 I = 0 2 t2 /2 /2 NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\02-Definite integration.p65 Illustration 15 : Prove that log(sin x)dx log(cos x)dx log 2 Solution : 2 0 0 /2 ........ (i) Let log(sin x)dx 0 /2 log sin x dx /2 log(cos x)dx then 0 2 0 I = ........ (ii) adding (i) and (ii), we get /2 /2 /2 2I = log sin x dx log cos x dx = (log sin x log cos x)dx 0 00 /2 x cos x)dx /2 log 2 sin x cos x dx 0 2 log(sin 0 2I = /2 log sin 2x dx /2 log(sin 2x)dx /2 (log 2)dx 0 2 0 0 = 36 E
JEE-Mathematics /2 = log sin 2x . dx log 2 x0 / 2 0 /2 log(sin 2x)dx log 2 2I = 2 ......... (iii) 0 /2 putting 2x = t, we get Let I1 = log(sin 2x)dx, 0 log sin t dt 1 1 /2 2 .2 log(sin t)dt 02 log sin tdt 20 0 I1 = = /2 I1 = log(sin x)dx 0 (iii) becomes ; 2I = I – log 2 2 /2 Hence log sin x dx log 2 Ans. 2 0 /2 Illustration 16 : (2 log sin x log sin 2x)dx equals - 0 (A) log 2 (B) – log 2 (C) (/2) log 2 (D) –(/2) log 2 Solution : /2 /2 (2 log sin x log 2 sin x cos x)dx (2 logsin x log 2 log sin x log cos x)dx Ans. (D) 00 /2 /2 /2 log sin xdx log 2dx log cos xdx = – (/2) log 2 0 00 Do yourself -4 : Evaluate : NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\02-Definite integration.p65 3 dx /2 ( i ) 3 (1 ex )(1 x2 ) (i i ) n sin2 x cos x dx (i ii ) / 2 sin x cos x dx 0 0 1 sin x cos x /2 (iv) cos x cos3 x dx / 2 nT T ( g ) f(x)dx n f(x) dx , (n I) ; where ‘T’ is the period of the function i.e. f(T + x) = f(x) 00 Tx T Note that : f(t)dt will be independent of x and equal to f(t)dt x0 bnT b ( h ) f(x)dx f(x)dx where f(x) is periodic with period T & n I. anT a nT T ( i ) f(x)dx (n m ) f(x)dx , (n, m I) if f(x) is periodic with period ‘T’. mT 0 E 37
JEE-Mathematics Illustration 17 : 4 Solution : Evaluate | cos x| dx 0 Note that |cos x| is a periodic function with period . Hence the given integral. 2 /2 | cos sin x 0 sin x /2 4 1 1 I = 4 = 4 cos cos xdx = 4 8 Ans. x| dx xdx 0 0 2 2[ x 14 ] Illustration 18 : x dx {x } 14]dx The value of x satisfying 2 , is equal to (where [.] and {.} denotes the [x 0 0 greatest integer and fractional part of (x) (A) [–14, –13) (B) (0, 1) (C) (–15, –14] (d) none of these 2[x14] x {x } 28 2[x] x {x } 0 2 Solution : dx [x 14]dx 0 2 dx (14 [x])dx 0 0 28 x 28 2[x] x 2 x 2[x] x 0 2 28 2 14 dx dx (1 4 [x]){x} 2 dx 2 dx (1 4 [ x ]){x } 0 0 nT T a nT nT {using f(x)dx n f(x)dx and f(x)dx f(x)dx where T is period of f(x)} a0 00 14 + [x] = (14 + [x]){x} (14 + [x])(1 – {x}) = 0 [x] = –14 x [–14, –13) Ans. (A) 16 / 3 Illustration 19 : Evaluate | sin x| dx 0 16 / 3 5 5/ 3 /3 Solution : | sin x| dx = | sin x| dx | sin x| dx = 5| sin x| dx | sin x| dx 0 0 5 00 = 5 cos x0 cos x0 / 3 = 10 + 1 1 21 Ans. 2 2 2n NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\02-Definite integration.p65 Illustration 20 : Evaluate : [sin x cos x]dx . Here [.] is the greatest integer function. 