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DETERMINANTS, , TOPIC-1-EVALUATION OF DETERMINANTS, , SIGN SYSTEM FOR EXPANSION OF DETERMINANT:, , INTRODUCTION, , 1. Consider two equations, a1 x  b1 y  0 .....(i) and, a 2 x  b 2 y  0 .....(ii), Multiplying (i) by b 2 and (ii) by b 1 and subtracting,, dividing by x, we get, a1 b 2  a 2 b1  0, The result a1 b 2  a 2 b1 is represented by, , a1, , b1, , a2 b 2, Which is known as determinant of order two and, a1 b 2  a 2 b1 is the expansion of this determinant. The, horizontal lines are called rows and vertical lines are, called columns., 2. Now let us consider three homogeneous linear, equations a1 x  b1 y  c1 z  0 , a 2 x  b 2 y  c 2 z  0, and a 3 x  b 3 y  c 3 z  0 Eliminated x, y, z from, above three equations we obtain, a1 (b 2 c 3  b 3 c 2 )  b1 (a 2 c 3  a 3 c 2 )  c1 (a 2 b 3  a 3 b 2 )  0, .....(iii), a1 b 1 c 1, The L.H.S. of (iii) is represented by a 2 b 2 c 2, a3 b 3 c 3, Its contains three rows and three columns, it is called a, determinant of third order., DETERMINANTS, DEFINITION:, , If M is the set of square matrices, K is the, set of numbers (real or complex) and f: M → K is, defined by f(A) = k, where A ∈ M and k ∈ K, then f(A), is called the determinant of A. It is also denoted by |A|, or det A or Δ., EXPANSION OF DETERMINANTS, , Unlike a matrix, determinant is not just a table of, numerical data but (quite differently) a short hand way, of writing algebraic expression, whose value can be, computed when the values of terms or elements are, known.(1) The value of the determinant is defined as, , a1, , b1, , a2, , b2, , a1 b1, ,  a1b2  a2b1 ., c1, , (2) a2 b2 c2  a1, , a3 b3 c3, , CHETHAN M G, , b2 c2, b3 c3, ,  b1, , a2 c2, a3 c3, ,  c1, , a2 b2, a3 b3, , The value of the determinant is the sum of products of, the elements of the row (or column) and the, corresponding determinant obtained by omitting the, row and the column of the element with a proper sign,, given by the rule (1)i j , where i and j are the number, of rows and the number of columns, respectively of the element of the row (or the column), chosen., Thus sign system for order 2, order 3, order 4,….. are, given by, , , , , ,  , ,   , , ,, ,   ,   , ,    , , ,, ,    ,    , , , ....., ,    , , EVALUATION OF DETERMINANTS., , If A is a square matrix of order 2, then its determinant, can be easily found. But to evaluate determinants of, square matrices of higher orders, we should always try, to introduce zeros at maximum number of places in a, particular row (column) by using the properties and, then we should expand the determinant along that row, (column).We shall be using the following notations to, evaluate a determinant :, 1. R i to denote i th row., 2. R i  R j to denote the interchange of i th and j th, rows., 3. If more than one operation like Ri  Ri   R j is, done in one step, care should be taken to see that a, row that is affected in one operation should not be, used in another operation., 4. R i ( ) to denote the multiplication of all element of, i th row by  .Similar notations are used to denote, column operations if R is replaced by C., SOLVED PROBLEMS, LEVEL-I, ,  2  3   1   3, 4, 3, 2, 1. If p  q  r  s  t    1 2     4 ,,   3   4 3, then value of t is, (a)16, (b)18, (c)17, (d)19, Solution:(b) Since it is an identity in  so satisfied, by every value of  . Now put   0 in the given, equation, we have, , [email protected], Phone number-8105418762 / 7019881906, , 1

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CONCEPTS OF MATHEMATICS FOR KCET -JEE MAIN AND ADVANCED, , 0, , 2. If a  b  c  0 ,then the solution of the, , 1, , 3, , 2, , 4  12  30  18, , 4, , 0, , t 1, 3, , 2. Evaluate, (a) 1, , cos150, , sin150, , sin 750, , Solution:(b)  , , x 1, , cos150, , sin150, , sin 750, , cos 750, , 3 2, 4 1, , (b) 4, (d) 2, , [KCET-17-1m][Only one option correct], [NCERT-DIRECT QUESTION], , 3 x, x 1, , , , 4 1, , 3  x  3  8  3  x  5,  8  x 2  x   8  2 2, 4., , log e e, , 5, , log10 10 5, (a) , , cx, , (d) 0,  a 2  b 2  c 2, Solution:(c)Trick: Put a  1, b  1 and c  0, so that they satisfy the condition a  b  c  0 ., Now the determinant becomes, , 1 x, , 0, , 0, , 1  x, , 1, , 1, , 3., , 5 , e, , (b) e, (c) 1, 1 5 , , (d) 0, , 1 0, x, , x 1, , x2, , x4, , x3, , x5, , x8 , , x7, , x  10, , x  14, (b) – 2, (d) None of these, , 2, , 3, , 5, , Trick: Put x  1 .Then 4, , 6, , 9  2, , e, , 8 11 15, , LEVEL-II, , 1, , Note: Since there is a option “None of these”, therefore, we should check for one more different value of x., Put x  1 ., , log x y log x z, , 1. The value of log y x, , 1, , log y z, , log z x log z y, , 1, , 1, , 4., , a) 0, b) 1, c) xyz, d) log xyz, Solution:(a),   1 1  log z y  log y z  log x y log y x  log z x  log y z, , , , 1, , (a) 2, 2, (c) x  2, Solution:(b), , 5  0 ( log a a  1), , 1 5, , , , , ,  log x z  log y x  log z y  log z x , ,   1  1  log x y  log y x  log y x   log x z  log z x  log z x , ,   000  0,  loga b  logb a  1 & loga m  log n a  log n m, 2, , a, , that option (c) gives the same values i.e., 0,  3 ., , , , Solution:(d)   1 5, , , , b, ,  0 is, , Now putting these in the options, we find, 2, , 5, , a, ,  (1  x){x(1  x)  1}  1(1  x)  0,  (1  x){x 2  x  1}  x  1  0  x( x 2  3)  0, , 3 2, , 2, , 1, , bx, , 3 2, (a  b 2  c 2 ), 2, 3 2 2 2, (c) 0, , (a  b  c ), 2, , then x is equal to, , (a) 2 2, (c) 8, , Solution:(a), , c, , (b) , ,   cos 750 cos150  sin 750 sin150,   cos  750  150   cos  900   0, , , , b, , (a) 0, , cos 750, (c) 2, (d) 3, , (b) 0, , 3 x, , c, , equation, , [KCET-15-1m][Only one option correct], , 3. If, , ax, , , , 1, , 1 , 1, , 2, , , , 1, ,  , 2, (b) 3 3i, , (a) 3 3i, (c) i 3, , (d) 3, , 1, , 1, , Solution:(a) 1 , , 1, , 2, , , , 1, ,   3(   2 ), 2, , [email protected], https://t.me/mcpmathematicskar_1986, Phone number-8105418762 / 701988190, , MCP-MATHEMATICS

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DETERMINANTS, ,  1  3i 1  3i ,  3, ,   3 3i ., 2 ,  2, 5. Let A  aij  , B  bij  are two 3  3 matrices, , sin x cos x tan x, 7. If f  x   x 3, , x2, , x, , 2x, , 1, , x, , a) 0, , such that bij   i  j 2 aij and B  81 .find A if, ,  3, (a) 1, 9, , (c) 1, 81, , (b) 3, ,  a11, 3a12, , B   3a21 32 a22, 32 a31 33 a32, , , b13 , i  j 2, aij , b23   bij  , b33 , , x 0, , 1, , 2, , 7, , [JEE-Main -18 online -4m ], , a) z, b) 1, c) 1, d)  z, 2, Solution:(d) w.k.t 1,  ,  are cube roots of unity and, , 1, ,   7   6   , , 3, , 2, 1, , 7, 1, ,    3k  3      3k, 0 2 , 0, , 2, , 2, , 4, , 2,  least positive integer is k      1    , , k  1 2  z, CHETHAN M G, , , , , , , , f  x, x2, , , , , , sin x, , ,  lim sin x  x  ,  cos x x 2  2  tan x   x , x 0, x, , , , f  x, x2, , , , , ,  0 1  2  0  1, , 1, , 8. If A, B, C be the angles of a triangle, then, , 1, , cos C, , cos C, , 1, , cos B, , cos A, , cos B, cos A , 1, , (a) 1, (b) 0, (c) cos A cos B cos C (d) cos A  cos B cos C, Solution:(b)Given, Angles of a triangle = A, B, and C. We know that as A + B + C = , therefore, A B   C, or cos( A  B)  cos(  C )   cos C, or cos A cos B  sin Asin B   cos C, cos A cos B  cos C  sin Asin B, and sin( A  B)  sin(  C)  sin C., Expanding the given determinant, we get, ,   (1  cos 2 A)  cos C (cos C  cos A cos B),  cos B(cos B  cos A cos C ),   sin 2 A  cos C (sin A sin B)  cos B(sin A sin C ),   sin 2 A  sin A(sin B cos C  cos B sin C ), , 1, , 1, 1  2  1  2  3k c1  c1  c2  c3, , 1, , , , x, 2, , , , lim, , where z  3 If 1   1  2  3k , then k is, , 1, , , , 1, 4, , , , x 0, , 2, , 3, , , , 1, , x 2 sin x  x    cos x x 2  2  tan x   x  , f  x, x, , , lim 2  lim , x 0, x 0, x, x2, f  x, , , 1, , lim 2  lim sin x  x    cos x x 2  2  tan x   x  , x 0, x, , 0, x, x, , , , , lim, , 1, , 1   2  0, 1, 1, 1, 2, 1   1  2  3k, 1, 2, 7, , , , x, , f  x   sin x x  x  cos x x  2 x  tan x x3  2 x3, , a13 , a23    81 9  A, a33 , B, 81, 1, B   81 9  A  A , , , 81 9 81 9 9, 6. Let  be a complex number such that 2 1  z, , 1, , x2, , 2x, , 32 a13 , , 33 a23 , 34 a33 , a12, a13 , 3a22 3a23 , 32 a32 32 a33 , , 1, , d) 1, , sin x cos x tan x, , 27, ,  a11, B  3  3  3a21, 32 a31,  a11 a12, 2 , B  27  3  3   a21 a22,  a31 a32, 2, , x2, , x 0, , c) 2, , Solution:(d) f  x   x 3, , (d) 1, , [JEE-MAIN-7/1/2020-SHIFT-II], [One option correct type-4M], ,  b11 b12, Solution:(a) B  b21 b22, , b31 b32, , b) 3, , f  x, , then lim, ,   sin 2 A  sin A sin( B  C )   sin 2 A  sin 2 A  0., , , , 0 cos x  sin x, , , 9. If S   x   0,2  : sin x, 0, cos x 0 , , , cos x sin x, 0, , , , , tan   x  is equal to, , 3 , xS, [email protected], Phone number-8105418762 / 7019881906, , then, , 3

