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M1-Allens Made Maths Theory + Exercise [II]

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JEE-Mathematics Illustration 16 : Evaluate : lim e tan x  e x x0 tan x  x Solution : lim e tan x  e x = lim ex  e (tan xx )  e x tan x  x tan x  x x0 x0 = lim ex (e tan xx  1) = lim ex (ey  1) y = tan x –x and lim ey 1 =1 tan x  x y where x0 x0 y0 y 0 y = e0 × 1 [as x  0, tan x – x  0] =1×1=1 Ans. Do yourself - 7 : (ii) Evaluate : lim 2x  1 ( i ) Evaluate : lim ex  ea x0 (1  x)1 / 2  1 xa x  a 1  x1 / x  e  Lim 1 1  x x  x  ( b ) (i) Lim  (Note : The base and exponent depends on the same variable.) x 0 In general, if Lim f(x)  0 , then Lim(1  f(x))1 / f(x)  e xa xa (ii) Lim n(1  x)  1 x0 x (iii) If Lim f(x)  1 and Lim (x)   , then ; Lim f(x )  (x)  ek where k  Lim  (x) [f(x)  1] x a xa x a xa Illustration 17 : Evaluate Lim (lo g 3 x )logx 3 x 1 3 Solution : Lim (log3 3 x )logx 3 = Lim (lo g 3  log3 x )logx 3 x 1 x 1 3 = Lim (1  log3 x )1 / log3 x = e  1 Ans. logb a  loga b Ans. x 1 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\01.LIMIT\\01.THEORY.p65 xn(1  2 tan x) Illustration 18 : Evaluate : Lim x0 1  cos x Solution : xn(1  2 tan x) xn(1  2 tan x) 2 tan x =4 Lim = Lim . x0 1  cos x x0 1  cos x .x2 2 tan x x2 Illustration 19 : Evaluate : lim  2x2  1  4x2 2 x  2x2  3  Solution : Since it is in the form of 1  2x2 4 x2 2 lim  2x2  1  2x2  3   2 x2 e x   (4x2 e–8 lim 1 = 2x2  3 + 2) = Ans.  3  x E9

JEE-Mathematics 1 Illustration 20 : lim  sin x x a ,a  n, n is an integer, equals -  sin a  xa (A) ecot a (B) e tan a (C) e sin a (D) ecos a sin x sin a 1 1 sin sin a x a  sin a 1  sin x sin a   x sin a  sin x  x  a 1 sin x sin a  x a  sin a    Solution : lim  sin a   lim  sin a   lim    xa xa xa  lim 2 cos  x  a sin  x  a  . 1 cos a  ecot a Ans. (A)   2   2  sin exa x a a  e sin a F IIllustration 21 : 1/ x ax  bx  cx limGH KJx0 3 = (A) abc (B) abc (C) (abc)1/3 (D) none of these F I F Ilim ax  bx  cx 1/ x 1 ax  bx  cx 3 1/ x  lim Solution : HG KJ HG KJx0 3 x0 3 LMMGF IJ POP lim a x 1 b x 1c x 1 NMH K PQx0 3x 3 (a x  1) (b x  1) (c x  1) (a x 1)(bx 1)(c x 1) 1   333 LM PO e1/3 lim a x  1  bx  1  cx  1 = e1/3 (log a + log b + log c) = elog (abc)1/3 = (abc)1/3 Ans. (C) NM PQx0 x x x Do yourself - 8 : ( i ) Evaluate : lim x{n(x  a)  nx} x     1 / x 1 1 pn q tan  4  n  (ii) Evaluate : lim   x (iii) Evaluate : lim  x0 n  1 (v) Evaluate : lim  x  6 x4  x  1   ( i v ) Evaluate : lim 1  tan2 x 2x x  Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\01.LIMIT\\01.THEORY.p65 x 0  ( c ) If Lim f(x)  A  0 & Lim (x)  B (a finite quantity) then ; Lim f(x)  (x)  eB ln A  A B xa xa xa x5 Illustration 22 : Evaluate : lim  7x2  1 1x3   x 5x2 1 Solution : Here f(x) = 7x2  1 , x5 = x 2 .x 3  x2 5x2 1 (x )  1  x3 1 1 1  x3 x3  lim f(x)  7 & lim (x)   x 5 x  lim (f(x))(x)  7   0 Ans.  5  x  E 10

JEE-Mathematics Do yourself - 9 : lim 1  5x2 x2 (i) Evaluate :   3x2  x  1 Miscellaneous Illustrations : Illustration 23 : Find lim 1  x 2 , where x = 1  xr , 0  r  (n –1), r I, n  N 0 2 n x1 x2 x3 x 4 .....x n r+1 Solution : Let x = cos  then x = 1  x0 = cos  0 1 22 x= 1  x1 = cos  ,.........x = cos  2 2 22 n 2n sin   Limits = cos  cos  cos  .....cos  , n  248 2n sin   = lim . sin  =  = cos–1x = lim 2n sin 2n n 2n 0 Ans. n sin   2n Illustration 24 : Evaluate : lim cos2 {1  cos2 (1  cos2 (............(1  cos2 (x))))} x0  x  4  2  sin   x   Solution : Let A= lim cos2 {1  cos2 (1  cos2 (............(1  cos2 (x))))} x0  x  4  2  x  sin    = lim cos2 {sin2 (sin2 (......(1  cos2 (x))))} = cos2 0 1 1 Ans. lim = x0   1 x0  ( x  4  4 )  sin  2 x 4 2. x  4  2  x  4  2  4 sin    x x  4  2   sin      Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\01.LIMIT\\01.THEORY.p65 x   1   1   1  n   2   22   2 n 1   Illustration 25 : Evaluate the following limits, if exist lim nn2 (n  1) n  n  ........... n  n   n   1   1   n  (n  1)  n  1  ......  n  1     2   2n 1     2  2 n 1   Solution : lim nn2 (n  1) n  ...... n  = lim    nn  n n   n  1 n n  1   n 1 n 1n 2 n 2.n 1 n n  . n 2    2n 1 n  = lim n   .......  = 1  . 1  1 2 ...... 1 1 2n 1 n  n 2n  2n1 n    lim     n  e = e11 1 1 1  2  4 ............ = e. e2 . e 4 ...... = 2 Ans. E 11

JEE-Mathematics Illustration 26 : Evaluate lim sin  . x0 x Solution : Again the function f(x) = sin(/x) is undefined at 0. Evaluating the function for some small values of x, we get f(1) = sin = 0, f  1   sin 2   0 , f(0.1) = sin10 = 0,  2  f(0.01) = sin100 = 0. On the basis of this information we might be tempted to guess that lim sin   0 but this time x0 x our guess is wrong. Note that although f(1/n) = sinn = 0 for any integer n, it is also true that f(x) = 1 for infinitely many values of x that approach 0. [In fact, sin(/x) = 1 when     2n x2 and solving for x, we get x = 2/(4n + 1)]. The graph of f is given in following figure y 1 y= sin(/x) –1 x 1 –1 The dashed line indicate that the values of sin(/x) oscillate between 1 and –1 infinitely often as x approaches 0. Since the values of f(x) do not approach a fixed number as x approaches 0,  lim sin  does not exist. x0 x ANSWERS FOR DO YOURSELF 1 : (i) (a) T (b) F (c) F (d) T (e) T (f) T (g) T (h) T (i) F (j) T 2: (i) 1 ENode-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\01.LIMIT\\01.THEORY.p65  3 3 : (i) q 2 1 p (ii) 3 3 (iii) 24 4 : (i) 1 (ii) 1 2  sin 2y (c ) 2asina + a2cosa 5 : (i) (a)  (b) 2y 1 1 6. (i) 6 (ii) 3 7 : (i) ea (ii) 2n2 8 : (i) a (ii) e2 (iii) ep 1 (v) e5 9 : (i) 0 (iv) e2 12

JEE-Mathematics CHECK YOUR GRASP PRINCIPLE OF MATHEMATICAL INDUCTION EXERCISE-I 1 . The sum of n terms of 12 + (12 + 22) + (12 + 22 + 32)  n  1 n  1 3 . If n is a natural number then  2   n! is true + .... is- n(n 1) (2n 1) n(n 1) (2n 1) when- (2) n  1 (3) n > 2 (4) Never (1) 6 (2) 6 (1) n > 1 11 1 4 . For natural number n, 2n (n – 1) ! < nn, if- (3) n(n + 1)2(n + 2) (4) n2(n + 1)2 12 12 2 . The greatest positive integer. which divides (1) n < 2 (2) n > 2 (3) n  2 (4) never (n + 16) (n + 17) (n + 18) (n + 19), for all n N, is- 1 5 . For every positive integer (1) 2 (2) 4 (3) 24 (4) 120 n n5 2n3 n 3 . Let P(n) : n2 + n is an odd integer. It is seen that n, 7 + 5 + 3 – 105 is- truth of P(n)  the truth of P(n + 1). Therefore, (1) an integer P(n) is true for all- (2) a rational number (1) n > 1 (2) n (3) a negative real number (3) n > 2 (4) None of these (4) an odd integer 4 . For every natural number n- 1 6 . For positive integer n, 3n < n! when- (1) n > 2n (2) n < 2n (3) n  2n (4) n  2n (1) n  6 (2) n > 7 (3) n  7 (4) n  7 5. If n  N, then x2n–1 + y2n–1 is divisible by- a 1 An (1) x + y (2) x – y (3) x2 + y2 (4) x2 + xy 17. If A =   , then for any n  N, equals -  0 a  6 . The inequality n! > 2n–1 is true- na n   a n na n 1    (1)  0 n a  (2)  0 an    (1) for all n > 1 (2) for all n > 2 (3) for all n  N (4) None of these na 1  an n  7 . 1.22 + 2.32 + 3.42 + ..... upto n terms, is equal to- (3)  0 n a  (4)  an    0  1 1 8 . The sum of n terms of the series (1) n(n + 1) (n + 2) (n + 3) 1.2 2.3 3.4 12 22 + 22 22 1 13 13 23 + 13 23  33 + ....... is- (2) n(n + 1) (n + 2) (n + 5) 12 1 1n n 1 n 1 NODE6 (E)\\DATA\\2014\\KOTA\\JEE-ADVANCED\\SMP\\MATHS\\UNIT#02\\ENG\\PART-2\\04-PRINCIPLES OF MATHEMATICAL INDUCTION\\02.EXE (3) n(n + 1) (n + 2) (3n + 5) (1) n(n 1) (2) n 1 (3) (4) n 2 n 12 (4) None of these 8 . The sum of the cubes of three consecutive natural 1 9 . For all n  N, 72n – 48n – 1 is divisible by- numbers is divisible by- (1) 25 (2) 26 (3) 1234 (4) 2304 (1) 2 (2) 5 (3) 7 (4) 9 2 0 . The nth term of the series 9 . If n  N, then 11n+2 + 122n+1 is divisible by- 4 + 14 + 30 + 52 + 80 + 114 + ..... is- (1) 113 (2) 123 (1) 5n – 1 (2) 2n2 + 2n (3) 3n2 + n (4) 2n2 + 2 (3) 133 (4) None of these 2 1 . If 10n + 3.4n+2 +  is exactly divisible by 9 for all 1 0 . If n  N, then 34n+2 + 52n+1 is a multiple of- n  N, then the least positive integral value of  is- (1) 14 (2) 16 (3) 18 (4) 20 (1) 5 (2) 3 (3) 7 (4) 1 1 1 . For each n  N, 102n+1 + 1 is divisible by- 2 2 . The sum of n terms of the series (1) 11 (2) 13 1 + (1 + a) + (1 + a + a2) + (1 + a + a2 + a3) +....., (3) 27 (4) None of these is- 1 2 . The difference between an +ve integer and its cube n a(1  an ) (2) n + a(1  an ) 1  a (1  a )2 is divisible by- (1) – (1  a )2 1a (1) 4 (2) 6 n a(1  an ) n a(1  an ) (3) 9 (4) None of these (3) + (1  a)2 (4) – + (1  a )2 1a 1a 30 E

