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LOGARITHMS

Logarithm: For ant two positive real numbers a, b, and a ≠ 1. If the real number x such then a^x = b, then x is called logarithm of b to the base a. it is denoted by

Standard formulae of logarithms:

Logarithmic Function:

Let a be a positive real number and a ≠ 1. The function f: (o, ∞) → R Defined by f(x) =

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PARTIAL FRACTIONS

Fractions:

If f(x) and g(x) are two polynomials, g(x) ≠ 0, then   is called rational fraction.

Ex:

   etc.  are rational fractions.

Proper Fraction:

A rational fraction is said to be a Proper fraction if the degree of g(x) is greater than the degree of f(x).

Ex:

  etc. are the proper fractions.

Improper Fraction:

A rational fraction is said to be an Improper fraction if the degree of g(x) is less than the degree of f(x).

Ex:

 etc. are the Improper fractions.

Partial Fractions:

Expressing rational fractions as the sum of two or more simpler fractions is called resolving a given fraction into a partial fraction.

∎ If R(x) =  is proper fraction, then

Case(i): – For every factor of g(x) of the form (ax + b) ⁿ, there will be a sum of n partial fractions of the form:

Case(ii): – For every factor of g(x) of the form (ax² + bx + c) ⁿ, there will be a sum of n partial fractions of the form:

∎ If R(x) = is improper fraction, then

Case (i): – If degree f(x) = degree of g(x),   where k is the quotient of the highest degree term of f(x) and g(x).

Case (ii): – If f(x) > g(x)

R(x) =

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 MATRICES AND DETERMINANTS

Matrix: A set of numbers arranged in the form of a rectangular array having rows and columns is called Matrix.

•Matrices are generally enclosed by brackets like

•Matrices are denoted by capital letters A, B, C, and so on

•Elements in a matrix are real or complex numbers; real or complex real-valued functions.

Oder of Matrix: A matrix having ‘m’ rows and ‘n’ columns is said to be of order m x n read as m by n.

Ex:

Types Of Matrices

Rectangular Matrix: A matrix in which the no. of rows is not equal to the no. of columns is called a rectangular matrix.                 

 Square Matrix: A matrix in which the no. of rows is equal to no. of columns is called a square matrix.

Principal diagonal (diagonal) Matrix: If A = [a _ij] is a square matrix of order ‘n’ the elements a₁₁, a₂₂, a₃₃, ………. A_{n n} is said to constitute its principal diagonal.

Trace Matrix: The sum of the elements of the principal diagonal of a square matrix A is called the trace of the matrix. It is denoted by Tr (A).

Diagonal Matrix: If each non-diagonal element of a square matrix is ‘zero’ then the matrix is called a diagonal matrix.

Scalar Matrix: If each non-diagonal element of a square matrix is ‘zero’ and all diagonal elements are equal to each other, then it is called a scalar matrix.

Identity Matrix or Unit Matrix: If each of the non-diagonal elements of a square matrix is ‘zero’ and all diagonal elements are equal to ‘1’, then that matrix is called unit matrix

Null Matrix or Zero Matrix: If each element of a matrix is zero, then it is called a null matrix.

Row matrix & column Matrix: A matrix with only one row s called a row matrix and a matrix with only one column is called a column matrix.

Triangular matrices:

A square matrix A = [a_ij] is said to be upper triangular if a_ij = 0   ∀ i > j

A square matrix A = [a_ij] is said to be lower triangular matrix a_ij = 0  ∀ i < j

 

Equality of matrices:

matrices A and B are said to be equal if A and B are of the same order and the corresponding elements of A and B are equal.

Addition of matrices:

If A and B are two matrices of the same order, then the matrix obtained by adding the corresponding elements of A and B is called the sum of A and B. It is denoted by A + B.

Subtraction of matrices:

If A and B are two matrices of the same order, then the matrix obtained by subtracting the corresponding elements of A and B is called the difference from A to B.

Product of Matrices:

Let A = [a_ik]_mxn and B = [b_kj]_nxp be two matrices, then the matrix C = [c_ij]_mxp   where

Note: Matrix multiplication of two matrices is possible when no. of columns of the first matrix is equal to no. of rows of the second matrix.

A _{m x n}  . B_{p x q} = AB _{mx q}; n = p

Transpose of Matrix: If A = [a_ij] is an m x n matrix, then the matrix obtained by interchanging the rows and columns is called the transpose of A. It is denoted by A^I or A^T.

Note: (i) (A^I)^I = A     (ii) (k A^I) = k . A^I    (iii)  (A + B )^T = A^T + B^T  (iv)  (AB)^T = B^TA^T

Symmetric Matrix: A square matrix A is said to be symmetric if A^T =A

If A is a symmetric matrix, then A + A^T is symmetric.

Skew-Symmetric Matrix: A square matrix A is said to be skew-symmetric if A^T = -A

If A is a skew-symmetric matrix, then A – A^T is skew-symmetric.

 Minor of an element: Consider a square matrix

the minor element in this matrix is defined as the determinant of the 2×2 matrix obtained after deleting the rows and the columns in which the element is present.

Ex: – minor of a₃ is     = b₁c₂ – b₂c₁

Minor of b₂ is   = a₁c₃ – a₃c₁

Cofactor of an element: The cofactor of an element in i ^th row and j ^th column of A_3×3 matrix is defined as its minor multiplied by (- 1) ^i+j.

Properties of determinants:

If each element of a row (column) f a square matrix is zero, then the determinant of that matrix is zero.

Ex: –    A =   ⇒ det(A) = 0

If A is a square matrix of order 3 and k is scalar then.

If two rows (columns) of a square matrix are identical (same), then Det. Of that matrix is zero.

Ex: –    A =     ⇒ det(A) = 0

If each element in a row (column) of a square matrix is the sum of two numbers then its determinant can be expressed as the sum of the determinants.

Ex: – 

If each element of a square matrix are polynomials in x and its determinant is zero when x = a, then (x-a) is a factor of that matrix.

For any square matrix A Det(A) = Det (A^I).

Det (AB) = Det(A). Det(B).

For any positive integer n Det (Aⁿ) = (DetA)ⁿ.

Singular and non-singular matrices:

A Square matrix is said to be singular if its determinant is is zero, otherwise it is said to be non-singular matrix.

Ex: –   A =      det(A) = 4 – 4 = 0

∴ A is singular matrix

B =

Det(B) = 4 + 4 = 8≠ 0

∴ B is non-singular

 Adjoint of a matrix: The transpose of the matrix formed by replacing the elements of a square matrix A with the corresponding cofactors is called the adjoint of A.

