EXERCISE - 03 JEE-Mathematics MISCELLANEOUS TYPE QUESTIONS TRUE / FALSE 0 1 . The value of the integral x.ex dx is not finite. 1 2 . If n is a positive integer then (nx)n dx (1)n n ! . 0 1 1 3 . 1 xp dx p 1 , where p R – {1} 4 . The average value of the function f(x) = sin2xcos3x on the interval [– ] is 0. sec x cos x sec2 x cot x cos ecx /2 1532 5 . If f(x) = cos2 x cos2 x cosec 2x . Then f(x) dx = – 60 [JEE 1987] 1 cos2 x cos2 x 0 2 x sin2n x 6 . dx [JEE 1996] For n > 0, sin2n x cos2n x 0 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 s t a t e m e n t i n C o l u m n - I c a n h a v e c o r r e c t m a t c h i n g w i t h O N E s t a t e m e n t i n C o l u m n - I I . 1 . Column-I Column-II 10 [x2 ]dx 1 (p) (A) = 100 4 [x2 28x 196] [x2 ] (q) 3 {where [.] denotes greatest integer function} 1 2| x| (r) (B) dx 3 1 x (s) 1 (C) lim 199 299 ..... n99 n NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\02-Definite integration.p65 n100 1 x200 dx 1 , 5050 1 (D) then = 2. Column-I Column-II (p) 7 E 1 3x2 1 (A) 1 1 4 tan x dx = (q) 8 sin x2dx 2 (r) 1 (B) 6 sin x2 sin(x 14)2 (s) 2 1 13 (C) [x]dx 156 1 {where [.] denotes greatest integer function} 1 0 n sin 2xdx n2 /2 (D) 53
JEE-Mathematics 3. Column-I Column-II (A) If [ ] denotes the greatest integer function and (p) 1 f(x) 3[x] 5| x| x 0 2 f(x)dx ; , then is equal to x 2 ; x 0 3 / 2 /2 cos x dx of is 11 (B) The value of (q) – /2 1 ex 2 (C) sin x cos ec 1 dx then the 3 If I = x2 dx and I = 1 x(x2 1) (r) 1 11 2 2 I1 I12 I2 value of e I1 I2 I22 1 , is I12 I22 1 1 (D) If f(x) and g(x) are two continuous functions defined on (s) 0 a R, then the value of {f(x) f(x)}{g(x) g(x)}dx, is a 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 : The equation 4x3 – 9x2 + 2x + 1 = 0 has atleast one real root in (0, 1). because b Statement-II : If 'f' is a continuous function such that f(x) 0 , then the equation f(x) = 0 has atleast one real a root in (a, b). (A) A (B) B (C) C (D) D 2. Statement-I : xdx tan x cos3 xdx . x tan x cos3 0 2 0 because b a b b NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\02-Definite integration.p65 xƒ (x)dx ƒ (x)dx Statement-II : . a 2a (A) A (B) B (C) C (D) D 3 . x ntdt > 0), f(x) = 1 Statement-I : If f(x) = 1 1 t t2 (x then – f x because Statement-II : If f(x) = x ntdt , then f(x) + f 1 1 nx2 . 1 t 1 x 2 (A) A (B) B (C) C (D) D 4 . Let f(x) = x – x2 + 1. Statement-I : g(x) = max{f(t) : 0 t x}, then 1 g(x)dx 29 0 24 because Statement-II : f(x) is increasing in 0, 1 and decreasing in 1 , 1 . 2 2 (A) A (B) B (C) C (D) D 54 E
JEE-Mathematics 5 . (sin m x. sin n x )dx Statement-I : Let m & n be positive integers. a = cos , if m n & b = cos (sin m x. sin n x )dx if m = n, then a + b = 2. because Statement-II : m x. sin n x )dx 0, m n , where m & n are positive integers. , m n (sin (A) A (B) B (C) C (D) D Statement-I 3 1 cosec99 1 . 6 . : 1/3 x x x dx 0 because a Statement-II : ƒ (x)dx 0 if ƒ (–x) = – ƒ (x). a (A) A (B) B (C) C (D) D n 1 1 r 1 n 1 r 7 . Statement-I : r0 n n 1 ( x 1) dx r1 n n 1 , n N. 0 because n 1 1 r 1 n 1 r r0 n n r 1 n n ƒ ( x )dx 0 Statement-II : If ƒ (x) is continuous and increasing in [0, 1], then ƒ ƒ , where n N (A) A (B) B (C) C (D) D COMPREHENSION BASED QUESTIONS Comprehension # 1 ƒ (t) 3 x Let g(x) = ƒ t dt , where ƒ is a function 2 1 0 whose graph is show adjacently. 0 1 23 4 5 6 7t On t he basis of above i nfor mat ion, a nswer t he fol low i ng que st ions : –1 1 . Maximum value of g(x) in x [0, 7] is - –2 (A) 3 (B) 9/2 –3 NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\02-Definite integration.p65 (C) 3/2 (D) 6 2 . Value of x at which g(x) becomes zero, is - (A) 3 (B) 4 (C) 5 (D) 6 3 . Set of values of x in [0, 7] for which g(x) is negative is - (A) (2, 7) (B) (3, 7) (C) (4, 6) (D) (5, 7) Comprehension # 2 The average value of a function f(x) over the interval, [a, b] is the number 1b µ f(x)dx b a a b 1 / 2 a The square root b 1 a f ( x )2 d x is called the root mean square of f on [a, b]. The average value of µ is attained if f is continuous on [a, b]. On the basis of above information, answer the following questions : 1. The average ordinate of y = sin x over the interval [0, ] is - E (A) 1/ (B) 2/ (C) 4/2 (D) 2/2 55
JEE-Mathematics 2 . The average value of the pressure varying from 2 to 10 atm if the pressure p and the volume v are related by pv3/2 = 160 is - 20 10 40 160 (B) 3 10 3 2 (A) 3 20 3 10 3 2 (C) 3 20 3 10 3 2 (D) 3 20 3 10 3 2 cos2 x 3 . The average value of f(x) = sin2 x 4 cos2 x on [0, /2] is - (A) /6 (B) 4/ (C) 6/ (D) 1/6 Comprehension # 3 m ax . ƒ t m in . ƒ t , 0 t 0 x 4 x 4 x 5 2 x5 Consider g(x) | x 5| | x 4| 6 x ta n sin 1 x2 12x 37 where ƒ (x) = x2 – 4x + 3. On the basis of above information, answer the following questions : 5 (B) 3 (C) 13/3 (D) 3/2 1 . g(x) dx is equal to 2 (A) 5/3 x2 2 . If h(x) g(t)dt , then complete set of values of x in the interval [0, 7] for which h(x) is decreasing, is - 0 (A) (6, 7] (B) (5, 7] (C) 6, 7 (D) 6, 7 3. lim g(x) g(2) is equal to - n(cos(4 x)) x 4 (A) 0 (B) 1 (C) 2 (D) does not exist MISCELLANEOUS TYPE QUESTION ANSWER KEY EXERCISE-3 NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\02-Definite integration.p65 True / False E 1. F 2. T 3. F 4. T 5. T 6. F Match the Column 1 . (A)(q), (B)(s), (C)(p), (D)(p) 2 . (A)(r), (B)(r), (C)(q), (D)(q) 3 . (A)(q), (B)(p), (C)(s), (D)(s) Assertion & Reason 1. A 2. C 3. D 4. A 5. D 6. A 7. A Comprehension Based Questions Comprehension # 1 : 1. B 2. C 3. D Comprehension # 2 : 1 . B 2. C 3. D Comprehension # 3 : 1 . B 2. D 3. A 56
EXERCISE - 04 [A] JEE-Mathematics CONCEPTUAL SUBJECTIVE EXERCISE 1 . Compute the integrals : 13 dx 1 1 xdx (a) 2 5 (3 x )4 (b) (ex 1)4 ex dx 3 xdx (d) 1 x 0 0 (c) sin2 x 2 dx 4 (e) 3 1 x2 (f) 2 x5 x2 1 1 dx 1 x2 3 dx (g) 0 (2x2 1) x2 1 2 x x 2. Prove that : (a) (x ) ( x) dx = 8 (b) dx = 2 dx where , > 0 x . d x = where < = (c) x (d) (x ) ( x) 2 (x ) ( x) 3 3 . Evaluate : (a) | (x 1)(x 2)| dx (b) | cos x| dx 0 0 x 2 , for 0 x 1 2 4 . = Given function f(x) . Evaluate f(x)dx x, for 1 x 2 0 2 1 5 . Evaluate : (a) [x2 ]dx (b) [cos1 x] dx , where [.] represents the greatest integer function 0 1 2t f(x) dx 6 . Evaluate : (a) log(1 cos x)dx (b) f(x) f(2t x) 0 0 4 2 x sin 2xdx (c) log(1 tan x)dx (d) 0 cos4 x sin4 x 0 NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\02-Definite integration.p65 7 . Prove that for any positive integer k, sin 2kx = 2[cosx + cos3x + ...... + cos(2k – 1)x] sin x /2 Hence prove that sin 2kx cot x dx = [JEE 1990] 2 0 8 . 1 x4 (1 x)4 dx Evaluate : 0 1 x2 / 2 a sin x b cos x 9 . Evaluate : dx 0sin x 4 2 dx 10. Evaluate : 2 sin 2x 0 Evaluate : 2 2 x7 3 x6 10 x5 7 x3 12 x2 x 1 11. x2 2 dx E 2 57
JEE-Mathematics 1 5 1 2 . Evaluate : (a) 1 1 x . dx (b) 2 x2 1 n 1 x 1 dx 1 x2 x 0 1 x x x2 x3 x4 1 ecos x [JEE 1999] 1 3 . Integrate : 0 ecos x e cos x dx 1 sin 1 x 1 4 . Evaluate : 0 x2 x 1 dx / 4 cos x sin x 15. Evaluate : 10 sin 2 x dx 0 x sin 2 x sin cos x 2 16. Evaluate dx [JEE 1991] 2x 0 1 7 . 2 ex cos x dx Evaluate : 0 4 2 2 (x2 1)dx u where u and v are in their lowest form. Find the value of (1000)u 18. 1 x3. 2x4 2x2 1 v v e Prove that if Jm n m , then = e – mJ integer). 1 9 . xdx J m – (m a positive m 1 1 2 0 . Prove the inequalities : 1 dx 2 2 1 2 dx 5 (c) (a) 8 (b) 2 e1/4 < e x2 x dx < 2e² 6 0 4 x2 x3 2 0 2 x2 6 0 2 1 . Suppose g(x) is the inverse of f(x) and f(x) has a domain x [a, b]. Given f(a) = and f(b) = , then find the value NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\02-Definite integration.p65 b of f(x)dx g(y )dy in terms of a, b, and a 1 2 2 . Let be the distinct positive roots of the equation tan x = 2x then evaluate (sin x.sin x)dx independent of 0 and . d sin x then compute the value of j(0) where 2 3 . Let h(x) = (fog)(x) + K where K is any constant. If (h(x)) = dx cos2 (cos x) j(x) = f(x) f(t) dt , where f and g are trigonometric functions. g(x) g(t) cos2 x tdt cos1 0 2 4 . (a) sin2 x t dt, x 0, determine f(x) (b) 2 f(x) sin1 0 f ( x ) e3 x tdt , x 0 find differential coefficient of f(x) w.r.t. nx when x = n2 ex nt 58 E
