Note : Unit matrix and null square matrices are also scalar matrices.
(9) Triangular Matrix : A square matrix [aij ] is said to be triangular matrix if each element above or below
the principal diagonal is zero. It is of two types
(i) Upper Triangular matrix : A square matrix [aij ] is called the upper triangular matrix, if aij = 0 when i > j .
3 1 2
Example : 0 4 3 is an upper triangular matrix of order 3×3.
0 0 6
(ii) Lower Triangular matrix : A square matrix [aij ] is called the lower triangular matrix, if aij = 0 when i< j.
1 0 0
0 is a lower triangular matrix of order 3×3.
Example : 2 3
4 5 2
Note : Minimum number of zeros in a triangular matrix is given by n(n − 1) where n is order of matrix.
2
Diagonal matrix is both upper and lower triangular.
A triangular matrix a = [aij ]n×n is called strictly triangular if aij = 0 for 1 ≤ i ≤ n
Example: 1 A square matrix A = [aij ] in which aij = 0 for i ≠ j and aij = k (constant) for i = j is called a [EAMCET 2001]
(a) Unit matrix (b) Scalar matrix (c) Null matrix (d) Diagonal matrix
Solution: (b) When aij = 0 for i ≠ j and aij is constant for i = j then the matrix [aij ]n×n is called a scalar matrix
Example: 2
Solution: (a) If A, B are square matrix of order 3, A is non singular and AB = 0, then B is a [EAMCET 2002]
Example: 3
(a) Null matrix (b) Singular matrix (c) Unit matrix (d) Non singular matrix
Solution: (c)
Example: 4 AB = 0 when B is null matrix.
Solution: (a) 2 5 − 7
The matrix 0
3 11 is known as
0 0 9
(a) Symmetric matrix (b) Diagonal matrix (c) Upper triangular matrix (d) Skew symmetric matrix
We know that if all the elements below the diagonal in a matrix are zero, then it is an upper triangular matrix.
In an upper triangular matrix n×n, minimum number of zeros is [Rajasthan PET 1999]
(a) n(n − 1) (b) n(n + 1) (c) 2n(n − 1) (d) None of these
2 2 2
As we know a square matrix A = [aij ] is called an upper triangular matrix if aij = 0 for all i>j
a11 a12 a13 a14 .... a1(n−2) a1(n−1) a1n
a2(n−1)
0 a22 a23 a24 .... a2(n−2) a2n
0 0 a33 a34 .... a3(n−2) a3(n−1) a3n
0 0 a44 .... a4(n−2) a4(n−1) a4n
0 (n − 1)n
A = − − − −− − − − . Number of zeros = (n − 1) + (n − 2) + ..... + 2 +1 = 2
− − − − − − − −
0 0 0 0 .... 0 a(n−1)(n−1) a(n−1)n
0 0 0 0 .... 0 0 ann
Example: 5 If A = [aij ] is a scalar matrix then trace of A is
∑ ∑(a) aij ∑(b) aij ∑(c) aij ∑(d) aii
ij i j i
Solution: (d) n
∑The trace of A = aii = Sum of diagonal elements.
i=1
Addition and Subtraction of Matrices.
If A = [aij ]m×n and B = [bij ]m×n are two matrices of the same order then their sum A+B is a matrix whose each
element is the sum of corresponding elements. i.e. A + B = [aij + bij ]m×n
5 2 1 5 5 + 1 2 + 5 6 7
Example : If A = 1 3 and B = 2 2 , then A + B = 1 + 2 3 + 2 = 3 5
1 3 3 4 + 3 1 + 3 7 4
4
Similarly, their subtraction A − B is defined as A − B = [aij − bij ]m×n
5 −1 2 − 5 4 − 3
i.e. in above example A − B = 1 − 2 3 − 2 = − 1
1 − 3 1 1
4 − 3
− 2
Note : Matrix addition and subtraction can be possible only when matrices are of the same order.
Properties of matrix addition : If A, B and C are matrices of same order, then
(i) A + B = B + A (Commutative law)
(ii) (A + B) + C = A + (B + C) (Associative law)
(iii) A + O = O + A = A, where O is zero matrix which is additive identity of the matrix.
(iv) A + (− A) = 0 = (− A) + A , where (− A) is obtained by changing the sign of every element of A, which is
additive inverse of the matrix.
(v) A+ B = A + C ⇒ B = C (Cancellation law)
B+ A = C + A
Scalar Multiplication of Matrices.
Let A = [aij ]m×n be a matrix and k be a number, then the matrix which is obtained by multiplying every
element of A by k is called scalar multiplication of A by k and it is denoted by kA.
2 4 10 20
Thus, if A = [aij ]m×n , then kA = Ak = [kaij ]m×n . Example : If A = 3 5 A = 15
1 , then 5
4
6 20 30
Properties of scalar multiplication:
If A, B are matrices of the same order and λ, µ are any two scalars then
(i) λ(A + B) = λA + λB (ii) (λ + µ)A = λA + µA
(iii) λ(µA) = (λµA) = µ(λA) (iv) (−λA) = −(λA) = λ (− A)
Note : All the laws of ordinary algebra hold for the addition or subtraction of matrices and their
multiplication by scalars.
Multiplication of Matrices.
Two matrices A and B are conformable for the product AB if the number of columns in A (pre-multiplier) is
same as the number of rows in B (post multiplier).Thus, if A = [aij ]m×n and B = [bij ]n×p are two matrices of order
n
∑m×n and n × p respectively, then their product AB is of order m × p and is defined as (AB)ij = airbrj
r =1
b1j
[ai1ai2...ain ]bbn2jj
= = (ith row of A)(jth column of B) .....(i), where i=1, 2, ..., m and j=1, 2, ...p
Now we define the product of a row matrix and a column matrix.
Let A = [a1a2 ....an ] be a row matrix and B = bb12 be a column matrix.
bn
Then AB = [a1b1 + a2b2 + .... + anbn ] …(ii). Thus, from (i),
(AB)ij = Sum of the product of elements of ith row of A with the corresponding elements of jth column of B.
Properties of matrix multiplication
If A,B and C are three matrices such that their product is defined, then
(i) AB ≠ BA (Generally not commutative)
(ii) (AB)C = A(BC) (Associative Law)
(iii) IA = A = AI , where I is identity matrix for matrix multiplication
(iv) A(B + C) = AB + AC (Distributive law)
(v) If AB = AC ⇒/ B = C (Cancellation law is not applicable)
(vi) If AB= 0
It does not mean that A= 0 or B = 0, again product of two non zero matrix
may be a zero matrix.
Note : If A and B are two matrices such that AB exists, then BA may or may not exist.
The multiplication of two triangular matrices is a triangular matrix.
The multiplication of two diagonal matrices is also a diagonal matrix and
diag (a1, a2 ,....an ) × diag (b1,b2 ,....bn ) = diag (a1b1, a2b2 ,....anbn )
The multiplication of two scalar matrices is also a scalar matrix.
If A and B are two matrices of the same order, then
(i) (A + B)2 = A2 + B2 + AB + BA (ii) (A − B2 ) = A2 + B2 − AB − BA
(iii) (A − B)(A + B) = A2 − B2 + AB − BA (iv) (A + B)(A − B) = A2 − B2 − AB + BA
(v) A(−B) = (− A)B = −(AB)
Positive Integral Powers of A Matrix.
The positive integral powers of a matrix A are defined only when A is a square matrix. Also then A2 = A.A ,
A3 = A.A.A = A2 A . Also for any positive integers m ,n.
(i) Am An = Am+n (ii) (Am )n = Amn = (An )m
(iii) I n = I, I m = I (iv) A0 = In where A is a square matrix of order n.
Matrix Polynomial .
Let f(x) = a0 x n + a1 x n−1 + a2 x n−2 + ... + an−1 x + an be a polynomial and let A be a square matrix of order n.
Then f(A) = a0 An + a1 An−1 + a2 An−2 + ... + an−1 A + an In is called a matrix polynomial.
Example : If f(x) = x 2 − 3x + 2 is a polynomial and A is a square matrix, then A2 − 3A + 2I is a matrix polynomial.
Example: 6 If A = cos α sin α , then A2 = [Rajasthan PET 2001]
− sin α cos α
Solution: (c)
Example: 7 (a) cos 2α sin 2α (b) cos 2α − sin 2α (c) cos 2α sin 2α (d) − cos 2α sin 2α
Solution: (b) − sin 2α
Example: 8 sin 2α cos 2α sin 2α cos 2α cos 2α − sin 2α − cos 2α
Solution: (a) Since A2 A.A = cosα sinα cosα sinα cos 2α sin 2α
Example: 9 = − sinα − = − sin 2α
Solution: (b) cosα sinα cosα cos 2α
Example: 10 If A = a b and A2 = α β then [AIEEE 2003]
Solution: (b) b a β α
(a) α = a2 + b2, β = ab (b) α = a2 + b2, β = 2ab (c) α = a2 + b2, β = a2 − b2 (d) α = 2ab, β = a 2 + b 2
A2 = α β = a b a b = a2 + b2 2ab . On comparing, we get, α = a2 + b2, β = 2ab
β α b b a
a 2ab a2 + b2
If A = i 0i ,n ∈ N , then A4n equals [AMU 1992]
0
(a) 1 0 (b) i 0 (c) 0 i (d) 0 0
0 1 0 0 0 0
i i
A2 = i 0 i 0 = −1 0 , A4 = A2. A2 = −1 0 −1 0 = 1 0 = I; (A4 )n = In = I = 1 0
0 0 − 1 1 − 1 0 1 0 1
i i 0 0 − 0
If A = 1 −−11, B = a 1 and (A + B)2 = A2 + B2 then value of a and b are [Kurukshetra CEE 2002]
2 b − 1
(d) a = 2,b = 4
(a) a = 4 , b = 1 (b) a = 1,b = 4 (c) a = 0, b = 4
We have (A + B)2 = A2 + B2 + A.B + BA
∵ AB + BA = 0 ∴ a −b 2 + a + 2 −a − 1 = 0
2a − b 3 b − 2 − b + 1
2a + 2 − b −a + 1 = 0 . On comparing, we get, −a + 1 = 0⇒ a =1 and 4−b =0 ⇒ b =4
2a − 2 4 − b
a h g x
The order of [x y z]h y
b f is [EAMCET 1994]
g f c z
(a) 3×1 (b) 1×1 (c) 1×3 (d) 3×3
Order will be (1× 3)(3 × 3)(3 ×1) = (1×1)
cos α − sin α 0
cos α 0 . Then F (α).F (α') is equal to
Example: 11 Let F(α) = sin α 1
0
0
(a) F (αα') (b) F (α / α' ) (c) F (α + α') (d) F (α − α' )
cos α − sin α 0 cosα′ − sinα′ 0
cos α 0 , cos α ′ 0
Solution: (c) We have F(α) = sin α F(α ′) = sin α ′ 1
Example: 12 0 0
Solution: (b) 0 1 0
Example: 13
cos α − sin α 0 cos α′ − sin α′ 0 cos(α + α ′) − sin(α + α ′) 0
cos α 0 cos α′ 0 cos(α + α ′) 0 = F (α + α′)
F (α ).F (α ′) = sin α sin α′ = sin(α + α ′) 1
0 0 0
0 1 0 1 0
1 1 0 [Kerala (Engg.) 2001]
For the matrix A = 1 2 1 , which of the following is correct
2 1 0
(a) A3 + 3A2 − I = 0 (b) A3 − 3A2 − I = 0 (c) A3 + 2A2 − I = 0 (d) A3 − A 2 + I = 0
1 1 0 1 1 0 2 3 1 2 3 1 1 1 0 7 9 3
A 2 = A.A = 1 2 1 1 2 1 = 5 6 2 , A3 = A.2 A = 5 6 2 1 2 1 = 15 19 6
2 1 0 2 1 0 3 4 1 3 4 1 2 1 0 9 12 4
7 9 3 6 9 3 1 0 0
A3 − 3.A2 = 15 19 6 − 15 18 6 = 0 1 0 = I ⇒ A3 − 3A2 − I = 0
9 12 4 9 12 3 0 0 1
If A = α 01, B = 1 0 , then the value of α for which A2 =B is [IIT Screening 2003]
1 5 1
(d) No real values
(a) 1 (b) –1 (c) 4
Solution: (d) A2 = α 0 α 0 = α2 0 ∵ A2 = B (given)
1 1 1 1 α + 1 1
Then α2 0 = 1 0 ⇒ α2 = 1 and α +1 = 5 . Clearly no real value of α
α + 1 1 5 1
Transpose of a Matrix.
The matrix obtained from a given matrix A by changing its rows into columns or columns into rows is called
transpose of Matrix A and is denoted by AT or A′ .
From the definition it is obvious that if order of A is m×n, then order of AT is n×m
a1 a2 a3 a1 b1
b1 b2 a 2
Example : Transpose of matrix b3 is a3 b 2
2×3 b3 3×2
Properties of transpose : Let A and B be two matrices then
(i) (AT )T = A
(ii) (A + B)T = AT + BT , A and B being of the same order
(iii) (kA)T = kAT , k be any scalar (real or complex)
(iv) (AB)T = BT AT , A and B being conformable for the product AB
(v) (A1 A2 A3 ..... An−1 An )T = AnT An−1T ....... A3 T A2T A1T
(vi) I T = I
Determinant of a Matrix .
a11 a12 a13
If A = a21
a 22 a 23 be a square matrix, then its determinant, denoted by |A| or Det (A) is defined as
a31 a32 a33
a11 a12 a13
| A|= a21 a22 a23
a31 a32 a33
Properties of determinant of a matrix
(i) | A|exists ⇔ A is square matrix (ii) | AB|=| A|| B|
(iii) | AT |=| A| (iv) |kA|= k n | A|, if A is a square matrix of order n
(v) If A and B are square matrices of same order then |AB|=|BA|
(vi) If A is a skew symmetric matrix of odd order then | A|= 0
(vii) If A = diag(a1, a2 ,.....an ) then | A|= a1a2 ...an (viii)| A|n =| An |, n ∈ N.
Example: 14 If A and B are square matrices of same order then
[Pb. CET 1992; Roorkee 1995; MP PET 1990; Rajasthan PET 1992, 94]
(a) (AB)′ = A′B′ (b) (AB)′ = B′A′
(c) AB = 0, if | A|= 0 or | B|= 0 (d) AB = 0, if | A|= I or B = I
Solution: (b) A = [aij ]n×n , B = [b jk ]n×n , AB = [aij ]n×n[bjk ]n×n = [cik ]n×n , where cik = aij b jk
(AB)′ = [cik ]′n×n = [cki]n×n = [bkj ]n×n [a ji ]n×n = B′ A′
Alternatively, Let A = 1 2 , B = 1 3 ; AB = 1 11
3 4 0 4 3 25
2×2 2×2
(AB)' = 1 3 …..(i) and B' A' = 1 0 1 3 = 1 3 …..(ii)
11 25 3 4 2 4 11 25
From (i) and (ii), (AB) = B′A′
Example: 15 If A,B are 3×2 order matrices and C is a 2×3 order matrix, then which of the following matrices not defined
[Rajasthan PET 1998]
(a) At + B (b) B + Ct (c) At + C (d) At + Bt
Solution: (a) Order of A is 3 × 2 and order of B is 3 × 2 and order of At is 2 × 3 then
= At + B = [At ]2×3 + [B]3×2 is not possible because order are not same.
Special Types of Matrices.
(1) Symmetric and skew-symmetric matrix
(i) Symmetric matrix : A square matrix A = [aij]is called symmetric matrix if aij = aji for all i, j or AT = A
a h g
Example : h
b f
g f c
Note : Every unit matrix and square zero matrix are symmetric matrices.
Maximum number of different elements in a symmetric matrix is n(n + 1)
2
(ii) Skew-symmetric matrix : A square matrix A = [aij]is called skew- symmetric matrix if aij = −aji for all i, j
0 h g
or AT = − A . Example : − h
0 f
− g − f 0
Note : All principal diagonal elements of a skew- symmetric matrix are always zero because for any
diagonal element. aij = −aij ⇒ aij = 0
Trace of a skew symmetric matrix is always 0.
Properties of symmetric and skew-symmetric matrices:
(i) If A is a square matrix, then A + AT , AAT , AT A are symmetric matrices, while A − AT is skew- symmetric matrix.