Let A = and     cofactor matrix of A =

Then adj (A) =

 Invertible matrix: Let A be a square matrix, we say that A is invertible if there exists a matrix B such that AB =BA = I, where I is a unit matrix of the same order as A and B.

1. (A^{– I})^{– I} = A
2. (A^I)^{– I} = (A^-I)^I
3. (AB)^-I = B^-I A^-I
4. A^-I =

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Compound Angles

The algebraic sum of two or more angles is called a ‘compound angle’. Thus, angles A + B, A – B, A + B + C etc., are Compound Angles

For any two real numbers A and B

⋇ sin (A + B) = sin A cos B + cos A Cos B

⋇ sin (A − B) = sin A cos B − cos A Cos B

⋇ cos (A + B) = cos A cos B − sin A sin B

⋇ cos (A − B) = cos A cos B + sin A sin B

⋇ tan (A + B) =

⋇ tan (A − B) =

⋇ cot (A + B) =

⋇ cot (A − B) =

⋇ tan ( + A) =

⋇ tan ( − A) =

⋇ cot (+ A) =

⋇ cot (− A) =

⋇ sin (A + B + C) = ∑sin A cos B cos C − sin A sin B sin C  

⋇ cos (A + B + C) = cos A cos B cos C− ∑cos A sin B sin C 

⋇ tan (A + B + C) =

⋇ cot (A + B + C) =

⋇ sin (A + B) sin (A – B) = sin² A – sin² B = cos² B – cos² A

⋇ cos (A + B) cos (A – B) = cos² A – sin² B = cos² B – sin² A

 

Multiple and Sub Multiple Angles

If A is an angle, then its integral multiples 2A, 3A, 4A, … are called ‘multiple angles ‘of A and the multiple of A by fraction like are called ‘submultiple angles.

∎ If  is not an add multiple of

 

PROPERTIES OF TRIANGLES

In ∆ABC,

Lengths AB = c; BC = a; AC =b

The area of the triangle is denoted by ∆.

Perimeter of the triangle = 2s = a + b + c

A = ∠CAB; B = ∠ABC; C = ∠BCA.

R is circumradius.

Sine rule:

In ∆ABC,

⟹ a = 2R sin A; b = 2R sin B; c = 2R sin C

Where R is the circumradius and a, b, c, are lengths of the sides of ∆ABC.

Cosine rule:

In ∆ABC,

Projection rule:

In ∆ABC,

a = b cos C + c cos B

b = a cos C + c cos A

c = a cos B + b cos A

Tangent rule (Napier’s analogy):

 Area of the triangle:

In ∆ABC, a, b, and c are sides

S = and area of the triangle

 

HYPERBOLIC FUNCTIONS

⨂

⨂ The function f: R→R defined by f(x) = ∀ x ∈ R is called the ‘hyperbolic sin’ function. It is denoted by Sinh x.

∴Sinh x =

Similarly,

Identities:

⨂ cosh²x – sinh² x = 1

cosh²x = 1 + sinh² x

sinh² x = cosh² x – 1

⨂ sech² x = 1 – tanh² x

tanh² x = 1 – sesh² x

⨂ cosech² x = coth² x – 1

coth² x = 1 + coth² x

Addition formulas of hyperbolic functions:

⨂ Sinh (x + y) = Sinh x Cosh y + Cosh x Sinh y

⨂ Sinh (x − y) = Sinh x Cosh y − Cosh x Sinh y

⨂ Cosh (x + y) = Cosh x Cosh y + Sinh x Sinh y  

⨂ Cosh (x − y) = Cosh x Cosh y − Sinh x Sinh y 

⨂ tanh (x + y) =

⨂ tanh (x − y) =

⨂ coth (x + y) =

⨂ sinh 2x = 2 sinh x cosh 2x =

⨂ cosh 2x = cosh²x + sinh² x = 2 cosh²x – 1 = 1 + 2 sinh²x =

⨂ tanh 2x =

 

Inverse hyperbolic functions:

⨂ Sinh⁻¹x =  ∀ x ∈ R

⨂ Cosh⁻¹x =   ∀ x ∈ (1, ∞)

⨂ Tanh⁻¹x =    ∀ < 1

 

COMPLEX NUMBERS

The equation x² + 1 = 0 has no roots in real number system.

∴ scientists imagined a number ‘i’ such that i² = − 1.

Complex number:

if x, y are any two real numbers then the general form of the complex number is

z = x + i y; where x real part and y is the imaginary part.

3 + 4i, 2 – 5i, – 3 + 2i are the examples for Complex numbers.

• z = x +i y can be written as (x, y).
• If z₁ = x₁ + i y₁, z₂ = x₂ + i y₂, then
• z_{1 +} z₂ = (x₁ + x₂, y₁ + y₂) = (x₁ + x₂) + i (y₁ + y₂)
• z_{1 −} z₂ = (x₁ − x₂, y₁ − y₂) = (x₁ − x₂) + i (y₁ − y₂)
• z_1∙   z₂ = (x₁ x₂ −y₁ y₂, x₁y₂ + x₂y₁) = (x₁x₂ −y₁ y₂) + i (x₁y₂ +x₂ y₁)
• z_1/ z_{2 = (}x₁x₂ + y₁ y₂/x₂² +y₂², x₂ y₁ – x₁y₂/ x₂² +y₂²)

= (x₁x₂ + y₁ y₂/x₂² +y₂²) + i (x₂ y₁ – x₁y₂/ x₂² +y₂²)

  Multiplicative inverse of complex number:

   The multiplicative inverse of the complex number z is 1/z.

z = x + i y then 1/z = x – i y/ x² + y²

Conjugate complex numbers:

The complex numbers x + i y, x – i y are called conjugate complex numbers.

Conjugate of z is denoted by

The sum and product of two conjugate complex numbers are real.

If z₁, z₂ are two complex numbers then

 Modulus and amplitude of complex numbers:

Modulus: – If z = x + i y, then the non-negative real number is called modulus of z and it is denoted by or ‘r’.

Amplitude: – The complex number z = x + i y represented by the point P (x, y) on the XOY plane. ∠XOP = θ is called amplitude of z or argument of z.

x = r cos θ, y = r sin θ

x² + y² = r² cos²θ + r² sin²θ = r² (cos²θ + sin²θ) = r²(1)

⇒ x² + y² = r²

⇒ r =  and = r.