0 Solution : 2n 2 Let I = [sin x cos x]dx n [sin x cos x]dx 00 ( [sinx + cosx] is periodic function with period 2] 1, 0x 2 x 3 0, 24 3 x 1, 4 x 3 [sin x cos x ] 2 2, 3 x 7 24 7 x 2 4 1, 0, 38 E
JEE-Mathematics / 2 3 / 4 3/2 7 / 4 2 Hence I = n 1dx 0dx 1dx 2dx 1dx 0dx 0 / 2 3 / 4 3/2 7 / 4 I = n 0 3 3 2 7 3 0 n Ans. 2 4 4 2 Do yourself -5 : 20 Evaluate : 3 10 (i i ) (sin x cos x)dx 20 ( i ) {2x}dx , where {.} denotes fractional part of x. 6 1.5 [x] 3x (i i i ) 0 3[x] dx , where [.] denotes greatest integer function. 4 . WALLI’S FORMULA : /2 /2 (n 1)(n 3 ).....(1 or 2) sin n cosn ( a ) x dx x dx n(n 2).....(1 or 2) K 00 where K = / 2 if n is even 1 if n is odd /2 x. co s m x dx [(n 1)(n 3)(n 5)....1 or 2][(m 1)(m 3)....1 or 2] K sinn ( b ) 0 (m n)(m n 2)(m n 4)....1 or 2 if both m an d n a re e ve n (m , n N ) Where K = 2 1 otherwise Illustration 21 : /2 sin4 x cos6 x dx / 2 3 3 3 3 (A) (B) 572 (C) (D) 64 256 128 NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\02-Definite integration.p65 / 2 /2 cos6 x.dx = 2 3.1 5.3.1 . 3 sin4 x 2 sin4 x Solution : cos6 x dx Ans. (C) /2 0 10.8.6.4.2 2 256 5 . DERIVATIVE OF ANTIDERIVATIVE FUNCTION (Newton-Leibnitz Formula) : d h(x) f(t)dt f[h(x)].h '(x) f[g(x)].g '(x) If h(x) & g(x) are differentiable functions of x then, dx g(x) x2 t2 5t 4 dt Illustration 22 : Find the points of maxima/minima of 2 et 0 Solution : Let f(x) = x2 t2 5 t 4 dt y E 0 2 et x 2 x4 5x2 4 (x 1)(x 1)(x 2)(x 2)2x –2 –1 0 1 f'(x) = 2 ex2 2x 0 2 ex2 Graph of f'(x) 39
JEE-Mathematics From the wavy curve, it is clear that f'(x) changes its sign at x = ± 2, ±1, 0 and hence the points of maxima are –1, 1 and of the minima are –2, 0, 2. d t3 1 Evaluate Illustration 23 : dx dt t2 log x Solution : d t3 1 dx 1 t3 . d (t3 ) 1 t2 . d (t2 ) = 3 3t2 2t t(t 1) Ans. t2 log x log dt log dt log t 2 log t log t dt Do yourself - 6 : x ( i ) If ƒ(x) sin t dt , then find ƒ '(1). 