JEE-Mathematics 2 5 . Given a function f(x) such that [JEE 1984] (a) it is integrable over every interval on the real line and aT (b) f(T + x) = f(x), for every x and a real T, then show that the integral f(x) dx is independent of a. a x2 x x2 2 sin xdx e dx lim 0 2 6 . Find the limits : (a) lim 0 (b) x 0 x3 x x e2 x2 dx 0 2 7 . Evaluate : (a) lim 1 1 ..... 1 [JEE 1981] n 1 n 2 6n n Limit 1 1 2 3n n n n 1 n 2 ..... 4 n (b) (c) Limit n ! 1 / n n nn CONCEPTUAL SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(A) NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\02-Definite integration.p65 1 . (a) 5 5 16 1 (b) 0.2 (e – 1)5 (9 4 3 ) 1 n 3 (e) 2 2 n 2 3 (c) (d) 2 36 2 2 2 3 1 2 1 7 3 1 11 1 4 2 1 (f) 32 2 8 (g) arc tan 2 3 . (a) (b) 2 4. 3 6 (b) t (c) log 2 5 . (a) 5 2 3 (b) cos1 + cos2 + cos3 + 3 6 . (a) log 2 8 2 (d) 8 8. 22 9. (a b) 2 11. 16 2 (b) n2 7 22 10. 1 2 . (a) 8 22 5 3 3 2 1 2 1 8 3 2 e2 1 13 . 14 . 1 5 . 3 arc tan 3 arc tan 3 16. 2 17. 5 2 63 1 8 .1 2 5 2 1 . b – a 22. 0 2 3 . sec(1) – 1 2 4 . (a) f(x) = /4 (b) 60 2 2 6 . (a) ; (b) 0 3 27. (a) log 6; (b) 3 n 4 (c) 1 e E 59
JEE-Mathematics BRAIN STORMING SUBJECTIVE EXERCISE EXERCISE - 04 [B] b xn1 (n 2)x2 (n 1)(a b)x nab bn 1 a n 1 1 . Prove that : dx (x a)2 (x b)2 2(a b) a If a , a and a are the three values of a which satisfy the equation /2 x)3 dx 4a /2 2 0 (sin 0 2 . x x cos 2 12 3 a cos x dx then find the value of 1000 a12 a 2 a 2 . 2 3 5 2 / 3 3 . Show that the sum of the two integrals e(x5)2 dx + 3 e9(x2 / 3)2 dx is zero. 4 1 / 3 if (0, ) if (, 2) dx 1 dx sin 0 2x cos 2x cos 2 4 . 2 Show that x2 1 = x2 1 = 0 sin 5. Evaluate : / 4 x2 (sin 2x cos 2x) dx (1 sin 2x ) cos2 x 0 1 6 . Comment upon the nature of roots of the quadratic equation x2 + 2x = k + t k dt depending on the value 0 of k R. 7 . If the derivative of f(x) wrt x is cos x then show that f(x) is a periodic function. f(x) 1 8 . Determine a positive integer n 5, such that ex (x 1)n dx = 16 6 e . 0 9. (a) If x < 1 prove that 1 2 x 2 x 4 x3 4 x3 8 x7 1 2 x . 1 x x2 1 x2 x4 1x4 x8 ...... 1 x x2 1 x1 x 1 x1 x (b) Prove the identity f(x) = tan x tan 22 tan ....... 2n1 tan cot 2 cot2x 22 22 2 n 1 2 n 1 2 n 1 NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\02-Definite integration.p65 1 1 0 . If f(x) = x + [xy² + x²y] f(y) dy where x and y are independent variables. Find f(x). 0 11. Given that U = {x(1 x)}n & n 2 prove that d2 U n = n (n 1) U 2 n(2n 1)U , n d x2 n2 n1 1 further if V = ex . U dx, prove that when n 2, V + 2n (2n 1).Vn1 n (n 1) V = 0 n n n n2 0 1 2 . Evaluate : Limit 1 1 22 32 ..... n2 1 / n n n2 1 n2 1 n2 1 (a) n 2 1 (b) For potitive integers n, let A = {(n + 1) + (n + 2) +.......(n + n)}, B = {(n + 1)(n + 2).......(n + n)}1/n. nn n If Lim A n ae where a, b N and relatively prime find the value of (a + b). n Bn b 60 E
JEE-Mathematics 1 m! n! (m n 1)! Prove that : (a) I = xm . (1 x)n dx = m, n 1 3 . m , n N. 0 1 n! (m 1)n1 (b) I = xm . (n x)n dx = (1)n m , n N. m, n 0 1 4 . Prove that the sum to (n +1) terms of C0 C1 C2 1 ....... equals x n1.(1 x)n1 dx n(n 1) (n 1)(n 2) (n 2)(n 3) 0 & evaluate the integral. 5 1 5 . Evaluate the integral : x 2 2x 4 x 2 2x 4 dx 3 xu x 16. Prove that f ( t ) dt du f ( u ).( x u)du 0 0 0 1 1 7 . Evaluate (tx 1 x)n dx , where n is a positive integer and t is a parameter independent of x. Hence show 0 that 1 dx = 1 for k = 0, 1, ........ n. [JEE 1981] [nCk (n + 1)] xk (1 x )n k 0 x 1 8 . If 'f' is a continuous function with f(t) dt as |x| , (0, 2) 0A then show that every line y = mx O (xp, 0) X [JEE 1991] B (0, –2) x intersects the curve y2 + f(t) dt = 2! 0 NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\02-Definite integration.p65 BRAIN STORMING SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(B) 2. 5250 5. 2 n2 6 . real & distinct k R 8 . n = 3 16 4 1 2 . (a) 2 e(1/2) ( 4) (b) 11 1 4 . n 1!(n 1)! 61 80 1 0 .f(x) = x + x + x² (2n 1)! 119 119 1 5 . 2 2 4 3 3 2 2 tn1 1 17. (t 1) (n 1) 3 E 61
JEE-Mathematics JEE-[MAIN] : PREVIOUS YEAR QUESTIONS EXERCISE - 05 [A] / 4 (3) /4 [AIEEE-2002] (4) n 1 . If In = tann x dx then the value of n(In–1 + In+1) is- 0 (1) 1 (2) /2 2 . 2x(1 sin x) = (2) 2/4 (3) /8 [AIEEE-2002] 1 cos2 x (4) 2/8 (1) 2 10 [AIEEE-2002] [AIEEE-2002] 3 . | sin x|dx = (1) 9 (2) 10 (3) 18 (4) 20 2 4 . [x2 ] dx is equal to (where [.] denotes greatest integer function) 0 (1) 2 – 1 (2) 2( 2 – 1) (3) 2 (4) none of these 1 5 . lim 1P 2P 3P .... nP equals - [AIEEE-2002] n nP 1 (3) P2 (4) P2 1 (1) 1 (2) P 1 d esin x 4 3 e sin x3 dx x 1 x 6 . Let F(x) = , x > 0. If dx = F(k) – F(1), then one of the possible values of k, is- [AIEEE-2003] (1) 64 (2) 15 (3) 16 (4) 63 b [AIEEE-2003] 7 . If f(a + b – x) = f(x), then x f(x) dx is equal to- a ab b ab b ab b ba b NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\02-Definite integration.p65 (1) f(a b x) dx (2) f(b x) dx (3) f(x) dx (4) f(x) dx 2 2 2 2 a a a a 1 [AIEEE-2003] 11 8 . The value of the integral I = x(1 x)n dx is- 0 (4) – n 1 n 2 11 1 1 (1) + (2) (3) n 1 n 2 n 1 n 2 x2 sec2 t dt 9 . The value of lim 0 x0 x sin x is - [AIEEE-2003] (1) 0 (2) 3 (3) 2 (4) 1 [AIEEE-2003] 10. lim (1)4 2 34 ... n4 – lim (1)3 23 33 .... n3 is equal to - n n5 n n5 (4) 1/4 (1) 1/5 (2) 1/30 (3) zero 62 E
JEE-Mathematics t [AIEEE-2003] 1 1 . If f(y) = ey, g(y) = y; y > 0 and F(t) = f(t y) g(y ) dy, then- 0 (1) F(t) = te–t (2) F(t) = 1 – e–1(1 + t) (3) F(t) = et – (1 + t) (4) F(t) = tet 12 . Let f(x) be a function satisfying f'(x) = f(x) with f(0) = 1 and g(x) be a function that satisfies 1 f(x) + g(x) = x2. Then the value of the integral f(x) g(x) dx is - [AIEEE-2003] 0 e2 3 e2 5 e2 5 e2 3 (4) e – – (1) e + + (2) e – – (3) e + – 22 22 22 22 lim n 1 er / n is- nn r 1 1 3 . [AIEEE-2004] (1) e (2) e – 1 (3) 1 – e (4) e + 1 3 [AIEEE-2004] 1 4 . The value of |1 x2 |dx is- (4) 1/3 2 [AIEEE-2004] (1) 28/3 (2) 14/3 (3) 7/3 (4) 3 (3) 2 [AIEEE-2004] / 2 (sin x cos x)2 (3) /4 1 5 . The value of I = dx is- (4) 2 0 1 sin 2x (1) 0 (2) 1 /2 1 6 . If x f(sin x) dx = A f (sin x) dx , then A is - 00 (1) 0 (2) ex f(a) f(a) I2 If f(x) = 1 e x I1 xg{x(1 x)}dx and I2 = g{x(1 x)}dx , then the value of f (a ) f (– a ) 1 7 . is- [AIEEE-2004] , I1 = (1) 2 (2) –3 (3) –1 (4) 1 18. lim 1 sec2 1 2 sec2 4 ... 1 se c2 1 equals- [AIEEE-2005] n2 n2 n2 n n n2 NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\02-Definite integration.p65 1 1 (3) tan1 1 (1) sec 1 (2) cosec 1 (4) tan1 2 2 2 112 2 1 9 . If I = 2 x2 dx, I = 2 x3 dx, I = 2 x2 dx and I = 2 x3 dx then- 1 2 3 4 [AIEEE-2005] 001 1 (1) I > I (2) I > I (3) I = I (4) I > I 2 1 1 2 3 4 3 4 1 lim f(x) 4t3 48 . Then 6 x 2 2 0 . x 2 Let f : R R be a differentiable function having f(2) = 6, f'(2) = dt equals - [AIEEE-2005] (1) 24 (2) 36 (3) 12 (4) 18 2 1 . The value of cos2 x dx, a > 0 is- [AIEEE-2005] 1 ax (4) 2 (1) a (2) (3) a 2 E 63
JEE-Mathematics 6x [AIEEE-2006] 2 2 . The value of the integral, dx is - 1 (4) 3 9x x 2 3 (2) 2 (3) 1 (1) 2 / 2 2 3 . [(x )3 cos2 (x 3)] dx is equal to- [AIEEE-2006] 3 / 2 (4) 4/32 (1) (4/32) + (/2) (2) /2 (3) (/4) – 1 2 4 . x f(sin x) dx is equal to- [AIEEE-2006] 0 /2 /2 (2) f (sin x)dx (1) f(sin x) dx 2 (3) f(cos x) dx (4) f(cos x)dx 0 0 0 0 a 2 5 . The value of [x]f '(x) dx , a > 1, where [x] denotes the greatest integer not exceeding x is- [AIEEE-2006] 1 (1) [a] f(a) – {f(1) + f(2) + ... + f([a])} (2) [a] f([a]) – {f(1) + f(2) + ... + f(a)} (3) a f([a]) – {f(1) + f(2) + ... + f(a)} (4) a f(a) – {f(1) + f(2) + ... + f([a])} 1 x log t [AIEEE-2007] 2 6 . Let F(x) = f(x) + f x , where f(x) = 1 1 t dt. Then F(e) equals- 1 (1) (2) 0 (3) 1 (4) 2 2 x dt 2 7 . The solution for x of the equation = is- [AIEEE-2007] 2 t t2 1 12 (1) 2 (2) (3) 3 /2 (4) 2 2 2 8 . Let I = 1 sin x dx and J = 1 cos x dx. Then which one of the following is true ? [AIEEE-2008] 0x 0x 2 2 2 2 (1) I > 3 and J > 2 (2) I < 3 and J < 2 (3) I < 3 and J > 2 (4) I > 3 and J < 2 z [AIEEE-2009] 2 9 . [cot x]dx , where [ . ] denotes the greatest integer function, is equal to - NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\02-Definite integration.p65 0 (1) –1 (4) 1 (2) – (3) 2 2 3 0 . Let p(x) be a function defined on R such that p'(x) = p'(1 – x), for all x [0, 1], p(0) = 1 and p(1) = 41. Then 1 (2) 21 (3) 41 [AIEEE-2010] (4) 42 p(x) dx equals :- 0 (1) 41 3 1 . The value of 1 8log(1 x) dx is :- [AIEEE-2011] 1 x2 0 E (2) log 2 (1) log 2 (3) log 2 (4) 8 log 2 64 2