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CONCEPTS OF MATHEMATICS FOR KCET -JEE MAIN AND ADVANCED, , (a) 2  3, , (b) 4  2 3, , (c) 4  2 3, , (d) 2  3, ,  A  b2  3, , [JEE Main online-17-4m][Only one option correct], , Solution:(c), , cos x  sin x, , 0, , cos x  0, , sin x, , 0, , cos x, , sin x, , , , 0, , , , , , , ,  cos x  cos2 x  sin x sin 2 x  0, cos x  sin x  0  tan x  1,  5, x ,, x   0, 2 , 4 4, , ,   ,   5 , tan   x   tan     tan   , , 3 , 3 4, 3 4 , xS,   ,   5 ,  tan     tan   , 3 4, 3 4 ,  ,  ,  ,  5 , tan    tan   tan    tan  , 3, 4 , 3,  4 , ,    ,     5 , 1  tan   tan   1  tan   tan  , 3 4, 3  4 , , , 3 1 3 1  3 1   3 1 1 3 , , ,  2 ,   2, 1 3 1 3,  1  3   1  3 1  3 ,  3  3 1 3  , ,   2  4  2 3 ,  2, ,   2 , 1  3, , ,  4  2 3, b, 1, 2, , , 2, 10. Let A  b b  1 b where, , ,  1, b, 2 , det  A , is, b  0. Then the minimum value of, b, (a)  3, (b) 2 3, (c) 3, (d) 2 3, 3, , 3, , 3, , , , , , , , , , , , , , , [JEE-MAIN-10/1/2019-SHIFT-II], [One option correct type-4M], , 2, , b, , 1, , Solution:(b) (i) A  b b  1 b, 2, , 1, , ,  A  2 b, , , , b, , 2, , , , , ,  A  2 2b2  2  b2  b  2b  b   b2  b2  1, , 4, , 2, , , , b2  3, 3, , b, b, b, b, w.k.t AM  GM, 3, b, b  b 3   b  3  2 3, ,  , b, 2, b, A, 2 3, b, sin , 1 ,  1, , 1, sin  ; then for all, 11. If A   sin , ,  1,  sin , 1 ,  3 5 ,  , ,,  , det  A lies in the interval,  4 4 , (ii), , A, , 3, (a)  , 3, 2 , , 5, (b)  , 4 , 2 , ,  5, , , , (d)  0, , 2, , , , , (c)  1, , 2, , 3, , , [JEE-MAIN-12/1/2019-SHIFT-II], [One option correct type-4M], , sin , , 1, , 1, , sin , ,  sin , , 1, , 1, Solution:(a) (i) A   sin , , 1, , , , , , , , A  1  sin 2   sin    sin   sin    1  sin 2 , , , , A  2 1  sin , (ii) Given, , 2, , , , , , 3, 5,  , 4, 4, , 3, 5,  sin   sin, 4, 4, 1, 1, ,  sin  , 2, 2, 1,  0  sin 2  , 2, 3,  1  1  sin 2    2  2 1  sin 2   3, 2, 3 ,  A   2, 3   , 3 , 2 , x, sin  cos , ,  sin, , , , 12. If 1   sin , , cos , , x, , 1, , 1, , x, , , , and, ,  2  b  1  A  2b  4  b  1, 2, , 2, , 2, , [email protected], https://t.me/mcpmathematicskar_1986, Phone number-8105418762 / 701988190, , MCP-MATHEMATICS

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a, , DETERMINANTS, , sin 2, , cos 2, ,  2   sin 2, , x, , 1, , cos 2, , 1, , x, , x, , , x  0; then for all, ,  ,    0, ,  2, (a) 1  2  2x3, , , , , , x, ,  1   x  x  x   x, , Denotes the sum of the diagonal entries of A. Then, (a) (P) is false and (Q) is true, (b) Both (P) and (Q) are false, (c) (P) is true and (Q) is false, (d) Both (P) and (Q) are true, [JEE-MAIN-2/9/2020-SHIFT-I], [One option correct type-4M], , a b ,  & a, b, c, d  0,1, c d , , , , Solution:(a) (i) Let A  , , 3, , Now A  ad  bc  0  either ad  1, bc  0, , 3, , or ad  0, bc  1, 0 1, 0 1 , ,  0  1  1, (ii) A  , , 1 0, 1 0 , , (ii) 1  2  2 x & 1  2  0, 13. The sum of the real roots of the equation, 3, , 1, , So (P) is false, (iii) If ad  1, bc  0 then A  1 . In this case, , x  3  0, is equal to, x2, (c)-4, , tr  A  2 So (Q) is true, , (d)0, , 15. Suppose the vectors x1 , x2 & x3 are the solutions, , [JEE-MAIN-10/4/2019-SHIFT-II], [One option correct type-4M], , x, , 6, , Solution:(d) 2, , 3 x, , x3  0, , 2x, , x2, , of the system of linear equations, Ax  b when the, vector b on the right side is equal to b1 , b2 & b3, respectively. If, , 1, , , ,  x  5x   6  5x  5  5x  0, ,  x 3x2  6 x  2 x2  6 x  6  4  2 x  3x  9   4 x  9 x   0, 2, ,  5 x3  30 x  30  5 x  0,  x3  7 x  6  0, , 1 , 0, 1, 0, 0, x1  1 , x2   2  , x3  0  , b1  0  , b2   2 , 0 ,  0 , 1, 1 , 1 , 0, &b3   0  ; then the determinant of A is equal to,  2 , (a), , CHETHAN M G, , , , Where I 2 denotes 2  2 identity matrix and tr  A, , , , 3, ,  5 x3  35 x  30  0, , 0, , x, ,  1   x3  x  x sin 2   cos2 , , 3, , -6, , 1, , , , (b) 6, , 1, , 1, ,  cos  sin   x cos 2 , , (a) 1, , 1, , , , x, ,  1   x3  x  x sin 2   sin  cos , , 2x, , -6, , (Q) If A  1, then tr  A   2, ,  cos    sin   x cos  , , 3, , 1, , statements, (P) If A  I 2 , then A  1, ,  1  x  x  1  sin    x sin   cos  , , 3 x, , 1, , cos , , cos , , 2, , 0, , sin , , Solution:(a) (i) 1   sin , , 6, , 6, , 14. Let A be a 2  2 real matrix with entries from, 0,1 and A  0. Consider the following two, , [JEE-MAIN-10/4/2019-SHIFT-I], [One option correct type-4M], , x, , -7, ,   x  1 x  3 x  2   0  x  1, 3, 2,  Sum of the real roots  1  3  2  0, , (d) 1  2  2 x3  x  1, , Similarly  2   x, , 0, , , , (c) 1  2  2x3, , , , 1, ,   x  1 x2  x  6  0, , (b) 1   2   x  cos 2  cos 4 , , 2, , 1, , 3, 2, , (b) 2, , (c), , 1, 2, , (d) 4, , [email protected], Phone number-8105418762 / 7019881906, , 5