1/x x y dy dx 3 sin2 tdt cos tdt 0 , then evaluate . (ii) /3 0 6 . DEFINITE INTEGRAL AS LIMIT OF A SUM : bb An alternative way of describing ƒ(x)dx is that the definite integral ƒ(x)dx is a limiting case of the summation a a of an infinite series, provided ƒ(x) is continuous on [a,b] b n 1 b a h i.e. ƒ( x ) dx lim ƒ(a rh) where h n . The converse is also true i.e., if we have an infinite series of the a n r 0 above form, it can be expressed as a definite integral. Step I : Express the given series in the form 1 f r n n Step II : Then the limit is its sum when n , i.e. lim 1 f r Step III : n n n Replace r by x and 1 by dx and lim by the sign of n n n Step IV : r NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\02-Definite integration.p65 The lower and the upper limit of integration are the limiting values of for the first and the last n term of r respectively. Some particular cases of the above are. ( a ) lim n 1 f r lim n 1 1 f r 1 ƒ( x )dx r 1 n n or r 0 n n 0 n n ( b ) lim pn 1 r ƒ( x )dx r 1 n n x r (as r = 1) where lim 0 x n r (as r = pn) and lin p n n 40 E
JEE-Mathematics Illustration 24 : Evaluate Lim 1 1 1 2 ......... 1 2n 2n 6n n 4n 1 4n 1 1 1 1 1 Solution : Let Sn = ......... = . 2n 1 2n 2 6n 2n r n2 r r 1 r 1 n Lim S 4 dx [n| 2 x| ]04 n6 n2 n3 02x n S = n Ans. Illustration 25 : n n n ....... 1 Evaluate Lim n 2 2 3 2 4 n 2 3 3 3 4 n 2 49n n 3 4 n n n lim n 2 2 3 2 4 n 2 ......... Solution : n Let p = 3 4 n 3 n 4 n 2 Analyzing the expression with the view of increasing integral value we get the expression in terms of r as n nn 11 dx r = lim 2 lim 2 x 3 x 42 n r 1 3 r 4 n n r 1 r r 0 n n 3 n 4 3 dx dt x Put 3 x 4 t, 2 Hence p = 2 7 dt 2 1 7 2 1 1 1 Ans. 3 4 t2 3 t 3 7 4 14 4 Do yourself - 7 : Evaluate : 1 1 1 1 n 1 1 (i) lim n 2.1 n 2.2 n 2.3 ....... 3n (ii) lim n r 0 n2 r2 n NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\02-Definite integration.p65 7 . ESTIMATION OF DEFINITE INTEGR AL : b ( a ) If f(x) is continuous in [a, b] and it’s range in this interval is [m, M], then m(b – a) f(x)dx M(b a) a 3 Illustration 26 : Prove that 4 3 x3 dx 2 30 1 Solution : Since the function f(x) = 3 x2 increases monotonically on the interval [1, 3], m = 2, M = 30 , b – a =2. Ans. 33 Hence, 2.2 3 x3 dx 2 30 4 3 x3 dx 2 30 11 bb ( b ) If f(x) (x) for a x b then f(x)dx (x) dx aa E 41
JEE-Mathematics Illustration 27 : 1 dx Solution : Prove that 6 0 4 x2 x3 4 2 Since 4 – x2 4 – x2 – x3 4 – 2x2 > 0 x [0, 1] 4 x2 4 x2 x3 4 2 x2 0 x [0, 1] 0 11 1 x [0 1] 4 x2 4 x2 x3 4 2x2 1 dx 1 dx 1 dx x [0, 1] 0 4 x2 0 4 x2 x3 0 4 2x2 sin 1 x 1 1 dx 1 sin 1 x 1 1 dx 0 4 x2 x3 2 2 6 0 4 x2 x3 4 2 0 Ans. 2 0 bb (c) f(x)dx f(x) dx . aa Illustration 28 : Prove that 19 sin x dx < 10–7 10 1 x8 19 sin x 19 sin x dx dx 10 1 x8 I = Solution : To find 10 1 x8 ....... (i) ....... (iii) Since | sin x| 1 for x 10 The inequality sin x 1 ....... (ii) 1 x8 |1 x8| also, 10 x 19 1 + x8 > 108 1 1 or 1 1 0 8 from (ii) and (iii) ; 1 x8 x | 1 8 | NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\02-Definite integration.p65 108 sin x 108 1 x8 19 sin x dx 19 10 1 x8 108 dx 10 19 sin x dx (19 10).108 107 Ans. 10 1 x8 Illustration 29 : If ƒ (x) is integrable function such that |ƒ (x) – ƒ (y) | < |x2 – y2|, x, y [a,b] then prove that b ƒ(x) ƒ(a) dx (a b)2 . xa 2 a 42 E
JEE-Mathematics Solution : b ƒ(x) ƒ(a) dx b ƒ(x) ƒ(a) dx Given, a xa a xa b x2 a2 b b (a b)2 x a dx | x a| dx (x a)dx 2 a aa b ( d ) If f(x) 0 on the interval [a,b], then f(x)dx 0 . a Illustration 30: 8 Solution : If ƒ (x) is a continous function such that ƒ (x) > 0 x [2,10] and ƒ(x) dx 0 , then find ƒ (6). 4 ƒ (x) is above the x-axis or on the x-axis for all x [2,10]. If ƒ(x) is greater than zero for any sub 88 interval of [4,8], then ƒ(x)dx must be greater than zero. But ƒ(x)dx 0 ƒ (x) = 0 x [4,8] 44 ƒ (6) = 0. Do yourself - 8 : 3 2 dx . ( i ) Prove that 4 3 x2 dx 4 3 (ii) Prove that 4 5 3 sin x 0 1 Show that 3 1 x4 (iii) 21 / 3 1 dx 1 5 0 (1 x6 )2 / 3 Miscelleneous Illustrations : x3 cos4 x sin2 x Illustration 31 : Evaluate : 0 (2 3x 3x2 ) dx Solution : x3 cos4 x sin2 x ........ (i) NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\02-Definite integration.p65 Let I = 0 (2 3x 3x2 ) dx ( x)3 cos4 ( x) sin2 ( x)dx = (By. Prop.) 2 3( x) 3( x)2 0 (3 x3 32 x 3x2 ) cos4 x sin2 x dx = ......... (ii) (2 3x 3x2 ) 0 Adding (i) and (ii) we have (3 32 x 3x2 ) cos4 x sin2 x 2I = dx (2 3x 3x2 ) 0 /2 2I = cos4 x sin2 xdx 2I = 2 cos4 x sin2 xdx 00 /2 I = cos4 x sin2 x dx 0 Using walli's formula, we get I = (3.1)(1) = 2 Ans. 6.4.2 2 32 E 43
JEE-Mathematics Illustration 32 : Let f be an injective function such that f(x) f(y) + 2 = f(x) + f(y) + f(xy) for all non negative real x and y with f(0) = 1 and f'(1) = 2 find f(x) and show that 3 f(x)dx x(f(x) 2) is a constant. Solution : We have f(x)f(y) + 2 = f(x) + f(y) + f(xy) Putting x = 1 & y = 1 then f(1)f(1) + 2 = 3f(1) we get f(1) = 1,2 f(1) 1 ( f(0) = 1 & function is injective) then f(1) = 2 1 Replacing y by x in (1) then f(x)f 1 2 f(x) f 1 f (1) f ( x)f 1 f(x) f 1 x x x x Hence f(x) is of the type f(x) = 1 ± xn f(1) = 2 f(x) = 1 + xn and f'(x) = nxn–1 f'(1) = n = 2 f(x) = 1 + x2 3 f(x)dx x(f(x) 2) 3 (1 x2 )dx x(1 x2 2) = 3 x3 x(3 x2 ) c = c = constant x 3 1 Illustration 33 : Evaluate : [x[1 sin x] 1]dx , [.] is the greatest integer function. 