JEE-Mathematics 3 2 . Let [.] denote the greatest integet function then the value of 1.5 xx2 dx is :- [AIEEE-2011] 0 5 (2) 0 3 3 (1) (3) (4) 4 2 4 x (3) g(x) + g() [AIEEE-2012] 3 3 . If g(x) cos 4t dt , then g(x + ) equals : (4) g(x) – g() 0 (1) g(x) . g() g(x) (2) g() / 3 dx 3 4 . tan x is equal to 6 . Statement-I : The value of the integral [JEE-MAIN-2013] /6 1 bb Statement-II : ƒ(x)dx ƒ(a b x)dx . aa (1) Statement-I is true, Statement-II is true; Statement-II is a correct explanation for Statement-I. (2) Statement-I is true, Statement-II is true; Statement-II is not a correct explanation for Statement-I. (3) Statement-I is true, Statement-II is false. (4) Statement-I is false, Statement-II is true. NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\02-Definite integration.p65 PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE-5 [A] Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Ans 1 1 3 1 2 1 3 4 4 1 3 4 2 1 3 Que. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Ans 2 1 4 2 4 2 1 2 3 1 1 1 2 2 2 Que. 31 32 33 34 Ans 3 4 3,4 4 E 65
JEE-Mathematics EXERCISE - 05 [B] JEE-[ADVANCED] : PREVIOUS YEAR QUESTIONS 1 . (a) The value of the integral e2 loge x dx is - xe 1 (A) 3/2 (B) 5/2 (C) 3 (D) 5 x (b) Let g(x) = f(t) dt, where f is such that 1 f(t) 1 for t (0, 1] and 0 f(t) 1 for t (1, 2]. 02 2 Then g(2) satisfies the inequality - (A) 3 g(2) 1 (B) 0 g(2) 2 35 (D) 2 g(2) 4 22 (C) 2 g(2) 2 ecos x .sin x for x 2 3 (c) If f(x) = . Then f(x)dx - 2 otherwise 2 (A) 0 (B) 1 (D) 3 (C) 2 [JEE 2000, Screening, 1+1+1+1M out of 35] x nt dt . Find the function f(x) + f(1/x) and show that, f(e) + f(1/e) = 1/2. ( d ) For x > 0, let f(x) = 11t [JEE 2000, (Mains) 5M out of 100] cos2 x dx, a The value of [JEE 2001] 2 . 1 ax 0 is - (A) (B) a (D) 2 (C) 2 [JEE 2001] x 3 . Let f : (0, ) R and F(x) = f(t)dt. If F(x2) = x2 (1 + x), then f(4) equals - 0 5 (B) 7 (C) 4 (D) 2 (A) 4 x NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\02-Definite integration.p65 4 . (a) Let f(x) = 2 t2 dt. Then the real roots of the equation x2 – f’ (x) = 0 are - 1 (A) ±1 1 1 (D) 0 and 1 (B) (C) 2 2 (b) Let T > 0 be a fixed real number. Suppose f is a continuous function such that for all x R f (x + T) = f(x). T 33T If I = f(x) dx then the value of f(2x) dx is - 03 (A) 3 I (B) 2I (C) 3I (D) 6I 2 1 2 1 x [ x ] 1 x (c) n dx equals - The integral 1 2 1 (B) 0 (C) 1 (D) 2 n 1 (A) 2 66 2 [JEE 2002 (Screening) 3+3+3M] E
JEE-Mathematics z1 5 . ( a ) If (m, n) = t m (1 t) n dt , then the expression for (m, n) in terms of (m + 1, n – 1) is - 0 (A) m (m + 1, n – 1) (B) n (m + 1, n – 1) n 1 m 1 2n n (m + 1, n – 1) 2n n (m + 1, n – 1) (C) (D) m 1 m 1 m 1 m 1 zx2 1 ( b ) If function f defined by f(x) = e -t2 dt increases in the interval - x2 (A) nowhere (B) x 0 (C) x [–2, 2] (D) x 0 [JEE 2003 (Screening) 3+3M] /2 /4 z z6 . If f(x) is an even function, then prove that f (cos 2x) cos x dx 2 f (sin 2x) cos x dx 00 [JEE 2003 (Mains) 2M out of 60] 1 1x 7 . ( a ) The value of the integral dx is - [JEE 2004] 0 1x (A) 1 (B) 1 (C) –1 (D) 1 2 2 t2 z GF JI( b ) If f(x) is differentiable and x f(x) dx 2 t 5 , then f 4 equals - H K5 25 0 2 (B) 5 (C) 1 5 (A) 5 2 (D) 2 [JEE 2004 (Screening)] y(x) x2 cos x. cos d , then find dy ( c )If 2 / 16 1 sin2 dx at x = . NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\02-Definite integration.p65 /3 4x3 dx ( d ) Evaluate : [JEE 2004 (Mains) 2+4M out of 60] | [JEE 2005 (Screening) 3+3M] / 3 2 cos x | 3 8. 1 1 is - 9. z GF JI( a ) If t2 (f (t)) dt (1 sin x) then f H Ksin x 3 E (A) 1/3 (B) 1 / 3 (C) 3 (D) 3 z0 ( b ) (x 3 3x 2 3x 3 (x 1) cos(x 1)) dx is equal to - 2 (A) – 4 (B) 0 (C) 4 (D) 6 z GFH HGF IKJ FHG KIJIJEvaluate e|cos x| 2 sin 1 cos x 3 cos 1 cos x sin x dx [JEE 2005, (Mains), 2M out of 60] K2 2 0 67
JEE-Mathematics 10 to 12 are based on the following Comprehension zb f(x) dx b a (f(a ) f(b)) 2 Suppose we define the definite integral using the following formula , for more accurate a result for c (a , b) F(c) c a (f(a) f(c)) b c (f(b) f(c)) . 22 zWhen b ba 4 c a b , f(x) dx (f(a) f(b) 2f(c)) 2 a z1 0 . sin x dx is equal to - /2 0 (A) (1 2) (B) (1 2 ) 8 4 (C) 8 2 (D) 4 2 [JEE 2006, 5M out of 184] 1 1 . If f''(x) 0, x (a , b) and c is a point such that a < c < b and (c, f(c)) is the point lying on the curve for which F(c) is maximum then f'(c) is equal to - f(b) f(a) 2(f(b) f(a)) 2(f(b) f(a)) (D) 0 (A) b a (B) b a (C) 2b a t [JEE 2006, 5M out of 184] z FG IJf(x) dx t a (f(t) f(a)) H K2 1 2 . If f(x) is a polynomial and if lim a 0 for all a, then the degree of f(x) can atmost be - (t a)3 ta (A) 1 (B) 2 (C) 3 (D) 4 z1 [JEE 2006, 5M out of 184] 5050 (1 x 50 )100 dx 1 3 . The value of z0 is. [JEE 2006, 6M] 1 (1 x 50 )101 dx 0 [JEE 2006, (1.5, +1.5)M out of 184] 1 4 . Match the following : Column-I Column-II NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\02-Definite integration.p65 (p) 4/3 /2 (q) 1 z l q(A) (sin x)cos x cos x cot x sin x . ln(sin x) dx 0 z z1 0 z z1 0 (r) 1 x dx 1 x dx (B) (1 y 2 ) dy (y 2 1) dy 0 1 01 sec2 x f(t)dt 1 5 . Lim 2 equals - [JEE 2007] x x2 2 4 16 8 2 (C) 2 f 1 (D) 4f(2) (A) f(2) (B) f(2) 2 68 E
JEE-Mathematics 1 6 . Match the integrals in Column-I with the values in Column-II [JEE 2007, 6M] Column-I Column-II 1 dx (p) 1 log 2 2 3 (A) 1 1 x2 (q) 2 lo g 2 1 dx 3 (B) 0 1 x2 (C) 3 dx 2 1 x2 (r) 3 (D) 2 dx (s) x2 1 1x 2 nn n 1 n 1 7 . [JEE 2008, 4M] Let Sn = n2 kn k2 and Tn = n2 kn k2 k 1 k 0 for n = 1, 2, 3, ........ Then, (A) Sn < 3 3 (B) Sn > 3 3 (C) Tn < 3 3 (D) Tn > 3 3 xx 1 8 . Let ƒ be a non-negative function defined on the interval [0, 1]. If 1 (ƒ '(t))2 dt ƒ(t)dt, 0 x 1, and 00 ƒ (0) = 0, then - [JEE 2009, 3M, –1M] 1 1 1 1 1 1 1 1 (A) ƒ 2 < 2 and ƒ 3 > 3 (B) ƒ 2 > 2 and ƒ 3 > 3 1 1 1 1 1 1 1 1 (C) ƒ 2 < 2 and ƒ 3 < 3 (D) ƒ 2 > 2 and ƒ 3 < 3 sin nx dx, n = 0, 1, 2, ..., then - [JEE 2009, 4M, –1M] 1 9 . If I = NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\02-Definite integration.p65 n 1 x sin x (A) I = I 10 10 (D) I = I n n+2 n n+1 (B) I2m1 10 (C) I2m 0 m 1 m 1 x 2 0 . Let f: R R be a continuous function which satisfies f(x) = f(t)dt .Then the value of f(ln 5) is................ 0 [JEE 2009, 4M, –1M] 2 1 . lim 1 x tn(1 t)dt [JEE 10, 3M, –1M] The value of x3 0 t4 4 is x 0 (A) 0 1 1 1 (B) (C) (D) 64 12 24 The of 1 x4 (1 x)4 dx is (are) 22. value(s) 0 1 x2 [JEE 10, 3M] E (A) 22 2 (C) 0 (D) 71 3 7 (B) 105 15 2 69
JEE-Mathematics 2 3 . For any real number x, let [x] denote the largest integer less than or equal to x. Let f be a real valued function defined on the interval [–10, 10] by f(x) x [x] x if [x] is odd, 1 [x] if [x]is even Then the value of 2 10 f(x) cos x dx is [JEE 10, 3M] 10 10 x 2 4 . Let f be a real-valued function defined on the interval (–1,1) such that ex f(x) 2 t4 1dt , for all 0 x (1,1), and let f–1 be the inverse function of f. Then (f–1)' (2) is equal to - [JEE 10, 5M, –2M] (A) 1 1 1 1 (B) 3 (C) (D) e 2 ln 3 x sin x2 dx is [JEE 2011, 3 (–1)M] 2 5 . The value of sin x2 sin(ln 6 x2 ) ln 2 (A) n 3 (B) n 3 (C) n 3 (D) n 3 42 22 2 62 2 6 . Let S be the area of the region enclosed by y ex2 , y = 0, x = 0, and x = 1. Then - [JEE 2012, 4M] (A) S 1 (B) S 1 1 e e (C) S 1 1 (D) S 1 1 1 1 4 1 e 2 e 2 2 7 . /2 x 2 ln x cos xdx is [JEE 2012, 3M, –1M] The value of the integral / 2 x 2 (D) (A) 0 (B) 2 4 (C) 2 4 2 2 2 2 8 . For a R (the set of all real numbers), a –1. lim (1a 2a ...... n a ) 1 NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\02-Definite integration.p65 n (n 1)a1 [(na 1) (na 2) ...... (na n )] 60 Then a = [JEE(Advanced) 2013, 3, (–1)M] (A) 5 (B) 7 15 17 (C) (D) 2 2 PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE-5 [B] 1 . (a) B (b) B (c) C (d) 1 n2 x 2. C 3 . C 4. (a) A; (b) C; (c) B 5 . (a) D (b) B 2 HGFe cos 1KJI 7 . (a) B (b) A (c) 2 (d) 4 tan 1 1 8 . (a) C (b) C 9. 24 HFG 1 IKJ e sin HFG 1 IKJ 32 5 2 2 2 1 0 . A 1 1 . A 12. A 13. 5051 14. (A)(q), (B)(p, r) 15. A 1 6 . (A)(s); (B)(s); (C)(p); (D)(r) 1 7 . A, D 18. C 19. A,B,C 20. 0 B 21. B 22. A 23. 4 24. B 25. A 26. A,B,D 27. B 28. 70 E
JEE-Mathematics EXERCISE - 01 CHECK YOUR GRASP SELECT THE CORRECT ALTERNATIVE (ONLY ONE CORRECT ANSWER) ab bc ca 1 . The value of determinant b c c a a b is equal to - ca ab bc (A) abc (B) 2abc (C) 0 (D) 4abc 2. sin 2x cos2 x cos 4x 3. 4. If cos2 x cos 2x sin2 x = a0 + a1 (sinx) + a2 (sin2x) +.......+ an (sinnx) then the value of a0 is - 5. cos4 x sin2 x sin 2x 6. 7. (A) –1 (B) 1 (C) 0 (D) 2 8. 9. a b c The value of the determinant a b c is equal to - E a b c (A) 0 (B) (a – b)(b – c)(c – a) (C) (a + b)(b + c)(c + a) (D) 4abc For any ABC, the value of determinant sin2 A cot A 1 is equal to - sin2 B cot B 1 sin2 C cot C 1 (A) 0 (B) 1 (C) sin A sin B sin C (D) sin A + sin B + sin C If Dp = p 15 8 p2 35 9 , then D1 + D2 + D3 + D4 + D5 is equal to - p3 25 10 (A) 0 (B) 25 (C) 625 (D) none of these E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\01.Determinants\\02.EXERCISES.p65 sin(A B C) sin B cos C If A + B + C = , then sin B 0 tan A is equal to - cos(A B) tan A 0 (A) 0 (B) 2 sin B tan A cos C (C) 1 (D) none of these The number of real values of x satisfying x 3x 2 2x 1 = 0 is - 2x 1 4x 3x 1 7x 2 17x 6 12x 1 (A) 3 (B) 0 (C) 1 (D) infinite log a p 1 is equal to - If a, b, c are pth, qth and rth terms of a GP, then log b q 1 log c r 1 (A) 0 (B) 1 (C) log abc (D) pqr If a , a ,.......a , a ,...... are in GP and a > 0 i, then log an log an2 log an4 is equal to - 12 n n+1 i log an6 log an8 log an10 log an12 log an14 log a n 16 (A) 0 (B) n log a (C) n(n + 1) log a (D) none of these n n 17