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CONCEPTS OF MATHEMATICS FOR KCET -JEE MAIN AND ADVANCED, , [JEE-MAIN-4/9/2020-SHIFT-II], [One option correct type-4M], ,  a1 a2 a3 , , , Solution:(b) (i) Let A  a4 a5 a6, , ,  a7 a8 a9 , And given Ax  b, (ii) Given Ax3  b3  a3  0, a6  0, a9  2., (iii) Given Ax2  b2,  2a2  a3  0  a2  0,, 2a5  a6  2  a5  1, 2a8  a9  0  a8  1, (iv) Given Ax1  b1,  a1  a2  a3  1  a1  1, a4  a5  a6  0  a4  1, a7  a8  a9  0  a7  1, 1, (v)  A  1, , 0, , 0, , 1, , 0 2, , 1 1 2, 16. Let  , , , 5, ,  cos ,   sin , , and A  , , B  A  A4 , then det  B , , sin  , . If, cos  , , (a) is one, , (b) lies in  2,3, , (c) is zero, , (d) lies in 1, 2 , , Solution:(d) (i), ,  cos  sin    cos  sin  , A2  , , ,   sin  cos     sin  cos  , , cos 2   sin 2 , sin  cos   sin  cos  , A2  , , cos 2   sin 2 ,   sin  cos   sin  cos , , 2, , (iii) B  A  A4, ,  cos , B,   sin , 6, , sin    cos 4, , cos     sin 4, , (iv) B   cos   cos 4    sin   sin 4 , 2, , 2, ,  B  cos 2   sin 2   cos 2 4  sin 4 4,  2  cos  cos 4  sin  sin 4 ,  B  2  2cos   4 , ,  B  2 1  cos  3   , Put  , , , 5, , ,  3  ,  B  2 1  cos ,    cos       cos  ,  5 , , ,  2  , 2 ,  B  2 1  cos ,    B  2  2sin  , 5,  5 , , 2, ,  10  2 5  , ,   sin 360  10  2 5 ,  B  4, ,  , , 4, 4, , ,  ,  10  2 5  5  5,  B  ,  1, 2 ,  , 4, 2, , , LEVEL-III, , 1. Which of the following is (are) Not the square of a, 3  3 matrix with real entries?, [JEE Advanced-P-1-17-4m][More than one option correct], , [JEE-MAIN-6/9/2020-SHIFT-II], [One option correct type-4M], ,  cos 2 sin 2 , A , ,   sin 2 cos 2 ,  cos 2 sin 2   cos 2, (ii) A4  , ,   sin 2 cos 2    sin 2,  cos 4 sin 4 ,  A4  , ,   sin 4 cos 4 , , sin   sin 4 ,  cos   cos 4, B, ,    sin   sin 4  cos   cos 4 , , sin 2 , cos 2 , , sin 4 , cos 4 , , 1 0 0 ,  1 0 0 , , , a) 0 1 0, b)  0 1 0 , , , , ,  0 0 1 ,  0 1 1, 1 0 0 , 1 0 0 , , , c) 0 1 0, d) 0 1 0 , , , , , 0 1 1, 0 1 1, Solution:(b,d) If some matrix, say X is the square of the, 2, matrix, say Y, then X  Y 2  det X   det Y   0 ., 1, , 0, , 0, , 1 0, , 0, , But 0 1 0  0 1 0  1 so these are not the, 0 1 1 0 1 1, square of a matrix., TOPIC-2-PROPERTIES OF DETRMINANT, PROPERTY-1: The value of determinant remains, , unchanged, if the rows and the columns are, a1 b 1 c 1, interchanged. If D  a 2 b 2 c 2 and, a3 b 3 c 3, , [email protected], https://t.me/mcpmathematicskar_1986, Phone number-8105418762 / 701988190, , MCP-MATHEMATICS

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DETERMINANTS, PROPERTY-7:, , a1, , a2, , a3, , D'  b 1, , b2, , b 3 .Then D'  D,, , c1, , c2, , c3, , D and D' are transpose of each other., Note: Since the determinant remains unchanged when, rows and columns are interchanged, it is obvious that, any theorem which is true for „rows‟ must also be true, for „columns‟., PROPERTY-2: If any two rows (or columns) of a, determinant be interchanged, the determinant is, unaltered in numerical value but is changed in sign, only., a1 b 1 c 1, a2 b 2 c 2, Let D  a 2 b 2 c 2 and D'  a1 b1 c 1 . Then, a3 b 3 c 3, a3 b 3 c 3, , D'  D, PROPERTY-3:, , If a determinant has two rows, (or columns) identical, then its value is zero., a1 b 1 c 1, Let D  a1 b1 c 1 . Then, D = 0, a2 b2 c 2, If all the elements of any row (or, column) be multiplied by the same number, then the, value of determinant is multiplied by that number., a1 b 1 c 1, ka1 kb1 kc1, Let D  a 2 b 2 c 2 and D   a 2, b2, c2 ., a3 b 3 c 3, a3, b3, c3, Then D'  kD, If each element of any row (or column), can be expressed as a sum of two terms, then the, determinant can be expressed as the sum of the, determinants.e.g., a1  x b1  y c1  z a1 b1 c1, x y z, PROPERTY-5:, , b2, , a3, , b3, , c2, ,  a2, , b2, , c2  a2, , b2, , c2, , c3, a3 b3 c3 a3 b3 c3, PROPERTY-6: The value of a determinant is not altered, by adding to the elements of any row (or column) the, same multiples of the corresponding elements of any, other row (or column), a1 b 1 c 1, e.g., D  a 2 b 2 c 2 and, a3 b 3 c 3, D' , , a1  ma 2, , b1  mb 2, , c1  mc 2, , a2, , b2, , c2, , a 3  na1, , b 3  nb 1, , c 3  nc 1, , 0, , 0, , c3, , a3 b3 c3 a3 b3 c3, PROPERTY-8: If a determinant D becomes zero on, putting x   , then we say that ( x   ) is factor of, x, 5 2, 2, 9 4 ., determinant. e.g. if D  x, 3, x, 16 8, At x  2, D  0, (because C 1 and C 2 are identical at x  2 )Hence, (x  2) is a factor of D., If A is of order n×n then |, , PROPERTY-9:, , PROPERTY-4:, , a2, , If all elements below leading diagonal or, above leading diagonal or except leading diagonal, elements are zero then the value of the determinant, equal to multiplied of all leading diagonal elements., a1 b1 c1, a1 0 0, a1 0 0, e.g. 0 b2 c2  a2 b2 0  a2 b2 0  a1b2c3, , |=, , | | ., , PROPERTY-10:, , |, , | =| | | | Provided AB is defined, , PROPERTY-11:, , |, , |=| |, , , n∈N., , PROPERTY-12:, , If A & B are square matrices of same, order then | | = | | ., , PROPERTY-13:, , Determinant of any triangular matrix and, diagonal matrix is the product of its diagonal elements., , PROPERTY-14:, , The determinant of a skew symmetric, matrix of odd order is always zero., Note: i) It should be noted that while applying, operations on determinants then at least one row, (or column) must remain unchanged. Or, Maximum, number of operations = order or determinant –1, ii) It should be noted that if the row (or column) which, is changed by multiplied a non-zero number, then the, determinant will be divided by that number., SHORT CUT METHODS, , 1. |, , |, , 1, , 1, , 1, , |= a, , b, , a2, , b2, , c =, c2, , |= (a-b) (b-c) (c-a), , . Then D'  D, 2., , |, , |=(x-y)(y-z)(z-x)= |, , |., , Note: It should be noted that while applying P-6 at, least one row (or column) must remain unchanged., CHETHAN M G, , [email protected], Phone number-8105418762 / 7019881906, , 7

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CONCEPTS OF MATHEMATICS FOR KCET -JEE MAIN AND ADVANCED, , SOLVED PROBLEMS, , 3. |, , |=(x-y)(y-z)(z-x)(x+y+z)., , 4. |, , |=(x-y)(y-z)(z-x)(xy+yz+zx)., , 5. |, 6., , LEVEL-I, , | = -(, , )., , |, , |=4abc., , 7. |, , |, , ., , 1. If , 1, , 2, , is the cube root of unity, then,  2, 2 1 =, 1 , , (b)0, (c) , (d) , 2, 1  , 1   2  2, Solution:(b)   2 1  1     2  2 1, 2 1  1   2 1 , 2, , (a)1, ,  2,  0 2 1  0 ., 0 1 , 0, , 8., , |, , 9., , |, , 10., , |, , ., |, , ., , x3  x, , 1, , 1, , 1, , 2. If x  a, , a, , b, , c, , x b, , 4, , 4, , 4, , a, , b, , c, , , , , ,   a  b  b  c  c  a   a 2  b 2  c 2   ab  bc  ca  , , 11. a b c  a 2  b 2  c 2  ab  bc  ca, , b a c, , 12. y, , 2, , z, , 2, , 1, 2, 2, 2,  a  b    b  c    c  a  , , 2, , x  a 2 x  a 2,  y  a 2  y  a 2, z  a 2 z  a 2, ,  4a x  y  y  z z  x , 3, , yz, , x, , y, , 13. z  x, , z, , x  ( x  y  z )( x  z ) 2, , x y, , y, , z, th, , 14. In an AP p term is x, qth term is y and rth term is, x y z, z , then p, , 1, , bx, , x  x c  x then, 2, , x c, , 0, , (a) f  1  0, , (b) f 1  0, , (c) f  2   0, , (d) f  0  0, , [KCET-2020-1m][Only one option correct], , 1 1 1, , x2, , ax, , q r  0., 1 1, , Solution:(d) By inspection method, x3  x a  x b  x, f  x   x  a x 2  x c  x Put x  0, , xb, , xc, , 0, , a, , b, , f  0   a, , 0, , c 0, , 0, , b c 0, Because the determinant of a skew symmetric matrix of, odd order is always zero., 3. If a1, a2, a3, .....a9, are in A.P. then the value of, a1, , a2, , a3, , a4, , a5, , a6 is, , a7, , a8, , a9, 9,  a1  a9 , 2, (d) loge  loge e , , (a) 1, , (b), , (c)  a1  a9 , , [KCET-2020-1m][Only one option correct], , Solution:(d), , a1, , a2, , a3, , 1 2 3, ,   a4, , a5, , a6  4 5 6, , a7 a8 a9 7 8 9, Here 1,2,3,4,5,6,7,8,9 are in AP, 8, , [email protected], https://t.me/mcpmathematicskar_1986, Phone number-8105418762 / 701988190, , MCP-MATHEMATICS