• Arg (z) = tan⁻¹(y/x)

• Arg (z₁.z₂) = Arg (z₁) + Arg (z₂) + nπ for some n ∈ { −1, 0, 1}

• Arg(z₁/z₂) = Arg (z₁) − Arg (z₂) + nπ for some n ∈ { −1, 0, 1}

Note:

∎ e^iθ = cos θ + i sin θ

∎ e^−iθ = cos θ − i sin θ

De- Moiver’s theorem

For any integer n and real number θ, (cos θ + i sin θ) ⁿ = cos n θ + i sin n θ.

→ cos α + i sin α can be written as cis α

→ cis α.cis β= cis (α + β)

→ 1/cisα = cis(-α)

→ cisα/cisβ = cis (α – β)

⟹ (cos θ + i sin θ) ^-n = cos n θ – i sin n θ

⟹ (cos θ + i sin θ) (cos θ – i sin θ) = cos²θ – i² sin²θ = cos²θ + sin²θ = 1.

→ cos θ + i sin θ = 1/ cos θ – i sin θ and cos θ – i sin θ = 1/ cos θ + i sin θ

⟹ (cos θ – i sin θ) ⁿ = (1/ (cos θ –+i sin θ)) ⁿ = (cos θ + i sin θ)^-n = cos n θ – i sin n θ

n^th root of a complex number: let n be a positive integer and z₀ ≠ 0 be a given complex number. Any complex number z satisfying z ⁿ = z₀ is called an n^th root of z₀. It is denoted by z₀^1/n or

⟹ let z = r (cos θ + i sin θ) ≠ 0 and n be a positive integer. For k∈ {0, 1, 2, 3…, (n – 1)}

let Then a₀, a₁, a_{2, …,} a_n-1 are all n distinct n^th roots of z and any n^th root of z is coincided with one of them.

n^th root of unity:  Let n be a positive integer greater than 1 and 

Note:

• The sum of the n^th roots of unity is zero.
• The product of n^th roots of unity is (– 1) ^{n – 1}.
• The n^th roots of unity 1, ω, ω², …, ω^n-1 are in geometric progression with common ratio ω.

Cube root of unity:      

x³ – 1 = 0 ⇒ x³ = 1

x =1^1/3

cube roots of unity are: 1    

ω =  , ω² =

ω² +ω + 1 = 0 and ω³ = 1

 

TRANSFORMATIONS

 

For A, B∈ R

⋇ sin (A + B) + sin (A – B) = 2sin A cos B

⋇ sin (A + B) −sin (A – B) = 2cos A sin B

⋇ cos (A + B) + cos (A – B) = 2 cos A cos B

⋇ cos (A + B) − cos (A – B) = − 2sin A sin B

For any two real numbers C and D

⋇ sin C + sin D = 2sin cos

 

⋇ sin C −sin D= 2cos  sin

⋇ cos C + cos D = 2 cos   cos

⋇ cos C − cos D = − 2sin   sin

If A + B + C = π or 180⁰, then

⋇ sin (A + B) = sin C; sin (B + C) = sin A; sin (A + C) = sin B

⋇ cos (A + B) = − cos C; cos (B + C) = −cos A; cos (A + C) = − cos B

If A + B + C = 90⁰ or then

⋇ sin   = cos;   sin   = cos ;  sin   = cos

⋇ cos  = sin; cos   = sin; cos   = sin

If    then

⋇ sin (A + B) = cos C; sin (B + C) = cos A; sin (A + C) = cos B

⋇ cos (A + B) = sin C; cos (B + C) = sin A; cos (A + C) = sin B

INVERSE TRIGONOMETRIC RATIOS

If A, B are two sets and f: A→ B is a bijection, then f^-1 is existing and f^-1: B → A is an inverse function.

Properties of Inverse Trigonometric functions:

Solutions of Simultaneous Equations

Matrix Inversion Method:

Let a system of simultaneous equations be

a₁ x + b₁ y + c₁z = d₁

a₂ x + b₂ y + c₂z = d₂

a₃ x + b₃ y + c₃z = d₃

The matrix form of the above equations is

Therefore, the matrix equation is AX = B

If Det A ≠ 0, A^-1 is exists

X = A^-1 B

By using above Condition, we get the values of x, y and z

This Method is called as Matrix Inversion Method

Cramer’s Method:

Let system of simultaneous equations be

a₁ x + b₁ y + c₁z = d₁

a₂ x + b₂ y + c₂z = d₂

a₃ x + b₃ y + c₃z = d₃

∆₁ is obtained by replacing the coefficients of x (1^st column elements of ∆) by constant values

∆₂ is obtained by replacing the coefficients of y (2^nd column elements of ∆) by constant values

∆₃ is obtained by replacing the coefficients of z (3^rd column elements of ∆) by constant values

Now

This method is called Cramer’s Method

Gauss-Jordan Method:

Let a system of simultaneous equations be

a₁ x + b₁ y + c₁z = d₁

a₂ x + b₂ y + c₂z = d₂

a₃ x + b₃ y + c₃z = d₃

Augmented matrix: The coefficient matrix (A) augmented with the constant column matrix (B) is called the augmented matrix. It is denoted by [AD].

[AD] =

This Matrix is reduced to the standard form ofby using row operations

1. Interchanging any two rows
2. Multiplying the elements of any two elements by a constant.
3. Adding to the elements of one row with the corresponding elements of another row multiplied by a constant.

∴ The solution of a given system of simultaneous equations is x = α, y = β, and z = γ.

Procedure to get the standard form:

1. Take the coefficient of x as the unity as a first equation.
2. If 1 is there in the first-row first column, then make the remaining two elements in the first column zero.
3. After that, if one element in R₂ or R₃ is 1, then make the remaining two elements in that column C₂ or C₃ as zeroes.
4. If any row contains two elements as zeros and only non-zero divide that row elements with the non-zero element to get unity and make the remaining two elements in that column as zeros.

 

 

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Errors and Approximations

Question 1

Find dy and ∆y for the following functions for the values of x and ∆x which are shown against each of the functions

(i) y = f(x) = x² + x at x = 10 when ∆x = 0.1.