(ii) If A is a symmetric matrix, then − A, KA, AT , An, A−1, BT AB are also symmetric matrices, where n ∈ N ,
K ∈ R and B is a square matrix of order that of A
(iii) If A is a skew-symmetric matrix, then
(a) A2n is a symmetric matrix for n ∈ N ,
(b) A2n+1 is a skew-symmetric matrix for n ∈ N ,
(c) kA is also skew-symmetric matrix, where k ∈ R ,
(d) BT AB is also skew- symmetric matrix where B is a square matrix of order that of A.
(iv) If A, B are two symmetric matrices, then
(a) A ± B, AB + BA are also symmetric matrices,
(b) AB − BA is a skew- symmetric matrix,
(c) AB is a symmetric matrix, when AB = BA .
(v) If A,B are two skew-symmetric matrices, then
(a) A ± B, AB − BA are skew-symmetric matrices,
(b) AB + BA is a symmetric matrix.
(vi) If A a skew-symmetric matrix and C is a column matrix, then CT AC is a zero matrix.
(vii) Every square matrix A can uniquelly be expressed as sum of a symmetric and skew-symmetric matrix i.e.
A = 1 (A + AT ) + 1 (A − AT ) .
2 2
(2) Singular and Non-singular matrix : Any square matrix A is said to be non-singular if | A|≠ 0, and a
square matrix A is said to be singular if | A|= 0 . Here | A|(or det(A) or simply det |A| means corresponding
determinant of square matrix A.
Example : A = 2 3 then| A|= 2 3 = 10 − 12 = −2 ⇒ A is a non singular matrix.
4 5 4 5
(3) Hermitian and skew-Hermitian matrix : A square matrix A = [aij ] is said to be hermitian matrix if
aij = a ji ∀i. j i.e. A = Aθ . Example : a b + ic , 3 4i 3 − 4i 5 + 2i
b − ic d 3 + 2i 5 − 2 + i are Hermitian matrices.
5 −
−2−i 2
Note : If A is a Hermitian matrix then aii = aii ⇒ aii is real ∀i, thus every diagonal element of a
Hermitian matrix must be real.
A Hermitian matrix over the set of real numbers is actually a real symmetric matrix and a square
matrix, A=|aij| is said to be a skew-Hermitian if aij = −a ji .∀i, j i.e. Aθ = − A .
Example : 0 − 2+ i , 3i − 3 + 2i −1−i
2 − 0 3 + 2i − 2i − 2 − 4i are skew-Hermitian matrices.
i 1 − i 2 − 4i
0
If A is a skew-Hermitian matrix, then aii = −aii ⇒ aii + aii = 0 i.e. aii must be purely imaginary or zero.
A skew-Hermitian matrix over the set of real numbers is actually a real skew-symmetric matrix.
(4) Orthogonal matrix : A square matrix A is called orthogonal if AAT = I = AT A i.e. if A−1 = AT
Example : A = cosα − sinα is orthogonal because A −1 = cosα sinα = AT
− sinα cosα − sinα
cosα
In fact every unit matrix is orthogonal.
(5) Idempotent matrix : A square matrix A is called an idempotent matrix if A2 = A .
Example : 1 / 2 1 / 2 is an idempotent matrix, because A2 = 1 / 4 +1/ 4 1/ 4 +1/ 4 = 1 / 2 1/ 2 = A.
1 / 2 1 / 2 1 / 4 +1/ 4 1/ 4 +1/ 4 1 / 2 1/ 2
Also, A = 1 0 and B= 0 0 are idempotent matrices because A2 = A and B2 = B.
0 0 0 1
In fact every unit matrix is indempotent.
(6) Involutory matrix : A square matrix A is called an involutory matrix if A2 = I or A−1 = A
Example : A = 1 0 is an involutory matrix because A2 = 1 0 = I
0 1 0 1
In fact every unit matrix is involutory.
(7) Nilpotent matrix : A square matrix A is called a nilpotent matrix if there exists a p ∈ N such that A p = 0
Example : A = 0 0 is a nilpotent matrix because A2 = 0 0 = 0 (Here P = 2)
1 0 0 0
(8) Unitary matrix : A square matrix is said to be unitary, if A' A = I since | A′|=| A|and
| A ' A|=| A '|| A|therefore if A′ A=I, we have | A '|| A|= 1
Thus the determinant of unitary matrix is of unit modulus. For a matrix to be unitary it must be non-singular.
Hence A′ A = I ⇒ A A′ = I
(9) Periodic matrix : A matrix A will be called a periodic matrix if Ak+1 = A where k is a positive integer. If,
however k is the least positive integer for which Ak+1 = A, then k is said to be the period of A.
(10) Differentiation of a matrix : If A= f(x) g(x) then dA = f ′(x) g′(x) is a differentiation of matrix A.
h(x) dx h′(x)
l(x) l′(x)
Example : If A = x2 sin x then dA = 2x cos x
2x 2 dx
2 0
(11) Submatrix : Let A be m×n matrix, then a matrix obtained by leaving some rows or columns or both, of
2 1 0 2 1 0 − 1
A is called a sub matrix of A. Example : If A' = 3 2 and 2 2 3
2 2 3 5 3 are sub matrices of matrix A = 2 2 2 4
5 5 3
1
(12) Conjugate of a matrix : The matrix obtained from any given matrix A containing complex number as
its elements, on replacing its elements by the corresponding conjugate complex numbers is called conjugate of A
1 + 2i 2 − 3i 3 + 4i 1 − 2i 2 + 3i 3 − 4i
and is denoted by A . Example : A = 4 − 5i 5 + 6i 6 − 7i then A = 4 + 5i 5 − 6i
7 + 8i 7 − 8i 6 + 7i
8
7 8 7
Properties of conjugates :
(i) (A) = A (ii) (A + B) = A + B
(iii) (αA) = αA,α being any number (iv) (AB) = A B, A and B being conformable for multiplication.
(13) Transpose conjugate of a matrix : The transpose of the conjugate of a matrix A is called transposed
conjugate of A and is denoted by Aθ . The conjugate of the transpose of A is the same as the transpose of the
conjugate of A i.e. (A′) = (A)′ = Aθ .
If A = [aij ]m×n then Aθ = [bji ]n×m where b ji = aij i.e. the ( j,i)th element of Aθ = the conjugate of (i, j)th element of A.
1 + 2i 2 − 3i 3 + 4i 1 − 2i 4 + 5i 8
Example : If A = 4 − 5i = 2 + 3i 5 − 6i 7 − 8i
5 + 6i 6 − 7i , then Aθ 6 + 7i
7
8 7 + 8i 7 3 − 4i
Properties of transpose conjugate
(i) (Aθ )θ = A (ii) (A + B)θ = Aθ + Bθ (iii) (kA)θ = KAθ , K being any number (iv) (AB)θ = Bθ Aθ
0 5 − 7
The matrix − 5
Example: 16 0 11 is known as [Karnataka CET 2000]
Solution: (b)
Example: 17 7 − 11 0 (d) Diagonal matrix
Solution: (a)
(a) Upper triangular matrix (b) Skew-symmetric matrix (c) Symmetric matrix [Kurukshetra CEE 2002]
Example: 18
Solution: (a) In a skew-symmetric matrix, aij = −a ji ∀ i, j = 1,2,3 and j = i , aii = −aii ⇒ each aii =0 (d) λ ≠ −3
Example: 19 Hence the given matrix is skew-symmetric matrix [ AT = − A] .
Solution: (b)
2 λ − 4
The matrix − 1 3
4 is non singular if
1 − 2 − 3
(a) λ ≠ −2 (b) λ ≠ 2 (c) λ ≠ 3
2 λ − 4
The given matrix A = − 1 3
4 is non singular If |A| ≠ 0
1 − 2 − 3
2 λ −4 1 λ+3 0
⇒ | A|= − 1 3 4 ≠ 0 ⇒ − 1 3 4 ≠ 0 [R1 → R1 + R2 ]
1 −2 −3 1 −2 −3
1 λ+3 0 R2 → R2 + R3
⇒ | A| = 0 1 R3 →
1 ≠ 0 , R3 − R1
0 −λ −5
− 3
⇒ 1(−3 + λ + 5) ≠ 0 ⇒ λ + 2 ≠ 0 ⇒ λ ≠ −2
1 1 2 2
3 1 − 2 is
The matrix A= 2 2 − 1 [Kurukshetra CEE 2002]
− 2 (d) Nilpotent
(a) Orthogonal (b) Involutary (c) Idempotent
1 122
3
Since for given A= 2 1 − 2 . For orthogonal matrix AAT = AT A = I(3×3)
−2 2
−1
1 1 2 2 1 2 − 2 1 9 0 0
3 1 − 2 2 1 3 0 9
⇒ AA T = 2 2 − 1 2 −2 2 = 0 0 0 = 3I . Similarly AT A = 3I . Hence A is orthogonal
− 2 − 1 9
If A = 4 3 x + 2 is symmetric, then x = [Karnataka CET 1994]
2x − x + 1
(d) 4
(a) 3 (b) 5 (c) 2
For symmetric matrix, A= AT ⇒ 4 2x − 3 = 4 x + 2 ⇒ 2x − 3 = x+2 ⇒ x =5
2x − x + 1
x + 2 x +1 3
Example: 20 If A and B are square matrices of order n×n, then (A − B)2 is equal to [Karnataka CET 1999; Kerala (Engg.) 2002]
(a) A2 − B2 (b) A2 − 2AB + B2 (c) A2 + 2AB + B2 (d) A2 − AB − BA + B2
Solution: (d) Given A and B are square matrices of order n×n we know that (A − B)2 = (A − B)(A − B) = A2 − AB − BA + B2
Example: 21 If A = cos θ − sinθ , then which of the following statement is not correct [DCE 2001]
cos θ
sin θ (d) A is not invertible
(a) A is orthogonal matrix (b) AT is orthogonal matrix (c) Determinant A = 1
Solution: (d) | A|= 1 ≠ 0, therefore A is invertible. Thus (d) is not correct
Example: 22 Matrix A is such that A2 = 2A − I where I is the identity matrix. Then for n ≥ 2, An = [EAMCET 1992]
(a) nA − (n − 1)I (b) nA − I (c) 2n−1 A − (n − 1)I (d) 2n−1 A − I
Solution: (a) We have, A2 = 2A − I ⇒ A2.A = (2A − I)A ; A3 = 2A2 − IA = 2[2A − I] − IA ⇒ A3 = 3A − 2I
Similarly A4 = 4 A − 3I and hence An = nA − (n − 1)I
0 0 − 1
Example: 23 Let A = 0 −1 0 , the only correct statement about the matrix A is [AIEEE 2004]
− 1 0 0
(a) A2 = I (b) A = (−1)I , where I is unit matrix
(c) A−1 does not exist (d) A is zero matrix
0 0 − 1 0 0 − 1 1 0 0
0 0 = I . Also, A−1 exists as | A|= 1
Solution: (a) A2 = A. A = 0 −1 0 0 −1 = 0 1
− 1 0 0 − 1 0 0 0 0 1
Adjoint of a Square Matrix.
Let A = [aij ] be a square matrix of order n and let Cij be cofactor of aij in A. Then the transpose of the matrix
of cofactors of elements of A is called the adjoint of A and is denoted by adj A
Thus, adj A = [Cij ]T ⇒ (adj A)ij = C ji = cofactor of a ji in A.
a11 a12 a13 C11 C12 C13 T C11 C21 C31
If A = a21 a23 , then adj A = C21 = C12 C32 ;
a 22 C22 C23 C22
a31 a32 a33 C31 C32 C33 C13 C23 C33
Where Cij denotes the cofactor of aij in A.
Example : A = p qs , C11 = s, C12 = −r, C21 = −q, C22 = p
r
∴ adj A = s − rT = s − q
− q − r
p p
Note : The adjoint of a square matrix of order 2 can be easily obtained by interchanging the diagonal
elements and changing signs of off diagonal elements.
Determinants-and-matrices
Properties of adjoint matrix : If A, B are square matrices of order n and In is corresponding unit matrix,
then (i) A(adj A) =| A| In = (adj A)A (Thus A (adj A) is always a scalar matrix)
(ii) |adj A|=| A|n−1 (iii) adj (adj A) =| A|n−2 A
(iv) | adj (adj A)|=| A|(n−1)2 (v) adj (AT ) = (adj A)T
(vi) adj (AB) = (adj B)(adj A) (vii) adj(Am ) = (adj A)m,m ∈ N
(viii) adj(kA) = k n−1(adj A),k ∈ R (ix) adj (In ) = In
(x) adj (O) = O (xi) A is symmetric ⇒ adj A is also symmetric.
(xii) A is diagonal ⇒ adj A is also diagonal. (xiii) A is triangular ⇒ adj A is also triangular.
(xiv) A is singular ⇒ |adj A|= 0
Inverse of a Matrix.
A non-singular square matrix of order n is invertible if there exists a square matrix B of the same order such
that AB = In = BA .
In such a case, we say that the inverse of A is B and we write A−1 = B
The inverse of A is given by A −1 = | 1 |.adj A
A
The necessary and sufficient condition for the existence of the inverse of a square matrix A is that | A|≠ 0
Properties of inverse matrix:
If A and B are invertible matrices of the same order, then
(i) (A−1 )−1 = A (ii) (AT )−1 = (A−1 )T
(iii) (AB)−1 = B−1 A−1 (iv) (Ak )−1 = (A−1)k ,k ∈ N [In particular (A2 )−1 = (A−1)2 ]
(v) adj(A−1) = (adj A)−1
(vi) | A −1 |= | 1 | =| A |−1
A
(vii) A = diag (a1a2 ...an ) ⇒ A −1 = diag (a1−1 a −1 ...a −1 ) (viii) A is symmetric ⇒ A−1 is also symmetric.
2 n (x) A is scalar matrix ⇒ A−1 is also scalar matrix.
(ix) A is diagonal, | A|≠ 0 ⇒ A−1 is also diagonal.
(xi) A is triangular, | A|≠ 0 ⇒ A−1 is also triangular. (xii) Every invertible matrix possesses a unique inverse.
Note : (Cancellation law with respect to multiplication)
If A is a non singular matrix i.e., if | A|≠ 0 ,then A−1 exists and AB = AC ⇒ A−1(AB) = A−1(AC)
⇒ (A−1 A)B = (A−1 A)C ⇒ IB = IC ⇒ B = C ∴ AB = AC ⇒ B = C ⇔| A|≠ 0
Example: 24 If A = 4 2 , then | adj A |is equal to [UPSEAT 2003]
Solution: (b) 3 4
Example: 25 (d) None of these
(a) 16 (b) 10 (c) 6
adj A = 4 −2
− 3
4
| adj A|= 4 −2 = 16 − 6 = 10
−3 4
If 3, – 2 are the Eigen values of non-singular matrix A and |A| = 4 . Then Eigen values of adj (A) are
(a) 3/4, –1/2 (b) 4/3, –2 (c) 12, –8 [Kurukshetra CEE 2002]
(d) –12, 8
Solution: (b) Since A−1 = adj A and if λ is Eigen value of A then λ−1 is Eigen value of A−1 , thus for adj (A) x = (A−1x)| A|
| A|
Example: 26
Solution: (d) =| A|.λ−1I
Example: 27
Solution: (b) adj(A) corresponding to Eigen value
Example: 28
Solution: (a) λ = 3 is = 4/3 and for λ = −2 is = 4 /− 2 = –2
Example: 29 3 2 4 A−1 = 1 adj (A) , then K is [UPSEAT 2002]
Solution: (d) If matrix A = 1 2 − 1 and K
Example: 30 1
0 1
(a) 7 (b) –7 (c) 1/7 (d) 11
We know that A−1 = adj(A) . We have A−1 = 1 adj(A) i.e. K =| A|
| A| K
32 4
and K = 1 2 − 1 = 3(3) − 2(1) + 4(1) = 9 − 2 + 4 = 11
01 1
The inverse of matrix A = a b is
c d
(a) d −b (b) 1 d −b (c) 1 1 0 (d) b −a [AMU 2001]
− c ad − bc − c | A| 0 1 d − c [AIEEE 2004]
a a
Here | A|= a b = ad − bc , adj (A) = d −b . Hence A−1 = 1 d −b
c d − c ad − bc − c
a a
1 − 1 1 4 2 2
Let A = 2 − 3 and 10.B = − 5
1 0 α . If B is the inverse of matrix A, then α is
1 1 1 1 − 2 3
(a) 5 (b) –1 (c) 2 (d) –2
1 −1 1 4 2 2
We have, A = − 2 1 − 3 , ∴ | A|= 1(4) + 1(5) + 1(1) = 10 and adj (A) = − 5 0 5
1 1 1 1 − 2 3
Then A−1 = 1 4 2 2
10 − 5 0 5
1 −2 3
According to question, B is the inverse of matrix A. Hence α = 5
1 0 − K
Matrix A = 2 1 3 is invertible for [UPSEAT 2002]
K 0 1
(a) K = 1 (b) K = −1 (c) K = 0 (d) All real K
1 0 −K
For invertible, | A|≠ 0 i.e., 2 1 3 ≠ 0
K0 1
⇒ 1(1) − K(−K) ≠ 0 ⇒| A|= K 2 + 1 ≠ 0 , which is true for all real K . [Orissa JEE 2002]
Let p be a non-singular matrix,
1 + p + p2 + ....... + pn = 0 (0 denotes the null matrix), then p−1 =
(a) pn (b) − pn (c) − (1 + p + ..... + pn) (d) None of these
Solution: (a) We have, 1 + p + p2 + − − − + pn = 0
Example: 31
Solution: (a) Multiplying both sides by p−1 , p−1 + I + Ip + ..... + pn−1I = 0.p−1
Example: 32 p−1 + I(1 + p + .... + pn−1) = 0 ⇒ p−1 = −I(1 + p + p2 + .... + pn−1) = − (−pn) = pn .