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DETERMINANTS, , 1, , 1 2 3, ,  R2  R2  R1 ,  3 3 3 , , R3  R3  R1 , , 6 6 6,   0 Because row-2&3 are in ratio, , 1, , equation., , 4. If 4 3i  1  x  i y , then, , 7. 1/ b, , 1/ c, , 20 3 i, , (d) x = 0, y=0, , 6i  3i 1, Solution:(d) 4 3i  1 Applying C2  C 2  3 i C3, , i, , 6i 0 1, 4 0  1 = 0 = 0+ 0 i, Equating real and imaginary, 20 0 i, parts x = 0, y = 0, , ab, , a  4b a  5b a  6b, (a) a  b  c  3abc, (c) 3a  5b, Solution:(d), a  b a  2b a  3b, 2, , 2, , 2, , (b) 3ab, (d) 0, , b, , b, , a  4b a  5b a  6b, , 2b, , 2b, , 2b, , by R2  R2  R1 & R3  R3  R2 , 6. One of the roots of the given equation, , c, , b, , xc, , a, , c, , a, , xb, ,  0 is, (b) (b  c), , (c)  a, , (d) (a  b  c), , xa, , b, , c, , b, , xc, , a, , c, , a, , xb, , CHETHAN M G, , ab, (b) 1/ abc, (d) 0, 1 a3 1, , abc, abc , abc, , 0, , abc, 1 b3 1  0, abc, 1 c3 1, , yz, , x, , y, , 8. If z  x, , z, , x y, , y, , x  k ( x  y  z )( x  z ) 2 , then k , z, , (a) 2xyz, , (b)1, (d) x 2 y 2 z 2, , zx, , z, , x  ( x  y  z )( x  z ) 2, , x y, , y, , z, , 0, , 9. If ax  bx  cx  dx  e  x  1, 4, , 3, , 2, , x 3, then e is equal to, a) 1, b) 0, Solution:(b), , c) 2, , x 1 x  3,  2x, , x4, , x4, , 3x, , d) -1, , x3  3x, , (a) ( a  b), , Solution:(d), , c, , [MP PET 2000; IIT 1998], , x 3  3x, , a  b a  2b a  3b, b, , b, , ca , , 2, , k 1, , a  2b a  3b a  4b , , xa, , b, , (c) xyz, Solution:(b) W.K.T, yz x y, , a  2b a  3b, , a  2b a  3b a  4b , , 5., , bc, , 2, , (a) abc, (c) ab  bc  ca, Solution:(d), 1 a3, 1/ a a 2 bc, 1, 1 b3, 1/ b b 2 ca , abc, 1 c3, 1/ c c 2 ab, , [One option correct type], , 20 3, , xb, , a, ,  x  (a  b  c) is one of the root of the, 1/ a a 2, , (c) x = 0, y=3, , 0 ,, , a, , (C1  C1  C2  C3 ), , 6i  3i 1, , (b) x =1, y=3, , c, ,  ( x  a  b  c) 1 x  c, , By inspection method option (d) is correct, loge  loge e   loge 1  0, , (a) x = 3, y=1, , b, , x 1, , x3, , ax 4  bx3  cx 2  dx  e  x  1, , 2 x, , x4, , x 3, , x4, , 3x, , Put, , x0 e, , 0, , 1, , 3, , 1, , 0, , 4  0  e  0, , 3, , 4, , 0, , The determinant of a skew symmetric matrix of odd, order is always zero., , [email protected], Phone number-8105418762 / 7019881906, , 9

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CONCEPTS OF MATHEMATICS FOR KCET -JEE MAIN AND ADVANCED, , 1 a bc, 10. The value of the determinant 1 b, , 1 c, (a) a  b  c, (c) 0, , c  a is, ab, , (b) (a  b  c), (d) 1  a  b  c, 1 a bc, 2, , Solution:(c)   1 b, , ca, , 1 c, , ab, , 1 1 ab, ,  0 ( C1  C2 ), 11. If |A| denotes the value of the determinant of the, square matrix A of order 3, then | 2 A |, (a) 8 | A |, (b) 8 | A |, (c) 2 | A |, (d) None of these, Solution:(a)We know that, det. ( A)  (1)n det A ,, where n is order of square matrix. If A is square matrix, of order 3, Then n  3 . Hence, | 2 A | (2)3 | A | 8 | A | ., 12. If I is a unit matrix of order 10, then the, determinant of I is equal to, (a)10, (b)1, (c)1/10, (d)9, Solution:(b) Determinant of unit matrix of any order, =1., 13.   100  i, , 0, , i  500, 1000  i is equal to, , 500  i i  1000, , 0, , (a)100 (b)500, (c)1000, (d)0, Solution:(d)This determinant of skew symmetric of, odd order, hence is equal to 0., 1 0 0 0 0, , 2 2 0 0 0, 14. The value of 4 4 3 0 0 is, , 5 5 5 4 0, 6 6 6 6 5, (a) 6, (b) 5!, (c) 1.22.3.43.54.64, (d) None of these, Solution:(b) The elements in the leading, diagonal are 1, 2, 3, 4, 5. On one side of the, leading diagonal all the elements are zero.,  The value of the determinant, = The product of the elements in the leading, diagonal, = 1.2.3.4.5 = 5!, 10, , ( x  1), , 2x, , x( x 1), , ( x 1) x, , then f (100) is equal to, (a)0, (b)1, (c)100, Solution:(a), , ,, , (d)–100, , 1, , x, , ( x  1), , 2x, , x( x  1), , ( x  1) x, , 3 x( x  1), ,  (a  b  c) 1 1 c  a (C2  C2  C3 ), , i  100, , x, , 3x( x 1) x( x 1)( x  2) ( x 1) x( x 1), , f ( x) , , 1 1 bc, , 0, , 15. If f ( x) , , 1, , x( x  1)( x  2) ( x  1) x( x  1), , Applying C3  C3  C2 , we get, , f ( x) , , 1, , x, , 1, , 2x, , x( x  1), , 2x, , 0., , 3x( x  1) x( x  1)( x  2) 3x( x  1), Hence f (100)  0, 16. Consider the following statements;, (a) If any two rows (or columns) of a determinant, are identical (all corresponding elements are, same), then value of determinant is zero., (b) The value of the determinant remains unchanged, if its rows and columns are interchanged., (c) If any two rows (or columns) of a determinant, are interchanged then sign of determinant, changes., Which of these are correct?, (a) ( a ) & (b), (b) (b) & (c), (d) (a), (b) &  c , , (c) (a) & (c), [NCERT-DIRECT QUESTION], , Solution:(d), , x2, 17. If a, b &c are in AP then x  4, , x3 xa, x5, , x6 x7, , x  b is, xc, , a) x   a  b  c , , b) 9x 2  a  b  c, , c) 0, , d) a  b  c, , [KCET-14-1m] [Only one option correct], , x2, , x3 xa, , Solution:(a)   x  4, , x5, , xb, , x6, , x7, , xc, , R1  R1  R2, , 2 x  8 2 x  10 2 x  a  c,   x4, , x5, , xb, , x6, , x7, , xc, , But 2b  a  c, , [email protected], https://t.me/mcpmathematicskar_1986, Phone number-8105418762 / 701988190, , MCP-MATHEMATICS