JEE-Mathematics 1 0 . If px4 + qx3 + rx2 + sx + t = x2 3x x 1 x3 then t is equal to - x 1 2x x 3 x 3 x4 3x (A) 33 (B) 0 (C) 21 (D) none 1 logx y logx z 1 1 . For positive numbers x, y and z, the numerical value of the determinant logy x 1 logy z is - logz x logz y 1 (A) 0 (B) log xyz (C) log(x + y + z) (D) logx logy logz (ax ax )2 (ax ax )2 1 1 2 . If a, b, c > 0 and x, y, z R, then the determinant (by by )2 (by by )2 1 is equal to - (cz cz )2 (cz cz )2 1 (A) axbycx (B) a–xb–yc–z (C) a2xb2yc2z (D) zero a2 b2 c c c b2 c2 a = abc, then the values of is - 1 3 . For a non-zero real a, b and c a a c2 a2 b b b (A) –4 (B) 0 (C) 2 (D) 4 (1 x)2 (1 x)2 (2 x2 ) (1 x)2 2x 1 x 1 1 4 . The equation 2x 1 3x 3x 2x 2x 1 5 x (1 x)2 2x 3 =0 x 1 3x 2 2 3x 1 2x (A) has no real solution (B) has 4 real solutions (C) has two real and two non-real solutions (D) has infinite number of solutions, real or non-real a bc px qy rz E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\01.Determinants\\02.EXERCISES.p65 1 5 . Let a determinant is given by A = p q r and suppose determinant A = 6. If B = a x b y c z xyz ap bq cr then - (B) det. B = –6 (C) det. B = 12 (D) det. B = –12 (A) det. B = 6 abc 1 6 . If a b c and a2 b2 c2 = 0 then - bc ca ab (A) a + b+ c = 0 (B) ab + bc + ca = 0 (C) a2 + b2 + c2 = ab + bc + ca (D) abc = 0 b2c2 bc b c 1 7 . If a, b, & c are nonzero real numbers, then c2a2 ca c a is equal to - a2b2 ab a b (A) a2b2c2(a + b + c) (B) abc(a + b + c)2 (C) zero (D) none of these 1 x x 1 x(x 1) 1 8 . If f(x) = 2x x(x 1)(x 2) (x 1)x , then f (100) is equal to - [JEE 98] (x 1)x (x 1) 3x(x 1) (B) 1 E (A) 0 (C) 100 (D) –100 18
JEE-Mathematics ab0 1 9 . The value of the determinant 0 a b is equal to - b0a (A) a3 – b3 (B) a3 + b3 (C) 0 (D) none of these 2 0 . An equilateral triangle has each of its sides of length 6 cm. If (x1, y1); (x2, y2) & (x3, y3) are its vertices then the x1 y1 1 2 value of the determinant, x2 y2 1 is equal to - x3 y3 1 (A) 192 (B) 243 (C) 486 (D) 972 2 1 . If the system of equations x + 2y + 3z =4, x + py + 2z = 3, x + 4y + z = 3 has an infinite number of solutions, then - (A) p = 2, µ = 3 (B) p =2, µ= 4 (C) 3p = 2µ (D) none of these SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS) aax 2 2 . If m m m 0 , then x may be equal to - bxb (A) a (B) b (C) a + b (D) m sin 2x ex sin x x cos x sin x x2 cos x 2 3 . If D(x) cos x sin x ex x 1 x2 , then the value of |n cos (Dx)| will be - ex cos x e2x ex (A) independent of x (B) dependent on x (C) 0 (D) non-existent 2 4 . The value of the determinant x n is x (A) independent of (B) independent of n (C) (x – )(x – ) (D)(x – )(x – n) E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\01.Determinants\\02.EXERCISES.p65 2 5 . If the system of linear equations x + ay + az = 0, x + by + bz = 0, x + cy + cz = 0 has a non-zero solution then (A) System has always non-trivial solutions. (B) System is consistent only when a = b = c (C) If a b c then x = 0, y = t, z=–t t R (D) If a = b = c then y = t1, z = t2, x = –a(t1+ t2) t1,t2 R 2 6 . If the system of equations x + y – 3 = 0, (1 + K ) x + (2 + K ) y – 8 = 0 & x – (1 + K) y + (2 + K) = 0 is consistent then the value of K may be - (A) 1 3 5 (D) 2 (B) 5 (C) – 3 CHECK YOUR GRASP ANSWER KEY EXERCISE-1 Que. 1 2 3 4 5 6 7 8 9 10 Ans. C A D A D A D A A C Que. 11 12 13 14 15 16 17 18 19 20 Ans. A D D D C A C A B D Que. 21 22 23 24 25 26 Ans. D A,B A,C B,C A,C,D A,C E 19
JEE-Mathematics BRAIN TEASERS EXERCISE - 02 SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS) 1 . Which of the following determinant(s) vanish(es) ? 1 bc bc(b c) 1 ab 1 1 (B) 1 ab (A) 1 ca ca(c a) 1 ab ab(a b) 1 bc 1 1 bc 11 ca ca 0 a b a c logx xyz logx y logx z 0 bc 1 logy z (C) b a (D) logy xyz c a cb 0 logz xyz logz y 1 mx mx p mx p 2 . If f'(x) = n np n p , then y = f(x) represents - mx 2n mx 2n p mx 2n p (A) a straight line parallel to x–axis (B) a straight line parallel to y–axis (C) parabola (D) a straight line with negative slope a2 a2 (b c)2 bc 3 . The determinant b2 b2 (c a )2 ca is divisible by - c2 c2 (a b)2 ab (A) a + b + c (B) (a + b) (b + c) (c + a) (C) a2 + b2 + c2 (D) (a – b)(b – c) (c – a) a b a b 4 . The determinant b c b c is equal to zero, if - a b b c 0 E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\01.Determinants\\02.EXERCISES.p65 (A) a, b, c are in AP (B) a, b, c are in GP (C) is a root of the equation ax2+bx+c=0 (D) (x– ) is a factor of ax2 + 2bx + c 1 sin2 x cos2 x 4 sin 2x 5 . Let f(x) = sin2 x 1 cos2 x 4 sin 2x , then the maximum value of f(x) = sin2 x cos2 x 1 4 sin 2x (A) 2 (B) 4 (C) 6 (D) 8 1 a a2 6 . The parameter on which the value of the determinant cos(p d)x cos px cos(p d)x does not depend upon sin(p d)x sin px sin(p d)x is- (A) a (B) p (C) d (D) x 1 a2 x (1 b2 )x (1 c2 )x 7 . If a2 + b2 + c2 = -2 and f(x) (1 a2 )x 1 b2 x (1 c2 )x , then f(x) is a polynomial of degree- (1 a2 )x (1 b2 )x 1 c2 x (A) 2 (B) 3 (C) 0 (D) 1 20 E
JEE-Mathematics p q px qy 8 . Given that q2 –pr < 0, p > 0, then the value of q r qx ry is- px qy qx ry 0 (A) zero (B) positive (C) negative (D) q2 + pr 9 . The value of lying between & and 0 A and satisfying the equation 42 2 1 sin2 A cos2 A 2 sin 4 sin2 A 1 cos2 A 2 sin 4 = 0 are - sin2 A cos2 A 1 2 sin 4 (A) A = , (B) A = 3 (C) A , (D) A , 3 48 8 58 6 8 1 0 . The set of equations x – y + 3z = 2, 2x – y + z = 4, x – 2y + z = 3 has - (A) unique solution only for = 0 (B) unique solution for 8 (C) infinite number of solutions of = 8 (D) no solution for = 8 E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\01.Determinants\\02.EXERCISES.p65 BRAIN TEASERS ANSWER KEY EXERCISE-2 Que. 1 2 3 45 6 7 8 9 10 Ans. A,B,C,D A A,C,D B,D C B A C A,B,C,D B,D E 21
JEE-Mathematics EXERCISE - 03 MISCELLANEOUS TYPE QUESTIONS TRUE / FALSE abc 1 . If a, b, c are sides of scalene triangle, then the value of b c a is positive. cab ahg 2 . If ax2 + 2hxy + by2 + 2gx + 2fy + c (1x + m1y + n1) (2x + m2y + n2), then h b f = 0. gfc 3 . If x = cy + bz, y = az + cx, z = bx + ay, where x, y, z are not all zero, then a2 + b2 + c2 + 2abc + 1 = 0. 333 3 33 x1 x2 x3 2 4 .2 2 2 If x i y i z i 1 and xi yi yizi zi xi 0 then y1 y2 y3 1 i 1 i 1 i 1 i 1 i 1 i 1 z1 z2 z3 5 . Consider the system of equations aix + biy + ciz = di where i = 1, 2, 3. a1 b1 c1 d1 b1 c1 a1 d1 c1 a1 b1 d1 If a2 b2 c2 = d2 b2 c2 = a2 d2 c2 = a2 b2 d2 = 0 a3 b3 c3 d3 b3 c3 a3 d3 c3 a3 b3 d3 then the system of equations has infinite solutions. MATCH THE COLUMN Following question 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 ap x uf (p) 3 (A) If the determinant b q m y v g E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\01.Determinants\\02.EXERCISES.p65 cr nz wh splits into exactly K determinants of order 3, (q) 8 (r) 5 each element of which contains only one term, then the values of K is (B) The values of for which the system of equations x + y + z = 6, x + 2y + 3z = 10 & x + 2y + z = 12 is inconsistent (C) If x, y, z are in A.P. then the value of the determinant a 2 a 3 a 2x is a 3 a 4 a 2y a 4 a 5 a 2z (D) Let p be the sum of all possible (s) 0 determinants of order 2 having 0, 1, 2 & 3 as their four elements E (without repeatition of digits). The value of 'p' is 22