JEE-Mathematics 2 3 . For all n  N, n4 is less than- 11 1 3 1 . 1.3 + 3.5 + 5.7 + ..... upto n terms is- (1) 10n (2) 4n (3) 1010 (4) None of these 1 n 1 2n (1) (2) (3) 2n –1 (4) 2 4 . For all n  N, n 2n 1 2n 1 3(n 1) (2n 1)2 (2n 1)2 3 2 . For positive integer n, 10n–2 > 81n when- (1) < (2) > (1) n < 5 (2) n > 5 (3) n  5 (4) n > 6 8 8 3 3 . If P is a prime number then np – n is divisible by p (2n 1)2 (3) = (4) None of these when n is a 8 (1) natural number greater than 1 2 5 . For all n  N, cos cos2 cos4 ...... cos 2n – 1 (2) odd number equals to- (3) even number sin 2n  sin 2n  (4) None of these (1) 2n sin  (2) 3 4 . 1 + 3 + 6 + 10 + ...... upto n terms is equal to- sin  cos 2n  cos 2n  1 1 (3) 2n cos 2 (4) 2n sin  (1) 3 n(n + 1)(n + 2) (2) 6 n(n + 1)(n + 2) 2 6 . For all positive integral values of n, 32n – 2n + 1 is 1 1 (3) n(n + 2)(n + 3) (4) n(n + 1)(n + 2) divisible by- 12 12 (1) 2 (2) 4 (3) 8 (4) 12 3 5 . A student was asked to prove a statement by 12 12 22 12  22  32 induction. He proved 2 7 . 1 + 1 2 + 1 2  3 + ... upto n terms is- (i) P(5) is true and (ii) Truth of P(n)  truth of p(n + 1), n  N 1 1 On the basis of this, he could conclude that P(n) is (1) 3 (2n + 1) (2) 3 n2 true for (1) no n  N (2) all n  N 1 1 (3) all n  5 (4) None of these (3) 3 (n + 2) (4) 3 n(n + 2) 3 6 . The sum of the series 2 8 . The smallest positive integer for which the 35 7 statement 3n+1 < 4n holds is- 12 + 12  22 + 12  22  32 + ..... upto n terms (1) 1 (2) 2 (3) 3 (4) 4 2n 3n 3n 6n (1) (2) (3) 2(n  1) (4) NODE6 (E)\\DATA\\2014\\KOTA\\JEE-ADVANCED\\SMP\\MATHS\\UNIT#02\\ENG\\PART-2\\04-PRINCIPLES OF MATHEMATICAL INDUCTION\\02.EXE 2 9 . Sum of n terms of the series n 1 n 1 n 1 11 1 1 3 7 15 1 + 1  2 + 1  2  3 + ...... is- 3 7 . 2 + 4 + 8 + 16 + ... upto n terms equal to- n 2 2n 2(n 1) 1 1 (1) n  1 (2) n(n  1) (3) n  1 (4) n  2 (1) n + 2n (2) 2n + 2n 3 0 . For every natural number n, n(n + 3) is always- 1 1 (3) n – 1 + 2n (4) n + 1 + 2n (1) multiple of 4 (2) multiple of 5 (3) even (4) odd ANSWER KEY Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Ans. 3 3 4 2 1 2 3 4 3 1 1 2 2 2 1 Que. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Ans. 3 2 2 4 3 1 1 1 1 1 1 4 4 3 3 Que. 31 32 33 34 35 36 37 Ans. 2 3 1 2 3 4 3 E 31

JEE-Mathematics PREVIOUS YEAR QUESTIONS PRINCIPLE OF MATHEMATICAL INDUCTION EXERCISE-II 1 . Let S(k) = 1 + 3 + 5 + ...... + (2k – 1) = 3 + k2, 4 . Statement –1 : For every natural number n  2 then which of the following is true ? [AIEEE-2004] (1) S(1) is true 11 1 (2) S(k)  S(k + 1) 1 + 2 +... + n > n (3) S(k)  S(k + 1) Statement –2 : For every natural number n  2 , (4) Principle of mathematical Induction can be used n n  1 < n+1. [AIEEE-2008] to prove that formula (1) Statement –1 is false, Statement –2 is true 2 . The sum of first n terms of the given series (2) Statement–1 is true, Statement–2 is false 12 + 2.22 + 32 + 2.42 + 52 + 2.62 + .... is n(n 1)2 , (3) Statement–1 is true, Statement–2 is true; 2 Statement–2 is a correct explanation for Statement–1 when n is even. When n is odd, then sum will be- (4) Statement–1 is true, Statement–2 is true; [AIEEE-2004] Statement–2 is not a correct explanation for n(n 1)2 1 Statement–1 (1) (2) n2(n + 1) 5 . Statement - 1: For each natural number n, 2 2 (n + 1)7 – n7 –1 is divisible by 7. (3) n(n + 1)2 (4) None of these Statement - 2: For each natural number n, n7 – n 1 0  1 0 is divisible by 7. [AIEEE-2011] 3. If A = 1 1  and I = 0 1  , then which one of (1) Statement-1 is false, statement-2 is true.   (2) Statement-1 is true, statement-2 is true; the following holds for all n  1, (by the principal Statement-2 is correct explanation for of mathematical induction) [AIEEE-2005] statement-1. (1) An = nA + (n – 1)I (2) An = 2n–1A + (n + 1)I (3) Statement-1 is true, statement-2 is tru e; (3) An = nA – (n – 1)I (4) An = 2n–1A – (n – 1)I Statement-2 is not a correct explanation for statement-1. (4) Statement-1 is true, statement-2 is false. Que. 1 2 3 4 5 ANSWER KEY NODE6 (E)\\DATA\\2014\\KOTA\\JEE-ADVANCED\\SMP\\MATHS\\UNIT#02\\ENG\\PART-2\\04-PRINCIPLES OF MATHEMATICAL INDUCTION\\02.EXE Ans. 2 2 3 3 2 32 E

JEE-Mathematics CHECK YOUR GRASP MATHEMATICAL REASONING EXERCISE-I 1 . The inverse of the statement (p  ~q)  r is- 1 3 If statements p, q, r have truth values T, F, T (1) ~(p ~q) ~r (2) (~p q) ~r respectively then which of the following statement (3) (~p q) ~r (4) None of these is true- 2 . (~p  ~q) is logically equivalent to- (1) (p q) r (2) (p q) ~r (1) p q (2) ~p q (3) p ~q (4) ~p ~q (3) (p q) (q r) (4) (p q) r 3 . The equivalent statement of (p  q) is- 1 4 . If statement p (q r) is true then the truth values of statements p, q, r respectively- (1) (p q) (p q) (2) (p q) (q p) (1) T, F, T (2) F, T, F (3) (~p q) (p ~q) (4) (~p q) (p ~q) (3) F, F, F (4) All of these 4 . If the compound statement p  (~p  q) is false 1 5 . Which of the following statement is a contradiction- then the truth value of p and q are respectively- (1) (p q) (~(p q)) (2) p (~p q) (1) T, T (2) T, F (3) F, T (4) F, F (3) (p q) p (4) ~p ~q 5 . The statement (p  ~p)  (~p  p) is- 1 6 . The negative of the statement \"If a number is (1) a tautology divisible by 15 then it is divisible by 5 or 3\" (2) a contradiction (1) If a number is divisible by 15 then it is not divisible (3) neither a tautology nor a contradiction by 5 and 3 (4) None of these (2) A number is divisible by 15 and it is not divisible by 5 or 3 6 . Negation of the statement (p  r)  (r  q) is- (3) A number is divisible by 15 or it is not divisible (1) ~(p r) ~(r q) (2) (~p ~r) (r q) by 5 and 3 (3) (p r) (r q) (4) (p r) (~r ~q) (4) A number is divisible by 15 and it is not divisible 7 . The dual of the statement ~p [~q (p q)  ~r] by 5 and 3 is- 1 7 . Which of the following is a statement- (1) ~p [~q (p q) ~r] (1) Open the door (2) p [q (~p ~q) r] (2) Do your home work (3) ~p [~q (p q) ~r] (3) Hurrah! we have won the match (4) ~p [~q (p q) ~r] (4) Two plus two is five 8 . Which of the following is correct- 1 8 . The negation of the statement \"2 + 3 = 5 and 8 < 10\" is- (1) (~p ~q) (p q) (1) 2 + 3  5 and 8  10 (2) 2 + 3 5 or 8 > 10 (2) (p q) (~q ~p) (3) 2 + 3 5 or 8  10 (4) None of these NODE6 (E)\\DATA\\2014\\KOTA\\JEE-ADVANCED\\SMP\\MATHS\\UNIT#02\\ENG\\PART-2\\03-MATHEMATICAL REASONING\\2.EXE (3) ~(p ~q) (p ~q) 1 9 . For any three simple statement p, q, r the statement (4) ~(p  q) (p q) (q p) (p q) (q r) is true when- 9 . The contrapositive of p  (~q  ~r) is- (1) p and r true and q is false (1) (~q r) ~p (2) (q r) ~p (2) p and r false and q is true (3) (q ~r) ~p (4) None of these (3) p, q, r all are false 1 0 . The converse of p  (q  r) is- (4) q and r true and p is false (1) (q ~r) p (2) (~q r) p 2 0 . Which of the following statement is a tautology- (3) (q ~r) ~p (4) (q ~r) p (1) (~p ~q) (p ~q) (2) (~p ~q) (p ~q) 1 1 . If p and q are two statement then (p  ~q) is true (3) ~p (~p ~q) (4) ~q (~p ~q) when- 2 1 . Which of the following statement is a contradiction- (1) p and q both are true (2) p and q both are false (1) (~p ~q) (p ~q) (2) (p q) (p ~q) (3) (~p q) (~q) (4) (~p q) (~q) (3) p is false and q is true (4) None of these 2 2 . The negation of the statement q  (p  ~r) is 1 2 . Statement (p  q)  p is- equivalent to- (1) a tautology (2) a contradiction (1) ~q (p r) (2) ~q ~(p r) (3) neither (1) nor (2) (4) None of these (3) ~q (~p r) (4) None of these E 23

JEE-Mathematics 2 3 . Which of the following is not a statement- 3 0 . If p is any statement, t is a tautology and c is a contradiction then which fo the following is not (1) every set is a finite set correct- (2) every square is a rectangle (3) The sun is a star (1) p (~c) p (4) Shut the window (2) p (~t) p 2 4 . The statement ~(p  q)  (~p  ~q) is- (3) t c p t (1) a tautology (4) (p t) (p c) (t c) (2) a contradiction 3 1 . If p, q, r are simple statement with truth values T, F, T respect ively then the tr uth value of (3) neither a tautology nor a contradiction ((~p q) ~r) p is- (4) None of these 2 5 . Which of the following is equivalent to (p  q) (1) True (2) False (1) p ~q (2) ~(~p  ~q) (3) True if r is false (4) True if q is true (3) ~(p  ~q) (4) None of these 3 2 . Which of the following is wrong- 2 6 . The dual of the following statement \"Reena is (1) p  ~p is a tautology healthy and Meena is beautiful\" is- (2) ~(~p)  p is a tautology (3) p  ~p is a contradiction (1) Reena is beaufiful and Meena is healthy (4) ((p  p)  q) p is a tautology 3 3 . The statement \"If 22 = 5 then I get first class\" is (2) Reena is beautiful or Meena is healthy logically equivalent to- (3) Reena is healthy or Meena is beutiful (4) None of these 2 7 . If p is any statement, t and c are a tautology and a (1) 22 = 5 and I donot get first class contradiction respectively then which of the following is not correct- (1) p t p (2) p c c (2) 22 = 5 or I do not get first class (3) p t c (4) p c p (3) 22  5 or I get first class 2 8 . If S*(p, q) is the dual of the compound statement (4) None of these S(p, q) then S*(~p, ~q) is equivalent to- (1) S(~p, ~q) (2) ~S(p, q) 3 4 . If statement (p ~r) (q r) is false and statement q is true then statement p is- (3) ~S*(p, q) (4) None of these 2 9 . Which of the following is a statement- (1) true (2) false (1) I am Lion (2) Logic is an interesting subject (3) may be true or false (4) None of these (3) A triangle is a circle and 10 is a prime number 3 5 . Which of the following statement are not logically NODE6 (E)\\DATA\\2014\\KOTA\\JEE-ADVANCED\\SMP\\MATHS\\UNIT#02\\ENG\\PART-2\\03-MATHEMATICAL REASONING\\2.EXE equivalent- (4) None of these (1) ~(p ~q) and (~p q) (2) ~(p q) and (p ~q) (3) (p q) and (~q ~p) (4) (p q) and (~p q) ANSWER KEY Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Ans. 3 3 4 2 2 4 3 2 1 1 3 1 4 4 1 Que. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Ans. 4 4 3 4 1 3 1 4 3 3 3 3 2 3 4 Que. 31 32 33 34 35 Ans. 1 4 3 3 4 24 E