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DETERMINANTS, , 2 x  8 2 x  10 2 x  2b,   x4, , x5, , xb, , x6, , x7, , xc, , a) k A, , x4, , x5, , xb, , x6, , x7, , xc, , Solution:(b), 20. If A is a square matrix of order 3  3, , then 5A is, equal to:, a) 5 A, ,  x  4   x  5  x  b ,  2 x4, , x5, , xb  0, , x6, , x7, , xc, , , , Solution:(b) 5 A  53 A  125 A, , log z, , log 3 y, , log 3 z, , (a) log  xyz , , (b) log  6xyz , , (c) 0, , (d) log  x  y  z , , log y, log 3 y, , log z, , log z, , log 3 y  log y, , d) 9 A, , c) k 3 A, ,   log 2 x  log x log 2 y  log y log 2 z  log z, log 3 x  log x, , c) 27 A, , 22. If A is a square matrix of order 3  3 , then kA is, equal to, a) k A, b) k 2 A, , log 3 z, , log y, , b), ,  If A   aij  & k  R  kA  k n A ,   nn, , , , R2  R2  R1 & R3  R3  R1, , log x, , 1, A, 3, , a) 3 A, , Solution:(c) 3 A  33 A  27 A, , Solution:(c)   log 2 x log 2 y log 2 z, , log 3 x, , 21. If A is a square matrix of order 3  3, then 3A is, , [KCET-16-1m] [Only one option correct], [NCERT-DIRECT QUESTION], , [KCET-16-1m][Only one option correct], , log x, ,  If A   aij  & k  R  kA  k n A ,   n n, , , , equal to:, , log 2 x log 2 y log 2 z is equal to, log 3 x, , d) 15 A, , [KCET-18-1m] [Only one option correct], [NCERT-DIRECT QUESTION], , 18. If x,y,z are not equal and  0,  1, then the value of, , log y, , b) 125 A, , c) 25 A, , R1  R2 , , log x, , d) 3k A, , c) k A, , [KCET-17-1m] [Only one option correct], , 2  x  4  2  x  5 2  x  b , , , b) k 3 A, , 2, , d) 3k A, , [KCET-17-1m] [Only one option correct], [NCERT-DIRECT QUESTION], , log 3 z  log z, , Solution:(c) If A   aij  & k  R  kA  k n A, nn, log x, , log y, , kA  k 3 A, , log z, ,  2y ,  2 x ,  2 z ,   log   log    log  ,  x ,  z ,  y ,  3 x , log  ,  x , log x, ,  3y , log   ,  y , log y, , 23. If x, y, z are all different and not equal to zero and, ,  3 z , log  ,  z , , log  3, , log  3, , log x log y log z,   log  2  log  3, , 1, , 1, , 1, , 1 y, , 1, , 1, , 1, , 1 z, ,  0 then the value of, , x 1  y 1  z 1 is equal to:, a) xyz, c)  x  y  z, , log z, ,   log  2  log  2  log  2 , log  3, , 1 x, , 1, , 1, , 1, , 1, , 1, , 1, , , , b) x 1 y 1 z 1, d) 1, , [KCET-2016-1M] [One option correct type], [NCERT-DIRECT QUESTION], , R2  R3 , , Solution:(d), ,   log  2  log  3 0  0, 19. If A is a square matrix of order 3  3, , then kA is, , 1 x, , 1, , 1, , 1 y, , 1, , 1, , 1, 1 0, 1 z, , R2  R2  R1 & R3  R3  R1, , equal to:, CHETHAN M G, , [email protected], Phone number-8105418762 / 7019881906, , 11

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CONCEPTS OF MATHEMATICS FOR KCET -JEE MAIN AND ADVANCED, , 5, A  6, 7, , 1 x 1 1, x, , y 0  0 C2  C2  C3, , x, , 0, , z, , 1 x, , 0, , 1, , x, , y, , 0 0, , x, , z z, , 0, , 0, ,  5,  6  6,  7  7, ,  1 1 1, xyz 1      0, z x y, , 1 1 1, 1    0, z x y, 1 1 1,    1, z x y, , 2, , 0, , 2, , 0, , 2, , xyz  xy  yz  xz  0, , 2, ,  50, ,  2, 2, A   2, 2,  2, , 1  x  yz  0  1 xz  xy   0, , , , 2, , 2 , , x, ,  5 x, , x, ,  6 x, , x, , a) 10, ,  7 x, , 2, , x, ,  5 x, , x, ,  6 x, , x, ,  7 x, , b) 12, , 2, , , , , , 2, , A, , B, , y, , 1 & 1  x, , y, , z, , 2, , 1, , zy, , C, , z , then, zx xy, , b) 1  2, , c) 1  , , d) 1  , , A, , B, , C, , Solution:(a) 1  x, , y, , z, , zy, , zx, , xy, , 2, , 2, , xA, xyz 2, 1 , x, xyz, 1, , 1, 1 is, 1, , B, , C, , y, , z, , zx, , xy, , x, , x, , 5, , x, ,  6 x, , x, ,  7 x, ,  5,  6,  7, 2, , 2, , 2, , x, , x, , 5, , x, ,  6 x, , x, ,  7 x, , yB, , zC, , 2, , z2, , y, , xyz, , xyz, , yB, , zC, , 2, , z2, , 1, , 1, , y, , Ax, , x2 1, , 1  By, , y2 1, , Cz, , z2 1, , 1  , , d) 0, , c) 1, , Put x  0 because x, y & z  R, , 12, , 1, , a) 1  , , xA, 1, 1 , x2, xyz, xzy, , [KCET-18-1m] [Only one option correct], , 5, Solution:(d) Let A   6, 7, , 2, , C1  xC1 , C2  yC2 , C3  zC3, , 2, , 2, , 1, , 2, , A, xyz, 1 , x, xyz, zy, , 3, ,  5,  6,  7, , 2, , [KCET-17-1m] [Only one option correct], ,   4  3   3 4    7, 2   7, 25. If x, y & z  R then value of the determinant, , 5, 6, 7, , 0, , 1, , x2 1, , Cz, , n, An  A , , , 2, , 2, , 0, , 0, , 2, , 0 1, , 26. Let   By, , A3  27,, 3, A  33 , , A 3, , 0, , , 6 , 7 ,  50, , 0 1 0, , [KCET-15-1m] [Only one option correct], , Solution:(c), , 0, , 0 1, , Ax, , x 1  y 1  z 1  1, 2, and A3  27, then  , , , b) 2, d)  5, , , 24. If A  , 2, a) 1, c)  7, , 0, , , , , , 2, , 2, , 2, , 1, 1, 1, , 1 3 ,  2 1, ,B  , ,  ,then ABB is, 4 2, 1 2 , a) 250, b) 100, c) 250, d) 50, , 27. If A  , , [KCET-19-1m] [Only one option correct], , Solution:(c), 2, 2, ABB  A B B  A B   2  12  4  1, , ABB   10  25  250, , [email protected], https://t.me/mcpmathematicskar_1986, Phone number-8105418762 / 701988190, , MCP-MATHEMATICS

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DETERMINANTS, , LEVEL-II, , sin , , sin , , 1., , cos , , 1  cos , , cos 2 , , 4sin 4, , 4sin 4, , 1  4sin 4, , 2, , 2, , 2, , 2, , sin 4 equal to, (a) 1/2, (b)1, (c) –1/2, (d)–1, 2, 1  sin , sin 2 , Solution:(c) cos 2  1  cos 2 , , 4sin 4, ,  0 then, , Solution:(a) 125 | A3 |  | A |3  | A |  5 and, , | A |  2  4  5   2  9    3, 4. If  ,  and  are the roots of the equations, x3  px  q  0 then value of the determinant,   ,    is,   , , sin , 2, , cos 2 , , 4sin 4, , 0, , 1  4sin 4, , Using C2  C2  C1 , C3  C3  C1, , (a) p, , cos , , 1, , 0 0, , x3  px  q  0 ,        0, , 4sin 4, , 0, , 1, ,   ,    ,   , , , , 2, , , ,  , , , ,  1  sin 2   cos2    4sin 4   0 ., , 1, 2, ,  sin 4 , , 1 x, , 1, , 1, , 1, , 1 x, , 1, , 1, , 1, , 1 x, ,            , 0 0 0, , , ,     0, , , ,   , ,  0 are, , (a) 0, – 3, (c) 0, 0, 0, – 3, , , , , , , , (b) 0, 0, – 3, (d) None of these, , 1 x, , 1, , 1, , 1, , 1 x, , 1, , 1, , 1, , 1 x, , 3x, , 0, , 1, , 3x, , x, , 1, , 3x, , (d) 0, , Applying R1  R1  R2  R3 , We get,, , 2. The roots of the equation, , , , (c) p 2  2q, , (b) q, , Solution:(d)Since  ,  ,  are the roots of, , 1  sin  1 1, 2, , Solution:(b), ,  2 , 3,  and | A | 125 then  , 2, , , , (a) 3, (b) 2, (c) 5, (d) 0, , 3. If A  , , 1  sin , 2, , 0, , C  C  C, 2, 3,  2, , 1, , 0, , 1, , ( x  3) 1, , x, , 1, , b, , a  b, , b, , c, , b  c  0 if, , a  b b  c, 0, (a) a, b, c are in A.P., (b) a, b, c are in G.P. or ( x   ) is a factor of, ax 2  2bx  c  0, (c) a, b, c are in H.P., (d)  is a root of the equation, Solution:(b)Applying R3  R3   R1  R2 , we get, ,  0  C1  C1  C 2  C 3 , , x 1 x, , 5. The determinant, , a, , , , , a  b, , a b, , b  c, , b c, , 0, , 0, , 0 0 a  b  b  c, 2, , 1 x 1 x, ,  (a 2  2b  c)(ac  b 2 )  0, , 1, ,  a 2  2b  c  0 or b 2  ac,  x   is a root of ax 2  2bx  c  0 or a, b, c are, , 0, , 1, ,  R2  R2  R1 , ( x  3) 0 x 0  0 , , , R3  R3  R1 , , 0 x x, , in G.P., ,  ( x   ) is a factor of ax 2  2bx  c  0 or a, b, c, , ( x  3) x 2  0  x  0, 0,  3 ., , Trick: Obviously the equation is of degree three,, therefore it must have three solutions. So, check for option (b)., CHETHAN M G, , are in G.P., , [email protected], Phone number-8105418762 / 7019881906, , 13