Solution: (b)
Example: 33 cos α − sin α 0
Solution: (a) cos α 0 , where α ∈ R ,then [ f(α)]−1 is equal to
Example: 34 Let f (α ) = sin α 1 [AMU 2000]
Solution: (a) 0
Example: 35 0
Solution: (d) (a) f(−α) (b) f(α −1) (c) f(2α) (d) None
Example: 36 cos α − sin α 0 cos α sin α 0
| f(α)|= sin α cos α 0 =1 , adj of f(α) = − sin α cos α 0
1 1
0 0 0 0
cos α sin α 0 cos α sin α 0
[ f(α)]−1 = − sin α cos α 0 ......(i) and f (−α) = − sin α cos α 0 ......(ii)
0 01 0 01
From (i) and (ii), [f(α ]) −1 = f[−α]
If I is a unit matrix of order 10, then the determinant of I is equal to [Kerala (Engg.) 2002]
(a) 10 (b) 1 (c) 1/10 (d) 9
Determinant of unit matrix of any order =1.
If A = α 2 and | A3 |= 125 then α = [IIT Screening 2004]
2 α
(a) ± 3 (b) ± 2 (c) ± 5 (d) 0
125 =| A3 |=| A|3 ⇒| A|= 5 and | A|= α 2 − 4 = 5 ⇒ α 2 = 9 ⇒ α = ± 3
If |A| denotes the value of the determinant of the square matrix A of order 3, then | −2A|= [MP PET 1987, 89, 92, 2000]
(a) −8| A| (b) 8| A| (c) −2| A| (d) None of these
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 |−2A|= (−2)3| A|= −8| A|.
3 −1+ x 2
−1 x + 2 singular
For how many value (s) of x in the closed interval [–4, –1] is the matrix 3 −1
2
x + 3
[Karnataka CET 2002]
(a) 2 (b) 0 (c) 3
(d) 1
3 x −1 2
3 −1 x + 2 = 0
x +3 −1 2
0 x −x [R1 → R1 − R2] , 0 x −x [R2 → R2 − R3]
3 −1 x + 2 = 0 −x 0 x =0
x +3 −1 2 x +3 −1 2
0 0 −x [C2 → C2 + C3]
−x x x =0
x+3 12
−x[(−x) − x(x + 3)] = 0 ⇒ x(x2 + 4 x) = 0 ⇒ x = 0, − 4
Hence only one value of x in closed interval [–4,–1] i.e. x = −4
Inverse of diagonal matrix (if it exists) is a
(a) Skew-symmetric matrix (b) Diagonal matrix (c) Non invertible matrix (d) None of these
Solution: (b) Let A = diag(d1, d2, d3......, dn)
As A is invertible, therefore det(A) ≠ 0 ⇒ d1, d2, d3......., dn ≠ 0 ⇒ di ≠ 0 for i = 1, 2, 3…..n
Here, cofactor of each non diagonal entry is 0 and cofactor of aii
= (−1)i+1 det [diag(d1, d2, d3....., di−1, di+1....., dn)] = d1, d2, d3.....di−1.di+1,......, dn = 1 [d1, d2, d3 ....., di −1, di , di+1....., dn] = | A |
di di
A−1 = | 1 |(adj A) = diag 1 , 1 ......., 1 , which is a diagonal matrix
A d1 d2 dn
Elementary Transformations or Elementary Operations of a Matrix.
The following three operations applied on the rows (columns) of a matrix are called elementary row (column)
transformations
(1) Interchange of any two rows (columns)
If ith row (column) of a matrix is interchanged with the ith row (column), it will be denoted by
Ri ↔ R j (Ci ↔ C j )
2 1 3 2 1 3
Example : A = − 1 1 , 2 4
2 then by applying R2 ↔ R3 , we get B = 3
3 2 4 − 1 2 1
(2) Multiplying all elements of a row (column) of a matrix by a non-zero scalar
If the elements of ith row (column) are multiplied by a non-zero scalar k, it will be denoted by
Ri → Ri(k) , [Ci → Ci(k)] or Ri → kRi, [Ci → kCi]
3 2 − 1 3 2 − 1
If A = 0 1 2 , then by applying R2 → 3R2 we obtain B = 0 3 6
− 1 2 − 3 − 1 2 − 3
(3) Adding to the elements of a row (column), the corresponding elements of any other row (column) multiplied
by any scalar k. If k times the elements of jth row (column) are added to the corresponding elements of the ith row
( )(column), it will be denoted by Ri → Ri + kR j Ci → Ci + kC j
2 1 3 1
If A = − 1 − 1 0 2, then the application of elementary operation R3 → R3 + 2R1 gives the matrix
0 1 3 1
2 1 3 1
B = − 1 − 1 0 2, If a matrix B is obtained from a matrix A by one or more elementary transformations,
4 3 9 3
then A and B are equivalent matrices and we write A ~ B
1 2 3 4 1 2 3 4
3, then − 1 1 − 1, applying R2 → R2 + (−1)R1
Let A= 2 1 4 A ~ 1
3 1 2 4 3 1 2 4
1 2 3 1
~ 1 − 1 1 − 2 , applying C4 → C4 + (−1)C3
1 1 2 2
An elementary transformation is called a row transformation or a column transformation according as it is
applied to rows or columns.
Elementary Martix.
A matrix obtained from an identity matrix by a single elementary operation (transformation) is called an
1 3 0 0 0 1 1 0 0
elementary matrix. Example : 0 1 0 0 1 0, 0 0 1 are elementary matrices obtained from I 3 by
0 0 1 1 0 0 0 1 0
subjecting it to the elementary transformations R1 → R1 + 3 R2, C1 ↔ C3 and R2 ↔ R3 respectively.
Theorem 1 : Every elementary row (column) transformation of an m×n matrix (not identity matrix) can be
obtained by pre-multiplication (post- multiplication) with the corresponding elementary matrix obtained from the
identity matrix Im (In ) by subjecting it to the same elementary row (column) transformation.
Theorem 2 : Let C = AB be a product of two matrices. Any elementary row (column) transformation of AB
can be obtained by subjecting the pre-factor A (post factor B) to the same elementary row (column) transformation.
Method of finding the inverse of a matrix by elementary transformations : Let A be a non singular
matrix of order n. Then A can be reduced to the identity matrix In by a finite sequence of elementary transformation
only. As we have discussed every elementary row transformation of a matrix is equivalent to pre-multiplication by
the corresponding elementary matrix. Therefore there exist elementary matrices E1, E2.....Ek such that
(Ek Ek−1... E2 E1 )A = I n
⇒ (Ek Ek−1...E2 E1 )AA−1 = In A−1 (post multiplying by A−1 )
⇒ (Ek Ek−1...E2 E1 ) In = A−1 ( In A−1 = A−1and AA−1 = In ) ⇒ A−1 = (Ek Ek−1...E2 E1 ) In
Algorithm for finding the inverse of a non singular matrix by elementary row transformations
Let A be non- singular matrix of order n
Step I : Write A = In A
Step II : Perform a sequence of elementary row operations successively on A on the LHS and the pre factor
In on the RHS till we obtain the result In = BA
Step III : Write A−1 = B
Note : The following steps will be helpful to find the inverse of a square matrix of order 3 by using
elementary row transformations.
Step I : Introduce unity at the intersection of first row and first column either by interchanging two rows or by
adding a constant multiple of elements of some other row to first row.
Step II : After introducing unity at (1,1) place introduce zeros at all other places in first column.
Step III Introduce unity at the intersection of 2nd row and 2nd column with the help of 2nd and 3rd row.
Step IV : Introduce zeros at all other places in the second column except at the intersection of 2nd row and
2nd column.
Step V : Introduce unity at the intersection of 3rd row and third column.
Step VI : Finally introduce zeros at all other places in the third column except at the intersection of third row
and third column.
Example: 37 3 − 1 − 2
Using elementary row transformation find the inverse of the matrix A = 2 0 − 1
3 − 5 0
− 5 / 8 5/4 1/8 1 5 −5 1 1 5 5 1
(a) − 3 / 8 3/4 8 3 −3 1 8 6
3/2 − 1 / 8 (b) 0 1 (c) 3 − 12 − 1 (d) None of these
− 5 / 4 3
1 / 4 10 2
Solution: (a) 3 − 1 − 2 1 0 0
We have A=IA ⇒ 2 0 − 1 = 0 1 0 A
3 − 5 0 0 0 1
1 − 1 − 1 1 − 1 0
Applying (R1 → R1 − R2) 2 0 − 1 = 0 1 0 A
3 − 5 0 0 0 1
1 − 1 − 1 1 − 1 0
Applying R2 → R2 − 2R1 and R3 → R3 − 3R1 , 0 2 − 1 = − 2 3 0 A
0 − 2 3 − 3 3 1
1 − 1 − 1 1 −1 0
Applying R2 → R2 / 2, 0 2 3 / 2 0 A
1 1 / = − 1
3 1
0 − 2 3 − 3
1 0 − 1 / 2 0 1/ 2 0
Applying R1 → R1 + R2 and R3 → R3 + 2R2 , 0 3/2 0 A
1 1/ 2 = − 1 6 1
0
0 4 − 5
1 0 − 1 / 2 0 1/ 2 0
0 3/2
Applying R3 → R3 / 4 , 0 1 1 / 2 = − 1 6/4 0 A
0 1 5 / 4 1 / 4
1 1 1 0 0 − 5 / 8 5/4 1/ 8
2 2 0 1 − 3/4 − 1 / 8 A
Applying R1 → R1 + R3 and R2 → R2 − R3 , 0 0 0 = 3 / 8 3/2
1 / 4
1 − 5 / 4
− 5 / 8 5 / 4 1/ 8
A−1 = − 3 / 8 3 / 4 − 1 / 8
− 5 / 4 3 / 2 1 / 4
Rank of Matrix .
Definition : Let A be a m×n matrix. If we retain any r rows and r columns of A we shall have a square sub-
matrix of order r. The determinant of the square sub-matrix of order r is called a minor of A order r. Consider any
13 4 5
matrix A which is of the order of 3×4 say, A = 1 2 6 7 . It is 3×4 matrix so we can have minors of order 3, 2
15 0 1
134
or 1. Taking any three rows and three columns minor of order three. Hence minor of order 3 = 1 2 6 = 0
15 0
Making two zeros and expanding above minor is zero. Similarly we can consider any other minor of order 3
and it can be shown to be zero. Minor of order 2 is obtained by taking any two rows and any two columns.
Minor of order 2= 1 3 = 2 − 3 = −1 ≠ 0 . Minor of order 1 is every element of the matrix.
1 2
Rank of a matrix: The rank of a given matrix A is said to be r if
(1) Every minor of A of order r+1 is zero
(2) There is at least one minor of A of order r which does not vanish
Note : If a minor of A is zero the corresponding submatrix is singular and if a minor of A is not zero then
corresponding submatrix is non-singular.
Here we can also say that the rank of a matrix A is said to be r if
(i) Every square submatrix of order r+1 is singular.
(ii) There is at least one square submatrix of order r which is non-singular.
The rank r of matrix A is written as ρ(A) = r
Working rule : Calculate the minors of highest possible order of a given matrix A. If it is not zero, then
the order of the minor is the rank. If it is zero and all other minors of the same order be also zero, then calculate
minor of next lower order and if at least one of them is not zero then this next lower order will be the rank. If,
however, all the minors of next lower orders are zero, then calculate minors of still next lower order and so on.
Note : The rank of the null matrix is not defined and the rank of every non-null matrix is greater than or
equal to 1.
The rank of a singular square matrix of order n cannot be n.
Echelon form of a Matrix.
A matrix A is said to be in Echelon form if either A is the null matrix or A satisfies the following conditions:
(1) Every non- zero row in A precedes every zero row.
(2) The number of zeros before the first non-zero element in a row is less than the number of such zeros in the
next row.
It can be easily proved that the rank of a matrix in Echelon form is equal to the number of non-zero row of the
0 3 2 1
matrix. Example : The rank of the matrix A = 0 0 2 5 is 2 because it is in Echelon form and it has two non-
0 0 0 0
0 2 5
zero rows. The matrix A = 0 0
1 is not in Echelon form, because the number of zeros in second row is not less
0 0 − 4
than the number of zeros in the third row. To reduce A in the echelon form, we apply some elementary row
0 2 5
transformations on it. Applying R3 → R3 + 4R2, we obtain A ~ 0 0 1 , which is in Echelon form and contains
0 0 0
2 non zero rows. Hence, r(A) = 2
Rank of a matrix in Echelon form : The rank of a matrix in Echelon form is equal to. the number of non-
zero rows in that matrix.
Algorithm for finding the rank of a matrix : Let A = [aij ] be an m×n matrix.
Step I : Using elementary row transformations make a11 = 1
Step II : Make a21, a31,...., am1 all zeros by using elementary transformations,
R2 → R2 − a21R1, R3 → R3 − a31R1,.....Rm → Rm − am1 R1
Step III : Make a22 = 1 by using elementary row transformations.
Step IV : Make a32 , a42 ,...., am2 all zeros by using R3 → R3 − a32 R2 , R4 → R4 − a42 R2 ,...Rm → Rm − am2 R2
The process used in steps III and IV is repeated upto (m − 1)th row. Finally we obtain a matrix in Echelon
form, which is equivalent to the matrix A. The rank of A will be equal to the number of non-zeros rows in it.
Example: 38 2 3 1 4 [Kurukshetra CEE 2002 ]
Solution: (a) The rank of the matrix A = 0 1 2 − 1 is
2 (d) Indeterminate
Example: 39 0 − 2 − 4
Solution: (b,d)
(a) 2 (b) 3 (c) 1
Example: 40
Solution: (c) 2 3 1 4 2 3 1
We have A = 0 1 2 − 1 , Considering 3×3 minor 0 1 2 its determinant is 0.