1 Solution : 1 01 Illustration 34 : Let I = [x[1 sin x] 1]dx = [x[1 sin x] 1]dx [x[1 sin x] 1]dx 1 1 0 Now [1 + sinx] = 0 if –1 < x < 0 [1 + sinx] = 1 if 0 < x < 1 NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\02-Definite integration.p65 01 1 I = 1.dx [x 1]dx = 1 + 1 dx = 1 + 1 = 2. Ans. 1 0 0 Find the limit, when n of 1 1 1 ..... 1 (2n 12 ) (4n 22 ) (6n 32 ) n Solution : Let 1 1 1 1 P = Lim .... n 2n 12 6n 32 n 4n 22 Lim 1 1 1 1 .... n 1(2n) 12 2(2n) 22 3(2n) 32 n(2n) n2 n 1 n 11 dx (2x x2 ) Lim Lim r 2 n r 1 r(2n) r2 n r n r 1 n 0 n. 2 44 E
JEE-Mathematics Put x = t2 dx = 2t dt 1 2tdt = sin 1 t 1 2 s in 1 1 2 2 t2 2 2 4 P = 0t 2 0 Hence P = /2. Ans. 1 x , x 1 5 Illustration 35 : If f(x) = x 1, x 1 , and g(x) = f(x – 1) + f(x + 1). Find the value of g(x ) dx . Solution : 3 Given, x 1, x 1 x, x 1 1 x0 1 x 1 0 0 x 1 f(x) = 1 x, 1 x 0 ; f(x–1) = x, 0 x 1 1 1x2 1 x, 0x 1 2 x 1 1 x2 x, x 1, x 1 x 2, Similarly x 2, x 1 1 x 2 1 x 1 0 2 x 1 f(x+1) = x 2, 0 x 1 1 1 x 0 x, x 1 1 x0 x, 2 x 2 x 2 2, 2 x 1 2 x, 1 x 0 g(x) = f(x – 1) + f(x + 1) = 2x, 0 x 1 2, 1 x 2 2x 2, 2 x Clearly g(x) is even, Now 5 g(x )dx = 3 + 5 = 1 dx 2 3 2 ) dx + 5 2) dx = 24 2 g(x) dx g(x) dx 22x 2 dx (2x (2x 3 0 3 0 1 2 3 NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\02-Definite integration.p65 ANSWERS FOR DO YOURSELF 1 : (i ) 31 (ii) 2 (iii) 2( 2 1) 9 2 : (i ) 6 (iv) 3 : (i ) 2 ( i i ) n2 4 : (i ) 2 (ii) /12 2 5 : (i ) (ii) – 3 n2 (iii) 0 4 3 2 (iv) 3 6 : (i ) 23 (ii) 3 1 2[x] 7 : (i ) 4 (i i i ) n3 E 3 sin1 (ii) dy 3 sin2 x 2 dx cos y 1 n3 2 (ii) 2 45
JEE-Mathematics CHECK YOUR GRASP EXERCISE - 01 SELECT THE CORRECT ALTERNATIVE (ONLY ONE CORRECT ANSWER) / 3 cos x 3 2 3 0 4 sin 3 then k is- 1 .If 3 x dx k log 1 1 1 1 (A) 2 (B) 3 (C) 4 (D) 8 (D) 1 + e eeee dx 2 . eee xnx.n(nx).n(n(nx)) equals - (A) 1 (B) 1/e (C) e – 1 3 . The value of the definite integral (e x1 e3x )1 dx is 1 (C) 1 tan 1 1 (A) 4e2 (B) e2 2 e (D) 2e2 (D) ee(e – 1) + e 4e e 4 . The value of the definite integral (x 1)ex .nx dx is - 1 (A) e (B) ee + 1 (C) ee(e – 1) 12 5 . Let a, b, c be non-zero real numbers such that ; (1 cos8 x)(ax2 bx c)dx (1 cos8 x)(ax2 bx c)dx , 00 then the quadratic equation ax2 + bx + c = 0 has - (A) no root in (0,2) (B) atleast one root in (0,2) (C) a double root in (0,2) (D) none x 1 1 2A 2 2 2 and f(x)dx , then the constant A and B are- 0 6 . If f(x) = A sin + B, f' 2 4 4 NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\02-Definite