JEE-Mathematics 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. a1 a2 a3 1 . Statement - I : Consider D = b1 b2 b3 c1 c2 c3 Let B1, B2, B3 be the co-factors of b1, b2, and b3 respectively then a1B1 + a2B2 + a3B3 = 0 Because Statement - II : If any two rows (or columns) in a determinant are identical then value of determinant is zero. (A) A (B) B (C) C (D) D 2 . Statement - I : Consider the system of equations, 2x + 3y + 4z = 5 x+y+z=1 x + 2y + 3z = 4 This system of equations has infinite solutions. Because Statement - II : If the system of equations is e1 : a1x + b1y + c1z – d1 = 0 & a1 b1 e2 : a2x + b2y + c2z – d2 = 0 a2 b2 e3 : e1 + e2 = 0, where R Then such system of equations has infinite solutions. (A) A (B) B (C) C (D) D 3 . Statement - I : If a, b, c R and a b c and x,y,z are non zero. Then the system of equations E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\01.Determinants\\02.EXERCISES.p65 ax + by + cz = 0 bx + cy + az = 0 cx + ay + bz = 0 has infinite solutions. Because Statement - II : If the homogeneous system of equations has non trivial solution, then it has infinitely many solutions. (A) A (B) B (C) C (D) D COMPREHENSION BASED QUESTIONS Comprehension # 1 x x3 x4 1 Let x, y, z R+ & D = y y3 y4 1 z z3 z4 1 On the basis of above information, answer the following questions : 1. If x y z & x, y, z are in GP and D = 0, then y is equal to - 2. (A) 1 (B) 2 (C) 4 (D) none of these E If x, y, z are the roots of t3 – 21t2 + bt – 343 = 0, b R, then D is equal to- (A) 1 (B) 0 (C) dependent on x, y, z (D) data inadequate 23
JEE-Mathematics 3 . If x y z & x, y, z are in A.P. and D = 0, then 2xy2z + x2z2 is equal to- (A) 1 (B) 2 (C) 3 (D) none of these Comprehension # 2 Consider the system of linear equations x + y + z = m x + y + z = n and x + y + z = p On the basis of above information, answer the following questions : 1 . If 1, – 2 then the system has - (A) no solution (B) infinte solutions (C) unique solution (D) trivial solution if m n p 2 . If = –2 & m + n + p 0 then system of linear equations has - (A) no solution (B) infinite solutions (C) unique solution (D) finitely many solution 3 . If = 1 & m p then the system of linear equations has - (A) no solution (B) infinite solutions (C) unique solution (D) unique solution if p = n MISCELLANEOUS TYPE QUESTION ANSWER KEY EXERCISE-3 E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\01.Determinants\\02.EXERCISES.p65 True / False 3. F 4. T 5. F E 1. F 2. T Match the Column 1. (A) (q); (B) (p); (C) (s); (D) (s) Assertion & Reason 1. A 2. A 3. A Comprehension Based Questions Comprehension # 1 : 1. A 2. B 3. C 3. A Comprehension # 2 : 1. C 2. A 24
EXERCISE - 04 [A] JEE-Mathematics CONCEPTUAL SUBJECTIVE EXERCISE 1 . Without expanding the determinant prove that : 0 b c 0 pq pr (a) b 0 a = 0 (b) q p 0 q r = 0 c a 0 rp rq 0 2 . Prove that : ax by cz a b c 1 a a2 b c (a) x2 y2 z2 = x y z (b) 1 b b2 c a = 0 1 1 1 yz zx xy 1 c c2 a b a2 2 a 2 a 1 1 3 . Prove that : 2 a 1 a 2 1 = (a 1)3 3 31 18 40 89 4 . Using properties of determinants or otherwise evaluate 40 89 198 . 89 198 440 abc bc ca ab 5 . If D = c a b and D = a b b c c a then prove that D = 2 D. bca ca ab bc 1 a2 b2 2 ab 2 b 6 . Prove that 2 ab 1 a2 b2 2 a = (1 + a² + b²)3 . 2 b 2 a 1 a2 b2 E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\01.Determinants\\02.EXERCISES.p65 a bc cb 7 . Prove that a c b c a (a b c)(a2 b2 c2 ) ab ba c x 2 2x 3 3x 4 8 . Solve for x, 2 x 3 3 x 4 4 x 5 = 0. 3 x 5 5 x 8 10 x 17 ax c b 9 . If a + b + c = 0 , solve for x : c b x a = 0. b a cx bc bc ' b 'c b 'c ' 1 0 . Prove that ca ca ' c 'a c 'a ' (ab ' a 'b)(bc ' b 'c)(ca ' c 'a) . ab ab ' a 'b a ' b ' 1 1 . Let the three digit numbers A28, 3B9, and 62C, where A, B, and C are integers between 0 and 9, be A36 divisible by a fixed integer k. Show that the determinant 8 9 C is divisible by k. 2B2 E 25
JEE-Mathematics n! (n 1)! (n 2)! D (n !)3 12. For a fixed positive integer n, if D = (n 1)! (n 2)! (n 3)! then show that 4 is divisible by n. (n 2)! (n 3)! (n 4)! 2r 1 2 3r 1 4 5r 1 n 1 3 . If Dr = x y z then prove that Dr = 0. 2n 1 3n 1 5n 1 r 1 1 1 (x y) z z2 z 1 x 1 14. Find the value of the determinant (y z) x 2y z x x2 xz y(x y) xz2 y (y z) x2z 4 2 1 1 5 . Prove that 4 2 1 = 64( ) ( )( ) ( ) ( ) ( ) 4 2 1 a1 l1 b1 m1 a1 l2 b1 m2 a1 l3 b1 m 3 1 6 . Show that a2 l1 b2 m1 a2 l2 b2 m 2 a2 l3 b2 m 3 = 0. a3 l1 b3 m1 a3 l2 b3 m2 a3 l3 b3 m3 1 7 . Solve the following sets of equations using Cramer’s rule and remark about their consistency. xy z60 x 2y z 1 x 3y z 2 7x 7 y 5 z 3 (a) 2 x y z 1 0 (b) 3 x y z 6 (c) 3 x y z 6 (d) 3 x y 5z 7 x y 2z 3 0 x 2y 0 5x y 3z 3 2 x 3y 5 z 5 1 8 . Investigate for what values of , the simultaneous equations x + y + z = 6 ; x + 2 y + 3 z = 10 & E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\01.Determinants\\02.EXERCISES.p65 x + 2 y + z = have : (a) A unique solution. (b) An infinite number of solutions. (c) No solution. 1 9 . Find the values of c for which the equations 2x +3y = 0 (c + 2) x + (c + 4)y = c + 6 (c + 2)2x + (c + 4)2 y = (c + 6)2 are consistent. Also solve above equations for these values of c. 20. Let 1, 2 and 1, 2 be the roots of ax2 + bx + c = 0 and px2 + qx + r = 0 respectively. If the system of b2 ac equations 1y + 2z = 0 and 1y + 2z = 0 has a non-trivial solution, then prove that . q2 pr CONCEPTUAL SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(A) 4. 1 8 . x = 1 or x = 2 9 . x = 0 or x = ± 3 a2 b2 c2 14. 0 2 1 7 . (a) x = 1 , y = 2 , z = 3; consistent (b) x = 2 , y = 1 , z = 1 ; consistent 13 7 35 (d) inconsistent (c) x = 3 , y = 6 , z = 6 ; consistent 1 8 . (a) 3 (b) = 3, = 10 (c) = 3, 10 10 E 1 9 . c = –6,–1, for c = –6, x = 0 = y & for c=–1, x = –5 , y = 3 26
JEE-Mathematics EXERCISE - 04 [B] BRAIN STORMING SUBJECTIVE EXERCISE 1 . Let a, b, c, d be real numbers in G.P. If u, v, w satisf y the system of equations u + 2v + 3w = 6, 4u + 5v + 6w = 12, 6u + 9v = 4, then show that the roots of the equations 1 1 1 x2 + [ (b – c)2 + (c – a)2 + (d – b)2 ] x + u + v + w = 0 and 20x2 +10 (a – d)2 x – 9 = 0 are reciprocals u v w of each other. [JEE 99] bc a2 c a b2 a b c2 2 . Prove that b c c a a b b c c a a b b c c a a b = 3 . (b c) (c a) (a b) (a + b + c) (ab + bc + ca) (a b) (a c) (b c) (b a) (c a) (c b) 3 . If a2 + b2 + c2 = 1 then show that the value of the determinant a2 (b2 c2 ) cos ba(1 cos ) ca(1 cos ) simplifies to cos2 ab(1 cos ) b2 (c2 a2 ) cos cb(1 cos ) ac(1 cos ) c2 (a2 b2 ) cos bc(1 cos ) cos(x y) cos(y z) cos(z x) 4 . Find the value of the determinant cos(x y ) cos(y z) cos(z x) . sin(x y) sin(y z) sin(z x) S0 S1 S2 5 . If Sr = r + r + r then show that S1 S2 S3 = ( )2 ( )2 ( )2 . 6 . If S2 S3 S4 ax1² + by1² + cz12 = ax22 + by22 + cz22 = ax32 + by32 + cz32 = d and ax x + by y + cz z = ax x + by y + cz z = ax x + by y + cz z = f, 2 3 23 23 3 1 31 31 1 2 12 12 then prove that x1 y1 z1 = (d f) d 2 f 1/2 (a , b , c 0) x2 y2 z2 abc x3 y3 z3 E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\01.Determinants\\02.EXERCISES.p65 7 . If u = ax2 + 2 bxy + cy2 , u = ax2 + 2 bxy + cy2, then prove that- y2 xy x2 ax by bx cy 1 u u a b c ax by . ax by bx cy y ax by a b c z a y a2x a3 0 8. Solve the system of equations : z by b2 x b3 0 where a b c. z cy c2 x c3 0 9 . If x,y,z are not all zero and if ax + by + cz = 0; bx + cy + az = 0; cx + ay + bz = 0 Prove that x : y : z = 1 : 1 : 1 or 1 : : 2 or 1 : 2 : . 1 0 . Prove that the system of equations in x and y ; ax + hy + g = 0, hx + by + ƒ = 0, ax2 + 2hxy + by2 + 2gx + 2ƒy ahg ah + c = t is consistent if t h b ƒ hb gƒc BRAIN STORMING SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(B) 4. 2sin(x – y) sin(y – z) sin(x – z) 8 . x = (a + b + c) , y = ab + bc + ca , z = abc E 27
JEE-Mathematics JEE-[MAIN] : PREVIOUS YEAR QUESTIONS EXERCISE - 05 [A] log a p 1 1 . If a, b, c are pth, qth and rth terms of a GP, and all are positive then log b q 1 is equal to- [AIEEE-2002] log c r 1 (1) 0 (2) 1 (3) log abc (4) pqr 1 n 2n [AIEEE-2003] 2 . If 1, , 2 are cube roots of unity and n 3p, p Z, then 2n 1 n is equal to- n 2n 1 (1) 0 (2) (3) 2 (4) 1 a a2 1 a3 3 . If b b2 1 b3 = 0 and vectors (1 , a , a2 ), (1 , b,b2) and (1 , c, c2) are non-coplanar, then the product abc c c2 1 c3 equals- [AIEEE-2003] (1) 1 (2) 0 (3) 2 (4) –1 log an log an2 log an4 4. If a , a ,.......a , a ,...... are in GP and a > 0 i, then log an6 log an8 log an10 is equal to- 12 n n+1 i log a n 12 log a n 14 log a n 16 [AIEEE-04,05] (1) 0 (4) none of these (2) n log a (3) n(n + 1) log a n n 1 a2 x (1 b2 )x (1 c2 )x 5 . If a2 + b2 + c2 = –2 and f(x) (1 a2 )x 1 b2 x (1 c2 )x , then f(x) is a polynomial of degree- [AIEEE 2005] (1 a2 )x (1 b2 )x 1 c2 x (1) 2 (2) 3 (3) 0 (4) 1 E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\01.Determinants\\02.EXERCISES.p65 6 . The system of equations x + y + z = – 1 x + y + z = – 1 x + y + z = – 1 has no solution, If is [AIEEE 2005] (1) 1 (2) not –2 (3) either -2 or 1 (4) –2 11 1 7 . If D = 1 1 x 1 for x 0, y 0 then D is- [AIEEE - 2007] 1 1 1y (1) Divisible by both x and y (2) Divisible by x but not y (4) Divsible by neither x nor y (3) Divisible by y but not x 5 5 [AIEEE - 2007] 8 . Let A = 0 5 , if |A2| = 25 then || equals- 0 0 5 (1) 5 (2) 52 (3) 1 (4) 1/5 9 . Let a, b, c be any real numbers. Suppose that there are real numbers x, y, z not all zero such that x = cy + bz, y = az + cx and z = bx + ay, then a2 + b2 + c2 + 2abc is equal to [AIEEE - 2008] (1) 2 (2) –1 (3) 0 (4) 1 28 E