0 − 2 − 4 2 3×4 0 − 2 − 4 3×3
2 3 4 2 1 4 3 1 4
Similarly considering , 0 1 0 2 − 1 − 1 , their determinant is 0 each rank can not be 3
−2 − 1 , −4 and 1 2
0 2
2 0 2 − 2 − 4
Then again considering a 2×2 minor, 2 − 3 , which is non zero. Thus, rank =2
0 2
− 1 2 5
−4 a − 4
The rank of the matrix 2 −2 a + 1 is [Roorkee 1988]
1
(a) 1 if a = 6 (b) 2 if a =1 (c) 3 if a = 2 (d) 1 if a = – 6
− 1 2 5 0 0 a+6 0 0 0
−4
Let A = 2 −2 a − 4 = 0 0 − a − 6 = 0 0 − a − 6
1 a + 1 1 − 2 a + 1 1 − 2 a + 1
00 0
When a = −6 , A = 0 0 0 , ∴ r(A) = 1
1 −2 −5
00 0 00 0 ,∴ r(A) = 2
When a = 1 , A = 0 0 − 7 ∴r (A) = 2 , When a = 6 , A = 0 0 − 12
1 −2 2 1 −2 7
00 0 , ∴ r(A) = 2
When a = 2 , A = 0 0 − 8
1 −2 3
x + a b c
x+b
The value of x so that the matrix a c has rank 3 is
b
a x + c
(a) x ≠ 0 (b) x = a + b + c
(c) x ≠ 0 and x ≠ −(a + b + c) (d) x = 0, x = a + b + c
Since rank is 3, | A|3×3 ≠ 0 , x+a b c
a x+b c ≠0
a x + c 3×3
b
x+a+b+c b c , Applying (C1 → C1 + C2 + C3)
x+a+b+c x+b c ≠0
x+a+b+c x+c
b
1 b c1 b c
(x + a + b + c ) 1 x+b c ≠ 0 ⇒ x +a+b+c ≠ 0, 1 x+b c ≠0
x+c 1 x+c
1 b b
0 −x c
x ≠ −(a + b + c) , 0 x − x ≠ 0 ⇒ x ≠ 0
1 b x+c
System of Simultaneous Linear Equations .
Consider the following system of m linear equations in n unknowns
a11 x1 + a12 x 2 + ... + a1n xn = b1
a21 x1 + a22 x 2 + ... + a2n xn = b2
... ... ... ... ...
... ... ... ... ...
am1 x1 + am2 x 2 + ... + amn xn = bm
The system of equations can be written in matrix form as a11 a12 ... a1n x1 = bb12 or AX = B,
a 21 a 22 ... a 2n x2
am 1 bn
... amn
am2 xn
Where A = a11 a12 ... a1n , X = x1 and B = b1
a21 a22 ... a2n x2 b2
am 1 bm
... amn m×n n×1 m×1
am2 xn
The m×n matrix A is called the coefficient matrix of the system of linear equations.
(1) Solution : A set of values of the variables x1, x 2......xn which simultaneously satisfy all the equations is
called a solution of the system of equations. Example : x = 2, y = −3 is a solution of the system of linear equations
3x + y = 3, 2x + y = 1, because 3(2) +(–3)=3 and 2(2)+(–3)=1
(2) Consistent system : If the system of equations has one or more solutions, then it is said to be a consistent
system of equations, otherwise it is an inconsistent system of equations. Example : the system of linear equation
2x + 3y = 5, 4 x + 6y = 10 is consistent, because x=1, y =1 and x = 2, y = 1 / 3 are solutions of it.
However, the system of linear equations 2x + 3y = 5, 4 x + 6y = 10 is inconsistent, because there is no set of
values of x, y which satisfy the two equations simultaneously.
(3) Homogeneous and non-homogeneous system of linear equations: A system of equations AX=B is
called a homogeneous system if B = 0 . Otherwise, it is called a non-homogeneous system of equations.
Example : The system of equations, 2x + 3y = 0, 3x − y = 5 is a homogeneous system of linear equations
whereas the system of equations given by 2x + 3y = 1, 3x − y = 5 is a non homogeneous system of linear equations.
Solution of a Non Homogeneous System of Linear Equations.
There are three methods of solving a non homogeneous system of simultaneous linear equations.
(1) Determinant Method (Cramer's Rule) (2) Matrix method (3) Rank method
We have already discussed the determinant method (Cramer's rule) in chapter determinants.
(1) Matrix method : Let AX = B be a system of n linear equations with n unknowns. If A is non-singular,
then A−1 exists. ∴ AX = B ⇒ A−1(AX) = A−1B , [pre-multiplying by A−1 ]
⇒ (A−1 A)X = A−1B , [by associativity]
⇒ In X = A−1B ⇒ X = A−1B . Thus, the system of equations AX = B has a solution given by X = A−1B .
Now, let X1 and X 2 be two solutions of AX = B . then, AX1 = B and AX2 = B
⇒ AX1 = AX2 ⇒ A−1(AX1) = A−1(AX2) ⇒ (A−1 A)X1 = (A−1 A)X 2 ⇒ In X1 = In X 2 ⇒ X1 = X 2 .
Hence, the given system has a unique solution.
Thus, if A is a non-singular matrix, then the system of equations given by AX = B has a unique solution given
by X = A−1B .
If A is a singular matrix, then the system of equations given by AX=B may be consistent with infinitely many
solutions or it may be inconsistent also.
Criterion of consistency : Let AX = B be a system of n-linear equations in n unknowns.
(i) If | A|≠ 0, then the system is consistent and has a unique solution given by X = A−1B
(ii) If | A|= 0 and (adj A) B = 0 , then the system is consistent and has infinitely many solutions.
(iii) If | A|= 0 and (adj A) B ≠ 0 , then the system is inconsistent
Algorithm for solving a non-homogeneous system of linear equations : We shall give the algorithm
for three equations in three unknowns. But it can be generalized to any number of equations.
Let AX = B be a non-homogenous system of 3 linear equations in 3 unknowns. To solve this system of
equations we proceed as follows
Step I : Write the given system of equations in matrix form, AX = B and obtain A, B.
Step II : Find | A|
Step III : If | A|≠ 0 , then write "the system is consistent with unique solution". obtain the unique solution by
the following procedure. Find A−1 by using A−1 = | 1A|adj A obtain the unique solution given by X = A−1B
Step IV : If | A|= 0, then write "the system is either consistent with infinitely many solutions or it is
inconsistent. To distinguish these two, proceed as follows: Find (adj A) B.
If (adj A)B ≠ 0 , then write "the system is inconsistent”.
If (adj A)B = 0 , then the system is consistent with infinitely many solution. To find these solutions proceed as
follows. Put z = k (any real number) and take any two equations out of three equations. Solve these equations for x
and y. Let the values of x and y be λ and µ respectively. Then x = λ , y = µ z = k is the required solution, where
any two of λ, µ,k are functions of the third.
(2) Rank method : Consider a system of m simultaneous linear equations in n unknowns x1, x2...., xn , given
by a11 x1 + a12 x 2 + a13 x 3 + ....... + a1n x = b1
a21 x1 + a22 x 2 + a23 x 3 + ....... + a2n xn = b2
am1 x1 + am2 x 2 + am3 x 3 + .... + amn xn = bm
This system of equations can be written in matrix form as
a11 a12 a13 .... a1n x 1 = b1
a 21 a 22 a 23 .... a 2n x 2 b2
a am 2 am 3 am n xn bm
m1 ....
or AX = B ,where a11 a12 a13 .... a1n , X = x1 and B = b1
a 21 a 22 a 23 .... a 2n x2 b2
a am
m1 .... m×n n×1 m×1
am2 am3 amn xn
The matrix A is called the coefficient matrix and the matrix
a11 a12 a13 .... a1n b1
a 21 a 22 a 23 .... a2n b2
[A : B] = is called the augmented matrix of the given system of
a .... amn
m1
am2 am3 bm
equations. This matrix is obtained by adding (n + 1) column to A. The elements of this column are
b1, b2...., bm
For example, the augmented matrix of the system of equation
2x − y + 3z = 1
x + y − 2z = 5
x + y + z = −1 is
2 −1 3 1
1 1 − 2 5
1 1 1 − 1
A non-homogeneous system of linear equations may have a unique solution, or many solutions or no solution
at all. If it has a solution (whether unique or not) the system is said to be consistent. Otherwise it is called an
inconsistent system. The following theorems tell us about the condition for consistency of a system of linear
equations
Theorem 1 : The system of linear equations AX = B is consistent iff the rank of the augmented matrix
[A : B] is equal to the rank of the coefficient matrix A.
Theorem 2 : Let AX = B be a system of m simultaneous linear equations in n unknowns.
Case I : If m > n , then
(i) if r (A) = r (A : B) = n, then system of linear equations has a unique solution.
(ii) if r (A) = r (A : B) = r < n, then system of linear equations is consistent and has infinite number of solutions.
In fact, in this case (n − r)variables can be assigned arbitrary values.
(iii) if r (A) ≠ r (A : B),then the system of linear equations is inconsistent i.e. it has no solution.
Case II : If m < n and r (A) = r(A : B) = r, then r ≤ m < n and so from (ii) in case I, there are infinite number of
solutions.
Thus, when the number of equations is less than the number of unknowns and the system is consistent, then
the system of equations will always have an infinite number of solutions.
Algorithm for solving a non-homogeneous system AX=B of linear equations by rank method
Step I: Obtain A, B.
Step II : Write the Augmented matrix [A : B].
Step III : Reduce the augmented matrix to Echelon form by applying a sequence of elementary row-
operations.
Step IV : Determine the number of non-zero rows in A and [A : B] to determine the ranks of A and [A :B]
respectively.
Step V: If r (A) ≠ r(A : B)then write "the system is inconsistent" STOP else write "the system is consistent", go
to Step VI
Step VI : If r (A) = r (A : B) = number of unknowns, then the system has a unique solution which can be
obtained by back substitution.
If r (A) = r (A : B) < number of unknowns, then the system has an infinite number of solutions which can also
be obtained by back substitution.
Example: 41 If x + y 2x + z = 4 7 then values of x, y, z, w are [Rajasthan PET 2002]
2z + w 0 10
x − y
(a) 2, 2, 3, 4 (b) 2, 3, 1,2 (c) 3, 3, 0, 1 (d) None of these
Solution: (a) We have x + y 2x + z = 4 7
Example: 42 2z + w 0 10
Solution: (a) x − y
x + y = 4, 2x + z = 7 , x − y = 0 and 2z + ω = 10 ⇒ x = 2 and y = 2, z = 3, w = 4
The system of linear equation x + y + z = 2 , 2x + y − z = 3, 3x + 2y + kz = 4 has unique solution if [EAMCET 1994]
(a) K ≠ 0 (b) −1 < K < 1 (c) −2 < K < 2 (d) K = 0
11 1
The given system of equation has a unique solution if 2 1 − 1 ≠ 0 ⇒ K ≠ 0
32 k
Cayley-Hamilton Theorem .
Every matrix satisfies its characteristic equation e.g. let A be a square matrix then | A − xI |= 0 is the
characteristics equation of A. If x 3 − 4 x 2 − 5x − 7 = 0 is the characteristic equation for A, then
A3 − 4 A2 + 5A − 7I = 0
Roots of characteristic equation for A are called Eigen values of A or characteristic roots of A or latent roots of A.
If λ is characteristic root of A, then λ−1is characteristic root of A−1 .
Geometrical Transformations .
(1) Reflexion in the x-axis: If P' (x', y') is the reflexion of the point P(x, y)on the x-axis, then the matrix
1 0 describes the reflexion of a point P(x, y)in the x-axis.
0 − 1
(2) Reflexion in the y-axis : Here the matrix is − 1 0
1
0
(3) Reflexion through the origin : Here the matrix is − 1 0
− 1
0
(4) Reflexion in the line y = x: Here the matrix is 0 1
1 0
(5) Reflexion in the line y = – x : Here the matrix is 0 − 1
− 1
0
(6) Reflexion in y = x tan θ : Here matrix is cos 2θ sin 2θ
sin 2θ − cos 2θ
(7) Rotation through an angle θ : Here matrix is cosθ − sinθ
sinθ cosθ
Matrices of Rotation of Axes .
We know that if x and y axis are rotated through an angle θ about the origin the new coordinates are given
by x = X cosθ − Y sinθ and y = X sinθ + Y cosθ
⇒ x = cosθ − sinθ X ⇒ cosθ − sinθ is the matrix of rotation through an angle θ .
y cosθ Y cosθ
sinθ sinθ
1 1 3
3 − 3
Example: 43 Characteristic equation of the matrix A= 1
Solution: (a)
− 2 − 4 − 4
Example: 44
Solution: (b) (a) A3 − 20 A + 8I (b) A3 + 20 A + 8I (c) A3 − 80 A + 20I (d) None of these
Example: 45 The characteristic equation is | A − λI |= 0 .
Solution: (c)
1−λ 1 3
So, 1 3−λ − 3 = 0 i.e. λ3 − 20λ + 8 = 0
−4 −4−λ
−2
By cayley-Hamilton theorem , A3 − 20 A + 8I = 0
The transformation due to the reflection of (x, y) through the origin is described by the matrix
(a) 0 0 (b) −1 0 (c) 0 −1 (d) 1 0
0 1 − 1 − 1 0 1
0 0
If (x′, y′) is the new position
x′ = (−1)x + 0.y, y′ = 0.x + (−1)yd
∴ x′ = −1 0 x
y′ 1 y
0 −
∴ Transformation matrix is −1 0
− 1
0
The rotation through 1800 is identical to (d) Identity transformation
(a) The reflection in x -axis (b) The reflection in y-axis (c) A point reflection
Rotation through 1800 gives x′ = −x
y′ = −y . Hence this a point reflection.
Differentiation and application of derivatives
Introduction.
The rate of change of one quantity with respect to some another quantity has a great importance. For example,
the rate of change of displacement of a particle with respect to time is called its velocity and the rate of change of
velocity is called its acceleration.
The rate of change of a quantity ‘y’ with respect to another quantity ‘x’ is called the derivative or differential
coefficient of y with respect to x.
Derivative at a Point.
The derivative of a function at a point x=a is defined by f ′(a) = lim f (a + h) − f (a) (provided the limit exists
h
h→0
and is finite)
The above definition of derivative is also called derivative by first principle.
(1) Geometrical meaning of derivatives at a point: Consider the curve y = f(x) . Let f(x) be differentiable
at x = c. Let P(c, f(c)) be a point on the curve and Q (x, f(x)) be a neighbouring point on the curve. Then,
Slope of the chord PQ = f(x) − f (c) . Taking limit as Q→ P , i.e., x → c, y
x − c [c, f(c)]P
ψ Q[x, f(x)]
we get lim (slope of the chord PQ) = lim f(x) − f(c) ……(i) f(x) – f(c)
x−c
Q→P x→c
As Q → P , chord PQ becomes tangent at P.
Therefore from (i), we have x–c
Slope of the tangent at P = lim f(x) − f(c) = df(x) . ψ x
x−c dx x 0
x→c
= c
Note : Thus, the derivatives of a function at a point x = c is the slope of the tangent to curve, y = f(x) at
point (c, f(c)).
(2) Physical interpretation at a point : Let a particle moves in a straight line OX starting from O towards X.
Clearly, the position of the particle at any instant would depend upon the time elapsed. In other words, the distance
of the particle from O will be some function f of time t.
PQ X
O t = t0 t = t0 + h
Let at any time t = t0 , the particle be at P and after a further time h, it is at Q so that OP = f(t0) and
OQ = f(t0 + h) . Hence, the average speed of the particle during the journey from P to Q is PQ , i.e.,
h
f (t0 + h) − f (t0 ) = f(t0, h) . Taking the limit of f(t0, h) as h → 0 , we get its instantaneous speed to be
h
lim f (t 0 + h) − f(t0 ) , which is simply f ′(t0) . Thus, if f (t) gives the distance of a moving particle at time t, then the
h
h→0
derivative of f at t = t0 represents the instantaneous speed of the particle at the point P, i.e., at time t = t0 .
Important Tips
• dy is d (y) in which d is simply a symbol of operation and not ‘d’ divided by dx.
dx dx dx
• If f ′(x0) = ∞, the function is said to have an infinite derivative at the point x0. In this case the line tangent to the curve of y = f(x) at the
point x0 is perpendicular to the x-axis
Example: 1 If f(2) = 4, f ′(2) = 1, then lim xf(2) − 2 f(x) = [Rajasthan PET 1995, 2000]
Solution: (b) x−2
x→2 (d) – 2
(a) 1 (b) 2 (c) 3
Given f(2) = 4, f ′(2) = 1
∴ lim xf (2) − 2f(x) = lim xf (2) − 2 f(2) + 2 f(2) − 2 f (x) = lim (x − 2) f (2) − lim 2 f (x) − 2 f(2)
x −2 x−2 x −2 x − 2
x→2 x→2 x→2 x→2
= f(2) − 2 lim f(x) − f (2) = f(2) − 2 f ′(2) = 4 − 2(1) = 4 − 2 = 2
x− 2
x→2
Trick : Applying L-Hospital rule, we get lim f(2) − 2 f ′(2) = 2.