integration.p65 (A) 2 and 2 (B) and 3 (C) 0 and (D) and 0 /4 7 . tann xdx lim If In = then n (In + In–2) = n 0 (A) 1 (B) 1/2 (C) (D) x tan 1 x 8 . dx 0 (1 x2 )2 (A) (B) (C) (D) 2 4 6 8 e f(x) 1 9 . dx , then the value of Suppose f, f' and f'' are continuous on [0, e] and that f'(e) = f(e) = f(1) = 1 and x2 2 1 e (B) 1 (C) 2 (D) none of these f ''(x)n x dx equals - 46 1 (A) 0 E
JEE-Mathematics 1 0 . 2 1 sin x 1 dx has the value equal to - /2 x x 1 (A) 0 3 5 (D) 2 (B) (C) (D) 4 4 4 (C) 2 4 logx 22 1 1 . lo g x 2 n2 dx = 2 (A) 0 (B) 1 sin x 3 sin 2x 1 2 . Suppose that F(x) is an antiderivative of f(x) = , x > 0 then dx can be expressed as - x1 x (A) F(6) – F(2) 1 1 (D) 2(F(6) – F(2)) (B) (F(6) – F(2)) (C) (F(3) – F(1)) 2 2 1 3 . f x 1 . nx dx 0 x x (A) is equal to zero (B) is equal to one 1 (D) can not be evaluated (C) is equal to 2 1 2 (D) 1 4 . Integral | sin 2x| dx is equal to - 1 (C) 0 (D) none 1 (A) 0 1 (C) (B) 5 3x dx x 15. (5 x) 2 1 1 (A) (B) 2 3 [JEE 1985] (D) none of these 1 6 . For any integer n the integral ecos2 x cos3 (2n 1) xdx has the value (D) 5/2 0 [JEE 1981] (A) (B) 1 (C) 0 (D) none of these NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\02-Definite integration.p65 1 7 .3 2x2 (x 2)2 53 dx is equal to - (D) 2 2 10x (D) 7 23 (A) 2 (B) 1 (C) 1/2 2 1 1 8 . The value of the definite integral (1 ex2 ) dx is - 0 (A) –1 (B) 2 (C) 1 + e–1 1 9 . (cos ax sin bx)2 dx where a and b are integer is equal to - (A) – (B) 0 (C) 2 0 . The value of (1 x2 ) sin x cos2 xdx is - (A) 0 (B) 3 (C) 2 – 3 3 E 47
JEE-Mathematics 2 2 1 . The value of [2 sin x]dx , where [ ] represents the greatest integer function is - (A) 5 (B) – 5 (D) –2 3 (C) 3 f(x) 2 2 . If t2 dt x cos x , then f'(9) 0 1 1 1 (D) is non existent (A) is equal to – (B) is equal to – (C) is equal to 9 3 3 ƒ( x) 2t 2 3 . lim x 1 dt is - Let ƒ : R R be a differentiable function and ƒ (1) = 4. Then the value of x 1 4 [JEE 1990] (A) 8ƒ '(1) (B) 4ƒ '(1) (C) 2ƒ '(1) (D) ƒ '(1) x [JEE 1997] g(x) 2 4 . If g(x) = cos4 t dt , then g(x + ) equals - (D) g() 0 (A) g(x) + g() (B) g(x) – g() (C) g(x)g() n V 1 cos 2 x dx 2 2 For n N, the value of the definite integral 0 2 5 . where V is - (A) 2n + 1 – cosV (B) 2n – sinV (C) 2n + 2 – sinV (D) 2n + 1 – sinV SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS) x 2 6 . dx 0 (1 x )(1 x2 ) (A) (B) 4 2 dx (D) cannot be evaluated (C) is same as 0 (1 x)(1 x2 ) 2 7 . Which of the following are true ? a x.f(sin x)dx a f (sin x)dx aa NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\02-Definite integration.p65 (A) a 2 . (B) f(x2 )dx 2. f(x2 )dx a a 0 n bc b (C) f(cos2 x)dx n. f(cos2 x)dx (D) f(x c)dx f(x)dx 00 0c x dt 2 8 . Let f(x) = and g be the inverse of f. Then the value of g'(0) is - 2 1 t4 (A) 1 (B) 17 (C) 17 (D) none of these x nt 2 9 . If f(x) = 1 1 t dt where x > 0 then the value(s) of x satisfying the equation, f(x) + f(1/x) = 2 is - (A) 2 (B) e (C) e–2 (D) e2 r 4 n n 3 0 . The value of Lim n r 1 is equal to - 2 r 3 r 4 n 1 1 1 1 (A) (B) (C) (D) 35 14 10 5 48 E
JEE-Mathematics 11 2 2 3 1 . If I1 2 x2 dx, I2 2x3 dx I3 2 x2 dx and I4 2x3 dx then - 1 1 00 (A) I3 > I4 (B) I3 = I4 (C) I1 > I2 (D) I2 > I1 nnn ........ 1 32. Let Sn = , then Lim Sn is - (n 1)(n 2) (n 2)(n 4) (n 3)(n 6) 6n n (A) n 3 (B) n 9 (C) greater than one (D) less than two 2 2 /2 cot x dx is- 3 3 . The value of the integral [JEE 1983] 0 cot x tan x 3/ 8 cot x /2 dx (C) dx (A) /4 (B) /2 cot x tan x (D) 0 1 tan3 x /8 kk I1 3 4 . Let ƒ be a positive function, let I1 x ƒ [x(1 x)]dx , I2 ƒ [x(1 x)]dx , where 2k – 1 > 0. Then I2 1k 1k is - [JEE 1997] (A) 2 (B) k 1 (D) less than 1 (C) 2 NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\02-Definite integration.p65 CHECK YOUR GRASP ANSWER KEY EXERCISE-1 Que. 1 234 5678 9 10 Ans. C AAD BDAD DA Que. 11 12 13 14 15 16 17 18 19 20 Ans. A AAD ACCD DA Que. 21 22 23 24 25 26 27 28 29 30 Ans. A AAA C A,C A,B,C,D C C,D C Que. 31 32 33 34 Ans. C A,D A,D C,D 49 E
JEE-Mathematics BRAIN TEASERS EXERCISE - 02 SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THEN ONE CORRECT ANSWERS) 1 2x2 3x 3 1 . The value of 0 (x 1)(x2 2x 2) dx is - (A) 2n2 tan 1 2 (B) 2n2 tan 1 1 (C) 2n2 cot1 3 (D) n4 cot1 2 4 43 4 If 2 .In 1 dx ; n N, then which of the following statements hold good ? 0 (1 x2 )n 1 1 5 (A) 2nIn+1 =2–n + (2n – 1)In (B) I2 8 4 (C) I2 8 4 (D) I3 16 48 3 . If a, b, c R and satisfy 3a + 5b + 15c = 0, the equation ax4 + bx2 + c = 0 has - (A) atleast one root in (–1, 0) (B) atleast one root in (0, 1) (C) atleast two roots in (–1, 1) (D) no root in (–1, 1) dx x2dx 4 . Let u = 0 x4 7x2 1 & v = 0 x4 7x2 1 then - (A) v > u (B) 6v = (C) 3u + 2v = 5/6 (D) u + v = /3 5 . Let f(x) be a function satisfying f'(x) = f(x) with f(0) = 1 and g be the function satisfying f(x) + g(x) = x2. The value 1 of the integral f(x)g(x)dx is - 0 (A) e – 1 e2 5 (B) e – e2 – 3 1 (D) e 1 e2 3 22 (C) (e 3) 22 2 For f(x) = x4 +|x|, let I1 = and I2 = /2 I1 I2 f(cos x)dx f(sin x)dx 0 0 6 . then has the value