Sol:

Given y = f(x) = x² + x at x = 10, ∆x = 0.1

∆y = f (x + ∆x) – f (x)

     = f (10 + 0.1) – f (10)

    = f (10.1) – f (10)

    = (10. 1)² + 10.1 – (10² + 10)

   = 102.01 + 10.1 – 100 – 10

    = 112.11 – 110

    = 2.11

dy = f’ (x) ∆x

     = (2x + 1) (0.1)

     = [2(10) + 1] (0.1)

    = 21 × 0.1

    = 2.1

(ii)  y = cos x at x = 60⁰ with ∆x = 1⁰ (1⁰ = 0.0174 radians)

Sol:

Given y = cos x, x = 60⁰ and ∆x = 1⁰

 ∆y = f (x + ∆x) – f (x)

      = cos (60⁰ + 1⁰) – cos 60⁰

      = cos (61⁰) – 0.5

      = 0.4848 – 0.5

      = – 0.0152

dy = f’ (x) ∆x

      = – sin x (1⁰)

      = – sin 60⁰ × 0.0174

      =– 0.8660 × 0.0174

      = – 0.0150

(iii)  y = x² + 3x + 6, x = 10 with ∆x = 0.01 

Sol:

   y = x² + 3x + 6

  ∆y = f (x + ∆x) – f (x)

       = f (10 + 0.01) – f (10)

       = f (10.01) – f (10)

      = (10.01)² + 3 (10.01) + 6 – (10² + 3 (10) + 6)

       = 100. 2001 + 30.03 + 6 – 100 – 30 – 6

       =130. 2301 – 130

       = 0.2301

dy = f’ (x) ∆x

     = (2x + 3 + 0) (0.01)

     = (2× 10 + 3) (0.01)

      = 23 × 0.01

      = 0.23

(iv)  y = , x = 8 and ∆x =0.02

Sol:

Question 2

The side of a square is increased from 3cm to 3.01cm find the approximate increase in the area of the square.

Sol:

Let x be the side of the square and area be A

Area of the square A = x²

 x = 3 and ∆x = 0.01

∆A = 2x × ∆x

      = 2(3) (0.01)

      = 6 × 0.01

      = 0.06

Question 3

If an increase in the side of a square is 2% then find the approximate percentage of increase in its area.

Sol:

Let x be the side of the square and A be its area

 = 2

 A = x²

∆A = 2x × ∆x

The approximate percentage error in area A

= 2 × 2 =4

Question 4

From the following. Find the approximations 

(i)

Sol:

Let f(x) =  , where x = 1000 and ∆x =– 1

f’ (x) =

Approximate value is

f (x + ∆x) = f(x) + f’ (x) ∆x

 

 = 10 – 0. 0033

 = 9.9967

(ii)

Sol:

(iii)

Sol:

(iv) Sin 62⁰

Sol:

Let f(x) = sin x, where x = 60⁰ and ∆x =2⁰

Approximate value is

 f (x + ∆x) = f(x) + f’ (x) ∆x

= sin 60⁰ + cos x (2⁰)

 = sin 60⁰+ cos 60⁰ (0.0348)

= 0.8660 + 0.5 × 0.0348

= 0.8660 + 0.0174

=0.8834

Question 5

The radius of a sphere is measured as 14cm. Later it was found that there is an error of 0.02cm in measuring the radius. Find the approximate error in the surface area of the sphere.

Sol:

Given r = 14 cm and ∆r =0.02cm

Surface area of sphere =A = 4π r²

∆A = 8π r ∆r

      = 8 ×3.14× 14 × 0.02

      = 7.0336

 

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Differentiation

Question 1 

Find f’ (x) for the following functions

(i) f(x) = (ax + b) (x > -b/a)

Sol:

Given f(x) = (ax + b) ⁿ

 f’ (x) = n (ax + b) ^{n – 1} (ax + b)  

            = n (ax + b) ^{n – 1} a

            = an (ax + b) ^{n –} 1

(ii)  f(x) = x² 2^x log x

Sol:

Given f(x) = x² 2^x log x

f’ (x) = (x²) 2^x log x + x² (2^x) log x + x² 2^x (log x).

          = 2×2^x log x +x² 2^x log a log x + x² 2^x (1/x)

          = x 2^x[log x² + x log x log 2 + 1]

(iii)  f(x) = (x > 0)

Sol:

Given f(x) =

f’ (x) =   . log 7    (x³ + 3x)

          =     log 7 (3x² + 3)

          =3 (x² + 1)  log 7

(iv) f(x) = log (sec x + tan x)

Sol:

Given, f(x) = log (sec x + tan x)

f’ (x) = (sec x + tan x)

         = (sec² x + sec x tan x)

         =  sec x (sec x + tan x)

        = sec x

Question 2

Find the derivative of  the following  functions

(i) f(x) = e^x (x² + 1)

Sol:

Given f(x) = e^x (x² + 1)

 f’ (x) = e^x (x² + 1) + (x² + 1)  (e^x)

          = e^x (2x + 0) + (x² + 1) e^x

            = e^x (x² + 2x + 1)

            = e^x (x + 1)²

(ii)

 Let y =

 

(iii) cos (log x + e^x) 

 

(iv) x = tan (e^-y)

e^-y = tan^-1 x

 

(v) cos [log (cot x)]

(vi) sin[tan^-1(e^x)]

(vii) cos^-1(4x³ – 3x)

let y = cos^-1(4x³ – 3x)

put x = cos θ ⟹ θ = cos^-1 x

y = cos^-1(4 cos ³ θ – 3cos θ)

   = cos^-1(cos 3θ)

= 3 θ

= 3 cos^-1 x

 = 3  (cos^-1 x)

      = 3

     =

(viii)

(ix)  

(x)

 

Question 3

Find f’ (x), If f(x) = (x³ + 6 x² + 12x – 13)¹⁰⁰.

Sol:

Given f(x) = (x³ + 6 x² + 12x – 13)¹⁰⁰

f’ (x) = 100(x³ + 6 x² + 12x – 13)⁹⁹ (x³ + 6 x² + 12x – 13)

          = 100(x³ + 6 x² + 12x – 13)⁹⁹ (3x² + 12 x + 12 – 0)

          =100(x³ + 6 x² + 12x – 13)⁹⁹ 3 (x² + 4 x + 4)

          = 300 (x + 2)² (x³ + 6 x² + 12x – 13)⁹⁹

Question 4

If f(x) = 1 + x + x² + x³ + …. + x¹⁰⁰, then find f’ (1).

Sol:

Given f(x) = 1 + x + x² + x³ + …. + x¹⁰⁰

           f’(x) = 0 + 1 + 2x + 3 x² + … 100 x⁹⁹

           f’(1) =  1 + 2 + 3 + … + 100

                   =

                   = 50 × 101

                  = 5050

Question 5

 From the following functions. Find their derivatives.