1
x→2
Example: 2 If f(x + y) = f(x).f(y) for all x and y and f(5) = 2, f ′(0) = 3, then f ′(5) will be
Solution: (c)
[IIT 1981; Karnataka CET 2000; UPSEAT 2002; MP PET 2002; AIEEE 2002]
Example: 3
Solution: (b) (a) 2 (b) 4 (c) 6 (d) 8
Let x = 5, y = 0 ⇒ f(5 + 0) = f(5).f(0)
⇒ f(5) = f(5)f(0) ⇒ f(0) = 1
Therefore, f ′(5) = lim f (5 + h) − f (5) = lim f(5)f(h) − f (5) = lim 2 f(h) − 1 { f(5) = 2}
h h h
h→0 h→0 h→0
= 2 lim f (h) − f (0) = 2 × f ′(0) = 2×3 = 6.
h
h→0
If f(a) = 3, f ′(a) = −2, g(a) = −1, g′(a) = 4, then lim g(x)f(a) − g(a)f(x) = [MP PET 1997]
x−a
x→a
(a) – 5 (b) 10 (c) – 10 (d) 5
lim g(x) f (a) − g(a) f (x) . We add and subtract g(a)f(a) in numerator
x − a
x→a
= lim g(x) f (a) − g(a) f (a) + g(a) f (a) − g(a) f (x) = lim f (a) g(x) − g(a) − lim g(a) f(x) − f (a)
x − a x − a x − a
x→a x→a x→a
= f (a) lim g(x) − g(a) − g(a) lim f(x) − f (a) = f(a)g' (a) − g(a)f ' (a) [by using first principle formula]
x − a x − a
x→a x→a
= 3.4 – (–1)(–2) = 12 – 2 = 10
Trick : lim g(x)f(a) − g(a)f(x)
x−a
x→a
Using L–Hospital’s rule, Limit = lim g' (x) f (a) − g(a) f ' (x) ;
1
x→a
Limit = g' (a) f(a) − g(a)f ' (a) = (4)(3) − (−1)(−2) = 12 – 2 = 10.
Example: 4 If 5 f(x) + 3 f 1 = x+2 and y = xf (x) then dy is equal to
Solution: (b) x dx
x =1
(a) 14 (b) 7 (c) 1 (d) None of these
8
5 f(x) + 3 f 1 = x + 2 ......(i)
x
Replacing x by 1 in (i), 5 f 1 + 3 f (x) = 1 + 2 ......(ii)
x x x
On solving equation (i) and (ii), we get, 16 f(x) = 5x − 3 + 4 , ∴ 16 f ′(x) = 5 + 3
x x2
y = xf(x) ⇒ dy = f(x) + xf ′(x) = 1 (5 x − 3 + 4) + x. 1 (5 + 3 )
dx 16 x 16 x2
at x = 1, dy = 1 (5 − 3 + 4) + 1 (5 + 3) = 7 .
dx 16 16 8
Some Standard Differentiation.
(1) Differentiation of algebraic functions
(i) d xn = nxn−1, x ∈ R,n ∈ R, x > 0 (ii) d ( x) = 1 (iii) d 1 = − n
dx dx 2x dx xn xn+1
(2) Differentiation of trigonometric functions : The following formulae can be applied directly while
differentiating trigonometric functions
(i) d sin x = cos x (ii) d cos x = − sin x (iii) d tan x = sec 2 x
dx dx dx
(iv) d sec x = sec x tan x (v) d cosec x = −cosec x cot x (vi) d cot x = −cosec 2x
dx dx dx
(3) Differentiation of logarithmic and exponential functions : The following formulae can be applied
directly when differentiating logarithmic and exponential functions
(i) d log x = 1 , for x > 0 (ii) d e x = ex
dx x dx
(iii) d a x = a x log a , for a > 0 (iv) d log a x = 1 , for x > 0, a> 0, a ≠ 1
dx dx log
x a
(4) Differentiation of inverse trigonometrical functions : The following formulae can be applied directly
while differentiating inverse trigonometrical functions
(i) d sin−1 x = 1 , for − 1 < x < 1 (ii) d cos−1 x = − 1 , for − 1 < x < 1
dx 1− x2 dx 1− x2
(iii) d sec −1 x = |x| 1 , for | x |> 1 (iv) d cosec −1x = −1 −1 , for | x |> 1
dx x2 −1 dx |x| x2
(v) d tan−1 x = 1 , for x∈R (vi) d cot −1 x = −1 , for x∈R
dx 1+ x2 dx 1+ x2
(5) Differentiation of hyperbolic functions :
(i) d sin h x = cos h x (ii) d cos h x = sin h x
dx dx
(iii) d tan h x = sec h2x (iv) d cot h x = − cosec h2 x
dx dx
(v) d sec h x = − sec h x tan h x (vi) d cosec h x = −cosec h x cot h x
dx dx
(vii) d sin h−1x =1/ (1 + x2) (viii) d cos h−1x =1/ (x2 − 1)
dx dx
(ix) d tan h−1x = 1 /(x2 − 1) (x) d cot h−1x = 1 /(1 − x2)
dx dx
(xi) d sec h−1x = −1 / x (1 − x2 ) (xii) d cosec h−1 x = −1 / x (1 + x 2 )
dx dx
(6) Differentiation by inverse trigonometrical substitution: For trigonometrical substitutions following
formulae and substitution should be remembered
(i) sin−1 x + cos−1 x = π / 2 (ii) tan−1 x + cot−1 x = π / 2
(iii) sec −1 x + cosec −1 x = π / 2 (iv) sin−1 x ± sin−1 y = sin−1 x 1− y2 ± y 1 − x 2
(v) cos−1 x ± cos−1 y = cos−1 xy (1 − x 2 )(1 − y 2 ) (vi) tan−1 x ± tan−1 y = tan−1 x±y
1 xy
(vii) 2 sin−1 x = sin−1(2x 1 − x 2 ) (viii) 2 cos−1 x = cos−1(2x2 − 1)
(ix) 2 tan −1 x = tan −1 2x = sin −1 2 x = cos −1 1 − x2
−x + x 1 + x2
1 2 1 2
(x) 3 sin−1 x = sin−1(3x − 4 x3) (xi) 3 cos−1 x = cos−1(4 x3 − 3x)
(xii) 3 tan−1 x = tan−1 3x − x3 (xiii) tan−1 x + tan−1 y + tan−1 z = tan −1 x+y+z− xyz
1− 3x2 1 − xy − yz − zx
(xiv) sin−1(−x) = − sin−1 x (xv) cos −1 (−x) = π − cos −1 x
(xvi) tan−1(−x) = − tan−1 x or π − tan−1 x (xvii) π − tan−1 x = tan −1 1 − x
(7) Some suitable substitutions 4 1 + x
S. N. Function Substitution S. N. Function Substitution
(i) a2 − x2 x = a sinθ or a cosθ (ii) x2 + a2 x = a tanθ or a cotθ
(iii) x2 − a2 a−x x = a cos 2θ
x = a sec θ or a cosecθ (iv) a+x
(v) a2 − x2 x = a sin 2 θ
a2 + x2 x 2 = a 2 cos 2θ (vi) ax − x2
(vii) x x = a tan 2 θ (viii) x x = a sin 2 θ
a+x a−x
(ix) (x − a)(x − b) x = a sec2 θ − b tan2 θ (x) (x − a)(b − x) x = a cos2 θ + b sin2 θ
Theorems for Differentiation.
Let f(x), g(x) and u(x) be differentiable functions
(1) If at all points of a certain interval. f′(x) = 0, then the function f(x) has a constant value within this interval.
(2) Chain rule
(i) Case I : If y is a function of u and u is a function of x, then derivative of y with respect to x is dy = dy du
dx du dx
dy du
or y = f (u) ⇒ dx = f '(u) dx
(ii) Case II : If y and x both are expressed in terms of t, y and x both are differentiable with respect to t then
dy = dy / dt .
dx dx / dt
(3) Sum and difference rule : Using linear property d ( f (x) ± g(x)) = d ( f (x)) ± d (g(x))
dx dx dx
(4) Product rule : (i) d ( f (x)g (x)) = f(x) d g(x) + g(x) d f(x) (ii) d (u.v.w.) = u.v. dw + v.w. du + u.w. dv
dx dx dx dx dx dx dx
(5) Scalar multiple rule : d (k f (x)) = k d f(x)
dx dx
(6) Quotient rule : d f(x) g(x) d ( f (x)) − f(x) d (g(x)) , provided
dx g(x) dx (g(x))2 dx
= g(x) ≠ 0
Example: 5 The derivative of f(x) =| x |3 at x = 0 is [Rajasthan PET 2001; Haryana CEE 2002]
Solution: (a)
Example: 6 (a) 0 (b) 1 (c) –1 (d) Not defined
Solution: (b)
f(x) = x3 , x≥0 and f ′(x) = 3x 2 , x≥0
Example: 7 − x3 , x<0 − 3x 2 , x<0
f ′(0+ ) = f ′(0−) = 0
The first derivative of the function ( sin 2x cos 2x cos 3x + log2 2x+3) with respect to x at x = π is [MP PET 1998]
(a) 2 (b) –1 (c) − 2 + 2π loge 2 (d) −2 + loge 2
f(x) = sin 2x.cos 2x.cos 3x + log 2 2x+3 , f(x) = 1 sin 4 x cos 3x + (x + 3)log 2 2, f(x) = 1 [sin 7x + sin x] + x + 3
2 4
Differentiate w.r.t. x,
f ′(x) = 1 [7 cos 7 x + cos x] + 1 , f ′(x) = 1 7 cos 7 x + 1 cos x + 1 , f ′(π ) = −2 + 1 = −1 .
4 4 4
If y =| cos x | + | sin x | then dy at x = 2π is
dx 3
(a) 1− 3 (b) 0 (c) 1 ( 3 − 1) (d) None of these
2 2
Solution: (c) Around x = 2π , |cos x |= − cos x and | sin x |= sin x
3
Example: 8
Solution: (b) ∴ y = − cos x + sin x ∴ dy = sin x + cos x
Example: 9 dx
Solution: (d)
Example: 10 At x = 2π , dy = sin 2π + cos 2π = 3 − 1 = 1 ( 3 − 1) .
Solution: (c) 3 dx 3 3 2 2 2
Example: 11 If f(x) = log x (log x), then f ′(x) at x = e is [IIT 1985; Rajasthan PET 2000; MP PET 2000; Karnataka CET 2002]
Solution: (a)
(a) e (b) 1/e (c) 1 (d) None of these
log(log x) f ′(x) = 1 − 1 log(log x) 1 − 0 1
log x x x e e
f(x) = log x (log x) = ⇒ (log x)2 ⇒ f ′(e) = 1 =
If f(x) =|log x |, then for x ≠ 1, f′(x) equals
(a) 1 (b) 1 (c) −1 (d) None of these
x |x| x
f(x) =| log x |= − log x , if 0 < x < 1 f ′(x) − 1 , if 0 < x < 1
if x ≥ 1 x .
log x , ⇒ = 1
x if x > 1
,
Clearly f ′(1−) = −1 and f ′(1+) = 1 , ∴ f ′(x) does not exist at x = 1
d e x − 2 3 / 4
dx log x + 2
x equals to [Rajasthan PET 2001]
(a) 1 (b) x2 + 1 (c) x2 − 1 (d) ex x2 − 1
x2 − 4 x2 − 4 x2 − 4
Let e x x − 2 3 / 4 ex log x − 2 3 / 4
log x + 2 x + 2
y = = log +
⇒ y = x + 3 [log(x − 2) − log(x + 2)] ⇒ dy =1+ 3 x 1 2 − x 1 = 1 + 3 4)
4 dx 4 − + 2 (x 2 −
⇒ dy x2 −1 .
dx = x2 − 4
If x = exp tan −1 y − x2 then dy equals [MP PET 2002]
x2 dx
(a) 2x [1 + tan (log x)] + x sec 2 (log x) (b) x [1 + tan (log x)] + sec 2 (log x)
(d) 2x [1 + tan (log x)] + sec 2 (log x)
(c) 2x [1 + tan (log x)] + x 2 sec 2 (log x)
x exp tan −1 y − x 2 log x tan −1 y − x 2
x x
= 2 ⇒ = 2
⇒ y − x2 = tan(log x) ⇒ y = x 2 tan(log x) + x 2 ⇒ dy = 2x . tan(log x) + x 2 . sec 2 (log x) + 2x
x2 dx x
⇒ dy = 2x tan(log x) + x sec 2 (log x) + 2x ⇒ dy = 2x[1 + tan(log x)] + x sec 2 (log x) .
dx dx
Example: 12 If y = sec −1 x + 1 + sin −1 x − 1 , then dy = [UPSEAT 1999; AMU 2002]
Solution: (a) x − 1 x + 1 dx
Example: 13 (d) None of these
Solution: (d) (a) 0 (b) 1 (c) 1
Example: 14 x +1
Solution: (b)
y = sec −1 x + 1 + sin−1 x − 1 = cos −1 x −1 + sin −1 x − 1 = π ⇒ dy =0 sin −1 x + cos −1 x = π
Example: 15 x − 1 x + 1 x +1 x + 1 2 dx
Solution: (a) 2
Example: 16 d tan −1 cos x − sin x [AISSE 1985, 87; DSSE 1982, 84; MNR 1985; Karnataka CET 2002; Rajasthan PET 2002, 03]
Solution: (a) dx cos x + sin x
Example: 17
(a) 1 (b) 1 (c) 1 (d) – 1
2(1 + x2) 1+ x2
d tan−1 cos x − sin x = d tan−1 tan π − x = −1 .
dx cos x + sin x dx 4
d 2 cot −1 1− x equals [MP PET 2002; EAMCET 1996]
dx sin 1+ x
(d) 1
(a) −1 (b) 1 (c) − 1
2 2
Let y = sin 2 cot −1 1 − x
1+ x
Put x = cosθ ⇒ θ = cos−1 x
⇒ y = sin 2 cot −1 1 − cosθ = sin 2 cot −1 tan θ ⇒y = sin 2 π − θ = cos 2 θ = 1 (1 + cosθ ) = 1 (1 + x)
1 + cosθ 2 2 2 2 2 2
∴ dy = 1
dx 2
If y = cos−1 5 cos x − 12 sin x , x ∈ 0, π , then dy is equal to
13 2 dx
(a) 1 (b) – 1 (c) 0 (d) None of these
Let cos α = 5 . Then sin α = 12 . So, y = cos−1{cos α. cos x − sin α. sin x}
13 13
∴y = cos−1{cos(x + α)} = x + α ( x + α is in the first or the second quadrant)
∴ dy = 1.
dx
d cosh−1(sec x) = [Rajasthan PET 1997]
dx
(a) sec x (b) sin x (c) tan x (d) cosecx
We know that d cosh−1 x = 1, d cosh −1 (sec x) = 1 sec x tan x = sec x tan x = sec x .
dx x2 −1 dx sec 2 x tan x
−1
d tan 2 2x − tan 2 x cot 3x [AMU 2000]
dx 1 − tan 2 2x tan 2 x
(a) tan 2x tan x (b) tan 3x tan x (c) sec2 x (d) sec x tan x
Solution: (c) Let y = tan 2 2x − tan 2 x = (tan 2x − tan x) (tan 2x + tan x) = tan(2x − x) tan(2x + x) = tan x tan 3x .
Example: 18 1 − tan 2 2x tan 2 x (1 + tan 2x tan x) (1 − tan 2x tan x)
Solution: (a)
∴ d [y. cot 3x] = d [tan x] = sec 2 x .