JEE-Mathematics a a 1 a 1 a 1 b1 c1 1 0 . Let a, b, c be such that b(a + c) 0. If b b 1 b 1 + a 1 b 1 c 1 = 0, c c 1 c 1 (1)n2 a (1)n1 b (1)n c then the value of n is :- [AIEEE - 2009] (1) Any odd integer (2) Any integer (3) Zero (4) Any even integer 1 1 . Consider the system of linear equations : x1 + 2x2 + x3 = 3 2x1 + 3x2 + x3 = 3 3x1 + 5x2 + 2x3 = 1 The system has [AIEEE - 2010] (1) Infinite number of solutions (2) Exactly 3 solutions (3) A unique solution (4) No solution 1 2 . The number of values of k for which the linear equations 4x + ky + 2z = 0 kx + 4y + z = 0 2x + 2y + z = 0 possess a non-zero solution is :- [AIEEE - 2011] (1) 1 (2) zero (3) 3 (4) 2 1 3 . If the trivial solution is the only solution of the system of equations x – ky + z = 0 kx + 3y – kz = 0 3x + y – z = 0 Then the set of all values of k is: [AIEEE - 2011] (1) {2, –3} (2) R – {2, –3} (3) R – {2} (4) R – {–3} E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\01.Determinants\\02.EXERCISES.p65 PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE-5 [A] Que. 1 2 3 456 7 8 9 10 114 1 441 Ans. 1 1 4 Que. 11 12 13 Ans. 4 4 2 E 29
JEE-Mathematics JEE-[ADVANCED] : PREVIOUS YEAR QUESTIONS EXERCISE - 05 [B] a2 a 1 1 . Solve for x the equation sin(n 1)x sin nx sin(n 1)x 0 cos(n 1)x cos nx cos(n 1)x [REE 2001, (Mains), 3 out 100] 2 . Test the consistency and solve them when consistent, the following system of equations for all values of x+y+z=1 x + 3y – 2z = 3x +( + 2)y – 3z = 2 +1 [REE 2001,(Mains), 5 out 100] 3 . Let a, b, c, be real numbers with a2 + b2 + c2 = 1, Show that the equation ax by c bx ay cx a represents a straight line. bx ay ax by c cy b 0 cx a ax by c cy b [JEE 2001,(Mains), 6 out 100] 4 . The number of values of k for which the system of equations (k +1) x + 8y = 4k kx + (k +3)y = 3k – 1 has infinitely many solutions is [JEE 2002,(Screening), 3] (A) 0 (B) 1 (C) 2 (D) infinite 5 . The value of for which the system of equations 2x – y – z = 12, x – 2y + z = –4, x + y + z = 4 has no solution is [JEE 2004 (Screening)] (A) 3 (B) –3 (C) 2 (D) –2 E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\01.Determinants\\02.EXERCISES.p65 6 . (a) Consider three point P = (–sin( – ), – cos), Q = (cos( – ), sin) and R = (cos( ), sin( )), where 0 < /4 (A) P lies on the line segment RQ (B) Q lies on the line segment PR (C) R lies on the line segment QP (D) P, Q, R are non collinear (b) Consider the system of equations x – 2y + 3z = –1; –x + y – 2z = k; x – 3y + 4z = 1. Statement-I : The system of equations has no solution for k 3. and 1 3 1 [JEE 2008, 3+3] Statement-II : The determinant 1 2 k 0, for k 3. 141 (A) Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I. E (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. 30
JEE-Mathematics 7 . The number of all possible values of , where 0 < , for which the system of equations (y + z)cos3 = (xyz)sin3 x sin 3 2 cos 3 2 sin 3 yz (xyz)sin3 = (y + 2z)cos3 + ysin3 have a solution (x , y, z) with yz 0, is [JEE 2010, 3] 0 0 0 00 E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\01.Determinants\\02.EXERCISES.p65 PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE-5 [B] 1 . x = n, n I 2 . If = 5, system is consistent with infinite solution given by z = K, y 1 (3K 4) and x 1 (5K 2) where K R 2 2 If 5 , system is consistent with unique solution given by z 1 (1 ); x 1 ( 2) and y = 0. 33 4. B 5. D 6. (a) D; (b) A 7. 3 E 31
JEE-Mathematics DETERMINANT 1. INTRODUCTION : If the equations a1x + b1 = 0, a2x + b2 = 0 are satisfied by the same value of x, then a1b2 – a2b1 = 0. The expression a b – a b is called a determinant of the second order, and is denoted by : 12 21 a1 b1 a2 b2 A determinant of second order consists of two rows and two columns. Next consider the system of equations a1x + b1y + c1 = 0, a2x + b2y + c2 = 0, a3x + b3y + c3 = 0 If these equations are satisfied by the same values of x and y, then on eliminating x and y we get. a1(b2c3 – b3c2) + b1(c2a3 – c3a2) + c1(a2b3 – a3b2) = 0 The expression on the left is called a determinant of the third order, and is denoted by a1 b1 c1 a2 b2 c2 a3 b3 c3 A determinant of third order consists of three rows and three columns. Illustration 1 : Eliminate , m, n from the equations a + cm + bn = 0, c + bm + an = 0, b + am + cn = 0 Solution : and express the result in the simplest form. The given set of equations can also be written as (if n 0) : a c m b 0 ; c b m a 0 ; b a m c 0 n n n n n n Then, let m y x; nn System of equations : ax + cy + b = 0 ...(i) cx + by + a = 0 ...(ii) bx + ay + c = 0 ...(iii) We have to eliminate x & y from these simultaneous linear equations. NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\01.Determinants\\01.Theory.p65 Since these equations are satisfied by the same values of x and y, then eliminating x and y we get, acb c b a 0 bac 2. VALUE OF A DETERMINANT : E a1 b1 c1 b2 c2 a2 c2 a2 b2 b3 c3 a3 c3 a3 b3 = a1(b2c3 – b3c2) – b1(a2c3 – a3c2) + c1(a2b3 – a3b2) D a2 b2 c2 a1 b1 c1 a3 b3 c3 Note : Sarrus diagram to get the value of determinant of order three : –ve –ve –ve a1 b1 c1 a1 b1 c1 a1 b1 D a2 b2 c2 = a2 b2 c2 a2 b2 = (a1b2c3 + a2b3c1 +a3b1c2) – (a3b2c1 + a2b1c3 + a1b3c2) a3 b3 c3 a3 b3 c3 a3 b3 +ve +ve +ve Note that the product of the terms in first bracket (i.e. a a a b b b c c c ) is same as the product of the terms 123 1 2 3123 in second bracket. 1
JEE-Mathematics Illustration 2 : 1 23 is - The value of 4 3 6 2 7 9 (A) 213 (B) – 231 (C) 231 (D) 39 1 23 36 4 6 4 3 Solution : 4 3 6 = 1 7 9 –2 2 3 7 92 2 7 9 = (27 + 42) – 2 (–36 –12) + 3 (28 – 6) = 231 Alternative : By sarrus diagram 1 2 3 1 2 31 2 4 3 6 = 4 3 6 4 3 2 7 9 2 7 9 2 7 = (27 + 24 + 84) – (18 – 42 – 72)= 135 – (18 – 114) = 231 Ans. (C) 3 . MINORS & COFACTORS : The minor of a given element of determinant is the determinant obtained by deleting the row & the column in which the given element stands. For example, the minor of a1 in a1 b1 c1 is b2 c2 & the minor of b2 is a1 c1 . a2 b2 c2 b3 c3 a3 c3 a3 b3 c3 Hence a determinant of order three will have “9 minors”. NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\01.Determinants\\01.Theory.p65 If M represents the minor of the element belonging to ith row and jth column then the cofactor of that element is ij given by : Cij = (–1)i + j. Mij Illustration 3 : 2 3 1 Find the minors and cofactors of elements '–3', '5', '–1' & '7' in the determinant 4 0 5 1 6 7 Solution : 45 Minor of –3 = 1 7 = 33 ; Cofactor of – 3 = –33 Minor of 5 = 2 3 9 ; Cofactor of 5 = –9 1 6 3 1 Minor of –1 = 0 5 15 ; Cofactor of –1 = –15 Minor of 7 = 2 3 12 ; Cofactor of 7 = 12 40 2 E
JEE-Mathematics 4 . EXPANSION OF A DETERMINANT IN TERMS OF THE ELEMENTS OF AN Y ROW OR COLUMN : a1 b1 c1 Let D = a2 b2 c2 a3 b3 c3 (i) The sum of the product of elements of any row (column) with their corresponding cofactors is always equal to the value of the determinant. D can be expressed in any of the six forms : a1A1 + b1B1 + c1C1, a1A1 + a2A2 + a3A3, a2A2 + b2B2 + c2C2, b1B1 + b2B2 + b3B3, a3A3 + b3B3 + c3C3, c1C1 + c2C2 + c3C3, where A ,B & C (i = 1,2,3) denote cofactors of a ,b & c respectively. ii i ii i (ii) The sum of the product of elements of any row (column) with the cofactors of other row (column) is always equal to zero. Hence, a A + b B + c C = 0, 21 21 21 b A + b A + b A = 0 and so on. 11 22 33 where A ,B & C (i = 1,2,3) denote cofactors of a ,b & c respectively.i ii ii i Do yourself -1 : 213 ( i ) Find minors & cofactors of elements '6', '5', '0' & '4' of the determinant 6 5 7 . 304 5 3 7 ( i i ) Calculate the value of the determinant 2 4 8 9 3 10 NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\01.Determinants\\01.Theory.p65 ab0 (i i i ) The value of the determinant 0 a b is equal to - b0a (A) a3 – b3 (B) a3 + b3 (C) 0 (D) none of these 120 ( i v ) Find the value of 'k', if 2 3 1 4 3k2 1 z y ( v ) Prove that z 1 x 1 x2 y2 z2 y x 1 5. PROPERTIES OF DETERMINANTS : ( a) The value of a determinant remains unaltered, if the rows & columns are inter-changed, E e.g. if a1 b1 c1 a1 a2 a3 D a2 b2 c2 b1 b2 b3 b3 c3 c1 c2 c3 a3 3