PREVIOUS YEAR QUESTIONS MATHEMATICAL REASONING JEE-Mathematics EXERCISE-II 1 . The statement p  (q  p) is equivalent 4 . Let S be a non-empty subset of R. (1) p  (p  q) [AIEEE-2008] Consider the following statement : (3) p  (p  q) p : There is a rational number x  S such that x > 0 (2) p  (p  q) which of the following statements is the negation of the statement p ? (4) p  p  q 2 . Let p be the statement “x is an irrational num- [AIEEE-2010] ber”, q be the statement “y is a trascendental num- ber”, and r be the statement “x is a rational num- (1) There is a rational number x  S such that x  0 ber iff y is a transcendental number”. [AIEEE-2008] (2) There is no rational number x  S such that x  0 (3) Every rational number x  S satisfies x  0 Statement –1 : r is equivalent to either q or p. (4) x  S and x  0  x is not rational Statement –2 : r is equivalent to p  ~ q  5 . Consider the following statements p : Suman is brilliant (1) Statement –1 is false, Statement –2 is true q : Suman is rich (2) Statement–1 is true, Statement–2 is false r : Suman is honest (3) Statement–1 is true, Statement–2 is true; The negation of the statement \"Suman is brilliant Statement–2 is a correct explanation for Statement–1 and dishonest if and only if Suman is rich\" can be (4) Statement–1 is true, Statement–2 is true; expressed as :- [AIEEE-2011] Statement–2 is not a correct explanation for (1) ~ q  ~ p  r (2) ~ (p  ~ r)  q (3) ~ p  (q  ~ r) Statement–1 (4) ~ (q  (p  ~ r)) 3 . Statement–1 : ~(p  ~ q) is equivalent to p  q. 6 . The only statement among the followings that is a tautology is : Statement–2 : ~(p  ~ q) is a tautology. [AIEEE-2011] (1) q  [p (p  q)] NODE6 (E)\\DATA\\2014\\KOTA\\JEE-ADVANCED\\SMP\\MATHS\\UNIT#02\\ENG\\PART-2\\03-MATHEMATICAL REASONING\\2.EXE[AIEEE-2009] >>(2) p (p q) (1) Statement–1 is true, Statement–2 is false. >>> > (2) Statement–1 is false, Statement–2 is true. (3) p (p q) (3) Statement–1 is true, Statement–2 is true ; (4) [p (p  q)]  q Statement–2 is a correct explanation for 7 . The negation of the statement Statement–1. \"If I become a teacher, then I will open a school\", (4) Statement–1 is true, Statement–2 is true ; is Statement–2 is not a correct explanation for [AIEEE-2012] statement–1. (1) I will not become a teacher or I will open a school. (2) I will become a teacher and I will not open a school. (3) Either I will not become a teacher or I will not open a school. (4) Neither I will become a teacher nor I will open a school. ANSWER KEY Que. 1 2 3 4 5 6 7 Ans. 2 1 1 3 2,4 4 2 E 25

JEE-Mathematics CHECK YOUR GRASP MATHEMATICAL REASONING EXERCISE-I 1 . The inverse of the statement (p  ~q)  r is- 1 3 If statements p, q, r have truth values T, F, T (1) ~(p ~q) ~r (2) (~p q) ~r respectively then which of the following statement (3) (~p q) ~r (4) None of these is true- 2 . (~p  ~q) is logically equivalent to- (1) (p q) r (2) (p q) ~r (1) p q (2) ~p q (3) p ~q (4) ~p ~q (3) (p q) (q r) (4) (p q) r 3 . The equivalent statement of (p  q) is- 1 4 . If statement p (q r) is true then the truth values of statements p, q, r respectively- (1) (p q) (p q) (2) (p q) (q p) (1) T, F, T (2) F, T, F (3) (~p q) (p ~q) (4) (~p q) (p ~q) (3) F, F, F (4) All of these 4 . If the compound statement p  (~p  q) is false 1 5 . Which of the following statement is a contradiction- then the truth value of p and q are respectively- (1) (p q) (~(p q)) (2) p (~p q) (1) T, T (2) T, F (3) F, T (4) F, F (3) (p q) p (4) ~p ~q 5 . The statement (p  ~p)  (~p  p) is- 1 6 . The negative of the statement \"If a number is (1) a tautology divisible by 15 then it is divisible by 5 or 3\" (2) a contradiction (1) If a number is divisible by 15 then it is not divisible (3) neither a tautology nor a contradiction by 5 and 3 (4) None of these (2) A number is divisible by 15 and it is not divisible by 5 or 3 6 . Negation of the statement (p  r)  (r  q) is- (3) A number is divisible by 15 or it is not divisible (1) ~(p r) ~(r q) (2) (~p ~r) (r q) by 5 and 3 (3) (p r) (r q) (4) (p r) (~r ~q) (4) A number is divisible by 15 and it is not divisible 7 . The dual of the statement ~p [~q (p q)  ~r] by 5 and 3 is- 1 7 . Which of the following is a statement- (1) ~p [~q (p q) ~r] (1) Open the door (2) p [q (~p ~q) r] (2) Do your home work (3) ~p [~q (p q) ~r] (3) Hurrah! we have won the match (4) ~p [~q (p q) ~r] (4) Two plus two is five 8 . Which of the following is correct- 1 8 . The negation of the statement \"2 + 3 = 5 and 8 < 10\" is- (1) (~p ~q) (p q) (1) 2 + 3  5 and 8  10 (2) 2 + 3 5 or 8 > 10 (2) (p q) (~q ~p) (3) 2 + 3 5 or 8  10 (4) None of these NODE6 (E)\\DATA\\2014\\KOTA\\JEE-ADVANCED\\SMP\\MATHS\\UNIT#02\\ENG\\PART-2\\03-MATHEMATICAL REASONING\\2.EXE (3) ~(p ~q) (p ~q) 1 9 . For any three simple statement p, q, r the statement (4) ~(p  q) (p q) (q p) (p q) (q r) is true when- 9 . The contrapositive of p  (~q  ~r) is- (1) p and r true and q is false (1) (~q r) ~p (2) (q r) ~p (2) p and r false and q is true (3) (q ~r) ~p (4) None of these (3) p, q, r all are false 1 0 . The converse of p  (q  r) is- (4) q and r true and p is false (1) (q ~r) p (2) (~q r) p 2 0 . Which of the following statement is a tautology- (3) (q ~r) ~p (4) (q ~r) p (1) (~p ~q) (p ~q) (2) (~p ~q) (p ~q) 1 1 . If p and q are two statement then (p  ~q) is true (3) ~p (~p ~q) (4) ~q (~p ~q) when- 2 1 . Which of the following statement is a contradiction- (1) p and q both are true (2) p and q both are false (1) (~p ~q) (p ~q) (2) (p q) (p ~q) (3) (~p q) (~q) (4) (~p q) (~q) (3) p is false and q is true (4) None of these 2 2 . The negation of the statement q  (p  ~r) is 1 2 . Statement (p  q)  p is- equivalent to- (1) a tautology (2) a contradiction (1) ~q (p r) (2) ~q ~(p r) (3) neither (1) nor (2) (4) None of these (3) ~q (~p r) (4) None of these E 23

JEE-Mathematics 2 3 . Which of the following is not a statement- 3 0 . If p is any statement, t is a tautology and c is a contradiction then which fo the following is not (1) every set is a finite set correct- (2) every square is a rectangle (3) The sun is a star (1) p (~c) p (4) Shut the window (2) p (~t) p 2 4 . The statement ~(p  q)  (~p  ~q) is- (3) t c p t (1) a tautology (4) (p t) (p c) (t c) (2) a contradiction 3 1 . If p, q, r are simple statement with truth values T, F, T respect ively then the tr uth value of (3) neither a tautology nor a contradiction ((~p q) ~r) p is- (4) None of these 2 5 . Which of the following is equivalent to (p  q) (1) True (2) False (1) p ~q (2) ~(~p  ~q) (3) True if r is false (4) True if q is true (3) ~(p  ~q) (4) None of these 3 2 . Which of the following is wrong- 2 6 . The dual of the following statement \"Reena is (1) p  ~p is a tautology healthy and Meena is beautiful\" is- (2) ~(~p)  p is a tautology (3) p  ~p is a contradiction (1) Reena is beaufiful and Meena is healthy (4) ((p  p)  q) p is a tautology 3 3 . The statement \"If 22 = 5 then I get first class\" is (2) Reena is beautiful or Meena is healthy logically equivalent to- (3) Reena is healthy or Meena is beutiful (4) None of these 2 7 . If p is any statement, t and c are a tautology and a (1) 22 = 5 and I donot get first class contradiction respectively then which of the following is not correct- (1) p t p (2) p c c (2) 22 = 5 or I do not get first class (3) p t c (4) p c p (3) 22  5 or I get first class 2 8 . If S*(p, q) is the dual of the compound statement (4) None of these S(p, q) then S*(~p, ~q) is equivalent to- (1) S(~p, ~q) (2) ~S(p, q) 3 4 . If statement (p ~r) (q r) is false and statement q is true then statement p is- (3) ~S*(p, q) (4) None of these 2 9 . Which of the following is a statement- (1) true (2) false (1) I am Lion (2) Logic is an interesting subject (3) may be true or false (4) None of these (3) A triangle is a circle and 10 is a prime number 3 5 . Which of the following statement are not logically NODE6 (E)\\DATA\\2014\\KOTA\\JEE-ADVANCED\\SMP\\MATHS\\UNIT#02\\ENG\\PART-2\\03-MATHEMATICAL REASONING\\2.EXE equivalent- (4) None of these (1) ~(p ~q) and (~p q) (2) ~(p q) and (p ~q) (3) (p q) and (~q ~p) (4) (p q) and (~p q) ANSWER KEY Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Ans. 3 3 4 2 2 4 3 2 1 1 3 1 4 4 1 Que. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Ans. 4 4 3 4 1 3 1 4 3 3 3 3 2 3 4 Que. 31 32 33 34 35 Ans. 1 4 3 3 4 24 E

PREVIOUS YEAR QUESTIONS MATHEMATICAL REASONING JEE-Mathematics EXERCISE-II 1 . The statement p  (q  p) is equivalent 4 . Let S be a non-empty subset of R. (1) p  (p  q) [AIEEE-2008] Consider the following statement : (3) p  (p  q) p : There is a rational number x  S such that x > 0 (2) p  (p  q) which of the following statements is the negation of the statement p ? (4) p  p  q 2 . Let p be the statement “x is an irrational num- [AIEEE-2010] ber”, q be the statement “y is a trascendental num- ber”, and r be the statement “x is a rational num- (1) There is a rational number x  S such that x  0 ber iff y is a transcendental number”. [AIEEE-2008] (2) There is no rational number x  S such that x  0 (3) Every rational number x  S satisfies x  0 Statement –1 : r is equivalent to either q or p. (4) x  S and x  0  x is not rational Statement –2 : r is equivalent to p  ~ q  5 . Consider the following statements p : Suman is brilliant (1) Statement –1 is false, Statement –2 is true q : Suman is rich (2) Statement–1 is true, Statement–2 is false r : Suman is honest (3) Statement–1 is true, Statement–2 is true; The negation of the statement \"Suman is brilliant Statement–2 is a correct explanation for Statement–1 and dishonest if and only if Suman is rich\" can be (4) Statement–1 is true, Statement–2 is true; expressed as :- [AIEEE-2011] Statement–2 is not a correct explanation for (1) ~ q  ~ p  r (2) ~ (p  ~ r)  q (3) ~ p  (q  ~ r) Statement–1 (4) ~ (q  (p  ~ r)) 3 . Statement–1 : ~(p  ~ q) is equivalent to p  q. 6 . The only statement among the followings that is a tautology is : Statement–2 : ~(p  ~ q) is a tautology. [AIEEE-2011] (1) q  [p (p  q)] NODE6 (E)\\DATA\\2014\\KOTA\\JEE-ADVANCED\\SMP\\MATHS\\UNIT#02\\ENG\\PART-2\\03-MATHEMATICAL REASONING\\2.EXE[AIEEE-2009] >>(2) p (p q) (1) Statement–1 is true, Statement–2 is false. >>> > (2) Statement–1 is false, Statement–2 is true. (3) p (p q) (3) Statement–1 is true, Statement–2 is true ; (4) [p (p  q)]  q Statement–2 is a correct explanation for 7 . The negation of the statement Statement–1. \"If I become a teacher, then I will open a school\", (4) Statement–1 is true, Statement–2 is true ; is Statement–2 is not a correct explanation for [AIEEE-2012] statement–1. (1) I will not become a teacher or I will open a school. (2) I will become a teacher and I will not open a school. (3) Either I will not become a teacher or I will not open a school. (4) Neither I will become a teacher nor I will open a school. ANSWER KEY Que. 1 2 3 4 5 6 7 Ans. 2 1 1 3 2,4 4 2 E 25