Example: 19 dx dx
Solution: (a)
Example: 20 If f(x) = cot −1 x x − x−x , then f '(1) is equal to [Rajasthan PET 2000]
Solution: (a) 2
(d) − log 2
(a) – 1 (b) 1 (c) log 2
f(x) = cot −1 xx − x−x
2
Put x x = tanθ , ∴ y= f(x) = cot −1 tan 2 θ − 1 = cot−1(− cot 2θ) = π − cot −1(cot 2θ )
2 tanθ
⇒ y = π − 2θ = π − 2 tan −1(x x ) ⇒ dy = −2 .x x (1 + log x) ⇒ f ′(1) = −1 .
dx 1 + x 2x
If y = (1 + x)(1 + x2)(1 + x4 ).......(1 + x2n ) then dy at x=0 is
dx
(a) 1 (b) – 1 (c) 0 (d) None of these
y = (1 − x)(1 + x)(1 + x2).....(1 + x 2n ) = 1 − x2n+1
1− x 1− x
∴ dy = − 2n+1.x 2n+1 −1(1 − x) + 1 − x 2n+1 , ∴ At x =0, dy = − 2 n+1 0 .1 + 1 − 0 =1.
dx (1 − x)2 dx 12
If f(x) = cos x . cos 2x . cos 4 x . cos 8x . cos 16x then f ′ π is
4
(a) 2 (b) 1 (c) 1 (d) None of these
2
f(x) = 2 sin x . cos x . cos 2x . cos 4 x. cos 8x . cos 16x = sin 32x
2 sin x 25 sin x
∴ f ′(x) = 1 . 32 cos 32x . sin x − cos x . sin 32x
32 sin2 x
32. 1 − 1 . 0
f ′ π 22
∴ 4 = 2 = 2.
1
32. 2
Relation between dy/dx and dx/dy .
Let x and y be two variables connected by a relation of the form f(x, y) = 0 . Let ∆x be a small change in x and
let ∆y be the corresponding change in y. Then dy = lim ∆y and dx = lim ∆x .
dx ∆x dy ∆y
∆x →0 ∆y→0
Now, ∆y . ∆x = 1 ⇒ ∆lxim→0 ∆y . ∆x = 1
∆x ∆y ∆x ∆y
⇒ lim ∆y . lim ∆x = 1 [ ∆x → 0 ⇔ ∆y → 0 ] ⇒ dy . dx = 1. So, dy = 1 .
∆x ∆y dx dy dx dx / dy
∆x →0 ∆y → 0
Methods of Differentiation.
(1) Differentiation of implicit functions : If y is expressed entirely in terms of x, then we say that y is an
explicit function of x. For example y = sin x, y = ex, y = x2 + x + 1 etc. If y is related to x but can not be
conveniently expressed in the form of y = f(x) but can be expressed in the form f(x, y) = 0 , then we say that y is
an implicit function of x.
(i) Working rule 1 : (a) Differentiate each term of f(x, y) = 0 with respect to x.
(b) Collect the terms containing dy / dx on one side and the terms not involving dy/dx on the other side.
(c) Express dy/dx as a function of x or y or both.
Note : In case of implicit differentiation, dy/dx may contain both x and y.
(ii) Working rule 2 : If f(x, y)= constant, then dy = − ∂f
dx ∂x
∂f
∂y
where ∂f and ∂f are partial differential coefficients of f (x, y) with respect to x and y respectively.
∂x ∂x
Note : Partial differential coefficient of f(x,y) with respect to x means the ordinary differential coefficient
of f(x,y) with respect to x keeping y constant.
Example: 21 If xe xy = y + sin2 x , then at x =0, dy = [IIT 1996]
Solution: (c) dx
Example: 22 (a) – 1 (b) – 2 (c) 1 (d) 2
Solution: (d)
We are given that xe xy = y + sin2 x
When x = 0 , we get y = 0
Differentiating both sides w.r.t. x, we get, e xy + xe xy x dy + y = dy + 2 sin x cos x
dx dx
Putting, x =0, y = 0 , we get dy = 1.
dx
If sin(x + y) = log(x + y) , then dy = [Karnataka CET 1993; Rajasthan PET 1989, 1992; Roorkee 2000]
dx
(a) 2 (b) – 2 (c) 1 (d) – 1
sin(x + y) = log(x + y)
Differentiating with respect to x, cos(x + y) 1 + dy = 1 y 1 + dy
dx x+ dx
cos(x + y) − 1 1 + dy = 0
+ dx
x y
cos(x + y) ≠ 1 for any x and y. So, 1 + dy =0, dy = −1 .
x+y dx dx
Trick: It is an implicit function, so dy = − ∂f / ∂x cos(x + y) − 1 = −1 .
dx ∂f / ∂y x+y
=−
cos(x + y) − 1
x+y
Example: 23 If ln(x + y) = 2xy, then y′(0) = [IIT Screening 2004]
Solution: (a)
(a) 1 (b) – 1 (c) 2 (d) 0
ln(x + y) = 2xy ⇒ (1 + dy / dx) = 2 x dy + y ⇒ dy = 1 − 2xy − 2y2 ⇒ y′(0) = 1− 2 = 1 , at x = 0, y =1.
(x + y) dx dx 2x2 + 2xy − 1 −1
(2) Logarithmic differentiation : If differentiation of an expression or an equation is done after taking log on
both sides, then it is called logarithmic differentiation. This method is useful for the function having following forms.
(i) y = [f(x)]g(x)
(ii) y= f1(x). f2(x)......... where gi(x) ≠ 0 (where i = 1, 2, 3,.....), fi(x) and gi(x) both are differentiable
g1(x).g2(x)........
(i) Case I : y = [ f(x]g(x) where f(x) and g(x)are functions of x. To find the derivative of this type of functions
we proceed as follows:
Let y = [ f(x)]g(x) . Taking logarithm of both the sides, we have log y = g(x). log f (x)
Differentiating with respect to x, we get 1 dy = g(x). 1 df (x) + log { f (x)}. dg(x)
y dx f(x) dx dx
∴ dy = y g(x) . df(x) + log[ f(x)]. dg(x) = [ f (x)g(x) g(x) df (x) + log[ f(x) dg(x)
dx f(x) dx dx f(x) dx dx
(ii) Case II : y= f1(x). f2(x)
g1(x).g2(x)
Taking logarithm of both the sides, we have log y = log[ f1(x)] + log[ f2(x)] − log[g1(x)] − log[g2(x)]
Differentiating with respect to x, we get 1 dy = f1′(x) + f2′ (x) − g1′ (x) − g ′2 (x)
y dx f1 (x) f2 (x) g 2 (x) g 2 (x)
dy = y f1′(x) + f2′(x) − g1′ (x) − g′2(x) = f1 (x).f2 (x) f1′(x) + f2′ (x) − g1′ (x) − g ′2 (x)
dx f1(x) f2(x) g1(x) g2(x) g1 (x).g 2 (x) f1 (x) f2 (x) g 1 (x)
g 2 (x)
Working rule : (a) To take logarithm of the function (b) To differentiate the function
Example: 24 If x myn = 2(x + y)m+n , the value of dy is [MP PET 2003]
Solution: (c) dx
(d) x − y
(a) x + y (b) x (c) y
y x
x myn = 2(x + y)m+n ⇒ m log x + n log y = log 2 + (m + n)log(x + y)
Differentiating w.r.t. x both sides
m + n dy = m + n 1 + dy ⇒ dy = y .
x y dx x + y dx dx x
Example: 25 If y = (sin x)tan x , then dy is equal to [IIT 1994; Rajasthan PET 1996]
Solution: (a) dx
(b) tan x .(sin x)tan x−1 . cos x
(a) (sin x)tan x .(1 + sec2 x . log sin x)
(c) (sin x)tan x ., sec2 x log sin x (d) tan x .(sin x)tan x−1
Given y = (sin x)tan x
log y = tan x . log sin x
Differentiating w.r.t. x, 1 . dy = tan x . cot x + log sin x . sec2 x
y dx
dy = (sin x)tan x[1 + log sin x . sec2 x] .
dx
(3) Differentiation of parametric functions : Sometimes x and y are given as functions of a single variable,
e.g., x = φ (t) , y = ψ (t) are two functions and t is a variable. In such a case x and y are called parametric functions
or parametric equations and t is called the parameter. To find dy in case of parametric functions, we first obtain
dx
the relationship between x and y by eliminating the parameter t and then we differentiate it with respect to x. But
every time it is not convenient to eliminate the parameter. Therefore dy can also be obtained by the following
dx
formula dy = dy / dt
dx dx / dt
To prove it, let ∆x and ∆y be the changes in x and y respectively corresponding to a small change ∆t in t.
Since ∆y = ∆y / ∆t , ∴ dy = lim ∆y = lim ∆y = dy = Ψ' (t)
∆x ∆x / ∆t dx ∆x ∆t dt φ' (t)
∆x →0 ∆t →0 dx
dt
lim ∆x
∆t
∆t →0
Example: 26 If x = a(cosθ + θ sinθ ) , y = a(sinθ − θ cos θ ), dy = [DCE 1999]
Solution: (b) dx
Example: 27
(a) cosθ (b) tanθ (c) sec θ (d) cosecθ
Solution: (d) (c)
dy = dy / dθ = a[cos θ − θ (− sinθ ) − cos θ ] = θ sin θ = tanθ . 1
Example: 28 dx dx / dθ a[− sin θ +θ cos θ + sinθ ] θ cos θ 1+ t2
If cos x = 1 and sin y = t , then dy = [MP PET 1994]
1+ t2 1+ t2 dx
(a) – 1 (b) 1−t (d) 1
1+ t2
Obviously x = cos−1 1 and y = sin−1 t
1+ t2 1+ t2
⇒ x = tan−1 t and y = tan−1 t ⇒ y=x ⇒ dy = 1.
dx
If x = 1 − t 2 and y = 1 2t 2 , then dy = [Karnataka CET 2000]
1 + t 2 +t dx
(a) −y (b) y (c) −x (d) x
x x y y
Solution: (c) x = 1 − t2 and y = 2t
1 + t2 1+ t2
Put t = tanθ in both the equations, we get x = 1 − tan 2 θ = cos 2θ and y = 2 tanθ = sin 2θ . Differentiating both the
1 + tan 2 θ 1 + tan 2 θ
equations, we get dx = −2 sin 2θ and dy = 2 cos 2θ .
dθ dθ
Therefore dy = − cos 2θ =− x .
dx sin 2θ y
(4) Differentiation of infinite series : If y is given in the form of infinite series of x and we have to find out
dy then we remove one or more terms, it does not affect the series
dx
(i) If y = f(x) + f(x) + f(x) + .......∞ , then y = f(x) + y ⇒ y 2 = f(x) + y
2y dy = f ′(x) + dy , dy f ′(x)
dx dx ∴ dx = 2y − 1
(ii) If y = f(x) f(x) f(x)f(x).....∞ then y = f(x)y
∴ log y = y log f(x)
1 dy = y.f ′(x) + log f (x). dy , ∴ dy = y 2 f ′(x) f (x)]
y dx f(x) dx dx f(x)[1 − y log
(iii) If y= f(x) + 1 then dy = yf ′(x)
1 dx 2y − f(x)
f(x) +
f(x) + ....∞
Example: 29 If y = x+ x+ x + ........to ∞ then dy = [Rajasthan PET 2002]
dx
Solution: (d)
Example: 30 (a) x (b) 2 (c) −1 (d) 1
2y − 1 2y − 1 2y − 1 2y − 1
Solution: (b)
Example: 31 y= x+ x+ x + ........to ∞ ⇒ y = x+y ⇒ y2 = x + y ⇒ 2y dy = 1+ dy ⇒ dy (2y − 1) = 1 ⇒ dy = 1
dx dx dx dx 2y − 1
If y = x x x.....∞ , then x(1 − y log e x) dy is [DCE 2000]
dx
(a) x2 (b) y2 (c) xy2 (d) None of these
y = x x x......∞ ⇒ y = xy ⇒ loge y = y loge x ⇒ 1 ⋅ dy = y + log e x dy ⇒ 1 − loge x dy = y ⇒ x(1 − y log e x) dy = y2
y dx x dx y dx x dx
If y = x2 + x2 1 , then dy =
1 dx
+ 1
x2
+ x2 + ......∞
(a) 2xy (b) xy (c) xy (d) 2x
2y − x2 y + x2 y − x2
2+ x2
y
Solution: (a) y = x2 + 1 ⇒ y2 = x2y + 1 ⇒ 2y dy = y.2x + x2 dy ⇒ dy 2xy
Example: 32 y dx dx dx = 2y − x2
Solution: (c) If x = e y+ey+........to ∞ , then dy is [AIEEE 2004]
dx
(a) 1+ x (b) 1 (c) 1− x (d) x
x x x 1+ x
x = ey+x
Taking log both sides, log x = (y + x) log e = y + x ⇒ y + x = log x ⇒ dy +1 = 1 ⇒ dy = 1 −1 = 1− x
dx x dx x x
(5) Differentiation of composite function : Suppose function is given in form of fog(x) or f[g(x)]
Working rule : Differentiate applying chain rule d f[g(x)] = f '[g(x)].g'(x)
dx
Example: 33 If f(x) =| x − 2| and g(x) = f(f(x)) , then for x>20, g′(x) equals
Solution: (b)
(a) –1 (b) 1 (c) 0 (d) None of these
Example: 34 (d) None of these
Solution: (c) For x > 20, we have
f(x) =| x − 2|= x − 2 and, g(x) = f(f(x)) = f(x − 2) = x − 2 − 2 = x − 4
∴ g′(x) = 1
If g is inverse of f and f ′(x) = 1 1 n , then g′(x) equals
+x
(a) 1 + xn (b) 1 + [ f(x)]n (c) 1 + [g(x)]n
Since g is inverse of f. Therefore,
fog(x) = x for all x ⇒ d { fog(x)} = 1 for all x
dx
⇒ f′(g(x)).g′(x) = 1 ⇒ f′{g(x)} = 1 ⇒ 1 1 f ′(x) = 1
g′(x) 1 + [g(x)]n = g′(x) 1+ xn
⇒ g′(x) = 1 + [g(x)]n
Differentiation of a Function with Respect to Another Function.
In this section we will discuss derivative of a function with respect to another function. Let u = f(x) and v = g(x)
be two functions of x. Then, to find the derivative of f(x) w.r.t. g(x) i.e., to find du we use the following formula
dv
du = du / dx
dv dv / dx
Thus, to find the derivative of f(x) w.r.t. g(x) we first differentiate both w.r.t. x and then divide the derivative of
f(x) w.r.t. x by the derivative of g(x) w.r.t. x.
Example: 35 The differential coefficient of tan−1 1 2x w.r.t. sin−1 1 2x 2 is
Solution: (a) − x2 +x
[Roorkee 1966; BIT Mesra 1996; Karnataka CET 1994; MP PET 1999; UPSEAT 1999, 2001]
(a) 1 (b) –1 (c) 0 (d) None of these
Let y1 = tan−1 1 2x and y2 = sin−1 2x
− x2 1+ x2
Putting x = tanθ
∴ y1 = tan−1 tan 2θ = 2θ = 2 tan−1 x and y2 = sin−1 sin 2θ = 2 tan−1 x
Again dy1 = d [2 tan−1 x] = 2 ........(i)
dx dx 1+ x2
and dy2 = d [2 tan−1 x] = 1 2 ........(ii)
dx dx + x2
Hence dy1 =1
dy2
Example: 36 The first derivative of the function cos−1 sin 1+ x + x x with respect to x at x =1 is [MP PET 1998]
2
(a) 3 (b) 0 (c) 1 (d) 1
4 2 −2
Solution: (a) f(x) = cos−1 cos π − 1+ x + x x = π − 1+ x + xx
2 2 2 2
∴ f′(x) = − 1. 1 + x x(1 + log x) ⇒ f ′(1) = − 1 +1 = 3
22 1+ x 4 4
Successive Differentiation or Higher Order Derivatives.
(1) Definition and notation : If y is a function of x and is differentiable with respect to x, then its derivative
dy can be found which is known as derivative of first order. If the first derivative dy is also a differentiable.
dx dx
function, then it can be further differentiated with respect to x and this derivative is denoted by d 2y / dx 2 which is
called the second derivative of y with respect to x further if d2y is also differentiable then its derivative is called third
dx 2
derivative of y which is denoted by d3y . Similarly nth derivative of y is denoted by dny . All these derivatives are
dx 3 dx n
called as successive derivative and this process is known as successive differentiation. We also use the following
symbols for the successive derivatives of y = f(x) :
y1, y2 , y3, ........., yn ,...... y I , y II , y III ........., y n ,......
Dy, D 2 y, D 3 y........., Dny,...... (where D = d ) dy , d2 y , d3 y , ....... dn y ,......... ..
dx dx dx dx dx
2 3 n
f ′(x), f ′′(x), f ′′′(x),........., f n(x),......