equal to - (A) 1 (B) 1/2 (C) 2 (D) 4 7 . x 28 4 3 x 1 3 2 x 1 Number of values of x satisfying the equation 8 t 2 t dt , is - 1 log (x 1) NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\02-Definite integration.p65 (A) 0 (B) 1 (C) 2 (D) 3 0 zez dz 8 . The value of definite integral 1 e2z (A) – n2 (B) n2 (C) –n2 1 2 2 (D) n 2 /4 9 . (cos 2x)3 / 2 .cos x dx 0 3 3 3 3 2 (A) (B) (C) (D) 16 32 16 2 16 1 0 . 1 n n 1 The value of 0 r 1 (x r ) k 1 x k dx equals (A) n (B) n! (C) (n+1)! (D) n.n! 50 E
JEE-Mathematics 2 e4 1 1 . If the value of the integral ex2 dx is , then the value of nx dx is - 1e (A) e4 – e – (B) 2e4 – e – (C) 2(e4 – e) – (D) 2e4 – 1 – dx r3 1 2 . The value of Lim dr is - x dx (r 1)(r 1) 3 (A) 0 (B) 1 1 (D) non existent (C) 2 1 3 . [2ex ]dx where [x] denotes the greatest interger function is - 0 (A) 0 (B) n2 (C) e2 (D) 2/e Let f(x) sin x , then /2 x = 1 4 . = x 0 f(x)f 2 dx 2 1 (A) f(x)dx (D) f(x)dx (B) f(x)dx (C) f(x)dx 0 0 0 0 1 1 2 f(x)dx If for a non-zero x, af(x) + bf x x 1 1 5 . 5 , where a b, then = (A) a2 1 b2 a log 2 5a 7b (B) a2 1 b2 a log 2 5a 7b 2 2 (C) a2 1 b2 a log 2 5a 7b (D) none of these 2 n 1 t bx)c / x 1 6 . t 0 dx If a, b and c are real numbers then the value of Lim (1 a sin equals - t0 (A) abc ab bc ca (B) c (C) a (D) b NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\02-Definite integration.p65 x x2 2 t2 ƒ (t)dt / 4 ƒ (x) x9 x3 x 1 dx 1 7 . 2 2 x / 4 cos2 x Let y = ƒ (x) be a differentiable curve satisfying ƒ (t)dt , then equals - (B) 1 (C) 2 (D) 4 (A) 0 1 8 . If y = ƒ(x) is a linear function satisfying the relation ƒ(xy) = ƒ(x).ƒ(y) x, y R , then the curve x y2 (sin t a2 t3 bt)dt , R cuts y = ƒ–1(x) at - 0 (A) no point (B) exactly one point (C) atleast two points (D) infinite points 48 1 9 . If ƒ (8 – t) = ƒ (t) and ƒ ()d 8 , then ƒ ()d is - 00 (A) 4 (B) 8 (C) 16 (D) 32 E 51
JEE-Mathematics t2 z 2 tan z 1 tan2 z t2 1 tan2 z 2 tan z z sec2 z 2z sec2 z If x e dz z 0 2 0 . y e dz 2 & . 0 Then the inclination of the tangent to the curve at t = is - 4 3 (A) (B) 3 (C) (D) 4 2 4 2 1 . The value of integral x ƒ(sin x)dx = 0 (A) ƒ(sin x)dx /2 /2 (D) ƒ(cos x)dx 2 0 2 0 (B) ƒ(sin x)dx (C) ƒ(cos x)dx 0 0 BRAIN TEASERS ANSWER KEY EXERCISE-2 NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\02-Definite integration.p65 Que. 1 2 3 4 5 6 789 10 D C B A,D C D Ans. A,C,D A,B A,B,C B,C,D 15 16 17 18 19 20 B A CCC D Que. 11 12 13 14 52 E Ans. B C B A Que. 21 Ans. A,B,C
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