JEE-Mathematics MATHEMATICAL REASONING 1 . STATEMENT : A sentence which is either true or false but cannot be both are called a statement. A sentence which is an exclamatory or a wish or an imperative or an interrogative can not be a statement. If a statement is true then its truth value is T and if it is false then its truth value is F For ex. (i) \"New Delhi is the capital of India\", a true statement (ii) \"3 + 2 = 6\", a false statement (iii) \"Where are you going ?\" not a statement beasuse it connot be defined as true or false Note : A statement cannot be both true and false at a time 2 . SIMPLE STATEMENT : Any statement whose truth value does not depend on other statement are called simple statement For ex. (i) \" 2 is an irrational number\" (ii) \"The set of real number is an infinite set\" 3 . COMPOUND STATEMENT : A statement which is a combination of two or more simple statements are called compound statement Here the simple statements which form a compound statement are known as its sub statements For ex. (i) \"If x is divisible by 2 then x is even number\" (ii) \"ABC is equilatral if and only if its three sides are equal\" 4 . LOGICAL CONNECTIVES : The words or phrases which combined simple statements to form a compound statement are called logical connectives. In the following table some possible connectives, their symbols and the nature of the compound statement formed by them S.N. Connectives symbol use operation 1. and  p q conjunction p q disjunction 2. or   p or p' negation NODE6 (E)\\DATA\\2014\\KOTA\\JEE-ADVANCED\\SMP\\MATHS\\UNIT#02\\ENG\\PART-2\\03-MATHEMATICAL REASONING\\1.THEORY Implication or conditional 3 . not  or ' p q or p q Equivalence or Bi-conditional 4 . If .... then ..... or  p q or p  q 5 . If and only if (iff)  or  Explanation : E (i) p  q  statement p and q (p  q is true only when p and q both are true otherwise it is false) (ii) p  q  statement p or q (p  q is true if at least one from p and q is true i.e. p  q is false only when p and q both are false) (iii) ~ p  not statement p (~ p is true when p is false and ~ p is false when p is true) (iv) p  q  statement p then statement q (p  q is false only when p is true and q is false otherwise it is true for all other cases) (v) p  q  statement p if and only if statement q (p  q is true only when p and q both are true or false otherwise it is false) 16

JEE-Mathematics 5 . TRUTH TABLE : A table which shows the relationship between the truth value of compound statement S(p, q, r ....) and the truth values of its sub statements p, q, r, ... is said to be truth table of compound statement S If p and q are two simple statements then truth table for basic logical connectives are given below Conjunction Disjunction Negation p q pq p q pq p (~ p) TT T TT T TF TF F TF T FT FT F FT T FF F FF F Conditional Biconditional p q pq TT T p q p  q q  p (p  q)  (q  p) or p  q TF F FT T TT T T T FF T TF F T F FT T F F FF T T T Note : If the compound statement contain n sub statements then its truth table will contain 2n rows. Illustration 1 : Which of the following is correct for the statements p and q ? (1) p  q is true when at least one from p and q is true (2) p  q is true when p is true and q is false (3) p  q is true only when both p and q are true (4) ~ (p  q) is true only when both p and q are false Solution : We know that p  q is true only when both p and q are true so option (1) is not correct we know that p  q is false only when p is true and q is false so option (2) is not correct we know that p   q is true only when either p and q both are true or both are flase so option (3) is not correct we know that ~(p  q) is true only when (p  q) is false i.e. p and q both are false So option (4) is correct 6.NODE6 (E)\\DATA\\2014\\KOTA\\JEE-ADVANCED\\SMP\\MATHS\\UNIT#02\\ENG\\PART-2\\03-MATHEMATICAL REASONING\\1.THEORY LOGICAL EQUIVALENCE : E Two compound statements S (p, q, r...) and S (p, q, r ....) are said to be logically equivalent or simply 12 equivalent if they have same truth values for all logically possibilities Two statements S1 and S2 are equivalent if they have identical truth table i.e. the entries in the last column of their truth table are same. If statements S1 and S2 are equivalent then we write S1  S2 For ex. The truth table for (p  q) and (~p  q) given as below p q (~ p) p  q ~ p  q TT F T T TF F F F FT T T T FF T T T We observe that last two columns of the above truth table are identical hence compound statements (p  q) and (~p  q) are equivalent i.e. p  q ~ p  q 17

JEE-Mathematics Illustration 2 : Equivalent statement of the statement \"if 8 > 10 then 22 = 5\" will be :- (1) if 22 = 5 then 8 > 10 (2) 8 < 10 and 22  5 (3) 8 < 10 or 22 = 5 (4) none of these Solution : We know that p  q  ~p  q   equivalent statment will 8  10 or 22 = 5 or 8  10 or 22 = 5 So (4) will be the correct answer. Do yourself - 1 : ( i ) Which of the following is logically equivalent to (p  q) ? (1) p ~q (2) ~p  ~ q (3) ~(p  ~q) (4) ~(~p  ~q) 7 . TAUTOLOGY AND CONTRADICTION : ( i ) Tautology : A statement is said to be a tautology if it is true for all logical possibilities i.e. its truth value always T. it is denoted by t. For ex. the statement p  ~ (p  q) is a tautology p q p  q ~ (p  q) p ~ (p  q) TT T F T TF F T T FT F T T FF F T T Clearly, The truth value of p  ~ (p  q) is T for all values of p and q. so p  ~ (p  q) is a tautology ( i i ) Contradiction : A statement is a contradiction if it is false for all logical possibilities. i.e. its truth value always F. It is denoted by c. For ex. The statement (p  q)  (~p  ~q) is a contradiction p q ~ p ~ q p  q (~ p ~ q) (p  q)  (~ p ~ q) TT F F T F F TF F T T F F FT T F T F F FF T T F T F NODE6 (E)\\DATA\\2014\\KOTA\\JEE-ADVANCED\\SMP\\MATHS\\UNIT#02\\ENG\\PART-2\\03-MATHEMATICAL REASONING\\1.THEORY Clearly, then truth value of (p  q)  (~p  ~q) is F for all value of p and q. So (p  q)  (~p  ~q) is a contradiction. Note : The negation of a tautology is a contradiction and negation of a contradiction is a tautology Do yourself - 2 : By truth table prove that : ( i ) p  q  ~p  ~q (ii) p  (~ p  q)  p  q (iii) p  (~ p  q ) is a tautology. 8 . ALGEBR A OF STATEMENTS : If p, q, r are any three statements then the some low of algebra of statements are as follow (i) Idempotent Laws : (a) p p p (b) p p p i.e. p p p p p p (p  p) (p  p) TT T FF F 18 E

JEE-Mathematics (ii) Comutative laws : (b) p q q p (a) p q q p p q (p  q) (q  p) (p  q) (q  p) TT T T T T TF F F T T FT F F T T FF F F F F (iii) Associative laws : (a) (p q) r p (q r) (b) (p q) r p (q r) p q r (p  q) (q  r) (p  q)  r p  (q  r) TTT T T T T TTF T F F F TFT F F F F TFF F F F F FTT F T F F FTF F F F F FFT F F F F FFF F F F F Similarly we can proved result (b) ( i v ) Distributive laws : (a) p (q r)  (p q) (p r) (c) p  (q r)  (p q)  (p  r) (b) p (q r) (p q) (p r) (d) p (q r)  (p q) (p r) p q r (q  r) (p  q) (p  r) p  (q  r) (p  q)  (p  r) TTT T T T T T TTF T T F T T TFT T F T T T TFF F F F F F FTT T F F F F FTF T F F F F NODE6 (E)\\DATA\\2014\\KOTA\\JEE-ADVANCED\\SMP\\MATHS\\UNIT#02\\ENG\\PART-2\\03-MATHEMATICAL REASONING\\1.THEORY FFT T F F F F FFF F F F F F Similarly we can prove result (b), (c), (d) ( v ) De Morgan Laws : (a) ~ (p  q)  ~p  ~q (b) ~(p q) ~p ~q p q ~ p ~ q (p  q) ~ (p  q) (~ p ~ q) TT F F T F F TF F T F T T FT T F F T T FF T T F T T Similarly we can proved resulty (b) ~(~p)  p ( v i ) Involution laws (or Double negation laws) : p ~ p ~ (~ p) TF T FT F E 19

JEE-Mathematics (vii) Identity Laws : If p is a statement and t and c are tautology and contradiction respectively then (a) p t p (b) p t t (c) p c c (d) p c p p t c (p  t) (p  t) (p  c) (p  c) TTF T T F T FTF F T F F (viii) Complement Laws : (a) p (~p) c (b) p (~p) t (c) (~t) c (d) (~c) t p ~ p (p ~ p) (p ~ p) TF F T FT F T ( i x ) Contrapositive laws : p  q  ~q  ~p p q ~ p ~ q p  q ~ q ~ p TT F F T T TF F T F F FT T F T T FF T T T T Illustration 3 : ~(p q) (~p  q) is equivalent to- (1) p (2) ~p (3) q (4) ~q (By Demorgan Law) Solution :  ~(p q) (~p q) (~p ~q) (~p q) Ans. (2) ~p (~q q) (By distributive laws) ~p t (By complement laws) ~p (By Identity Laws) Do yourself - 3 : (2) a contradiction ( i ) Statement (p  ~q)  (~p  q) is (4) None of these (1) a tautology (3) neither a tautology not a contradiction 9 . NEGATION OF COMPOUND STATEMENTS : NODE6 (E)\\DATA\\2014\\KOTA\\JEE-ADVANCED\\SMP\\MATHS\\UNIT#02\\ENG\\PART-2\\03-MATHEMATICAL REASONING\\1.THEORY If p and q are two statements then (i) Negation of conjunction : ~(p  q)  ~p  ~q p q ~ p ~ q (p  q) ~ (p  q) (~ p ~ q) TT F F T F F TF F T F T T FT T F F T T FF T T F T T ( i i ) Negation of disjunction : ~(p  q)  ~p  ~q p q ~ p ~ q (p  q) (~ p  q) (~ p ~ q) TT F F T F F TF F T T F F FT T F T F F FF T T F T T 20 E

JEE-Mathematics Note : (i) the connectives  and  are also called dual of each other. (ii) If S*(p, q) is the dual of the compound statement S(p, q) then (a) S*(~p, ~q)  ~S(p, q) (b) ~S*(p, q)  S(~p, ~q) Illustration 6 : (iii) ~(p q) [p ~(q ~s)] The duals of the following statements (i) (p q) (r s) (ii) (p t) (p c) Solution : (i) (p q) (r s) (ii) (p c) (p t) (iii) ~(p q) [p ~(q ~s)] 1 1 . CON VERSE, IN VERSE AND CONTR APOSITIVE OF THE CONDITIONAL STATEMENT (p  q) : ( i ) Converse : The converse of the conditional statement p  q is defined as q  p ( i i ) Inverse : The inverse of the conditional statement p  q is defined as ~p  ~q ( i i i ) Contrapositive : The contrapositive of conditional statement p  q is defined as ~q  ~p Illustration 7 : If x = 5 and y = –2 then x – 2y = 9. The contrapositive of this statement is- (1) If x – 2y 9 then x 5 or y –2 (2) If x – 2y 9 then x 5 and y –2 (3) If x – 2y = 9 then x = 5 and y = –2 (4) None of these Solution : Let p, q, r be the three statements such that p : x = 5, q : y = –2 and r : x – 2y = 9 Here given statement is (p  q)  r and its contrapositive is ~r  ~(p  q) i.e. ~r (~p ~q) i.e. if x – 2y 9 then x 5 or y –2 Ans. (1) Do yourself - 5 : ( i ) If S*(p, q, r) is the dual of the compound statement S(p, q, r) and S(p, q, r) = ~p  [~(q  r)] then NODE6 (E)\\DATA\\2014\\KOTA\\JEE-ADVANCED\\SMP\\MATHS\\UNIT#02\\ENG\\PART-2\\03-MATHEMATICAL REASONING\\1.THEORY S*(~p, ~q, ~r) is equivalent to - (1) S(p, q, r) (2) ~S(~p, ~q, ~r) (3) ~S(p, q, r) (4) S*(p, q, r) ( i i ) Contrapositive of the statement (2) (r s) (~p q) (4) None (p  q)  (r ~ s) will be :- (1) ~(p  q)  ~(r  ~s) (3) (~r ~ s) (p ~q) 1. (i) 3 (ii) 3 ANSWERS FOR DO YOURSELF 2. (i) 2 22 4. (i) 3 5. (i) 3 E