If y = f(x) , then the value of the nth order derivative at x = a is usually denoted by
dn y or (yn )x=a or (yn )x=a or f n (a)
dx
n
x =a
(2) nth Derivatives of some standard functions :
(i) (a) dn sin(ax + b) = an sin nπ + ax + b (b) dn cos(ax + b) = an cos nπ + ax + b
dx n 2 dx n 2
(ii) dn (ax + b)m = m! an (ax + b)m−n , where m>n
dx n (m − n)!
Particular cases : (ii) When a = 1, b = 0 , then y = x n
(i) (a) When m = n
∴ Dn (x m ) = m(m − 1).......(m − n + 1)x m−n = m! x m−n
Dn {(ax + b)n } = an .n! (m − n)!
(b) When m < n, Dn{(ax + b)m} = 0 (iv) When m = −1, y = 1 b)
(iii) When a = 1, b = 0 and m = n, (ax +
then y = xn Dn(y) = an(−1)(−2)(−3)........(−n)(ax + b)−1−n
∴ Dn(xn) = n! = an(−1)n(1.2.3......n)(ax + b)−1−n = an(−1)nn !
(ax + b)n+1
(3) dn log(ax + b) = (−1)n−1 (n − 1)! a n (4) dn (e ax ) = a ne ax
dx n (ax + b)n dx n
(5) d n (a x ) = a x (log a)n (6) (i) dn e ax sin(bx + c) = r ne ax sin(bx + c + nφ)
dx n dx n
where r = a2 + b2 ; φ = tan −1 b , y = e ax sin(bx + c)
a
(ii) dn e ax cos(bx + c) = r ne ax cos(bx + c + nφ)
dx n
Example: 37 If y = x + 1+ x2 n , then (1 + x 2 ) d2 y + x dy is [AIEEE 2002]
Solution: (a) (a) n2y dx dx
2
Example: 38
Solution: (c) (b) − n2y (c) −y (d) 2x 2y
Example: 39 y = (x + 1 + x2 )n ⇒ dy = n(x + 1 + x2 )n−11 + x ⇒ dy = n(x + 1 + x2 )n ⇒( 1+ x2 ) dy = n x + 1 + x2 n
dx 1 + x2 dx 1+ x2 dx
⇒ d2y . 1+ x2 + dy x x2 = n2 x + 1 + x2 n−1 1 + x
dx 2 dx 1+ 1 + x2
⇒ (1 + x2). d2y + x. dy = n2(x + 1+ x 2 )n ⇒ (1 + x 2 ) d2y + x. dy = n2y .
dx 2 dx dx 2 dx
If f(x) = xn, then the value of f(1) − f ' (1) f ' ' (1) f ' '' (1) + ...... + (−1)n f n(1) is [AIEEE 2003]
1! + 2! − 3 ! n!
(a) 2n (b) 2n−1 (c) 0 (d) 1
f(x) = xn ⇒ f(1) = 1 , f′(x) = nxn−1 ⇒ f′(1) = n
f′′(x) = n(n − 1)xn−2 ⇒ f′′(1) = n(n − 1) …..
f n(x) = n! ⇒ f n(1) = n! , ∴ f(1) − f ′(1) + f ′′(1) ...... + (−1)n f n(1)
1! 2! n!
= 1 − n + n(n − 1) − n(n − 1)(n − 2) + ..... + (−1)n n! =nC0 − nC1 +nC2 −nC3 + ...... + (−1)n nCn =0.
1! 2! 3! n!
log e
x2
If f(x) = tan−1 + tan−1 3 + 2 log x , then dny is (n ≥ 1)
log(ex 2 ) 1 − 6 log x dxn
(a) tan−1{(log x)n} (b) 0 (c) 1/2 (d) None of these
Solution: (b) We have y = tan −1 log e − log x 2 + tan −1 3 + 2 log x = tan −1 1 − 2 log x + tan −1 3 + 2 log x
log e + log x 2 1 − 6 log x 1 + 2 log x 1 − 6 log x
Example: 40
Solution: (b) = tan−1 1 − tan−1(2 log x) + tan−1 3 + tan−1(2 log x) ⇒ y = tan−1 1 + tan−1 3 ⇒ dy = 0 ⇒ dny =0.
dx dxn
If f(x) = (cos x + i sin x)(cos 3x + i sin 3x)..... (cos(2n − 1)x + i sin(2n − 1)x), then f′′(x) is equal to
(a) n2 f(x) (b) − n4 f(x) (c) − n2 f(x) (d) n4 f(x)
We have, f(x) = cos(x + 3x + .... + (2n − 1)x) + i sin(x + 3x + 5x + .... + (2n − 1)x) = cos n2x + i sin n2x
⇒ f ′(x) = −n2 (sin n2 x) + n2 (i cos n2 x) ⇒ f ′′(x) = −n4 cos n2 x − n4i sin n2 x
⇒ f ′′(x) = −n4 (cos n2 x + i sin n2 x) ⇒ f′′(x) = −n4 f(x)
nth Derivative using Partial fractions.
For finding nth derivative of fractional expressions whose numerator and denominator are rational algebraic
expression, firstly we resolve them into partial fractions and then we find nth derivative by using the formula giving
the nth derivative of 1 b .
ax +
Example: 41 If y = x4 , then for n > 2 the value of yn is equal to
Solution: (a) x2 − 3x + 2
(a) (−1)n n![16(x − 2)−n−1 − (x − 1)−n−1 ] (b) (−1)n n![16(x − 2)−n−1 + (x − 1)−n−1 ]
(c) n![16(x − 2)−n−1 + (x − 1)−n−1 ] (d) None of these
y= x2 x4 + 2 = x2 + 3x + 7 + 15x − 14 = x2 + 3x + 7 − 1 + 16
− 3x (x − 1)(x − 2) (x − 1) (x − 2)
∴ yn = Dn (x 2 ) + Dn (3x) + Dn (7) − Dn[(x − 1)−1 ] + 16Dn[(x − 2)−1 ]
= (−1)n n![−(x − 1)−n−1 + 16(x − 2)−n−1 ] = (−1)n n![16(x − 2)−n−1 − (x − 1)−n−1 ] .
Differentiation of Determinants.
Let ∆(x) = a1(x) b1 (x) . Then ∆ ′(x) = a1′ (x) b1′ (x) + a1(x) b1 (x)
a 2 (x) b2 (x) a 2 (x) b2 (x) a′2 (x) b′2 (x)
If we write ∆(x) = | C1C2C3 |. Then ∆′(x) =|C1′ C2 C3 |+|C1 C′2 C3 |+|C1 C2 C3′ |
R1 R1′ R1 R1
Similarly, if ∆(x)= R2 , then ∆′(x) = R2 + R2′ + R2
R3 R3 R3 R3′
Thus, to differentiate a determinant, we differentiate one row (or column) at a time, keeping others unchanged.
Example: 42 If fr (x), gr (x), hr (x), r = 1,2,3 are polynomials in x such that fr (a) = gr (a) = hr (a),r = 1,2,3 and
f1 (x) f2 (x) f3 (x) [IIT 1985]
F(x) = g1(x) g 2 (x) g 3 (x) , then find F' (x) at x = a
h2 (x) h3 (x)
h1 (x)
(a) 0 (b) f1(a)g 2 (a)h3 (a) (c) 1 (d) None of these
f1' (x) f2' (x) f3' (x) f1(x) f2 (x) f3 (x) f1(x) f2 (x) f3 (x)
F'(x) = g1(x) g 2 (x) g 3 (x) + g1' (x) ' ' g 2 (x) g3 (x)
Solution: (a) h2 (x) h3 (x) h1(x) g 2 (x) g 3 (x) + g1(x) h2' (x) h3' (x)
h1 (x)
Example: 43 h2 (x) h3 (x) h1′ (x)
Solution: (d)
f1' (a) f2' (a) f3' (a) f1 (a) f2 (a) f3 (a) f1(a) f2 (a) f3 (a)
∴ F ′(a) = g1(a) g 2 (a) ' g′2 (a) ' g 2 (a) g 3 (a)
h2 (a) g 3 (a) + g 1 (a) h2 (a) g 3 (a) + g 1 (a) h2' (a) h3' (a)
h1 (a)
h3 (a) h1 (a) h3 (a) h1' (a)
=. 0 + 0 + 0 = 0 [ fr (a) = gr (a) = hr (a), r = 1, 2, 3]
x3 sin x cos x where p is a constant. Then d3 [ f(x)] at x =0 is [IIT 1997]
Let f(x) = 6 dx 3
−1 0
1 p2 p3
(a) p (b) p + p2 (c) p + p3 (d) Independent of p
x3 sin x cos x
Given f(x) = 6 0 , 2nd and 3rd rows are constant, so only 1st row will take part in differentiation
−1
1 p2 p3
d3 x3 d3 sin x d3 cos x
dx 3 dx 3 dx 3
d3
∴ dx 3 f(x) = 6 −1 0
1 p2 p3
We know that dn xn = n ! , dn sin x = sin(x + nπ ) and dn cos x = cos(x + nπ )
dx n dx n 2 dx n 2
d3 3! sin x + 3π cos x + 3π
dx 3 2 2
Using these results, f(x) = 6 −1 0
1
p2 p3
d3 f(x) = 6 −1 0
dx 3 6
at x =0 1 − 1 0 = 0 i.e., independent of p.
p2 p3
Differentiation of Integral Function.
If g1(x)and g 2 (x) both functions are defined on [a, b] and differentiable at a point x ∈ (a, b) and f(t) is
continuous for g1(a) ≤ f(t) ≤ g2(b)
d∫Then g2 ( x ) f (t)dt = f[g2(x)]g′2(x) − f[g1(x)]g1′ (x) = f[g 2 (x)] d g 2(x) − f [g 1 (x)] d g1 (x) .
dx dx dx
g1 ( x )
Example: 44 x3 [MP PET 2001]
∫If F(x) = log t dt (x > 0) , then F' (x) =
x2
(a) (9x 2 − 4 x)log x (b) (4 x − 9x 2 )log x (c) (9x 2 + 4 x)log x (d) None of these
Solution: (a) Applying formula we get F'(x) = (log x 3 )3x 2 − (log x 2 )2x
= (3 log x)3x 2 − 2x(2 log x) = 9x 2 log x − 4 x log x = (9x 2 − 4 x)log x .
Example: 45 ∫y 1 dt , then d2y is
Solution: (b) 1 + 4t 2 dx 2
If x =
0
(a) 2y (b) 4y (c) 8y (d) 6 y
∫y 1 dt ⇒ dx 1 ⇒ dy = 1+ 4y2 ⇒ d2y 4y dy ⇒ d2y = 4y ⋅ 1+ 4y2 = 4y
1+ 4t2 dy = 1+ 4y2 dx dx2 = 1+ 4y2 dx dx 2 1+ 4y2
x=
0
Leibnitz’s Theorem.
G.W. Leibnitz, a German mathematician gave a method for evaluating the nth differential coefficient of the
product of two functions. This method is known as Leibnitz’s theorem.
Statement of the theorem – If u and v are two functions of x such that their nth derivative exist then
Dn (u.v.) =nC0 (Dnu)v + nC1 Dn−1u.Dv + nC2 Dn−2u.D 2v + ............ + nCr Dn−r u.Dr v + ......... + u.(Dnv).
Note : The success in finding the nth derivative by this theorem lies in the proper selection of first and
second function. Here first function should be selected whose nth derivative can be found by
standard formulae. Second function should be such that on successive differentiation, at some stage,
it becomes zero so that we need not to write further terms.
Example: 46 If y = x 2e x , then value of yn is
Solution: (b)
Example: 47 (a) {x 2 − 2nx + n(n − 1)}e x (b) {x 2 + 2nx + n(n − 1)}e x
Solution: (b)
(c) {x 2 + 2nx − n(n − 1)}e x (d) None of these
Applying Leibnitz’s theorem by taking x 2 as second function. We get , Dny = Dn(e x .x 2 )
= n C0 Dn (e x )x 2 +nC1Dn−1(e x ).D(x 2 ) +nC2 Dn−2 (e x ).D 2 (x 2 ) + ........... = e x .x 2 + ne x .2x + n(n − 1) e x .2 + 0 + 0 + ..........
2!
yn = {x 2 + 2nx + n(n − 1)}e x .
If y = x 2 log x, then value of yn is
(a) (−1)n−1(n − 3)! (b) (−1)n−1(n − 3)! .2 (c) (−1)n−1(n − 2)! (d) None of these
xn−2 xn−2 xn−2
Applying Leibnitz’s theorem by taking x 2 as second function, we get, Dny = Dn (log x.x 2 )
= nC0Dn(log x).x2 +nC1Dn−1(log x).D(x2) +nC2Dn−2(log x)D2(x2) + ...........
= (−1)n−1(n − 1)! .x2 + n. (−1)n−2(n − 2)! .2x + n(n − 1) (−1)n−3(n − 3)! .2 + 0 + 0.........
xn x n−1 2! xn−2
= (−1)n−1(n − 1)! + 2n(−1)n−2(n − 2)! + n(n − 1)(−1)n−3(n − 3)!
xn−2 xn−2 xn−2
= (−1)n−1(n − 3)! × {(n − 1)(n − 2) − 2n(n − 2) + n(n − 1)} =. (−1)n−1 (n − 3)! .2
xn−2 x n−2
Differentiation and application of derivatives
Velocity and Acceleration in Rectilinear Motion.
The velocity of a moving particle is defined as the rate of change of its displacement with respect to time and the
acceleration is defined as the rate of change of its velocity with respect to time. s V V+δV
Let a particle A moves rectilinearly as shown in figure. O ∆s
A (t) B (t+∆t)
Let s be the displacement from a fixed point O along the path at time t; s is considered to be positive on right of
the point O and negative on the left of it.
Also, ∆s is positive when s increases i.e., when the particle moves towards right.
Thus, if ∆s be the increment in s in time ∆t . The average velocity in this interval is ∆s
∆t
And the instantaneous velocity i.e., velocity at time t is v = lim ∆s = ds
∆t dt
∆t →0
If the velocity varies, then there is change of velocity ∆v in time ∆t .
Hence, the acceleration at time t = lim ∆v = dv
∆t dt
∆t →0
Example: 1 The distance travelled s (in metre) by a particle in t second is given by s = t 3 + 2t 2 + t . The speed of the particle after
Solution: (a)
1 sec. will be [UPSEAT 2003]
Example: 2
Solution: (b) (a) 8 cm/sec. (b) 6 cm/sec. (c) 2 cm/sec (d) None of these
Example: 3 s = t3 + 2t2 + t , v= ds = 3t2 + 4t + 1
Solution: (d) dt
Example: 4
Speed of the particle after 1 second
v(t =1) = ds = 3 × 12 + 4 ×1 + 1 = 8cm / sec .
dt (t
=1)
A particle moves in a straight line in such a way that its velocity at any point is given by v2 = 2 − 3x , where x is measured
from a fixed point. The acceleration is [MP PET 1992]
(a) Zero (b) Uniform (c) Non-uniform (d) Indeterminate
Velocity, v2 = 2 − 3x
Differentiating with respect to t, we get
2v dv = −3. dx ⇒ 2v dv = −3v ⇒ dv 3
dt dt dt dt = − 2
Hence, acceleration is uniform.
The position of a point in time 't' is given by x = a + bt − ct2 , y = at + bt2 . Its acceleration at time 't' is [MP PET 2003]
(a) b − c (b) (b + c) (c) 2b − 2c (d) 2 b2 + c2
Acceleration in x-direction = d2x = −2c and acceleration in y-direction = d2y = 2b
dt 2 dt 2
Resultant acceleration is = (−2c)2 + (2b)2 = 2 b2 + c2 [SCRA 1996]
If the path of a moving point is the curve x = at y = b sin at , then its acceleration at any instant
(a) Is constant (b) Varies as the distance from the axis of x
(c) Varies as the distance from the axis of y (d) Varies as the of the point from the origin
Solution: (c) dx = vx = a ⇒ d2x = 0 = ax
dt dt 2
Example: 5
Solution: (b) ax is acceleration in x-axis
d2y = −ba 2 sin at ⇒ ay = −a2y
dt 2
Hence, ay changes as y changes.