Question 6 

If y = , find

Sol:

Given y =

Question 7

If y = log (cosh 2x), find

Sol:

Given y = log (cosh 2x)

Question 8

If x = a cos³ t, y = a sin³ t, find

Sol:

Given If x = a cos³ t, y = a sin³ t

Question 9

Differentiate f(x) with respect to g(x) for the following.

(i) f(x) = e^x, g(x) =

f’ (x) = e^x and g’ (x) =

derivative of f(x) with respect to g(x) =

 

(ii )   

  put x = tan θ ⟹ θ = tan^-1 x

 

Question 10

if y = then prove that

Sol:

Given y =

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Inter Maths – 1B two marks questions and solutions are very useful in IPE examinations.

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Limits

Question 1

Find

Sol:

= 9

Question 2

Compute

Sol:

= a + a = 2a

Question 3

compute  

Sol:

Question 4

Show that = 1and  = –1

Sol:

we know that if x > 0

                            = –x if x < 0

As x → 0+ ⟹ x > 0

⟹  = x

=  1

As x → 0+ ⟹ x < 0

⟹  = –x

  = –1

Question 5

If f (x) = , then find and . Does  exist?

Sol:

Question 6

Show that = –1

Sol:

As x → 2– ⟹ x < 2

   x – 2 < 0

 ⟹   < 0

 

Question 7

Compute and

Sol:

As x → 2+ ⟹ x > 2

  ⟹  = 2

=                          

As x → 2– ⟹ x < 2

  ⟹  = 1

Question 8

Find

Sol:

Question 9

Compute

Sol:

Question 10

Show that

Sol:

Question 11

Compute

Sol:

Question 12

Compute

Sol:

Question 13

Evaluate

Sol:

let y = x – 1 ⟹ x = y + 1

then as x → 1, y → 0

Question 14

Compute

Sol:

Question 15

Compute

Sol:

Question 16

Compute

Sol:

 

     As x → ∞,  →0

                      = ∞ (3/4)

                      = ∞

Question 17

Compute

Sol:

We know that – 1 ≤ sin x ≤ 1

x² – 1 ≤ x² – sin x ≤ x² + 1

Question 18

Find

Sol:

Question 19

Find

Sol:

Question 20

Find

Sol:

Question 21

Find

Sol:  

Question 22

Compute

Sol:  

Question 23

Compute

Sol:

 

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Inter Maths – 1A two marks questions and solutions are very useful in IPE examinations.

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Hyperbolic Functions

Question 1

Prove that for any x∈ R, sinh (3x) = 3 sinh x + 4 sinh³ x

Sol:

sinh (3x) = sinh (2x + x)

                  = sinh 2x cosh x + cosh 2x sinh x

                 = (2 sinh x cosh x) cosh x + (1 + 2 sinh² x) sinh x

                 = 2sinh x cosh² x + sinh x + 2 sinh³ x

                 = 2 sinh x (1 + sinh² x) + sinh x + 2 sinh³ x

                 = 2 sinh x + 2 sinh³ x+ sinh x + 2 sinh³ x

                 = 3 sinh x + 4 sinh³ x

Question 2

If cosh x = , find the values of (i) cosh 2x and (ii) sinh 2x

Sol:

Given cosh x =

 Cosh 2x = 2 cosh² x – 1

                 = 2.  – 1

                 =

Sinh² 2x = cosh² 2x – 1

                 = – 1

                 =  – 1

       =

  Sinh² 2x   =

Question 3

If cosh x = sec θ then prove that tanh²= tan²

Sol:

tanh²  =

              =

             =

             =

             = tan²

Question 4

If sinh x = 5, then show that x =

Sol:

Given, sinh x = 5

      ⟹ x = sinh^-15    

We know that sinh^-1x =  

        ⟹ x =       

               x  = 

Question 5

Show that tanh^-1 =   log3

Sol:

Given tanh^-1

We know that tanh^-1 x =    

 tanh^-1  =   

                 =    

                 =   log3

Question 6

For x, y ∈ R prove that sinh (x + y) = sinh (x) cosh (y) + cosh (x) sinh (y)

Sol:

R.H.S = sinh (x) cosh (y) + cosh (x) sinh (y)

=  

= sinh (x + y)

Question 7

For any x∈ R, prove that cosh⁴ x – sinh⁴ x = cosh 2x

Sol:

cosh⁴ x – sinh⁴ x = (cosh² x)² – (sinh² x)²

                                 = (cosh² x + sinh² x) (cosh² x – sinh² x)

                                 = 1. cosh 2x

                                 = cosh 2x

Question 8

Prove that = cosh x + sinh x

Sol:

 

= cosh x + sinh x

Question 9

If sin hx = ¾ find cosh 2x and sinh 2x.

Sol:

Given sin hx = ¾

We know that cosh² x = 1 + sinh² x

                                           = 1 + (3/4)²

                                           = 1 + 9/16

                                           = 25/16

cos hx = 5/4

cosh 2x = 2cosh² x – 1

                = 2(25/16) – 1

                = 25/8 – 1

                = 17/8

Sinh 2x = 2 sinh x cosh x

                = 2 (3/4) (5/4)

                = 15/8

Question 10

Prove that (cosh x – sinh x) ⁿ = cosh nx – sinh nx

Sol:

 

∴ (cosh x – sinh x) ⁿ = cosh nx – sinh nx

 

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Trigonometric Ratios Up to Transformations

 Question 1

Find the value of sin²(π/10) + sin²(4π/10) + sin²(6π/10) + sin²(9π/10)

Sol:

 sin²(π/10) + sin²(4π/10) + sin²(6π/10) + sin²(9π/10)

= sin²(π/10) + sin²(π/2 – π/10) + sin²(π/2+ π/10) + sin²(π – π/10)

= sin²(π/10) + cos²(π/10) + cos²(π/10) + sin²(π/10)

= 1 + 1 = 2

 Question 2

If sin θ = 4/5 and θ not in the first quadrant, find the value of cos θ

Sol:

Given sin θ = 4/5 and θ not in the first quadrant

⇒ θ in the second quadrant

⇒ cos θ < 0

    cos²θ = 1 – sin² θ

              =1 – (4/5)²

             = 1 – 16/25

∴cos θ   = – 3/5 (∵cos θ < 0)

 Question 3

If 3sin θ + 4 cos θ = 5, then find the value of 4 sin θ – 3cos θ

Sol:

Given, 3sin θ + 4 cos θ = 5

let 4 sin θ – 3cos θ = x

  (3sin θ + 4 cos θ )² + (4 sin θ – 3cos θ)² = 5² + x²

 9 sin² θ + 16 cos² θ + 12 sin θ cos θ + 16 sin² θ + 9 cos² θ – 12sin θ cis θ = 25 + x²

25 sin² θ + 25 cos² θ = 25 + x²

25 = 25 + x²

⇒ x² = 0

 x = 0

∴ 4 sin θ – 3cos θ = 0

 Question 4

If sec θ + tan θ =, find the value of sin θ and determine the quadrant in which θ lies

Sol:

Given, sec θ + tan θ =  ———— (1)

 We know that sec² θ – tan² θ = 1

⇒ (sec θ + tan θ) (sec θ – tan θ) = 1

 sec θ – tan θ =

⇒ sec θ – tan θ = ———— (2)

 (1) + (2)

⇒ (sec θ + tan θ) + (sec θ – tan θ) =

2sec θ =

 sec θ =

(1) – (2)

⇒ (sec θ + tan θ) – (sec θ – tan θ) =

2 tan θ =  ⇒ tan θ =

Now sin θ = tan θ ÷ sec θ =

     Sin θ =

Since sec θ positive and tan θ is negative θ lies in the 4^th quadrant.

 Question 5

Prove that cot (π/16). cot (2π/16). cot (3π/16).… cot (7π/16) = 1

Sol:

cot (π/16). cot (2π/16). cot (3π/16).… cot (7π/16)

= cot (π/16). cot (2π/16). cot (3π/16). cot (4π/16). cot (5π/16) cot (6π/16) cot (7π/16)

= cot (π/16). cot (2π/16). cot (3π/16). cot (π/4). cot (π/2 – 3π/16) cot (π/2 – 2π/16) cot (π/2 – π/16)

= cot (π/16). cot (2π/16). cot (3π/16). cot (π/4). tan (3π/16) tan (2π/16) tan (π/16)

= [cot (π/16). tan (π/16)] [cot (2π/16). tan (2π/16)] [cot (3π/16). tan (3π/16]. cot (π/4)

= 1.1.1.1

 =1

 Question 6

If cos θ + sin θ = cos θ, then prove that cos θ – sin θ =  sin θ

Sol:

Given, cos θ + sin θ = cos θ

Sin θ =  cos θ – cos θ

           = (  – 1) cos θ

(  + 1) sin θ = (  + 1) (  – 1) cos θ

 sin θ + sin θ = cos θ

∴ cos θ – sin θ =  sin θ

 Question 7

Find the value of 2(sin⁶ θ + cos⁶ θ) – 3 (sin⁴ θ + cos⁴ θ)

Sol:

2(sin⁶ θ + cos⁶ θ) – 3 (sin⁴ θ + cos⁴ θ)

= 2[(sin² θ)³ + (cos² θ)³] – 3[(sin² θ)² + (cos²)²

= 2[(sin² θ + cos² θ)³ – 3 sin² θ cos² θ (sin² θ + cos² θ)] – 3[(sin² θ + cos² θ)² – 2 sin² θ cos² θ]

= 2[1 – 3 sin² θ cos² θ] – 3 [1 – 2 sin² θ cos² θ]

= 2 – 6 sin² θ cos² θ – 3 + 6 sin² θ cos² θ

= – 1

 Question 8

If tan 20⁰ = λ, then show that    

Sol:

Given tan 20⁰ = λ

 =

                               =

                              =

                              =

 Question 9

If sin α + cosec α = 2, find the value of sinⁿ α + cosecⁿ α, n∈ Z

Sol:

Given sin α + cosec α = 2

 ⇒ sin α + 1/ sin α = 2

 ⇒  = 2

      sin² α + 1= 2 sin α

       sin² α – 2 sin α + 1= 0

  (sin α – 1 )² = 0

⇒ sin α – 1 = 0

sin α = 1 ⇒ cosec α = 1

 sinⁿ α + cosecⁿ α = (1)ⁿ + (1)ⁿ =1 + 1 =2

∴ sinⁿ α + cosecⁿ α = 2

 Question 10

Evaluate sin² + cos²   – tan²  

Sol:

 

 Question 11

Find the value of sin 330⁰. cos 120⁰ + cos 210⁰. Sin 300⁰

Sol:

 sin 330⁰. cos 120⁰ + cos 210⁰. Sin 300⁰

=sin (360⁰ – 30⁰). cos (180⁰ – 60⁰) + cos (180⁰ + 30⁰). sin (360⁰ – 60⁰)

= (– sin 30⁰). (– cos 60⁰) + (– cos30⁰). (– sin60⁰)

= sin 30⁰.  cos 60⁰ + cos30⁰.  Sin60⁰

= sin (60⁰ + 30⁰) = sin 90⁰

=1

 Question 12

Prove that cos⁴ α + 2 cos² α = (1 – sin⁴ α)

Sol:

cos⁴ α + 2 cos² α

= cos⁴ α + 2 cos² α (1 – cos² α)

= (cos² α)² + 2 (1 – sin² α) (sin² α)

= (1 – sin² α)² + 2 sin² α – 2sin⁴ α

= 1 + sin⁴ α – 2 sin² α + 2 sin² α – 2sin⁴ α

= 1 – sin⁴ α

 Question 13

Eliminate θ from x = a cos³ θ and y = b sin³ θ

Sol:

Given x = a cos³ θ and y = b sin³ θ

 cos³ θ = x/a and sin³ θ = y/b

 cos θ = (x/a)^1/3 and sin θ = (y/b)^1/3

we know that sin² θ + cos² θ = 1

 ⇒ [(y/b)^1/3]² + [(x/a)^1/3]² = 1

   (x/a)^2/3 + (y/b)^2/3 = 1

 Question 14

Find the period of the following functions

Sol:

(i) f(x) = tan 5x

we know that period of tan kx =

⇒ period of tan 5x =

(ii) f(x) =

we know that period of =

period of =

                                      =

(iii) f(x) = 2 sin + 2 cos

period of sin  =  = 8

period of cos  =  = 6

period of given function is = LCM (8, 6) = 24

 

(iv) f(x) = tan (x + 4x + 9x +…. + n²x)

f(x) = tan (x + 4x + 9x +…. + n²x)

       = tan (1 + 4 + 9 + … + n²) x

= tanx

we know that period of tan kx =

Period of tan  =

                        =

 Question 15

Prove that sin²(52 ½)⁰ – sin² (22 ½)⁰ =

Sol:

We know that sin² A – sin²B = sin (A +B) sin (A – B)

 ⇒ sin²(52 ½)⁰ – sin² (22 ½)⁰

= sin (52 ½+ 22 ½) sin (52 ½ – 22 ½)

 = sin 75⁰ sin 30⁰

 =       

∴ sin²(52 ½)⁰ – sin² (22 ½)⁰ =

 Question 16

Prove that tan 70⁰ – tan20⁰ = 2 tan 50⁰

Sol:

50⁰ = 70⁰ – 20⁰

Tan 50⁰ = tan (70⁰ – 20⁰)

We know that tan (A –B) =

  ⇒ Tan 50⁰ =

 ⇒ tan 70⁰ – tan 20⁰ = tan 50⁰ (1 + tan70⁰ tan 20⁰)

     tan 70⁰ – tan 20⁰ = tan 50⁰ [1 + tan70⁰ cot (90⁰ – 20⁰)]

     tan 70⁰ – tan 20⁰ = tan 50⁰ [1 + tan70⁰ cot 70⁰]

     tan 70⁰ – tan 20⁰ = tan 50⁰ [1 + 1]

∴ tan 70⁰ – tan20⁰ = 2 tan 50⁰

 Question 17 

If sin α = , sin β =  and α, β are acute, show that α + β =

Sol:

Given sin α =                                               sin β =

 tan α = 1/3                                                          tan β = ½

tan (α + β) =

       

  tan (α + β) = 1

∴ α + β =

 Question 18

Find tan in terms of tan A

Sol:

 tan  =

                        =

 Question 19

Prove that = cot 36⁰

Sol:

  =

(on dividing numerator and denominator by cos 9⁰)

      =

   = tan (45⁰ + 9⁰)

    = tan 54⁰

 = tan (90⁰ – 36⁰)

 = cot 36⁰

 ∴  = cot 36⁰      

 Question 20

Show that cos 42⁰ + cos 78⁰ + cos 162⁰ = 0

Sol:

cos 42⁰ + cos 78⁰ + cos 162⁰

= cos (60⁰ – 18⁰) + cos (60⁰ + 18⁰) + cos (180⁰ – 18⁰)

=cos 60⁰ cos18⁰ + sin 60⁰ sin 18⁰ + cos 60⁰ cos 18⁰ – sin 60⁰ sin 18⁰ – cos 18⁰

= 2 cos 60⁰ cos 18⁰ – cos 18⁰

= 2 (1/2) cos 18⁰ – cos 18⁰

 = cos 18⁰ – cos 18⁰

= 0

 Question 21

Express sin θ + cos θ as a single of an angle

Sol:

sin θ + cos θ = 2(  sin θ + cos θ)

                                = 2(cos 30⁰ sin θ + sin 30⁰ cos θ)

                                = 2 sin (θ + 30⁰)

 Question 22

Find the maximum and minimum value of the following functions

(i) 3 sin x –4 cos x

a= 3, b = –4 and c = 0

  

                                     = 5

∴ minimum value = –5 and maximum value = 5

(ii) cos (x + ) + 2  sin (x + ) – 3

a= 1, b = 2  and c = – 3

 ∴ minimum value = –6 and maximum value = 0

 Question 23

Find the range of the function f(x) = 7 cos x – 24sin x + 5

Sol:

Given f(x) = 7 cos x – 24sin x + 5

a= 7, b = –24 and c = 5

∴ Range = [–20, 30]    

 Question 24

Prove that sin²α + cos² (α + β) + 2 sin α sin β cos (α + β) is independent of α

Sol:

sin²α + cos² (α + β) + 2 sin α sin β cos (α + β)

= sin²α + cos (α + β) [ cos (α + β) +2 sin α sin β]

= sin²α + cos (α + β) [ cos α cos β – sin α sin β +2 sin α sin β]

=sin²α + cos (α + β) [ cos α cos β + sin α sin β]

=sin²α + cos (α + β) cos (α –β)

= sin² α + cos² α – sin² β

=1 – sin² β

= cos² β

 Question 25

Simplify

Sol:

   =

                 =

                 = tan θ

Question 26

For what values of x in the first quadrant is positive?

Sol:

 > 0 ⟹ tan 2x > 0

⟹ 0 < 2x < π/2 (∵ x is in first quadrant)

⟹ 0 < x < π/4

Question 27

If cos θ = and π < θ < 3π/2, find the value of tan θ/2.

Sol:

cos θ =

π < θ < 3π/2 ⟹ π/2 < θ/2 < 3π/4

tan θ/2 < 0

tan θ/2 =

               =–  (tan θ/2 < 0)

              =–

           = – 2

Question 28

If A is not an integral multiple of π/2, prove that cot A – tan A = 2 cot 2A.

Sol:

cot A – tan A =

                         =

                          =

                          =

                           =

                           = 2 cot 2A

Question 29

Evaluate 6 sin 20⁰ – 8sin³ 20⁰

Sol:

6 sin 20⁰ – 8sin³ 20⁰ = 2 (3 sin 20⁰ – 4sin³ 20⁰)

                                       = 2 sin 3(20⁰)

                              = 2 sin 60⁰

                              = 2 

                              =

Question 30

Express cos⁶ A + sin⁶ A in terms of sin 2A.

Sol:

cos⁶ A + sin⁶ A

= (sin² A)³ + (cos² A)³

= (sin² A + cos² A)³ – 3 sin² A cos² A (sin² A + cos² A)

= 1 – 3 sin² A cos² A

=1 – ¾ (4 sin² A cos² A)

 = 1 – ¾ sin²2 A

Question 31

If 0 < θ < π/8, show that  = 2 cos (θ/2)

Sol:

 =2 cos (θ/2)

Question 32

Find the extreme values of cos 2x + cos²x

Sol:

cos 2x + cos²x = 2cos² x– 1 + cos² x

                              =3cos² x – 1

We know that – 1 ≤ cos x ≤ 1

 ⟹ 0 ≤ cos² x ≤ 1

      3×0 ≤ 3×cos² x ≤ 3×1

      0– 1 ≤3 cos² x – 1≤ 3– 1

   – 1≤3 cos² x – 1≤ 2

     Minimum value = – 1

     Maximum value = 2

Question 33

Prove that = 4

Sol:

= 4

Question 34

Prove that sin 78⁰ + cos 132⁰ =

Sol:

 sin 78⁰ + cos 132⁰ = sin 78⁰ + cos (90⁰ + 42⁰)