JEE-Mathematics EXERCISE - 01 CHECK YOUR GRASP SELECT THE CORRECT ALTERNATIVE (ONLY ONE CORRECT ANSWER) 1. 1 5  and 2A – 3B = 2 5  , then matrix B is equal to - If A – 2B = 3 7   0 7     4 5   0 6 2 1 6 1 (A) 6 7  (B) 3 7  (C) 3 2  (D) 0 1       cos  sin   , then equal - 2. If A =  sin  AA is to cos    (A) A+ (B) A (C) A (D) none of these 3 . If number of elements in a matrix is 60 then how many different order of matrix are possible - (A) 12 (B) 6 (C) 24 (D) none of these 4 . Matrix A has x rows and x + 5 columns. Matrix B has y rows and 11 – y columns. Both AB and BA exist, then - (A) x = 3, y = 4 (B) x = 4, y = 3 (C) x = 3, y = 8 (D) x = 8, y = 3 5 . If A2 = A, then(I + A)4 is equal to - (A) I + A (B) I + 4A (C) I + 15A (D) none of these 6. If the product of n matrices 1 1 1 2  1 3  ...... 1 n is equal to the matrix 1 378 then the value 0 1  0 0 1  0 1  0  1  1      of n is equal to - (A) 26 (B) 27 (C) 377 (D) 378 7. If A = 0 1 and (aI +bA)2 = A , then - 1 2 0   (A) a = b = 2 (B) a = b = 1/ 2 (C) a = b = 3 (D) a = b = 1/ 3 8 . If A is a skew symmetric matrix such that ATA = I, then  A4n–1 n N is equal to - (A) – AT (B) I (C) – I (D) AT E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\02.MATRIX\\02.EXERCISES.p65 9 . If AAT = I and det(A) = 1, then - (A) Every element of A is equal to it's co-factor. (B) Every element of A and it's co-factor are additive inverse of each other. (C) Every element of A and it's co-factor are multiplicative inverse of each other. (D) None of these 1 0 . Which of the following is an orthogonal matrix - 6 / 7 2/7 3 / 7 6 / 7 2/7 3/7  3/7 3 / 7 (A) 2 / 7 6 / 7 6 / 7  (B) 2 / 7 6/7 6 /7  3 / 7  3 / 7  2 / 7  2 / 7 6 / 7 2 / 7 3 / 7  6/7 2 / 7 3/7  3/7 2/7 (C)  2 / 7 6/7 6/7  (D)  2 / 7 2/7 3 / 7      3 / 7 2 / 7  6 / 7 3 / 7  1 1 . If A is an orthogonal matrix & | A | = –1, then AT is equal to - (A) –A (B) A (C) –(adj A) (D) (adj A) 50 E

JEE-Mathematics 1 1 1   4 2 2 12. Let A = 2 1 3  and 10B = 5 0  . If B is the inverse of matrix A, then  is -  1 1 1   1 2 3  (A) –2 (B) –1 (C) 2 (D) 5 3 2  an d B  3 1 then the value of Det.(2A9B–1), is - Let the matrix A and B be defined as A = 2 1  7 13.  3   (A) 2 (B) 1 (C) –1 (D) –2 2 1  A 3 2  1 0 If 7 4  0 14.   5 3  1  , then matrix A equals -     7 5 2 1   7 1  5 3 (A) 11 8  (B) 5 3  (C) 34 5  (D) 13 8      15. 0 5  and (x) = 1 + x + x2 + ...... + x16, then (A) = If A = 0 0   1 5  1 5  0 5  (A) 0 (B) 0 (C) 0 (D) 1 1  0  1     16. 1 2 and M2 –M – I = O , then  equals - If M = 2 3  2  (A) –2 (B) 2 (C) –4 (D) 4 1 2  1 4 4 8 If A = 3 3 17. 0  , B =  2 3  and ABC = 7  , then C equals -     1 72 32  1 54 110  1 54 110  1 72 32 (A) 66 57 (B) (C) (D) 66 57 29  66  3 11  66  3 1 1  29       1 8 . If P is a two-rowed matrix satisfying PT = P–1, then P can be -  cos   sin   cos  sin     cos  sin  (B)  sin  cos  (A)  sin  cos   (C)  sin   cos  (D) none of these   E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\02.MATRIX\\02.EXERCISES.p65 a 0 0 19. If A   0 a 0  , then | A | | Adj A | is equal to -    0 0 a  (A) a25 (B) a27 (C) a81 (D) none of these (D) A 2 3  A  5 19A–1 20. If 2  , then is equal to -  (A) AT (B) 2A (C) 1A 2 SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS) 2 1 . If A and B are square matrices of same order, then which of the following is correct - (A) A + B = B + A (B) A + B = A – B (C) A – B = B – A (D) AB = BA 2 2 . A square matrix can always be expressed as a (A) sum of a symmetric matrix and skew symmetric matrix of the same order (B) difference of a symmetric matrix and skew symmetric matrix of the same order (C) skew symmetric matrix (D) symmetric matrix E 51

JEE-Mathematics 2 3 . Choose the correct answer : (A) every scalar matrix is an identity matrix. (B) every identity matrix is a scalar matrix (C) transpose of transpose of a matrix gives the matrix itself. (D) for every square matrix A there exists another matrix B such that AB = I = BA. 2 4 . Let aij denote the element of the ith row and jthcolumn in a 3 × 3 matrix and let aij = –aji for every i and j then this matrix is an - (A) orthogonal matrix (B) singular matrix (C) matrix whose principal diagonal elements are all zero (D) skew symmetric matrix 2 5 . Let A be an invertible matrix then which of the following is/are true : (A) |A–1| = |A|–1 (B) (A2)–1 = (A–1)2 (C) (AT)–1 = (A–1)T (D) none of these 1 9 7  26. If A   i n 8  , where i 1 and   is complex cube root of unity, then tr(A) will be-   1 6 2n  (A) 1, if n = 3k, k  N (B) 3, if n = 3k, k  N (C) 0, if n  3k, k  N (D) –1, if n  3k, n  N 2 7 . If A is a square matrix, then - (A) AAT is symmetric (B) AAT is skew-symmetric (C) ATA is symmetric (D) ATA is skew symmetric. a b satisfies the equation x2 + k = 0, then - 28. If A = c d   (A) a + d = 0 (B) k = –|A| (C) k = a2+ b2+ c2+ d2 (D) k = |A| 2 9 . If A and B are invertible matrices, which one of the following statement is/are correct - (A) Adj(A) = |A|A–1 (B) det(A–1) =|det(A)|–1 (C) (A + B)–1= B–1 + A–1 (D) (AB)–1 = B–1A–1 a b a  b 30. Matrix b c b  c   is non invertible if - E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\02.MATRIX\\02.EXERCISES.p65  2 1 0  (A)  = 1/2 (B) a, b, c are in A.P. (C) a, b, c are in G.P. (D) a, b, c are in H.P. CHECK YOUR GRASP ANSWER KEY EXERCISE-1 Que. 1 2 3456 7 8 9 10 Ans. A A ACCB B D A A Que. 11 12 13 14 15 16 17 18 19 20 Ans. C D DABD B B D D Que. 21 22 23 24 25 26 27 28 29 30 Ans. A A,B B,C B,C,D A,B,C B,C A,C A,D A,B,D A,C 52 E

JEE-Mathematics EXERCISE - 02 BRAIN TEASERS SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS) 1 . If A and B are square matrices of same order and AAT = I then (ATBA)10 is equal to - (A) AB10AT (B) ATB10A (C) A10B10(AT)10 (D) 10ATBA 2 . If A is a invertible idempotent matrix of order n, then adj A is equal to - (A) (adj A)2 (B)  (C) A–1 (D) none of these x 3 2 3. Matrix A = 1 y 4  , if xyz = 60 and 8x + 4y + 3z = 20, then A (adj A) is equal to -  2 2 z  64 0 0  88 0 0  68 0 0  34 0 0  (A)  0 64 0  (B)  0 88 0  (C)  0 68 0  (D)  0 34 0           0 0 64   0 0 88   0 0 68   0 0 34  2 1 3 4 and C =  3 –4  4. Let three matrices A = 4 1; B = 2 then 3   –2 3     tr (A)  tr  ABC   tr  A (B C )2   tr  A (B C )3   ......     2   4   8  (A) 6 (B) 9 (C) 12 (D) none of these 5 . Let A, B, C, D be (not necessarily square) real matrices such that AT = BCD ; BT = CDA; CT = DAB and DT = ABC for the matrix S = ABCD, then which of the following is/are true (A) S3 = S (B) S2 = S4 (C) S = S2 (D) none of these 1 tan x let function (ATA–1) 6. If A =  tan x  then us define a f(x) = det then which of the following can be the value 1  of f(f(f(f...........f(x)))) (n  2) n times (A) fn(x) (B) 1 (C) fn–1(x) (D) nf(x) 1 50 1 2r 1 E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\02.MATRIX\\02.EXERCISES.p65 For a matrix A = 0 2r  1 7. 1  , the value of r 1 0 1  is equal to -   1 100  1 4950  1 5050  1 2500  (A) 0 (B) 0 (C) 0 (D) 0 1  1  1  1      8 . If A and B are two invertible matrices of the same order, then adj (AB) is equal to - (A) adj (B) adj (A) (B) |B||A| B–1 A–1 (C) |B||A| A–1 B–1 (D) |A||B|(AB)–1 0 1 2 1 / 2 1 / 2 1 / 2 9. If A = 1 2 3  , A 1   4 3 c  , then -   3 a 1  5 / 2 3 / 2 1 / 2 (A) a = 1, c = –1 1 (C) a = –1, c = 1 11 (B) a = 2, c = – (D) a = , c = 2 22 1 0 . If A and B are different matrices satisfying A3 = B3 and A2B = B2A, then which of the following is/are incorrect- (A) det (A2 + B2) must be zero (B) det (A – B) must be zero (C) det (A2 + B2) as well as det (A – B) must be zero (D) At least one of det (A2 + B2) or det (A – B) must be zero E 53

JEE-Mathematics 0 0 1  11. If A  0 1 0  , then-  1 0 0  0 0 1 (A) AdjA is zero matrix (B) Adj A   0 1 0    1 0 0  (C) A–1 = A (D) A2 = I 1 2 . If A and B are square matrices of the same order such that A2 = A, B2 = B, AB = BA, then which one of the following may be true- (A) A(B)2 = O (B) (A + B)2 = A + B (C) (A – B)2 = A – B (D) none of these 1 3 . If B is an idempotent matrix and A = I – B, then- (A) A2 = A (B) A2 = I (C) AB = O (D) BA = O a11 a12 a13 1 4 . Let 0 = a21 a22 a23 (where 0  0) and let 1 denote the determinant formed by the cofactors of a31 a32 a33 elements of 0 and 2 denote the determinant formed by the cofactor at 1 and so on n denotes the determinant formed by the cofactors at n–1 then the determinant value of n is - (A)  2 n (B)  2 n (C)  n2 (D) 20 0 0 0 BRAIN TEASERS ANSWER KEY EXERCISE-2 E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\02.MATRIX\\02.EXERCISES.p65 Que. 1 2 3 4 5 6 7 8 9 10 Ans. B A,B,C C A A,B A,B,C D A,B,D A A,B,C Que. 11 13 14 Ans. B,C,D 12 A,C,D B 54 E A,B,C

JEE-Mathematics EXERCISE - 03 MISCELLANEOUS TYPE QUESTIONS TRUE / FALSE 1 . Let A, B be two matrices such that they commute, then A2 – B2 = (A – B)(A + B) 2 . If A is a periodic matrix with period 2 then A6 = A. 3 . Let A, B be two matrices such that they commute, then (AB)n = AnBn. 4 . All positive odd integral powers of skew symmetric matrix are symmetric. 5 . Let A, B be two matrices, such that AB = A and BA = B, then A2 = A and B2 = B. 6 . If A & B are symmetric matrices of same order then AB – BA is symmetric. 7 . If A and B are square matrices of order n, then A and B will commute, iff A – I and B – I commute for every scalar . MATCH THE COLUMN Following questions contains statements given in two columns, which have to be matched. The statements in Column-I are labelled as A, B, C and D while the statements in Column-II are labelled as p, q, r and s. Any given statement in Column-I can have correct matching with ONE statement in Column-II. 1 . Column-I Column-II Matrix Type of matrix (p) Idempotent  2 2 4  (q) Involutary (A) 1 3 4  (r) Nilpotent   1 2 3  5 8 0  (B)  3 5 0     1 2 1 1  1 2 2  3 (C) 2 1 2  2 2  1 1 1 3  (D)  5 2 6  (s) Orthogonal   2 1 3 E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\02.MATRIX\\02.EXERCISES.p65 2 . Column-I Column-II (A) If A is a square matrix of order 3 and (p) 6 (q) 5 det A = 162 then det  A  = (r) 0  3  (s) 9 (B) If A is a matrix such that A2 = A and (I + A)5 = I + A then 2 1 7 (C) 4 3 and A2 – xA + yI = 0 If A = 2 5   then y – x =  9 10 11 12 (D) If A = 13 14 15 16  and  17 18 19 20  1 3 5 7  3 3 10 10   B =  5 10 5 0  then (AB)23  7 10 0 7  E 55