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CONCEPTS OF MATHEMATICS FOR KCET -JEE MAIN AND ADVANCED, , Solution:(a)Applying C1  C1  C2  C3 , we get, , 6. The value of, , 1, , 1, (3  3 ), , x 2, , (3  3 ), , x, , (2  2 ), x, , 1, , x 2, , (2  2 ), x, , x, , x 2, , x 2, , (5  5 ), , x 2, , (5  5 ), , x, , x 2, , x, , 1, ,   2.2 .2.2, x, , x, , x, , x, , 2.3 .2.3, , x 2, , (2  2 ), x, , 1, , x, , 2.5 .2.5, , x 2, , (3  3 ), x, , 1, , 1, , 1, , 1, , (2 x  2 x ) 2, , (3x  3 x )2, , (5x  5 x )2, , 4, , cos x cos x, , cos x, , sin x, , (c) 1, , cos x, , 0, , 0, , 0, , sin x  cos x, , 0, ,  (2cos x  sin x)(sin x  cos x)  0  tan x  2,1, 2, ,   , But tan x  2 in   ,  .,  4 4, , Hence tan x  1  x , 4, 8. If n  3k and 1,  ,, , 1, unity, then   , , , value, (a)0, 14, , (b) , ,  are the cube roots of, , , 2n, n, , 1, , , , 2n, , x4, , 2x, , 2x, , 2x, , x4, , 2x, , 2x, , 2x, , x4, , 10. If  an  is an arithmetic sequence, then, am, , an, , ap, ,  m, , n, , p, , 1, , 1, , 1, , (c) , , Equals, , ,  n has the, , a  (m  1) d, , a  (n  1)d, , a  ( p  1) d, , m, , n, , p, , 1, , 1, , 1, , , , 2n, , a, , a, ,  m n, , 1, , 1, 2, , d)  4, 5 , , (a)1, (b)–1, (c)0, (d)None of these, Solution:(c), Let a be the first term and d the, common difference. Then ar  a  (r  1) d, , 2, , n, , b)  4, 5 , , 2, 2,  (5 x  4)  x  4    2 x   4 x  x  4  , , , 2, 2, 2,  (5 x  4)  x  4    2 x   (2 x)  16 x , , , 2,  (5 x  4)  x  4  , , , So A  4 , B  5   A, B)  (4,5) , , sin x, , (2 cos x  sin x) 0 sin x  cos x, , 2, , 2, 2, 2, 2,  (5 x  4)  x  4    2 x    2 x   2 x  x  4    2 x   2 x  x  4  , , , , (d) 3, , cos x, , x4, , c a b, , Applying, R2  R2  R1 and R3  R3  R1, , 1, , 2x, , 2x, , b c a  a  b  c  a 2  b 2  c 2  ab  bc  ca , , cos x  0, , 1 cos x, , 2x, , a b c, , 1 cos x cos x, (2 cos x  sin x) 1 sin x, , x4, , [JEE Main ofline-18-4m], , sin x, , (b) 2, , ( 1   n   2 n  0if n is not multiple of 3), , Solution:(d) W.K., , , ,   x  is, 4, 4, (a) 0, Solution:(c), ,  2n, , c)  4, 3, , cos x  0 in the interval, , cos x cos x, , 1   , , 2n, , a)  4,5 , , 0, , [ R1 and R2 are identical], Trick: Putting x  0 ,we get option (a) is correct, 7. The number of distinct real roots of, , sin x, , 1, ,  2n, 0  n  2n, n  0 1 n  0, 1, 0  2n, 1, , 2 x   A  Bx  x  A , then, 2x, 2x x  4, the order pair ( A, B ) is equal to:, , (5 x  5 x ) 2, , 1, ,   1   n   2n, , 9. If, , x, , 1, , n, , n, , (a) 0 (b) 30 x, (c) 30  x (d) None of these, Solution:(a) Applying R2  R2  R3, , 1, , 1   n   2n, , 1, , a, , m 1 n 1 p 1, , p d, , m, , n, , p, , 1, , 1, , 1, , 1, , (d)1, , [email protected], https://t.me/mcpmathematicskar_1986, Phone number-8105418762 / 701988190, , MCP-MATHEMATICS

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DETERMINANTS, , 1, , 1, , 1, , a m n, 1, , 1, , m n, , p, , p d m n, , p  a .0  d .0  0, , 1, , 1, , 1, , 1, , log an, 1,   log an 3, 2, log an  6, , 11. If 3n is a factor of the determinant, , 1, n, n, , c1, , 1, , 1, , n 3, , n6, , n 3, , c2, , c1, , n6, , c2, , c1, , , , then the maximum value, , c2, ,   n c1, n, , , , 1, , 1, , n 3, , n6, , n 3, , c2, , c1, , c2, , n6, , c1, , 1, , 2, , 1, , n, , n3, , n6, , 1, , 2, , 1, , 1, , n, , n3, , n6, , 3, , Solution:(b) yp  z, , y, , z, , xp  y, , yp  z, x, , y, , 0, , y, , z, ,  xp  py  yp  z, , xp  y, , yp  z, , log an  7, , , ,  xz  y  0  xz  y, , 6, , log an 5 is, , 0, , 2, , Hence x, y, z are in G.P, 10, , C4, , 10, , C5, , 11, , C6, , 11, , C7, , 12, , C8, , 12, , C9, , 13, , 12, , log an  4, , , , 2, , n  3  3n  3  3  9, 12. If a1 , a2 , a3 ,..., a n,... are in G.P.then the value of the, , log an  6, , , , , , 3, 9, 27 are the factors of  .Hence, , determinant log an 3, , 0, , 0, , 14. The value of  , , log an  2, , yp  z, ,  xp 2  py  yp  z xz  y 2  0, , 0, , log an 1, , xp  y, , y, , Expansion along C, ,   27, , log an, , z, , x, ,  n  3 n  2   n  6  n  5, 0, ,  0 if, , y, , b) x,y,z are in G.P, d) xy,yz,zx are in A.P., , 2, , n  n  1 6n  6 12n  30, , equal to zero, where m is, (a) 6, (b) 4, (c) 5, 10, C4 10 C5, Solution:(c)   11 C6 11 C7, 12, , 10, , , , log an 8, ,  an21  an .an2  2log an1  log an  log an 2, , an24  an3.an5  2 log an  4  log an 3  log an 5, an27  an6 .an8  2 log an 7  log an 6  log an 8, Putting these values in the second column of the given, determinant, we get, , 12, , C8, , C9, , 11, , Cm, , Cm 2 is, Cm  4, , (d) None of these, Cm, 12, Cm  2 = 0, 11, , 13, , Cm  4, , Applying C2  C1  C2, , (a) –2, (b) 1, (c) 2, (d) 0, Solution:(d)  a1 , a2 , a3 ,......., an , ...... are in G.P., , CHETHAN M G, , 1, (0)  0 [ C2 = C1 + C3 ], 2, xp  y, x, y, , xp  y, , 2, , 1, , n, , log an 8, , 13. The determinant yp  z, ,  n  3 n  2   n  6  n  5, , 1, , log an  6  log an 8, , C1  C1   pC2  C3, , 1, , n  n  1, , log an 5, , 0, , c2, , 2, 2, C2  C2  C1 , C3  C3  C1, , log an 3  log an 5, , a) x,y,z are in A.P, c) x,y,z are in H.P, , d) 4, , 1, n  n  1, , log an  2, , 0, , of n is, a) 5, b) 2, c) 3, Solution:(c) Expansion along C1, , 1, , log an  log an  2, , 11, 12, , , , C4, , 10, , C4 10 C5, , C6, , 11, , C8, , 12, , 10, , C4, , 11, , C5, , 11, , C6, , 12, , C7, , 12, , C8, , 13, , C9, , 11, , Cm, , C6  C7, , 12, , Cm  2  0, , C8  C9, , 13, , Cm  4, , 11, , 12, , 11, , Cm, , 12, , Cm  2 = 0, 13, Cm  4, , Clearly m  5 satisfies the above result, , [ C2 , C3 will beidentical], , [email protected], Phone number-8105418762 / 7019881906, , 15