A stone thrown vertically upwards from the surface of the moon at velocity of 24 m/sec. reaches a height of
s = 24t − 0.8t 2m after t sec. The acceleration due to gravity in m/sec2 at the surface of the moon is [MP PET 1992]
(a) 0.8 (b) 1.6 (c) 2.4 (d) 4.9
ds = velocity = 24 = 24 – 1.6 t
dt
So acceleration at t, is d2s = −1.6
dt 2
As stone is thrown upwards, so acceleration due to gravity (which acts downwards) = 1.6.
Derivative as the Rate of Change.
If a variable quantity y is some function of time t i.e., y = f(t), then small change in time ∆t have a
corresponding change ∆y in y.
Thus, the average rate of change = ∆y
∆t
When limit ∆t → 0 is applied, the rate of change becomes instantaneous and we get the rate of change with
respect to t.
i.e., lim ∆y = dy
∆t dt
∆t →0
Hence, it is clear that the rate of change of any variable with respect to some other variable is derivative of first
variable with respect to other variable.
Note : The differential coefficient of y with respect to x i.e, dy is nothing but the rate of increase of y
dx
relative to x.
Example: 6 The rate of change of the surface area of a sphere of radius r when the radius is increasing at the rate of 2cm/sec is
proportional to [Karnataka CET 2003]
(a) l (b) l (c) r (d) r 2
r r2
Solution: (c) Surface area s = 4πr 2 and dr =2
dt
∴ ds = 4π × 2r dr = 8πr × 2 = 16πr ⇒ ds ∝ r .
dt dt dt
Example: 7 If the volume of a spherical balloon is increasing at the rate of 900 cm2/sec. then the rate of change of radius of balloon at
Solution: (c)
instant when radius is 15 cm [in cm/sec] [Rajasthan PET 1996]
Example: 8
Solution: (b) (a) 22 (b) 22 (c) 7 (d) None of these
7 22
Example: 9
Solution: (a) V = 4 πr 3
3
Differentiate with respect to t
dV = 4 π 3r 2 . dr ⇒ dr ⇒ 1 . dV
dt 3 dt dt 4πr 2 dt
dr = 4 ×π 1 × 900 = 1 = 7 .
dt × 15 × 15 π 22
A man of height 1.8 m is moving away from a lamp post at the rate of 1.2 m/sec. If the height of the lamp post be 4.5
meter, then the rate at which the shadow of the man is lengthening
(a) 0.4 m/sec (b) 0.8 m/sec. (c) 1.2 m/sec. (d) None of these
dy = 1.2 According to the figure, A
dt
x = 2 y P
3 1.8
4.5
dx 2 dy
⇒ dt = 3 . dt
⇒ Rate of length of shadow dx = 0.8 m / s . C xQ y B
dt
A 10 cm long rod AB moves with its ends on two mutually perpendicular straight lines OX and OY. If the end A be moving
at the rate of 2 cm/sec. then when the distance of A from O is 8 cm, the rate at which the end B is moving, is [SCRA 1996]
(a) 8 cm/sec (b) 4 cm / sec (c) 2 cm / sec . (d) None of these
3 3 9
By figure, x 2 + y2 = 100 ......(i)
.....(ii)
dx dy y
dt dt
⇒ 2x + 2y = 0
x=8 B
y
dy 16 8 10 cm
dt =− 6 3
Therefore by (i) and (ii), = − cm / sec .
∴ B is moving at the rate 8 cm / sec . A x
3 Ox
Differentiation and application of derivatives
Slope of the Tangent and Normal.
(1) Slope of the tangent : If tangent is drawn on the curve y = f(x) at point P(x1, y1) and this tangent makes
an angle ψ with positive x-direction then,
dy tanψ = slope of the tangent y Tangent
dx (x1, y1) Normal
=
Note : If tangent is parallel to x-axis ψ = 0 ⇒ dy = 0
dx (
x1 , y1 )
ψ
π dy O x
2 dx (
If tangent is perpendicular to x-axis ψ = ⇒ = ∞
x1 , y1 )
(2) Slope of the normal : The normal to a curve at P(x1, y1) is a line perpendicular to the tangent at P and
passing through P and slope of the normal = −1 = −1 = − dx
Slope of tangent dy
dy P ( x1, y1 )
dx
P ( x1, y1 )
Note : If normal is parallel to x-axis
⇒ − dx ( =0 or dx ( =0
dy dy
x1 , y1 ) x1 , y1 )
If normal is perpendicular to x-axis (for parallel to y-axis)
⇒ − dy =0
dx (
x1 , y1 )
Example: 1 The slope of the tangent to the curve x2 + y2 = 2c2 at point (c, c) is [AMU 1998]
Solution: (b)
(a) 1 (b) – 1 (c) 0 (d) 2
Example: 2
Solution: (a) Given x2 + y2 = 2c2
Differentiating w.r.t. x, 2x + 2y dy =0
dx
⇒ 2y dy = −2x ⇒ dy = −x ⇒ dy = −1
dx dx y dx (c,
c )
The line x + y = 2 is tangent to the curve x2 = 3 − 2y at its point [MP PET 1998]
(a) (1, 1) (b) (–1, 1) (c) ( 3, 0) (d) (3, – 3)
Given curve x2 = 3 − 2y
diff. w.r.t. x, 2x = − 2dy ; dy = −x
dx dx
Slope of the line = – 1
dy = −x = −1 ; x = 1
dx
∴ y = 1 point (1, 1)
Example: 3 The tangent to the curve y = 2x2 − x + 1 at a point P is parallel to y = 3x + 4, the co-ordinate of P are [Rajasthan PET 2003]
Solution: (b)
(a) (2, 1) (b) (1, 2) (c) (– 1, 2) (d) (2, – 1)
Given y = 2x2 − x + 1
Let the co-ordinate of P is (h, k) then dy = 4h −1
dx (h,
k )
Clearly 4h − 1 = 3
h = 1 ⇒ k = 2 . P is (1, 2).
Equation of the Tangent and Normal.
(1) Equation of the tangent : We know that the equation of a line passing through a point P(x1, y1) and
having slope m is y − y1 = m(x − x1)
Slope of the tangent at (x1, y1) is = dy
dx (x1, y1)
The equation of the tangent to the curve y = f(x) at point P(x1, y1) is
y − y1 = dy (x − x1)
dx
( x1 , y1 )
(2) Equation of the normal : Slope of the Normal = −1
dy ( x1 , y1 )
dx
Thus equation of the normal to the curve y = f(x) at point P(x1, y1)
y − y1 = −1 (x − x1)
dy ( x1 , y1 )
dx
Note : If at any point P(x1, y1) on the curve y = f(x) , the tangent makes equal angle with
the axes, then at the point P, ψ = π or 3π . Hence, at P tanψ = dy = ±1 .
4 4 dx
Example: 4 The equation of the tangent at (−4, − 4) on the curve x2 = −4y is [Karnataka CET 2001]
Solution: (d)
(a) 2x + y + 4 = 0 (b) 2x − y − 12 = 0 (c) 2x + y − 4 = 0 (d) 2x − y + 4 = 0
x2 = −4y ⇒ 2x = −4 dy ⇒ dy = −x ⇒ dy = 2 .
dx dx 2 dx (−4, −4 )
We know that equation of tangent is (y − y1 ) = dy (x − x1 ) ⇒ y+4 = 2(x + 4) ⇒ 2x − y + 4 = 0 .
dx (x1,
y1 )
Example: 5 The equation of the normal to the curve y = sin πx at (1, 1) is [AMU 1999]
Solution: (b) 2
Example: 6 (d) y − 1 = −2 (x − 1)
Solution: (d) (a) y = 1 (b) x = 1 (c) y = x π
Example: 7 y = sin πx ⇒ dy = π cos π x ⇒ dy =0
Solution: (d) 2 dx 2 2 dx (1,1)
Example: 8
Solution: (d) ∴ Equation of normal is y − 1 = 1 (x − 1) ⇒ x = 1 . [Karnataka CET 2002]
0
Example: 9
Solution: (c) The equation of the tangent to the curve y = be−x / a at the point where it crosses y-axis is
(a) ax + by = 1 (b) ax − by = 1 (c) x − y =1 (d) x + y =1
a b a b
Curve is y = be−x / a
Since the curve crosses y-axis (i.e., x = 0 ) ∴ y = b
Now dy = −b e−x /a . At point (0, b), dy = −b
dx a dx (0, a
b)
∴ Equation of tangent is y − b = −b (x − 0) ⇒ x + y = 1.
a a b
If the normal to the curve y = f(x) at the point (3, 4) makes an angle 3π with the positive x-axis then f′(3) is equal to
4
[IIT Screening 2000; DCE 2001]
(a) – 1 (b) −3 (c) 4 (d) 1
4 3
Slope of the normal = −1 ⇒ tan 3π = −1
dy / dx 4
dy
dx (3, 4)
∴ dy = 1; f′(3) = 1 .
dx (3, 4)
The point (s) on the curve y 3 + 3x 2 = 12y where the tangent is vertical (parallel to y-axis), is are [IIT Screening 2002]
(a) 4 ,− 2 (b) ± 11 ,1 (c) (0,0) (d) ± 4 , 2
± 3 3 3
y3 + 3x 2 = 12y
⇒ 3y 2 . dy + 6x = 12. dy ⇒ dy (3y 2 − 12) + 6x = 0 ⇒ dy 6x ⇒ dx = 12 − 3y2
dx dx dx dx = 12 − 3y2 dy 6x
Tangent is parallel to y-axis, dx =0 ⇒ 12 − 3y2 = 0 or y = ±2. Then x=± 4 , for y=2
dy 3
y = −2 does not satisfy the equation of the curve, ∴ The point is ± 4 , 2
3
At which point the line x + y = 1 touches the curve y = be−x / a [Rajasthan PET 1999]
a b
(d) (b, 0)
(a) (0, 0) (b) (0, a) (c) (0, b)
Let the point be (x1, y1) ∴ y1 = be−x1 / a ......(i)
Also, curve y = be− x / a ⇒ dy = −b e− x / a
dx a
dy = −b e−x1 / a = −y1 (by (i))
dx ( a a
x1 , y1 )
Now, the equation of tangent of given curve at point (x1, y1) is y − y1 = −y1 (x − x1) ⇒ x + y = x1 +1
a a y1 a
Comparing with x + y =1, we get, y1 = b and 1+ x1 =1 ⇒ x1 = 0
a b a
Hence, the point is (0, b).
Example: 10 The abscissa of the point, where the tangent to curve y = x 3 − 3x 2 − 9x + 5 is parallel to x-axis are [Karnataka CET 2001]
Solution: (d)
(a) 0 and 0 (b) x = 1 and −1 (c) x = 1 and −3 (d) x = −1 and 3
y = x3 − 3x2 − 9x + 5 ⇒ dy = 3x2 − 6x − 9 .
dx
We know that this equation gives the slope of the tangent to the curve. The tangent is parallel to x-axis dy =0
dx
Therefore, 3x 2 − 6x − 9 = 0 ⇒ x = −1, 3 .
Angle of Intersection of Two Curves.
The angle of intersection of two curves is defined to be the angle between the tangents to the two curves at their
point of intersection.
We know that the angle between two straight lines having slopes m1 and m2 y y = f2xy = f1x
O
φ = tan−1 m1 − m2
1 + m1m2
Also slope of the tangent at P(x1,y1) P
φ
m1 = dy , m2 = dy x
dx dx
1( x1 , y1 ) 2( x1 , y1 )
Thus the angle between the tangents of the two curves y = f1 (x) and y = f2(x)
dy − dy
dx 1(x1, y1) dx 2(x1, y1)
tanφ =
dy dy
1+ dx 1(x1, y1) dx 2(x1, y1)
Orthogonal curves : If the angle of intersection of two curves is right angle, the two curves are said to intersect
orthogonally. The curves are called orthogonal curves. If the curves are orthogonal, then φ = π
2
m1m2 = −1 ⇒ dy dy = −1
dx 1 dx
2
Example: 11 The angle between the curves y2 = x and x 2 = y at (1, 1) is [Karnataka CET 1993]
Solution: (b)
(a) tan −1 4 (b) tan −1 3 (c) 90o (d) 45o
3 4
Given curve y2 = x and x 2 = y
Differentiating w.r.t. x, 2y dy =1 and 2x = dy
dx dx
dy = 1 and dy = 2
dx (1,1) 2 dx (1,1)
Angle between the curve
tanφ 2 − 1 tanφ 3 tan −1 3 .
2 4 4
⇒ = 1 ⇒ = ⇒φ =
2
1 + .2
Example: 12 If the two curves y = a x and y = b x intersect at α , then tanα equal [MP PET 2001]
Solution: (a)
(a) log a − log b (b) log a + log b (c) log a − log b (d) None of these
Example: 13 1 + log a log b 1 − log a log b 1 − log a log b
Solution: (b)
Clearly the point of intersection of curves is (0, 1)
Now, slope of tangent of first curve, m1 = dy = a x log a ⇒ dy = m1 = log a
dx dx (0,1)
Slope of tangent of second curve, m2 = dy = b x log b ⇒ m2 = dy = log b
dx dx (0,1)
∴ tanα = m1 − m2 = log a − log b .
1 + m1m2 1 + log a log b
The angle of intersection between curve xy = 6 and x 2y = 12
(a) tan −1 3 (b) tan −1 3 (c) tan −1 11 (d) 0o
4 11 3
The equation of two curves are xy = 6 and x 2y = 12 from (i) we obtain y = 6 putting this value of y in equation (ii) to
x
obtain x 2 6 = 12 ⇒ 6x = 12 ⇒ x=2
x
Putting x = 2 in (i) or (ii) we get, y = 3. Thus, the two curves intersect at P(2, 3)
Differentiating (i) w.r.t. x, we get x dy + y = 0 ⇒ dy = −y ⇒ dy = 3 = m1
dx dx x dx (2, −2
3)
Differentiating (ii) w.r.t. x, we get x2 dy + 2xy = 0 ⇒ dy = −2y
dx dx x
⇒ dy = −3 = m2 ⇒ tan θ = m1 − m2 = −3 + 3 1 + −3 (−3) = 3 ⇒θ = tan−1 3 .
dx (2, 1 + m1m2 2 2 11 11
3)
Length of Tangent, Normal, Subtangent and Subnormal .
Let the tangent and normal at point P(x, y) on the curve y = f(x) meet the x-axis at points A and B
respectively. Then PA and PB are called length of tangent and normal respectively at point P. If PC be the
perpendicular from P on x-axis, the AC and BC are called length of subtangent and subnormal respectively at P. If
PA makes angle ψ with x-axis, then tanψ = dy from fig., we find that y
dx
dy 2 Normal Tangent
dx
1 +
(1) Length of tangent PA = ycosecψ = y dy P (x, y)
dx
ψ
ψ C x
OA B
(2) Length of normal PB = y secψ =y 1 + dy 2
dx
(3) Length of subtangent AC = y cotψ = y
dy
dx
(4) Length of subnormal BC = y tanψ = y dy
dx
Example: 14 The length of subtangent to the curve x 2y2 = a4 at the point (−a, a) is [Karnataka CET 2001]
Solution: (c)
(a) 3a (b) 2a (c) a (d) 4a
Example: 15
Solution: (a) Equation of the curve x 2y2 = a4 .
Differentiating the given equation,
x 2 2y dy + y2 2x = 0 ⇒ dy = −y ⇒ dy = − a = 1
dx dx x dx (−a, −a
a)
Therefore, sub-tangent = y =a.
dy
dx
For the curve yn = an−1x, the sub-normal at any point is constant, the value of n must be [Karnataka CET 1999]
(a) 2 (b) 3 (c) 0 (d) 1
yn = an−1x ⇒ nyn−1 dy = an−1 ⇒ dy = an−1
dx dx nyn−1
∴ Length of the subnormal = y dy = ya n −1 = a n −1y 2 −n
dx nyn−1 n
We also know that if the subnormal is constant, then an−1 .y2−n should not contain y.
n
Therefore, 2 − n = 0 or n = 2 .
Length of Intercept made on Axis by the Tangent.
Equation of tangent at any point (x1, y1) to the curve y = f(x) is y − y1 = dy (x − x1) ......(i)
dx (x1, y1)
Equation of x-axis y = 0 ......(ii) Y
and Equation of y-axis x = 0 ......(iii) R
P (x1 y1)
Solving (i) and (ii) we get x = x1 − dy y1
dx
( x1 , y1 ) O X
Q
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Free Flip-Book Mathematics class 12th by Study Innovations. 494 Pages
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