JEE-Mathematics ( b ) If any two rows (or columns) of a determinant be interchanged, the value of determinant is changed in sign only. e.g. a1 b1 c1 a2 b2 c2 Let D a2 b2 c2 & D1 a1 b1 c1 Then D1 = – D. a3 b3 c3 a3 b3 c3 ( c ) If all the elements of a row (or column) are zero, then the value of the determinant is zero. ( d ) If all the elements of any row (or column) are multiplied by the same number, then the determinant is multiplied by that number. a1 b1 c1 Ka1 Kb1 Kc1 e.g. If D = a2 b2 c2 and D1 = a2 b2 c2 Then D1 = KD a3 b3 c3 a3 b3 c3 ( e ) If all the elements of a row (or column) are proportional (or identical) to the element of any other row, then the determinant vanishes, i.e. its value is zero. a1 b1 c1 a1 b1 c1 e.g. If D = a1 b1 c1 D = 0 ; If D1 ka1 kb1 kc1 D1 0 a3 b3 c3 a3 b3 c3 abc ybq Illustration 4 : Prove that x y z x a p pqr zc r Solution : abc axp (By interchanging rows & columns) D= x y z= b y q pqr cz r xap (C C) NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\01.Determinants\\01.Theory.p65 = y b q 1 2 zcr ybq (R R ) =x a p 12 zcr a2 ab ac Illustration 5 : Find the value of the determinant ab b2 bc ac bc c2 a2 ab ac abc abc Solution : D = ab b2 bc = a ab b2 bc = abc a b c = 0 ac bc c2 ac bc c2 abc Since all rows are same, hence value of the determinant is zero. 4 E
JEE-Mathematics Do yourself -2 : ap r n c ( i ) Without expanding the determinant prove that b q m q m b 0 cr n pa (ii) 2 2 is equal to - If D , then 2 2 (A) D (B) 2D (C) 4D (D) 16D pq r (i ii ) If D x y z , then KD is equal to - mn Kp q r pq r p Kx Kp Kx K (A) x Ky z (C) q Ky m (D) Kq Ky Km (B) x y z m Kn K Km Kn r Kz n Kr Kz Kn ( f ) If each element of any row (or column) is expressed as a sum of two (or more) terms, then the determinant can be expressed as the sum of two (or more) determinants. a1 x b1 y c1 z a1 b1 c1 xyz e.g. a2 b2 c2 a2 b2 c2 a2 b2 c2 a3 b3 c3 a3 b3 c3 a3 b3 c3 ƒ(r) g(r) h(r) Note that : If Dr a b c a1 b1 c1 where r N and a,b,c, a1, b1,c1 are constants, then NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\01.Determinants\\01.Theory.p65 n n n ƒ(r) g(r ) h(r) n r 1 r 1 r 1 Dr a b c ar 1 b1 c1 1 ( g ) Row - column operation : The value of a determinant remains unaltered under a column (C ) operation i of the form C C + Cj + Ck (j, k i) or row (R ) operation of the form R R + Rj + Rk (j, k i). i i i i i In other words, the value of a determinant is not altered by adding the elements of any row (or column) to the same multiples of the corresponding elements of any other row (or column) e.g. Let a1 b1 c1 D = a2 b2 c2 a3 b3 c3 a1 a2 b1 b2 c1 c2 (R1 R1 + R2; R3 R3 + R2) D = a2 b2 c2 a3 a1 b3 b1 c3 c1 Note : (i) By using the operation R xR + yR + zR (j, k i), the value of the determinant becomes x i i j k times the original one. (ii) While applying this property ATLEAST ONE ROW (OR COLUMN) must remain unchanged. E5
JEE-Mathematics Illustration 6 : r r3 2 If D r n n3 n n(n 1) 2 n(n 1) 2 2n , find Dr . 2 r 0 2(n 1) n n n n(n 1) n(n 1) 2 2(n 1) 2n = 0 r r3 2 2 2(n 1) n r 0 r 0 r0 2 Solution : Dr n n3 2n n n3 Ans. r 0 n(n 1) 2 2(n 1) n(n 1) n(n 1) 2 2 n(n 1) 2 2 2 a b c 2a 2a Illustration 7 : Prove that 2b b c a 2b (a b c)3 2c 2c c a b a b c 2a 2a Solution : D 2b b c a 2b 2c 2c c a b abc abc a b c (R R + R + R) D 2b b c a 2b 1 1 2 3 2c 2c c a b 11 1 D (a b c) 2b b c a 2b 2c 2c c a b 1 0 0 (C3 C – C; C C – C) NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\01.Determinants\\01.Theory.p65 D (a b c) 2b (a b c) 0 3 1 2 2 1 (a b c) 2c 0 D = (a + b + c)3 Illustration 8 : Determinant a b nc (n 1)a (n 1)b (n 1)c b c na (n 1)b is equal to - (n 1)c (n 1)a c a nb (A) (a + b + c)3 (B) n (a + b + c)3 (C) (n – 1) (a + b + c)3 (D) none of these Solution : Applying C C + (C + C) 1 1 2 3 1 (n 1)a (n 1)b D = n(a + b + c) 1 b c na (n 1)b 1 (n 1)a c a nb 1 (n 1)a (n 1)b R2 R2 R1 D = n(a + b + c) 0 a b c 0 R 3 R3 R 1 0 0 abc Ans. (B) = n(a + b + c)3 E 6
JEE-Mathematics 32 k 42 32 3 k Illustration 9 : If 42 k 52 42 4 k = 0, then the value of k is- 52 k 62 52 5 k (A) 2 (B) 1 (C) –1 (D) 0 (R R – R ; R R – R ) Solution : Applying (C C – C ) 3 31 3 3 22 2 1 32 k 42 3 D 42 k 52 4 0 52 k 62 5 9 k 16 3 Ans. (B) 7 9 1 0 9 11 1 k – 1 = 0 k = 1 Do yourself - 3 : x 20 ( i i ) Solve for x : 2 x 5 1 0 53 106 159 ( i ) Find the value of 52 65 91 . 5x 1 2 102 153 221 1 bc a(b c) (iii) Using row-column operations prove that (b) 1 ca b(c a) 0 xa xb xc 1 ab c(a b) (a) y a y b y c = 0 za zb zc 2r 1 n n ( i v ) Dr . If D = 1 2 3 , then find the value of r 3 21 r 1 NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\01.Determinants\\01.Theory.p65 ( h ) Factor theorem : If the elements of a determinant D are rational integral functions of x and two rows (or columns) become identical when x = a then (x – a) is a factor of D. Note that if r rows become identical when a is substituted for x, then (x – a)r–1 is a factor of D. aax Illustration 10 : Prove that m m m m (x a)(x b) bxb Solution : Using factor theorem, Put x = a aaa D= m m m=0 bab Since R and R are proportional which makes D = 0, therefore (x – a) is a factor of D. 12 Similarly, by putting x = b, D becomes zero, therefore (x – b) is a factor of D. E7
JEE-Mathematics aax ..........(i) D = m m m (x a)(x b) bxb To get the value of put x = 0 in equation (i) aa0 m m m ab b0b amb = ab = m D = m(x – a)(x – b) (x a)2 (x b)2 (x c)2 = 2(x – y) (y – z) (z – x) (a – b) (b – c) (c – a) Illustration 11 : Prove that (y a)2 (y b)2 (y c)2 (z b)2 (z c)2 (z a)2 Solution : (x a)2 (x b)2 (x c)2 D = (y a)2 (y b)2 (y c)2 (z b)2 (z c)2 (z a)2 Using factor theorem, Put x = y (y a)2 (y b)2 (y c)2 D (y a)2 (y b)2 (y c)2 (z b)2 (z c)2 (z a)2 R and R are identical which makes D = 0. Therefore, (x–y) is a factor of D. NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\01.Determinants\\01.Theory.p65 12 Similarly (y – z) & (z – x) are factors of D Now put a = b (x b)2 (x b)2 (x c)2 D (y b)2 (y b)2 (y c)2 (z b)2 (z c)2 (z b)2 C and C become identical which makes D = 0. Therefore, (a–b) is a factor of D. 12 Similarly (b–c) and (c–a) are factors of D. Therefore, D = (x – y) (y – z) (z – x) (a – b) (b – c) (c – a) To get the value of put x = –1 = a, y = 0 = b and z = 1 = c 014 D 1 0 1 (1)(1)(2)(1)(1)(2) 410 4 = 8 = 2 E D = 2(x – y) (y – z) (z – x) (a – b) (b – c) (c – a) 8
JEE-Mathematics Do yourself - 4 : 1 a bc ( i ) Without expanding the determinant prove that 1 b ca (a b)(b c)(c a) 1 c ab 1 4 20 ( i i ) Using factor theorem, find the solution set of the equation 1 2 5 0 1 2x 5x2 6 . MULTIPLICATION OF T WO DETERMINANTS : a1 b1 l1 m1 a1 l1 b1 l2 a1 m1 b1 m 2 a2 b2 l2 m 2 a2 l1 b2 l2 a2 m1 b2 m 2 Similarly two determinants of order three are multiplied. ( a ) Here we have multiplied row by column. We can also multiply row by row, column by row and column by column. (b ) If D1 is the determinant formed by replacing the elements of determinant D of order n by their corresponding cofactors then D1 = Dn–1 (a x)2 (b x)2 (c x )2 (1 ax )2 (1 bx)2 (1 cx)2 Illustration 12 : If a, b, c x, y, z R, then prove that (a y )2 (b y)2 (c y )2 (1 ay )2 (1 by )2 (1 cy )2 (b z)2 (c z)2 (1 az)2 (1 bz)2 (1 cz)2 (a z)2 Solution :NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\01.Determinants\\01.Theory.p65 (a x)2 (b x)2 (c x)2 a2 2ax x2 b2 2bx x2 c2 2cx x2 (b y)2 (c y)2 a2 2ay y2 b2 2by y2 c2 2cy y2 E L.H.S. = (a y)2 (b z)2 (c z)2 a2 2az z2 b2 2bz z2 c2 2az z2 (a z)2 1 x x2 a2 2a 1 = 1 y y2 b2 2b 1 (Row by Row) 1 z z2 c2 2c 1 1 x x2 a2 2a 1 = 1 y y2 (1) b2 2b 1 1 z z2 c2 2c 1 1 x x2 1 2a a2 = 1 y y2 (1)(1) 1 2b b2 (C C) 1 3 1 z z2 1 2c c2 9