JEE-Mathematics 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 : If a, b, c are distinct real number and x, y, z are not all zero given that ax + by + cz = 0, bx + cy + az = 0, cx + ay + bz = 0, then a + b + c  0 Statement - II : If a, b, c are distinct positive real number then a2 + b2 + c2  ab + bc + ca. (A) A (B) B (C) C (D) D 2 . Statement - I : If A is skew symmetric matrix of order 3 then its determinant should be zero Statement - II : If A is square matrix, then det A = det A' = det(–A') (A) A (B) B (C) C (D) D 3 . Statement-I : If A is a non-singular symmetric matrix, then its inverse is also symmetric. Because Statement-II : (A–1)T = (AT)–1, where A is a non-singular symmetric matrix. (A) A (B) B (C) C (D) D 4. Statement - I : There are only finitely many 2 × 1 2  . 2 matrices which commute with the matrix 1 1   Because Statement - II : If A is non-singular, then it commutes with I, adj A and A–1. (A) A (B) B (C) C (D) D E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\02.MATRIX\\02.EXERCISES.p65  3 1 3 1  3 1 1 3     2 2 22 2 2 22 5. Statement-I : If x = 1  3 A  3 1 & if A is idempotent matrix then x is also idempotent    3  1  3  1    2 2 2 2   2 2 2 2  matrix. Because Statement-II : If P is an orthogonal matrix & Q = PAPT, then Qn = PAnPT. (A) A (B) B (C) C (D) D 6 . Statement-I : The determinants of a matrix A = [aij]5 × 5 where aij + aji = 0 for each i and j is zero. Because Statement-II : The determinant of a skew symmetric matrix of odd order is zero. (A) A (B) B (C) C (D) D 56 E

JEE-Mathematics COMPREHENSION BASED QUESTIONS Comprehension # 1 Let P(x, y) be any point and P'(x1, y1) be its image in x-axis then x1 = x y1 = –y This system of equation is equivalent to the matrix equation. x1   A x    y   y1  where A is a square matrix of order 2 x2  x x3  x  x4  x    B ,   C y ,   D y  Similarly  y   y   y    y       2 3 4 represents the reflection of point (x, y) in y-axis, origin and the line y = x respectively. On the basis of above information, answer the following questions : 1 . The value of A + B + C + D is - 1 1  1 1  1 0  0 1  (A)  0 0  (B)  1 1  (C) 0 1  (D) 1 0        2 . Let X be a square matrix given by X = A + AD2 + AD4 +.........+AD2n – 2, then X is - n 0  n 0 n 0  n 0 (A)  0 n  (B) 0 n  (C) 0 n  (D)  0 n        3. Let P(a, b) be a point & x   DC B A a  then Q(x, y) represents the reflection of point P(a, b) in - y  b  (A) x-axis (B) y-axis (C) origin (D) line y = x Comprehension # 2 a1 a2 a3  Matrix A is called orthogonal matrix if AAT = I = ATA. Let A = b1  b2 b3  be an orthogonal matrix. Let E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\02.MATRIX\\02.EXERCISES.p65   a1ˆi  a2ˆj  a3 kˆ ,   b1ˆi  b2ˆj  b3kˆ ,   c1ˆi  c2ˆj  c 3 kˆ . Then c1 c2 c3 1 &    0 i.e. a b c | a|| b|| c| a.b  b.c  c.a   a, b & c forms mutually perpendicular triad of unit vectors. a b c  If abc = p and Q = c a b  , where Q is an orthogonal matrix. Then.  b c a  On the basis of above information, answer the following questions : 1. The values of a + b + c is - 2. 3. (A) 2 (B) p (C) 2p (D) ±1 4. The values of ab + bc + ca is - E (A) 0 (B) p (C) 2p (D) 3p The value of a3 + b3 + c3 is - (A) p (B) 2p (C) 3p (D) None of these The equation whose roots are a, b, c is - (A) x3 – 2x2 + p = 0 (B) x3 – px2 + px + p = 0 (D) x3 ± x2 – p = 0 (C) x3 – 2x2 + 2px + p = 0 57

JEE-Mathematics Comprehension # 3  2 2 4  4 3 3 If A 0  1 3 4  and B0   1 0 1      1 2 3   4 4 3  B = adj(B ), n  N and I is an identity matrix of order 3. n n–1 On the basis of above information, answer the following questions : 1. det. (A0  A 2 B 2  A 3  A 4 B 4  .......10 terms) is equal to - 0 0 0 0 0 (A) 1000 (B) –800 (C) 0 (D) –8000 2 . B + B + ....... + B is equal to - 12 49 (A) B (B) 7B (C) 49B (D) 49I 0 0 0 3 . For a variable matrix X the equation A X = B will have - 00 (A) unique solution (B) infinite solution (C) finitely many solution (D) no solution MISCELLANEOUS TYPE QUESTION ANSWER KEY EXERCISE-3 E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\02.MATRIX\\02.EXERCISES.p65  True / False 1. T 2. F 3. T 4. F 5. T 6. F 7. T  Match the Column 1. (A)  (p), (B)  (q), (C)  (s), (D)  (r) 2. (A)  (p), (B)  (s), (C)  (q), (D)  (r)  Assertion & Reason 1. D 2. C 3. A 4. D 5. A 6. A  Comprehension Based Questions Comprehension # 1 : 1. B 2. C 3. D Comprehension # 2 : 1. D 2. A 3. D 4. D Comprehension # 3 : 1. C 2. C 3. D 58 E

EXERCISE - 04 [A] JEE-Mathematics CONCEPTUAL SUBJECTIVE EXERCISE 1 . Find the value of x and y that satisfy the equations - 3 2  y 3 3  y 3 0    3 y 3 y  2 4   x x  10   10  2 . A is a square matrix of order n.  = maximum number of distinct entries if A is a triangular matrix m = maximum number of distinct entries if A is a diagonal matrix p = minimum number of zeroes if A is a triangular matrix If  + 5 = p + 2m, find the order of the matrix. 1 2  a b 3. If the matrices A = 3 4  and B = c d    (a, b, c, d not all simultaneously zero) commute, find the value of db . Also show that the matrix which acb commutes with A is of the form    2 / 3       1 2  5  4. Consider the two matrices A and B where A = 4 3 ; B = 3  . Let n(A) denotes the number of elements in  A. When the two matrices X and Y are not conformable for multiplication then n(XY) = 0  n(C ) (| D|2 n(D)  If C = (AB)(B'A); D = (B'A)(AB) then, find the value of  n(A )  n(B )  . 0 1  (A8 A6 A4 A2 0  5. Define A = 3 0  . Find a vertical vector V such that + + + + I)V = 1 1    E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\02.MATRIX\\02.EXERCISES.p65 (where I is the 2 × 2 identity matrix). 6 . If A is an idempotent matrix and I is an identity matrix of the same order, find the value of n, n N, such that (A + I)n = I + 127 A. 11 11 7 . If the matrix A is involutary, show that (I + A) and (I – A) are idempotent and (I + A). (I –A) = 0. 22 22 0 1 1 8. Let X be the solution set of the equation Ax = I, where A = 4 3 4  and I is the corresponding unit matrix  3 3 4  and x  N then find the minimum value of (cosx   sin x ),  R. 1 2 5  9. Express the matrix  2 3  6  as a sum of a lower triangular matrix & an upper triangular matrix with zero   1 0 4  in leading diagonal of upper triangular matrix. Also express the matrix as a sum of a symmetric and a skew symmetric matrix. E 59

JEE-Mathematics 1 1 1 2 3  1 0 1 Given A = 2 4 3 4  0 1 10. 3 1  , B =  . Find P such that BPA = 0  . 2   1   sin   cos  0 1 1 . If A = cos  sin  0  then find |AT| and |A–1|.   0 0 1  1  tan    1 tan  1 cos   sin   2   2  12. Show that, = .  tan  1   tan  1   sin  cos    2  2    cos x  sin x 0 13. If F(x) =  sin x cos x 0  then show that F(x).F(y) = F(x + y). Hence prove that [F(x)]–1 = F(–x).    0 0 1  1 0 2  14. If A = 0 2 1  , then show that the matrix A is a root of the polynomial f(x) = x3 – 6x2 + 7x + 2.  2 0 3 1 5 . Use matrix to solve the following system of equations x  y  z 3 x  y  z 6 x  y z 3 x  y z 3 (a) x  2y  3z  4 (b) x  y  z  2 (c) x  2y  3z  4 (d) x  2y  3z  4 x  4y  9z  6 2x  y  z  1 2x  3y  4z  7 2x  3y  4z  9 CONCEPTUAL SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(A) E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\02.MATRIX\\02.EXERCISES.p65 0 1 . x = 3/2, y = 2 2 . 4 3 . 1 4 . 650 5. V   1  6. n=7   11   1 0 0  0 2 5  1 2 2   0 0 3  4 7 7 0  2    8. 2 9.  2 3 0  + 0  6  ; 3 3    0 0 3  10.  3 5 5  11. 1, 1     1 5 . (a) 1 0 4  0 0 0  2 3 4  3 3 0  (c) x = 2, y = 1, z = 0 (b) x = 1, y = 2, z = 3 x = 2 + k, y = 1  2k, z = k where k  R (d) inconsistent, hence no solution 60 E

EXERCISE - 04 [B] JEE-Mathematics BRAIN STORMING SUBJECTIVE EXERCISE 1. a b satisfying the matrix equation A2 fA + gI = O are equal If A = c then prove that value of f and g + to d   – t (A) and determinant of A respectively. Given a, b, c, d are non zero reals and I = 1 0  ; O = 0 0 . r 0 1  0 0    2 . A is a matrix such that |A|=a, B = (adj A) such that |B|= b. Find the value of (ab2 + a2b + 1)S where 3×3 1 a a2 a3 up to , and a = 3. S= b  b3  b5  ....... 2 3 . Find the number of 2 × 2 matrix satisfying : (a) a is 1 or –1 ; (b) a121  a122  a 2  a 2 2 ; (c) a a + a a = 0 ij 21 22 11 21 12 22 4 . If A is a skew symmetric matrix and I + A is non singular, then prove that the matrix B = (I – A)(I + A)–1 is an orthogonal matrix. Use this to find a matrix B given A = 0 5  . 5 0   2 1 9 3 5. Given A = 2 1 ; B = 3 1  . I is a unit matrix of order 2. Find all possible matrix X in the following cases.  (a) AX = A (b) XA = I(c) XB = O but BX  O. 3 2 1  x  b  6. Determine the values of a and b for which the system 5 8 9  y   3     2 1 a  z  1 (a) has a unique solution ; (b) has no solution and (c) has infinitely many solutions 7 . If A is an orthogonal matrix and B = AP where P is a non singular matrix then show that the matrix PB–1 is also orthogonal. 1 2 a n 1 18 2007 8. If 0 1 4   0 1 36  then find a + n.   E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\02.MATRIX\\02.EXERCISES.p65 0 0 1  0 0 1  9. Let A  a b and P  p   0  . Such that AP = P and a + d = 5050. Find the value of (ad – bc). c 0  d  q     BRAIN STORMING SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(B) 2. 225 3. 8 4. 1 12 5  1 3  5 12    5 . (a) X = a b for a, b  R ; (b) X does not exist ; (c) X= a 3a  a, c  R and 3a + c  0 2  2a c 1  2 b  3c    6 . (a) a  –3 ; b  R ; (b) a = –3 and b  1/3 ; (c) a = –3, b = 1/3 8 . 2 0 0 9. 5049 E 61