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CONCEPTS OF MATHEMATICS FOR KCET -JEE MAIN AND ADVANCED, , 1, , 4, , 15. The roots of the equation 1 2, , 5  0 are, , 1 2x 5x, (a) 1, 2, , (b) 1, 2, , (c) 1, 2, , (d) 1, 2, , 1, , 4, , , , 6, ,  x 1   2, x 1   2, , ,  x 1 x  2, , 1, , 1, , 3.2.5. 0 (1  x) 1  x 2  0, , 2, , 1, , x , , 2, , , x  2, , 2, , 1, , x , , 1, , , (C1  C1  C2  C3 ), , 1, , ( 1     2  0), , 1, x , , 1, ,  x [1{( x   ) ( x   )  1}  {1  ( x   )}, 2, ,  2{1  ( x   2 )}],  x( x 2   x   2 x   3  1     x   2,  2   2 x   4 ), , x2, ,  x3 ,, (  3  1) ., 18. If a, b, c are unequal what is the condition that the, , (1  x) 0 1 1  x  0, , value of the following determinant is zero, , x, , 0, , 1, , 1, , 1, x2, , x, , x  1  0 or x  2  0  x  1, 2 ., Trick: Obviously by inspection, x  1, 2 satisfy the, equation., b 2  ab b  c bc  ac, , , 2, 16. ab  a, , a  b b 2  ab , , bc  ac c  a ab  a 2, (a) abc(a  b  c), (c) 0, , (b) 3a b c, (d) None of these, , c, , Solution:(c)   (b  a )(b  a ). a, , a b b, , c, , ca a, , b b, , c, ,  (a  b) a a b  0 , [by C2  C2  C3 ] ., 2, , c, , c, , a, , 17. If  is a cube root of unity, then, x 1, , 2, , , 2, , x  2, , 1, , 1, , x , , , , a a2, , a3  1, ,   b b2, , b3  1, , c2, , c3  1, , c, , (a) 1  abc  0, (b) a  b  c  1  0, (c) (a  b)(b  c)(c  a)  0, (d) None of these, Solution:(a)Splitting the determinant into, Two determinants, we get, , 1 a a2, , 2 2 2, , b bc, , 16, , 1, , , x  2, , 2, , 1, , 2, ,  R1  R1  R2 , 0 2  2 x 5(1  x 2 )  0 , , R2  R2  R3 , , 2, 1, 2x, 5x, , 1, , , x 1   2, , 15, , 0, , , , , 5 0, , 1 2x 5x, , Solution:(d)   , , 2, , 20, , Solution:(b) 1 2, , 0, , x 1, , 20, , 1 a a2, ,   1 b b 2  abc 1 b b 2  0, c2, , 1 c, , 1 c, , = (1  abc)[(a  b)(b  c)(c  a )]  0, Because a, b, c are different, the second factor, cannot be zero. Hence, option (a), 1  abc  0 , is, correct., 19. If a, b and c are non-zero numbers, then, , b2c 2,  c a, 2, , 2, , 2 2, , ab, , bc b  c, ca c  a is equal to, ab a  b, , 3, (b) x  , , 3, 2, (c) x  , , 3, (d) x, , 2 2 2, , (b) a b c, (d) None of these, Solution:(d)Multiplying R1 by a, R2 by b and R 3, (a) abc, (c) ab  bc  ca, , 3, (a) x  1, , c2, , [email protected], https://t.me/mcpmathematicskar_1986, Phone number-8105418762 / 701988190, , MCP-MATHEMATICS

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DETERMINANTS, , (c) 3, , ab c abc ab  ac, 1 2 2, by c , we have  , a bc abc bc  ab, abc 2 2, a b c abc ac  bc, 2 2, , , , 2 2 2, , bc 1 ab  ac, , 3, , bc 1 ab, , 22. If, ,  abc.ab ca 1 1  0 , [Since C2  C3 ]., , , , Solution:(a), , 1, , , 1, 2, , 1 x  0, x   2, , (b) x  1, (d)None of these, 1, , , 2 1 x  o, x   2, , , Check at x  0, we get , 1, , a bc, , 2a, , 2a, , 2b, , bca, , 2b, , 2c, , 2c, , c a b, , Solution:(c)  , , 2b, , bca, , 2b, , 2c, , 2c, , c a b, 1, , 1, 2b, c a b, , 2c, , 0, ,    a  b  c  2b   a  b  c , 2c, , 0, , 0, 0,  a  b  c, ,    a  b  c  a  b  c   1, 2, , 1     0, 2, , , , Given    a  b  c  x  a  b  c    2, 2, , From (1) & (2), ,  a  b  c  a  b  c    a  b  c  x  a  b  c , 2, 2,  a  b  c   x  a  b  c,  x  2  a  b  c , 2, , determinant 1 1   2, , CHETHAN M G, , (c) 2  a  b  c  (d) 2  a  b  c , , 1, , 1, 3, i, .Then the value of the, 2, 2, 1, 1, 1, , (a) 3, , c a b, , C2  C2  C1 & C3  C3  C1, , = 0, if x  0 ., , , , 2c, , 2c, , by C1  C1  C2  C3, , 1, , 2c, ,    a  b  c  2b b  c  a, , , 1,   1   2  x 2 1 x, 1   2  x x   2, , 21. Let   , , 2b, , 1, , 1   2  x, , , , bca, , [JEE-MAIN-11/1/2019-SHIFT-II], [One option correct type-4M], , , ,   2 (   )   (1  1)  0  0 Or, , , 1, 2,  x , 1 x ,, x x   2, , 2b, , abc a bc a bc, ,   2 ( 4   )   ( 3  1)  1( 2   2 ), , x, , 2a, , R1  R1  R2  R3, ,  1, 2 1,  2, , 2, , 2a, , 2, , Trick: Put a  1, b  2, c  3 and check it., , (a) x  0, (c) x  1, , a bc, ,   a  b  c  x  a  b  c  , x  0 & a  b  c  0, then x is equal to, (a) abc, (b)   a  b  c , , ab 1 1, , , 1, ,  2 (C1  C1  C2  C3 ), , , ( 1     2  0),  3[.   4 ]  3( 2   )  3 (  1) ., , bc 1 1, , 20. At what value of x, will, , 1, 2, , 2, , 0, , {by C3  C3  C1 }, , x  2, , 1, , Solution:(b)   0 1  , , abc, ac 1 bc  ab  abc ac 1  ab, abc, ab 1 ac  bc, ab 1  ab, , x  2, , (d) 3 (1   ), , 2, , 2, ,  2 is, 4, (b) 3 (  1), , 23. Let a1 , a2 , a3 ,..., a10 be in GP. With, , 2, , ai  0, , for i  1, 2, 3...,10 and S be the set of pairs, ,  r , k  , r , k  N (the set of natural numbers) for, [email protected], Phone number-8105418762 / 7019881906, , 17

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CONCEPTS OF MATHEMATICS FOR KCET -JEE MAIN AND ADVANCED, , log e a a, , log e a3r a4k, ,  A   sin   2   sin   2  2d   d  2sin   d , , Which log e a a, , log e a a, , log e a a  0, ,  A  2sin   4  4d  sin 2   2sin , , log e a a, , log e a a, , log e a a, , r k, 1 2, r k, 4 5, r k, 7 8, , log e a a, , r, 2, r, 5, r, 8, , k, 3, k, 6, k, 9, , r k, 6 7, r k, 9 10, , Then the numbers of elements in S, is, (a) 2 (b) 10, (c) 4, (d) infinitely many, [JEE-MAIN-10/1/2019-SHIFT-II], [One option correct type-4M], ,   log e a r  k R 3r  4 k, , log e a r  k R 4 r 5k, , log e a r  k R 5r  6 k, , log e a r  k R 6 r  7 k, , log e a r  k R 7 r 8k, , log e a r  k R8r 9 k, , C2 = C2 - C1 & C3 = C3 - C1, ,    log e a, log e a, , r k, r k, , R, R, , 3r  4 k, 6r 7 k, , log e R r  k, , log e R 2 r  k , , r k, , 2 r  k , , log e R, log e R, , log e R, , r k, , log e R, , 2 r  k , , 2, , (ii) Given A min  8, ,   d  2 1  8, , , ,  , , , , max sin 2   1, ,   d  2  9   d  2   3,  d  1 or d  5, 2, , 25. Let the numbers 2, b, c be in an AP and, , 1 1, A   2 b,  4 b 2, , 1, c  If det  A   2,16 ,, c 2 , , then c lies in the interval, (a)  2,3, , (b)  4, 6, , (c) 3, 2  2, , (d) 2  23/4 , 4, , 3/4, , , , , , , , [JEE-MAIN-8/4/2019-SHIFT-II], [One option correct type-4M], , C1 & C 2 are in the ratio,    0 r , k  N, , Solution:(b), , 1, , 1, , 1, , So S has infinitely many pairs  r , k  r , k  N, , (i) A  2, , b, , c C2  C2  C1 & C3  C3  C1, c2, , 24. Let d  R, and, , 4 b, sin   2, , 4d,  2, , A 1, sin   2,  5 2sin   d, , , , d, ,  sin   2  2d ,   0, 2 . If the minimum value of det  A is 8,, then a value of d is, (a) 2 2  2, , , , , , (d) 2, , , , 4d, , sin   2, , (i) A  1, , sin   2, , d, , 5, , 2sin   d, ,  sin   2  2d, , R1  R1  R3  2 R2, ,  A1, , 1,  A2, , 0, , 0, , b2, , c2, , 4 b2  4 c2  4, , , , , , , ,  A   b  2  c 2  4   c  2  b2  4, , , ,  A   b  2  c  2  c  b , c2, c2,  c  2  ,  A  c  2 ,  2  c ,  b, , 2 , 2 ,  2, , ,  c  2 .,  c  2  c  2 ,  A  c  2 , ,  A , 4,  2  2 , (ii) Given 2  A  16, 3, , Solution:(c), , 1, , , , 2 1, , [JEE-MAIN-10/1/2019-SHIFT-I], [One option correct type-4M], , 2, , 2, ,  A   b  2  c  2  c  2  b  2, , (b) 7, , (c) 5, ,  c  2, 2, , 3, ,  16, , 4, , 0, , 0, , sin   2, ,  8   c  2   64  2   c  2   4, , d, ,  4  c  6  c   4,6, , 5 2sin   d, 18, ,  A   d  2  sin 2 , 2, , Solution:(d) (i) Let a1 , a2 , a3 ,..., a10 be in GP., Let a be the first term & R be the common ratio of the, 2, 9, GP. a1  a, a2  aR, a3  aR ,..., a10  aR ., (ii), log e a r  k R k, log e a r  k R r  2 k log e a r  k R 2 r 3k, , log e a r  k R k, ,  2d sin   2d sin   d 2, ,  sin   2  2d, , 3, , [email protected], https://t.me/mcpmathematicskar_1986, Phone number-8105418762 / 701988190, , MCP-MATHEMATICS