JEE-Mathematics 1 x x2 1 2a a2 1 y y2 1 2b b2 1 z z2 1 2c c2 Multiplying row by row 1 2ax a2x2 1 2bx b2x2 1 2cx c2x2 = 1 2ay a2y2 1 2by b2y2 1 2cy c2y2 1 2bz b2z2 1 2cz c2z2 1 2az a2z2 (1 ax )2 (1 bx)2 (1 cx)2 (1 ay )2 (1 by )2 (1 cy )2 (1 bz)2 (1 cz)2 (1 az)2 = R.H.S. Illustration 13 : Let be the roots of equation ax2 + bx + c = 0 and S = n + n for n 1. Evaluate the value n 3 1 S1 1 S2 of the determinant 1 S1 1 S2 1 S3 . 1 S2 1 S3 1 S4 3 1 S1 1 S2 1 11 1 1 2 2 D = 1 S1 1 S2 1 2 2 1 3 3 Solution : 1 S3 1 S3 = 1 1 3 3 1 4 4 1 S2 1 S4 1 2 2 11 1 1 1 1 1 1 12 = 1 2 1 = 1 2 =[(1 – )(1 – )( – )]2 1 2 1 2 2 1 2 D = ()2 ( 1)2 NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\01.Determinants\\01.Theory.p65 are roots of the equation ax 2 + bx + c = 0 b & c b2 4ac a a a (b2 4ac) a b c 2 (b2 4ac)(a b c)2 D = a2 a = Ans. a4 y5z6 (z3 y3 ) x4z6 (x3 z3 ) x4y5 (y3 x3 ) x y2 z3 Illustration 14 : If D1 y2 z3 (y6 z6 ) xz3 (z6 x6 ) xy2 (x6 y 6 ) and D2 x 4 y5 z6 . Then D D is equal to - y2z3 (z3 y3 ) xz3 (x3 z3 ) 12 xy2 (y3 x3 ) x7 y8 z9 (A) D 3 (B) D 2 (C) D 4 (D) none of these 2 2 2 Solution : The given determinant D is obtained by corresponding cofactors of determinant D . 12 Hence D = D 2 D1D2 D 2 D 2 D 3 Ans. (A) 12 2 2 E 10
JEE-Mathematics Do yourself - 5 : 11 1 100 (i) If the determinant D = 2 2 2 and D1 0 , then find the determinant D 2 2 2 2 0 D such that D = . 2 D1 ab2 ac2 bc2 a2b a2c b2c 111 ( i i ) If D1 ac ab ab bc bc ac & D2 a b c , then D D is equal to - cb a c 12 b a bc ac ab (A) 0 (B) D12 (C) D 2 (D) D23 2 7. SPECIAL DETERMINANTS : (a) Cyclic Determinant : The elements of the rows (or columns) are in cyclic arrangement. abc b c a (a3 b3 c3 3abc) = –(a + b + c) (a2 + b2 + c2 – ab – bc – ac) ca b NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\01.Determinants\\01.Theory.p65 1 (a b c) {(a b)2 (b c)2 (c a )2 } 2 = – (a + b + c) (a + b + c2) (a + b2 + c), where ,2 are cube roots of unity (b) Other Important Determinants : 0 b c (i) b 0 a 0 c a 0 111 11 1 (ii) a b c a b c (a b)(b c)(c a) bc ac ab a2 b2 c2 11 1 (iii) a b c (a b)(b c)(c a)(a b c) a3 b3 c3 111 (iv) a2 b2 c2 (a b) (b c)(c a) (ab bc ca) a3 b3 c3 11 1 (v) a b c (a b) (b c) (c a) (a2 b2 c2 ab bc ca) a4 b4 c4 E 11
JEE-Mathematics 1 2 Illustration 15 : Prove that 2 1 = –(1– 3)2. 2 1 Solution : This is a cyclic determinant. 1 2 2 1 = – (1 + 2)(1 + 2 + 4 – – 2 – 3) 2 1 = – (1 + + 2)(– + 1 – 3 + 4) = – (1 + + 2)(1 – )2(1 + + 2) = – (1 – )2(1 + + 2)2 = –(1 – 3)2 Do yourself - 6 : ka k2 a2 1 ( i ) The value of the determinant kb k2 b2 1 is kc k2 c2 1 (A) k(a + b)(b + c)(c + a) (B) kabc(a2 + b2 + c2) (C) k(a – b)(b – c)(c – a) (D) k(a + b – c)(b + c – a)(c + a – b) a2 b2 a2 c2 a2 c2 ( i i ) Find the value of the determinant a2 0 c2 a2 . b2 c2 b2 abc NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\01.Determinants\\01.Theory.p65 (i i i ) Prove that bc ca ab (a b c)(a b)(b c)(c a) bc ca ab 8 . CR A MER'S RULE (SYSTEM OF LINE AR EQUATIONS) : Simultaneous linear equations Consistent Inconsistent (at least one solution) (no solution) Exactly one solution Infinite solutions or Unique solution Trivial solution Non trivial solution All variable At least one zero is the non zero variable only solution satisfies the system 12 E
JEE-Mathematics (a) Equations involving two variables : (i) Consistent Equations : Definite & unique solution (Intersecting lines) (ii) Inconsistent Equations : No solution (Parallel lines) (iii) Dependent Equations : Infinite solutions (Identical lines) Let, a1x + b1y + c1 = 0 a2x + b2y + c2 = 0 then : (1) a1 b1 Given equations are consistent with unique solution a2 b2 (2) a1 b1 c1 Given equations are inconsistent a2 b2 c2 (3) a1 b1 c1 Given equations are consistent with infinite solutions a2 b2 c2 (b) Equations Involving Three variables : Let a x + b y + c z = d ............ (i) 1 111 a x + b y + c z = d ............ (ii) 2 222 a x + b y + c z = d ............ (iii) 3 333 Then, x = D1 , y = D2 , z = D3 . DDD a1 b1 c1 d1 b1 c1 a1 d1 c1 a1 b1 d1 Where D = a2 b2 c2 ; D = d2 b2 c2 ; D = a2 d2 c2 & D = a2 b2 d2 1 2 3 a3 b3 c3 d3 b3 c3 a3 d3 c3 a3 b3 d3 Note : (i) If D 0 and atleast one of D , D , D 0, then the given system of equations is consistent 1 2 3 and has unique non trivial solution. NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\01.Determinants\\01.Theory.p65 (ii) If D0 & D = D = D = 0, then the given system of equations is consistent and has trivial 123 solution only. (iii) If D = D = D = D = 0, then the given system of equations is consistent and has infinite solutions. 123 a1x b1y c1z d1 Note that In case a1x b1y c1z d 2 (Atleast two of d1 , d2 & d3 are not equal) a1x b1y c1z d 3 D = D1= D2 = D3 = 0. But these three equations represent three parallel planes. Hence the system is inconsistent. (iv) If D = 0 but atleast one of D , D , D is not zero then the equations are inconsistent and have 123 no solution. (c) Homogeneous system of linear equations : Let a1x + b1y + c1z = 0 ............ (i) a x + b y + c z = 0 ............ (ii) 2 22 a x + b y + c z = 0 ............ (iii) 3 33 D =D =D =0 123 E 13
JEE-Mathematics The system always possesses atleast one solution x = 0, y = 0, z = 0, which is called Trivial solution, i.e. this system is always consistent. Check value of D D 0 D = 0 Unique Trivial solution Trivial & Non-Trivial solutions (infinite solutions) Note that if a given system of linear equations has Only Zero solutions for all its variables then the given equations are said to have TRIVIAL SOLUTION. Also, note that if the system of equations a x + b y + c = 0; a x + b y + c = 0; a x + b y + c = 0 1 11 2 22 3 33 a1 b1 c1 is always consistent then a2 b2 c2 0 but converse is NOT true. a3 b3 c3 9 . APPLICATION OF DETERMINANTS IN GEOMETRY : ( a ) The lines : a x + b y + c = 0 ........ (i) 1 11 a x + b y + c = 0 ........ (ii) 2 22 a3x + b3y + c3 = 0 ........ (iii) a1 b1 c1 are concurrent if a2 b2 c2 = 0. a3 b3 c3 This is the condition for consistency of three simultaneous linear equations in 2 variables. ( b ) Equation ax² + 2 hxy + by² + 2 gx + 2 fy + c = 0 represents a pair of straight lines if : ahg abc + 2 fgh af² bg² ch² = 0 = h b f gfc (c) Area of a triangle whose vertices are (x , y ) ; r=1,2,3 is D= 1 x1 y1 1 NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\01.Determinants\\01.Theory.p65 rr 2 x2 y2 1 x3 y3 1 If D = 0 then the three points are collinear. x y1 ( d ) Equation of a straight line passing through points (x1 , y1) & (x2 , y2) is x1 y1 1 = 0 x2 y2 1 Illustration 16 : Find the nature of solution for the given system of equations : x + 2y + 3z = 1; 2x + 3y + 4z = 3; 3x + 4y + 5z = 0 Solution : 123 D= 2 3 4 =0 345 123 Now, D = 3 3 4 = 5 1 045 D = 0 but D 0 1 Hence no solution. Ans. 14 E
JEE-Mathematics Illustration 17 : Find the value of , if the following equations are consistent : Solution : x + y – 3 = 0; (1 + )x + (2 + )y – 8 = 0; x – (1 + )y + (2 + ) = 0 The given equations in two unknowns are consistent, then = 0 1 1 3 i.e. 1 2 8 0 1 (1 ) 2 Applying C C – C and C C + 3C 2 2 1 3 3 1 10 0 1 1 3 5 0 1 2 5 (5 ) (3 5)(2 ) 0 32 2 5 0 1, 5 / 3 Illustration 18 : If the system of equations x + y + 1 = 0, x + y + 1 = 0 & x + y + = 0. is consistent then find the value of . Solution : For consistency of the given system of equations 1 1 (–1)2 ( + 2) = 0 = 1 or =–2 Ans. D 1 1 0 11 3 = 1 + 1 + 3 or 3 – 3 + 2 = 0 Illustration 19 : If x, y, z are not all simultaneously equal to zero, satisfying the system of equations sin(3) x – y + z = 0; cos(2)x + 4y + 3z = 0; 2x + 7y + 7z = 0, then find the values of (0 2) . NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\01.Determinants\\01.Theory.p65 Solution : Given system of equations is a system of homogeneous linear equations which posses non-zero solution set, therefore D = 0. sin3 1 1 sin 3 1 0 (C C + C) D = cos 2 4 3 3 3 2 D = cos2 4 7 2 77 2 7 14 sin 3 1 0 (R R – R3 ) D = cos2 1 0.5 0 2 2 2 2 7 14 D = 14 sin 3 cos 2 1 2 D=0 sin3 + 2cos2 – 2 = 0 3sin – 4sin3 = 4sin2 (sin)(4sin2 + 4sin – 3) = 0 E 15
JEE-Mathematics (sin)(2sin – 1)(2sin + 3) = 0 sin = 0 ; sin 1 ; 3 sin = – 22 sin = 0 = 0, , 2; sin 1 , 5 ; 3 no solution. 2 66 sin 2 0, , 5 , , 2 Ans. 66 Do yourself -7 : ( i ) Find nature of solution for given system of equations 2x + y + z = 3; x + 2y + z = 4 ; 3x + z = 2 ( i i ) If the system of equations x + y + z = 2, 2x + y – z = 3 & 3x + 2y + kz = 4 has a unique solution then (A) k 0 (B) –1 < k < 1 (C) –2 < k < 1 (D) k = 0 (i i i ) The system of equations x + y + z = 0, –x + y + z = 0 = 0 & –x – y + z = 0 has a non-trivial solution, then possible values of are - (A) 0 (B) 1 (C) –3 (D) 3 ANSWERS FOR DO YOURSELF NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\01.Determinants\\01.Theory.p65 1 . ( i ) minors : 4, –1, –4, 4 ; cofactors : –4, –1, 4, 4 (ii) –98 (iii) B (iv) 0 2. (ii) C (iii) B,C 3 . (i) 0 (ii) 2 (iv) 0 4 . ( i i ) x = –1, 2 5. (i) 11 1 (ii) D 1 1 6. (i) C (ii) 0 7 . ( i ) infinite solutions (ii) A (iii) A 16 E
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