JEE-Mathematics JEE-[MAIN] : PREVIOUS YEAR QUESTIONS EXERCISE - 05 [A] a b and A2 =   1. If A = b  then a    [AIEEE 2003]   (1)  = a2 + b2,  = a2 – b2 (2)  = a2 + b2,  = ab (4)  = 2ab,  = a2 + b2 (3)  = a2 + b2,  = 2ab  0 0 1 2. If A =  0 1 0  then- [AIEEE 2004]   (4) A = (–1)  1 0 0  (1) A–1 does not exist (2) A2 =  (3) A = 0 1 1 1   4 2 2 3. If A = 2 1 3  and 10B = 5 0  where B = A–1, then  is equal to- [AIEEE 2004]  1 1 1   1 2 3  (1) 2 (2) –1 (3) –2 (4) 5 4 . If A2 – A + I = 0, then the inverse of A [AIEEE 2005] (1) I – A (2) A – I (3) A (4) A + I NML OQP LNM OQP5. 10 and I = 1 0 , then which one of the following holds for all n1, (by the principal of If A = 1 1 01 mathematical induction) [AIEEE-2005] (1) An = nA – (n–1) I (3) An = nA + (n–1) I (2) An = 2n-1A+ (n–1) I (4) An = 2n-1A– (n–1) I 6 . If A and B are square matrices of size n × n such that A2 – B2 = (A – B) (A + B), then which of the following will be always true- [AIEEE- 2006] (1) AB = BA (2) Either of A or B is a zero matrix (3) Either of A or B is an identity matrix (4) A = B 7. Let A =  1 2 and B= a 0 , a, b  N. Then- [AIEEE- 2006] E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\02.MATRIX\\02.EXERCISES.p65  3   0 b  4  (1) there exist more than one but finite number of B's such that AB = BA (2) there exist exactly one B such that AB = BA (3) there exist infinitely many B's such that AB =BA (4) there cannot exist any B such that AB = BA 5 5   [AIEEE- 2006] 8 . Let A = 0  5 If |A2| = 25, then || equals- 0 0 5  (1) 52 (2) 1 (3) 1/5 (4) 5 9 . Let A be a 2 × 2 matrix with real entries. Let I be the 2 × 2 identity matrix. Denoted by tr(A), the sum of diagonal entries of A. Assume that A2= I. Statement –1: If A  I and A  I , then det A = –1 [AIEEE- 2008] Statement –2 : If A  I and A  I , then tr(A)  0. E (1) Statement –1 is false, Statement –2 is true. (2) Statement–1 is true, Statement–2 is true; Statement–2 is a 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 62

JEE-Mathematics 1 0 . Let A be a square matrix all of whose entries are integers. Then which one of the following is true? [AIEEE- 2008] (1) If det A = ± 1, then A–1 exists but all its entries are not necessarily integers (2) If det A  ±1, then A–1 exists and all its entries are non–integers (3) If det A = ±1, then A–1 exists and all its entries are integers (4) If det A = ±1, then A–1 need not exist 1 1 . Let A be a 2 × 2 matrix [AIEEE- 2009] Statement–1 : adj (adj A) = A Statement–2 : | adj A | = |A| (1) Statement–1 is true, Statement–2 is false. (2) Statement–1 is false, Statement–2 is true. (3) Statement–1 is true, Statement–2 is true; Statement–2 is a correct explanation for Statement–1. (4) Statement–1 is true, Statement–2 is true; Statement–2 is not a correct explanation for statement–1. 1 2 . The number of 3 × 3 non-singular matrices, with four entries as 1 and all other entries as 0, is :- [AIEEE-2010] (1) Less than 4 (2) 5 (3) 6 (4) At least 7 13 . Let A be a 2 × 2 matrix with non-zero entries and let A2 = I, where I is 2 × 2 identity matrix. Define Tr(A) = sum of diagonal elements of A and |A| = determinant of matrix A. [AIEEE-2010] Statement–1 : Tr(A) = 0. Statement–2 : |A| = 1. (1) Statement–1 is true, Statement–2 is true; Statement–2 is a correct explanation for Statement–1. (2) Statement–1 is true, Statement–2 is true; Statement–2 is not a correct explanation for statement–1. (3) Statement–1 is true, Statement–2 is false. (4) Statement–1 is false, Statement–2 is true. 1 4 . Let A and B be two symmetric matrices of order 3. Statement-1 : A(BA) and (AB)A are symmetric matrices. Statement-2 : AB is symmetric matrix if matrix multiplication of A with B is commutative. [AIEEE-2011] E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\02.MATRIX\\02.EXERCISES.p65 (1) Statement-1 is true, Statement-2 is false. (2) Statement-1 is false, Statement-2 is true (3) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1 (4) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1. 1 5 . Statement-1 : Determinant of a skew-symmetric matrix of order 3 is zero. Statement-1 : For any matrix A, det(AT) = det(A) and det(–A) = –det(A). Where det(B) denotes the determinant of matrix B. Then : [AIEEE-2011] (1) Statement-1 is true and statement-2 is false (2) Both statements are true (3) Both statements are false (4) Statement-1 is false and statement-2 is true. H   0 16. If   1 is the complex cube root of unity and matrix  , then H70 is equal to: [AIEEE-2011]  0  (1) H (2) 0 (3) –H (4) H2 17. Let P and Q be 3 × 3 matrice s with P  Q. If P3 = Q3 and P2Q = Q2P, then determi nant of E (P2 + Q2) is equal to : [AIEEE-2012] (1) –1 (2) –2 (3) 1 (4) 0 63

JEE-Mathematics 1 0 0 1 0    1  18. Let A   2 1 0  . If u1 and u2 are column matrices such that Au1   0  and Au2   , then u1 + u2 is 2    3 1   0   0  equal to : [AIEEE-2012] 1   1   1   1    (1)  1  (2)  1  (3)  1  (4)  1         1   0   1   0  PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE-5 [A] E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\02.MATRIX\\02.EXERCISES.p65 Que. 1 2 3 4 5 6 7 8 9 10 Ans. 3 2 4 1 1 1 3 3 4 3 Que. 11 12 13 14 15 16 17 18 Ans. 4 4 3 4 1 1 4 1 64 E

EXERCISE - 05 [B] JEE-Mathematics JEE-[ADVANCED] : PREVIOUS YEAR QUESTIONS a b c  1. If matrix A = b c a  where a,b,c are real positive numbers, abc = 1 and AT A = I, then find the value of  c a b a3 + b3 + c3. [JEE 2003, Mains 2M out of 60] 2. If A   2 and |A3| = 125, then  is equal to - [JEE 2004 (Screening)] 2  (A) ±3 (B) ±2 (C) ±5 (D) 0 3 . If M is a 3 × 3 matrix, where MTM = I and det (M) = 1, then prove that det (M–I) = 0. [JEE 2004 (Mains), 2M out of 60] a 1 0 a 1 1 f   a2  x b d 4. A   1 b d  , B   0 g c  , U   g , V   0  , X =  y             1 c   f h   h   0  z  If AX = U has infinitely many solutions, then prove that BX = V cannot have a unique solution. If further afd  0, then prove that BX = V has no solution [JEE 2004 (Mains), 4M out of 60] 1 0 0  1 0 0  A 1  1 (A 2  cA  d) , then the value of c and 0  6 5. A  0 1 1  ,   1 0  and d are - 0 2  0 4  0 1 [JEE 2005 (Screening)] (A) –6, –11 (B) 6, 11 (C) –6, 11 (D) 6, –11 3 1   6. If P  2 2 , 1 1 and Q = PAPT and x = PT Q2005 P, then x is equal to - A  0 1 3 1     2 2 E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\02.MATRIX\\02.EXERCISES.p65 [JEE 2005 (Screening)] 1 2005 4  2005 3 6015  (A) 0 (B)   1    2005 4  2005 3  1 2  3 1 1 2005 2  3  (C)   (D)   4 1 4 2  3 2005  2  3  Comprehension (3 questions) 1 0 0 1  2  2  3  3  7. A  2 1 0  , if U1, U2 and U3 are columns matrices satisfying. AU1  0  , AU2   , AU3   and   E 0 0 1  3 2 1  U is 3×3 matrix whose columns are U1, U2, U3 then answer the following questions - (a) The value of |U| is - (A) 3 (B) –3 (C) 3/2 (D) 2 (C) 1 (D) 3 (b) The sum of the elements of U–1 is - (A) –1 (B) 0 65

JEE-Mathematics 3  (c) The value of 3 2 0 U 2  is -  0  (A) [5] (B) [5/2] (C) [4] (D) [3/2] [JEE 2006, 5 marks each] 8 . Match the Statement / Expressions in Column I with the Statements / Expressions in Column II and indicate your answer by darkening the appropriate bubbles in the 4 × 4 matrix given in the ORS. Column I Column II (A) The minimum value of x2  2x  4 is (P) 0 (Q) 1 x 2 (B) Let A and B be 3 × 3 matrices of real numbers, where A is symmetric, B is skew-symmetric, and (A+B)(A–B) = (A – B) (A + B). If (AB)t = (–1)k AB, where (AB)t is the transpose of the matrix AB, then the possible values of k are (C) Let a = log3 log3 2. An integer k satisfying 1  2(k3a )  2 , (R) 2 must be less than (D) If sin = cos, then the possible values of 1        are (S) 3   2  [JEE 2008, 6] 9 . Let A be the set of all 3 × 3 symmetric matrices all of whose entries are either 0 or 1. Five of these entries are 1 and four of them are 0. (a) The number of matrices in A is - (A) 12 (B) 6 (C) 9 (D) 3 (b) The number of matrices A in A for which the system of linear equations x  1  A  y   0     E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\02.MATRIX\\02.EXERCISES.p65 z  0 has a unique solution, is - (A) less than 4 (B) at least 4 but less than 7 (C) at least 7 but less than 10 (D) at least 10 (c) The number of matrices A in A for which the system of linear equations x  1  A  y   0     z  0 is inconsistent, is - (B) more than 2 (C) 2 (D) 1 (A) 0 [JEE 2009, 4+4+4] x  1  10. (a) The number of 3  3 matrices A whose entries are either 0 or 1 and for which the system A y   0    z  0 has exactly two distinct solutions, is (A) 0 (B) 29 – 1 (C) 168 (D) 2 66 E

JEE-Mathematics (b) Let k be a positive real number and let 2k  1 2 k 2 k   0 2k 1 k     A   2 k 1 2k  and B  1  2k 0 2 k  .   2 k 2k 1     k 2 k 0  If det (adj A) + det(adj B) = 106, then [k] is equal to [Note : adj M denotes the adjoint of a square matrix M and [k] denotes the largest integer less than or equal to k]. ( c ) Let p be an odd prime number and Tp be the following set of 2  2 matrices : Tp  A  a b  : a, b, c  0,1, 2,....., p  1  c a    (i) The number of A in Tp such that A is either symmetric or skew-symmetric or both, and det(A) divisible by p is - (A) (p – 1)2 (B) 2 (p – 1) (C) (p – 1)2 + 1 (D) 2p –1 (ii) The number of A in Tp such that the trace of A is not divisible by p but det (A) is divisible by p is - [Note : The trace of a matrix is the sum of its diagonal entries.] (A) (p – 1) (p2 – p + 1) (B) p3 – (p – 1)2 (C) (p – 1)2 (D) (p – 1) (p2 – 2) (iii) The number of A in Tp such that det (A) is not divisible by p is - (A) 2p2 (B) p3 – 5p (C) p3 – 3p (D) p3 – p2 [JEE 2010, 3+3+3+3+3] 1 1 . Let M and N be two 3 × 3 non-singular skew-symmetric matrices such that MN = NM. If PT denotes the transpose of P, then M2N2(MTN)–1 (MN–1)T is equal to - [JEE 2011, 4] E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\02.MATRIX\\02.EXERCISES.p65 (A) M2 (B) –N2 (C) –M2 (D) MN  1 a b 12. Let 1 be a cube root of unity and S be the set of all non-singular matrices of the form   1 c  ,   2  1 where each of a,b and c is either  or 2. Then the number of distinct matrices in the set S is- (A) 2 (B) 6 (C) 4 (D) 8 [JEE 2011, 3, (–1)] 1 3 . Let M be 3 × 3 matrix satisfying 0 1  1   1  1  0  M 1    2  , M 1   1  and M 1    0          0  3   0  1 1 12 Then the sum of the diagonal entries of M is [JEE 2011, 4] 1 4 . Let P =[aij] be a 3 × 3 matrix and let Q = [bij], where bij = 2i+jaij for 1 < i, j < 3. If the determinant of P is 2, then the determinant of the matrix Q is - [JEE 2012, 3M, –1M] (A) 210 (B) 211 (C) 212 (D) 213 E 67

JEE-Mathematics 1 5 . If P is a 3 × 3 matrix such that PT = 2P + I, where PT is the transpose of P and I is the 3 × 3 identity matrix, then x 0 there exists a column matrix X  y   0  such that [JEE 2012, 3M, –1M]  z  0 0  (A) PX  0  (B) PX = X (C) PX = 2X (D) PX = –X  0  1 4 4 16. If the adjoint of a 3 × 3 matrix P is 2 1 7  , then the possible value(s) of the determinant of P  2 1 3 is (are) - [JEE 2012, 4M] (A) –2 (B) –1 (C) 1 (D) 2 PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE-5 [B] E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#02\\ENG\\Part-1\\02.MATRIX\\02.EXERCISES.p65 1. 4 2. A 5. C 6. A 7. (a) A, (b) B, (c) A 8 . (A) R (B) Q,S (C) R,S (D) P,R 9 . (a) A, (b) B, (c) B 1 0 . (a) A, (b) 4; (c) (i) D, (ii) C, (iii) D 1 1 . Bonus 12. A 13. 9 14. D 15. D 16. A,D 68 E