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DETERMINANTS, , 26. Let  &  be the root of the equation, , 1  cos , , sin , , 4 cos 6, , cos , , 1  sin , , 4 cos 6, , cos , , sin , , 1  4 cos 6, , 2, , x  x  1  0. Then for y  0 in R, 2, , y 1, , , , , , , , , y, , 1, , 1, , y , , , , (a) y y  3, 2, , Solution:(a), , is equal to:, , , , , , , , , , , , cos , , cos , , 1, , y , , R1  R1  R2  R3, , y 1   2, , y 1   2, , y 1   2, , , 2, , y  2, , 1, , 1, , y , , , ,  1     2  0, y, y, y, 2,    y , 1, 2, , 1, y , 1,  y , , 1, , 1, , y  2, , 1, , 1, , y , , 2, , , , 0, , 0, , 2, , 4 cos 6, , (a) f  50   501, , (b) f  50   1, , (c) f  50   1, , (d) f  50   501, , xa, ,  , , xc, , x4 x3, , , , R1  R1  R3  2 R2, , 1, , sin 2 , , 4 cos 6, , cos 2 , , 1  sin 2 , , 4 cos 6, , cos 2 , , sin 2 , , 1  4 cos 6, , , , 7, 36, , , 9, , (b), , 18, , (c), , [JEE-MAIN-12/4/2019-SHIFT-II], [One option correct type-4M], , CHETHAN M G, , (d), , xb, ,  0 is, , 7, 24, , 0, , 0, ,  f  x  x  b, , 0, , x  3 x  2  a  2b  c  1, x4 x3, , xc, , 1  cos 2 , , (a), , a  c  2b,  f  x , ,   y  y  1  1  y  y   y,  , 27. A value of    0,  for which,  3, 3, , x 1, , x3 x2, ,   y  y 2   y   2 y   3  1   y     2 y   , 2, , x2, , Solution:(b) f  x   x  b, ,   y  y   2  y    1   y   2   2     2 y   4 , , , 2, , 0, , [JEE-MAIN-9/1/2020-SHIFT-II], [One option correct type-4M], , , , , , 1  4 cos 6, ,  2  8cos6  4cos6  0,  2  4cos6  0  1  2cos6  0, 1, 2,  cos 6    cos 6  cos, 2, 3, 2, ,  6 ,  , 3, 9, 28. Let a  2b  c  1. If, x  a x  2 x 1, f  x   x  b x  3 x  2 , then, xc x4 x3, , 2, , 1, ,  0 C2  C2  C1, , 1 1  4 cos 6, , 2, , 2, , , , sin , 2, ,  cos 2 , , equation x  x  1  0. So    &    2, are cube roots of unity., , 4 cos 6, , 2, , 2, , Solution:(c) (i) Given  &  are the roots of, , , , 0, , 0, ,  cos  1  sin , 2, , 1, , , y  2, , 2, , 1, , 1, , (b) y  1, 3, , [JEE-MAIN-9/4/2019-SHIFT-I], [One option correct type-4M], , 2, , 2, , R1  R1  R2, , (d) y y 2  1, , (ii)   , , 2, 2, , (c) y 3, , y 1, , 2, , 0, , x3 x2, , xc, , x4, , x3, ,  f  x    x  3   x  2 x  4 , 2, ,  f  x   x2  9  6x  x2  4x  2x  8,  f  x  1,  f  50   1 (Because constant function), [email protected], Phone number-8105418762 / 7019881906, , 19

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CONCEPTS OF MATHEMATICS FOR KCET -JEE MAIN AND ADVANCED, ,  sin 2 , , 1  sin 2 , , 1, , Solution:(d) (i) f     cos , , 1  cos , , 1, , 10, , 2, , 29. If, , x2, , 2x  3, , 3x  4, ,   2 x  3 3x  4, , 2, , 4 x  5  Ax3  Bx 2  Cx  D,, , 12, , 3x  5 5 x  8 10 x  17, Then B  C is equal to, (a)-1, (b)1 (c) -3, (d) 9, , C3 = C3 - C1 + C2, , [JEE-MAIN-12/4/2019-SHIFT-II], [One option correct type-4M], , x2, , 4 x  5 R2  R2  R1, , 3x  5 5 x  8 10 x  17, 2x  3, , 3x  4, ,    x 1, , x 1, , x 1, ,  f     cos , , 1  cos , , 0, , 10, , 4, , 12, , sin 2   sin 2  cos 2   cos 2  ,  f    4 , , 2, 2,   sin  cos , , , (ii) Given, , x2, , 2x  3, , 3x  4, , 1, , 1, , 1, , C2 = C2 - C1 & C3 = C3 - C1, x2, , x 1, , 2x  2, , 1, , 0, , 0, , , , , , 4, , cos 2 x, ,      x  1  x  1 7 x  12   2 x  3 2 x  2 , ,    1  x  7 x2  12 x  7 x  12  4 x2  4 x  6 x  6, ,    1  x  3x2  9 x  6, ,    3x 2  9 x  6  3x3  9 x 2  6 x,    3x3  12 x 2  15x  6  1, But Given   Ax3  Bx 2  Cx  D   2 , From (1) & (2)  B  C  12 15  3, 30. If the minimum and the maximum values of the,   , function f :  ,   R, defined by, 4 2, ,  sin 2 , , 1  sin 2 , , 1, , f     cos 2 , , 1  cos 2 , , 1 are, , 10, , 2, , m & M respectively, then the ordered pair,  m, M  is equal to, , , , (a) 0, 2 2, , , , (c)  0, 4 , [JEE-MAIN-05/9/2020-SHIFT-I], [One option correct type-4M], , 20, , , , 2, ,   m, M    1, 0 , 31. Let m & M be respectively the minimum and, maximum values of, , 3x  5 2 x  3 7 x  12, , 12, ,  , ,  2    1  cos 2  0, 2,  4  4cos 2  0  4  f    0, , , , 3x  5 5 x  8 10 x  17, ,     x  1, , 0, , 2, ,  f    4 cos2   sin 2    4cos 2, , 3x  5 5 x  8 10 x  17, ,     x  1, , 1  sin 2 , , 3x  4, , Solution:(c)   2 x  3 3 x  4, , x2, ,  sin 2 , 2, , 2x  3, , 2, , (b)  4, 4 , (d)  4,0 , , 1  sin 2 x, , sin 2 x, , 2, , 1  cos x, , sin x, , sin 2 x, , cos 2 x, , sin 2 x, , 1  sin 2 x, , 2, , . Then, the, , ordered pair  m, M  is equal to, (a)  3,3, , (b)  3, 1, , (c)  4, 1, , (d) 1,3, , [JEE-MAIN-06/9/2020-SHIFT-I], [One option correct type-4M], , Solution:(b), , cos 2 x, , 1  sin 2 x, , (i)   1  cos x, 2, , sin 2 x, , 2, , sin 2 x, , 2, , 1  sin 2 x, , sin x, , 2, , cos x, , sin x, , R2  R2  R1 & R3  R3  R1, , cos 2 x 1  sin 2 x sin 2 x, , , 1, , 1, , 0, , 0, , 1, , 1, , , , , ,     cos2 x  1  sin 2 x  sin 2 x,     cos x  1  sin x  sin 2 x,    2  sin 2 x    2  sin 2 x , 2, , 2, , (ii) w.n.t. 1  sin 2x  1, , [email protected], https://t.me/mcpmathematicskar_1986, Phone number-8105418762 / 701988190, , MCP-MATHEMATICS

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DETERMINANTS, ,  1  2  sin 2x  3,  3    2  sin 2 x   1  3    1, ,   m, M    3, 1, 32. If a  x  b  y  c  z  1, where, a, b, c, x, y , z are non-zero distinct real numbers, x a y, then y, , b y, , z, , c y, , xa, y  b is equal to, zc, , (a) y  b  a , , (b) y  a  b , (d) y  a  c , , (c) 0, [JEE-MAIN-05/9/2020-SHIFT-II], [One option correct type-4M], , x a y, Solution:(b)   y, , b y, , z, , c y, , xa, y  b C3 = C3 - C1, zc, , x a y a,    y b  y b C2 = C2 - C3, z c y c, x, , y a, ,  y, , y b R2  R2  R1 , R3  R3  R1 ,, y c, , z, , x, , y, , a, ,    yx 0 ba, zx, , 0 ca, ,     y  y  x  c  a    b  a  z  x , ,  a  x  b y  y  x  a b, ,  & a  x  c  z  1  z  x  a  c  1, , ,     y  a  b  c  a    b  a  a  c  1,     y  a  b  c  a  a  c  1,     y  a  b   1  y  a  b , , CHETHAN M G, , [email protected], Phone number-8105418762 / 7019881906, , 21