JEE-Mathematics MAXIMA-MINIMA 1. INTRODUCTION : Some of the most important applications of differential calculus are optimization problems, in which we are required to find the optimal (best) way of doing something. Here are examples of such problems that we will solve in this chapter  What is the shape of a vessel that can with-stand maximum pressure ?  What is the maximum acceleration of a space shuttle ? (This is an important question to the astronauts who have to withstand the effects of acceleration)  What is the radius of a contracted windpipe that expels air most rapidly during a cough ? These problems can be reduced to finding the maximum or minimum values of a function. Let's first explain exactly what we mean by maxima and minima. (a) Maxima (Local/Relative maxima) : absolute maximum relative maximum No greater value of f. A function f(x) is said to have a A maximum at x = a if there exist a near by neighbourhood (a – h, a + h) – {a} such that Q f(a) > f(x)  x  (a – h, a + h) – {a} y = f(x) (b) Minima (Local/Relative minima): A function f(x) is said to have a minimum at x = a if there exist a neighbourhood (a – h, a + h) – {a} such that f(a) < f(x)  x  (a – h, a + h) – {a} NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\04-Maxima-minima.p65 B PR relative minimum absolute minimum No smaller value of f. No smaller value of f. near by Also a relative minimum x=a x=b (c) Absolute maximum (Global maximum) : A function f has an absolute maximum (or global maximum) at c if f(c)  f(x) for all x in D, where D is the domain of f. The number f(c) is called the maximum value of f on D. (d) Absolute minimum (Global minimum) : A function f has an absolute minimum at c if f(c)  f(x) for all x in D and the number f(c) is called the minimum value of f on D. The maximum and minimum values of f are called the extreme values of f. Note that : (i) the maximum & minimum values of a function are also known as local/relative maxima or local/relative minima as these are the greatest & least values of the function relative to some neighbourhood of the point in question. (ii) the term 'extremum' (or extremal) or 'turning value' is used both for maximum or a minimum value. (iii) a maximum (minimum) value of a function may not be the greatest (least) value in a finite interval. (iv) a function can have several maximum & minimum values & a minimum value may even be greater than a maximum value. (v) local maximum & local minimum values of a continuous function occur alternately & between two consecutive local maximum values there is a local minimum value & vice versa. E 79

JEE-Mathematics 2 . DERIVATIVE TEST FOR ASCERTAINING MA XIMA/MINIMA : (a) First derivative test : If ƒ '(x) = 0 at a point (say x = a) and (i) If f'(x) changes sign from positive to negative while graph of the function passes through x = a then x = a is said to be a point local maxima. (ii) If f'(x) changes sign from negative to positive while graph of the function passes through x = a then x = a is said to be a point local minima. YY dy dy >0 <0 dx dx dy dy <0 >0 dx dx O x=a X O x=a X Note : If f '(x) does not change sign i.e. has the same sign in a certain complete neighbourhood of a, then f(x) is either strictly increasing or decreasing throughout this neighbourhood implying that f(a) is not an extreme value of f. 1 Illustration 1 : Let f(x) = x + x ; x 0. Discuss the local maximum and local minimum values of f(x). 1 x2  1 (x  1)(x  1) +–+ Solution : Here, f'(x) = 1 – x2 = x2  –1 1 x2 Using number line rule, f(x) will have local maximum at x = –1 and local minimum at x = 1  local maximum value of f(x) = –2 at x = –1 and local minimum value of f(x) = 2 at x = 1 Ans. 3 x 2  12x  1, 1  x  2 , then  Illustration 2 : If f(x) =  37  x, 2  x  3, (A) f(x) is increasing on [–1, 2) (B) f(x) is continuous on [–1, 3] (C) f'(x) does not exist at x = 2 (D) f(x) has the maximum value at x = 2 Solution : 3x2  12x  1, 1  x  2 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\04-Maxima-minima.p65 Given,f(x) =   37  x, 2x3 6 x  12, 1  x  2 f'(x) =    1, 2x3 (A) which shows f'(x) > 0 for x  [–1, 2) So, f(x) is increasing on [–1, 2) Hence, (A) is correct. (B) for continuity of f(x). (check at x = 2) RHL = 35, LHL = 35 and f(2) = 35 So, (B) is correct (C) Rf'(2) = –1 and Lf'(2) = 24 so, not differentiable at x = 2. Hence, (C) is correct. 80 E

JEE-Mathematics (D) we know f(x) is increasing on [–1, 2) and decreasing on (2, 3], Ans. Thus maximum at x = 2, Hence, (D) is correct.  (A), (B), (C), (D) all are correct. Do yourself - 1 : ( i ) Find local maxima and local minima for the function f(x) = x3 – 3x. ( i i ) If function f(x) = x3 – 62x2 + ax + 9 has local maxima at x = 1, then find the value of a. (b) Second derivative test : If f(x) is continuous and differentiable at x = a where f'(a) = 0 and f''(a) also exists then for ascertaining maxima/minima at x = a, 2nd derivative test can be used - (i) If f''(a) > 0  x = a is a point of local minima (ii) If f''(a) < 0  x = a is a point of local maxima (iii) If f''(a) = 0  second derivative test fails. To identify maxima/minima at this point either first derivative test or higher derivative test can be used. Illustration 3 : If f (x) = 2x3 – 3x2 – 36x + 6 has local maximum and minimum at x = a and x = b respectively, then ordered pair (a, b) is - (A) (3, –2) (B) (2, –3) (C) (–2, 3) (D) (–3, 2) x = –2, 3 Solution : f(x) = 2x3 – 3x2 – 36x + 6 f'(x) = 6x2 – 6x – 36 & f''(x) = 12x – 6 Now f'(x) = 0  6(x2 – x – 6) = 0  (x – 3) (x + 2) = 0  f''(–2) = –30  x = –2 is a point of local maximum f''(3) = 30  x = 3 is a point of local minimum Hence, (–2, 3) is the required ordered pair. Ans. (C) Illustration 4 : Find the point of local maxima of f(x) = sinx (1+cosx) in x  (0, /2). NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\04-Maxima-minima.p65 Solution : 1 Let f(x) = sinx ( 1+ cosx)  sinx + sin 2x 2  f '(x) = cos x + cos 2x f ''(x) = – sin x – 2sin 2x Now f '(x) = 0  cos x + cos2x = 0  cos 2x = cos ( –x)  x = /3 Also f ''(/3) = – 3 / 2 – 3  0  f(x) has a maxima at x = /3 Ans. ex  ex in [–loge 2, loge 7]. Illustration 5 : Find the global maximum and global minimum of f(x) = 2 Solution : f(x) = ex  ex is differentiable at all x in its domain. 2 Then f '(x) = ex  ex , f ''(x) = ex  ex 22 ex  ex f '(x) = 0  = 0  e2x = 1  x = 0 2 E 81

JEE-Mathematics f''(0) = 1  x = 0 is a point of local minimum Points x = – loge2 and x = loge7 are extreme points. Now, check the value of f(x) at all these three points x = –loge2, 0, loge7 e  e loge 2  loge 2 5 25 Y  f(–loge2) =  7 24 f(0) = e0  e0 1 5 4 2 1 e  eloge 7  loge 7 25 –loge2 loge7 X 0 Ans. f(loge7) =  27 E  x = 0 is absolute minima & x = loge7 is absolute maxima Hence, absolute/global minimum value of f(x) is 1 at x = 0 25 and absolute/global maximum value of f(x) is 7 at x = loge7 Do yourself - 2 : nx ( i ) Find local maximum value of function f(x) = x ( i i ) If f(x) = x2e–2x (x > 0), then find the local maximum value of f(x). (b ) nth derivative test : Let f(x) function such that f'(a) = f''(a) = f'\"(a) = .......... = fn–1 (a) = 0 & fn(a)  0, then (i) n = even (1) fn(a) > 0  Minima (2) fn(a) < 0  Maxima (ii) n = odd Neither maxima nor minima at x = a Illustration 6 : Identify a point of maxima/minima in f(x) = (x + 1)4. Solution : f(x) = (x + 1)4 f'(x) = 4(x +1)3 f''(x) = 12(x +1)2 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\04-Maxima-minima.p65 f'''(x) = 24(x +1) f''''(x) = 24 Now f'(x) = 0  x = –1 f''(–1) = 0, f'''(–1) = 0, f''''(–1) = 24 > 0  at x = 1 f(x) has point of minima. Illustration 7 : Find point of local maxima and minima of f(x) = x5 – 5x4 + 5x3 – 1 Solution : f(x) = x5 – 5x4 + 5x3 – 1 f'(x) = 5x4 – 20x3 + 15x2 = 5x2 (x2 – 4x + 3) = 5x2 (x – 1)(x – 3) f'(x) = 0  x = 0, 1, 3 f''(x) = 10x(2x2 – 6x + 3) Now f''(1) < 0  Maxima at x = 1 f''(3) > 0  Minima at x = 3 82

JEE-Mathematics and f''(0) = 0  IInd derivative test fails so, f'''(x) = 30(2x2 – 4x + 1) f'''(0) = 30  Neither maxima nor minima at x = 0. Do yourself - 3 : ( i ) Identify the point of local maxima/minima in f(x) = (x – 3)10. 3 . SUMM ARY OF WORKING RULE FOR SOLVING REAL LIFE OPTIMIZATION PROBLEM : First :When possible, draw a figure to illustrate the problem & label those parts that are important in the problem. Constants & variables should be clearly distinguished. Second : Write an equation for the quantity that is to be maximized or minimized. If this quantity is denoted by ‘y’, it must be expressed in terms of a single independent variable x. This may require some algebraic manipulations. Third : If y = f (x) is a quantity to be maximum or minimum, find those values of x for which dy/dx = f (x)= 0. Fourth : Using derivative test, test each value of x for which f (x) = 0 to determine whether it provides a maximum or minimum or neither. Fifth : If the derivative fails to exist at some point, examine this point as possible maximum or minimum. Sixth : If the function y = f(x) is defined only for x  [a,b] then examine x = a & x = b for possible extreme values. Illustration 8 : Determine the largest area of the rectangle whose base is on the x-axis and two of its vertices lie on the curve y = ex2 . Solution : Area of the rectangle will be A = 2a. ea2 Y For max. area, dA d (2aea2 ) = ea2 [2  4a2 ] (–a, e–a2) (a, e–a2) y=e–x2  (–a,0) (a,0) X da da dA 0  a = 1 da 2 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\04-Maxima-minima.p65 sign of dA changes from positive to negative da 1 2 . e 1 2 = 2 sq units . Ans.  x  2 is a point of maxima  Amax = 2  2 e1/ 2 Illustration 9 : A box of maximum volume with top open is to be made by cutting out four equal squares from four corners of a square tin sheet of side length a ft, and then folding up the flaps. Find the side of the square base cut off. Solution : Volume of the box is, V = x(a – 2x)2 i.e., squares of side x are cut out then we will get a box with a square base of side (a – 2x) and height x. xx a xx xx xx Removed x x a a–2x a–2x E 83

JEE-Mathematics  dV = (a – 2x)2 + x·2(a – 2x)(–2) dx dV = (a – 2x) (a – 6x) dx dV For V to be extremum dx = 0  x = a/2, a/6 But when x = a/2; V = 0 (minimum) and we know minimum and maximum occurs alternately in a continuous function. Hence, V is maximum when x = a/6. Ans. Illustration 10 : A Conical vessel is to be prepared out of a circular sheet of gold of unit radius. How much sectorial area is to be removed from the sheet so that vessel has maximum volume. Solution : Lateral height of cone = Radius of circle = 1 Lateral area of cone = Area of circle with sector removed i.e. r(1) = (1)2 (2  2) O 2 2 A B i.e. r =  (here r is radius of cone)  Height 'h' of cone = 12  r2 Volume of cone V = 1 r2h = 1     2  12     2 3 3       upon maximizing V, we get     2  2  3   =  1  3  Area of sector removed = 1 (1)2 (2) =  2 Ans. 2  1  3  Do yourself - 4 : NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\04-Maxima-minima.p65 ( i ) Find the two positive numbers x & y such that their sum is 60 and xy3 is maximum. ( i i ) If from a wire of length 36 metre, a rectangle of greatest area is made, then find its two adjacent sides in metre. Important note : (i) If the sum of two real numbers x and y is constant then their product is maximum if they are equal. 1 i.e. xy = [(x + y)2 – (x – y)2] 4 (ii) If the product of two positive numbers is constant then their sum is least if they are equal. i.e. (x + y)2 = (x – y)2 + 4xy 4 . SHORTEST DISTANCE BETWEEN TWO CURVES : SD Shortest distance between two non-intersecting curves always lies along the common normal. (Wherever defined) 84 E

Willington Island

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