                                      = sin 78⁰ – sin 42⁰

                                      = 2 cos  sin

                                     = 2 cos 60⁰ sin 18⁰

                                      = 2

                                      =

Question 35

Find the value of sin 34⁰ + cos 64⁰ – cos4⁰

Sol:

sin 34⁰ + cos 64⁰ – cos4⁰

= sin 34⁰ –2 sin  sin

= sin 34⁰ – 2sin 34⁰ sin 30⁰

= sin 34⁰ – 2 sin 34⁰ (1/2)

=sin 34⁰ – sin 34⁰

=0

Question 36

Prove that 4(cos 66⁰ + sin 84⁰) =

Sol:

4(cos 66⁰ + sin 84⁰)  

=4(cos 66⁰ + sin (90⁰ – 6⁰)  

=4(cos 66⁰ + cos (6⁰)  

= 4[ 2 cos  cos ]

=4[ 2 cos  cos ]

=8 cos 36⁰ cos 30⁰

= 8

=

Question 35

Prove that (tan θ + cot θ)² = sec² θ + cosec² θ = sec² θ. cosec² θ

Sol:

tan θ + cot θ =   

                         =

                         =

                         = sec θ. cosec θ

(tan θ + cot θ)²   = sec² θ. cosec² θ

sec² θ + cosec² θ =

                               

                              = sec² θ. cosec² θ

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Inter Maths – 1A two marks questions and solutions are very useful in IPE examinations.

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Product Of Vectors

Question 1

If a = 6i +2 j +3 k, b = 2i – 9 j+ 6k, then find the angle between the vectors a and b

Sol:

Given vectors are a = 6i +2 j +3 k, b = 2i – 9 j+ 6k

 If θ is the angle between the vectors a and b, then cos θ = 

 a .b = (6i +2 j +3 k). (2i – 9 j+ 6k) = 6(2) + 2 (– 9) + 3(6)

          = 12 – 18 + 18 = 12

      

              = 7

      

              = 11

  ⟹ cos θ =

 θ =

Question 2

If a = i +2 j –3 k, b = 3i – j+ 2k, then show that a + b and a – b are perpendicular to each other.

Sol:

Given vectors are a = i +2 j –3 k, b = 3i – j+ 2k

a + b = (i +2 j –3 k) + (3i – j+ 2k) = 4i + j – k

a – b = (i +2 j –3 k) – (3i – j+ 2k) = –2i +3 j – 5k

(a + b). (a – b) = (4i + j – k). (–2i +3 j – 5k)

                            = – 8 + 3 + 5

                           = 0

∴ a + b and a – b is perpendicular to each other.

Question 3

If a and b be non-zero, non-collinear vectors. If , then find the angle between a and b

Sol:

Given

 Squaring on both sides

     

 (a + b) (a + b) = (a – b) (a – b)

  a² + 2 a. b + b² = a² – 2 a.b + b²

 ⟹ 4 a.b = 0

  a.b = 0

∴ the angle between a and b  is 90⁰

Question 4

If = 11, = 23 and  = 30, then find the angle between the vectors a and b and also find

Sol:

Given = 11,   = 23 and = 30

        = 30

 = 900

 = 900

(11)² – 2 ×11×23 cos θ + (23)² = 900

121 – 506 cos θ + 529   = 900

 650 – 506 cos θ = 900

cos θ =  ⟹ θ =

   

                = (11)² + 2 ×11×23 cos θ + (23)²

                = 121 + 2 ×11×23 ×  + 529

               = 400

 = 20

Question 5

If a = i – j – k and b = 2i – 3j + k, then find the projection vector of b on a and its magnitude.

Sol:

Given vectors are a = i – j – k and b = 2i – 3j + k

 a.b = (i – j – k). (2i – 3j + k) = 2 + 3 – 1 = 4

   =

 The projection vector of b on a =

                  =  (i – j – k)

  The magnitude of the projection vector =  =

Question 6

If the vectors λ i – 3j + 5k and 2λ i – λ j – k are perpendicular to each other, then find λ

Sol:

let a = λ i – 3j + 5k, b = 2λ i – λ j – k

Given, that a and b are perpendicular to each other

⟹ a.b = 0

(λ i – 3j + 5k). (2λ i – λ j – k) = 0

2 λ² + 3 λ – 5 = 0

2 λ² + 5 λ – 2 λ – 5 = 0

λ (2 λ + 5) – 1 (2 λ + 5) = 0

(2 λ + 5) ((λ – 1) = 0

λ = 1 or λ = -5/2

Question 7

Find the Cartesian equation of the plane passing through the point (– 2, 1, 3) and perpendicular to the vector 3i + j + 5k

Sol:

let P (x, y, z) be any point on the plane

 ⟹ OP = xi + yj + zk

 OA = – 2i +j +3k

AP = OP – OA = (xi + yj + zk) – (– 2i +j +3k)

 AP = (x + 2) i + (y – 1) j + (z – 3) k

 AP is perpendicular to the vector 3i + j + 5k

 ⟹ 3 (x + 2) + (y – 1) + 5(z – 3) = 0

  ⟹ 3x + 6 + y – 1 + 5z – 15 = 0

 ∴   3x + y + 5z – 10 = 0 is the required Cartesian equation of the plane

Question 8

Find the angle between the planes 2x – 3y – 6z = 5 and 6x + 2y – 9z = 4

Sol:

Given plane equations are: 2x – 3y – 6z = 5,6x + 2y – 9z = 4

 Vector equations of the above planes are: r. (2i – 3j – 6k) = 5 and r. (6i + 2j – 9k) = 4

 ⟹ n₁ = 2i – 3j – 6k and n₂ = 6i + 2j – 9k

 If θ is the angle between the planes r. n₁ = d₁ and r. n₂ = d₂, then

 Cos θ =

  ⟹ Cos θ =

 ⟹ θ =

Question 9

a = 2i – j + k, b = i – 3j – 5k. Find the vector c such that a, b, and c form the sides of a triangle.

Sol:

Given a = 2i – j + k, b = i – 3j – 5k

 If a, b, and c form the sides of a triangle, then a + b + c = 0

 ⟹ a + b = – c

⟹ c = – (a + b)

       = – [(2i – j + k) +( i – 3j – 5k)]

      = – (3i –4 j –4k)

     ∴ c = – 3i +4 j + 4k     

Question 10

Find the equation of the plane through the point (3, –2, 1) and perpendicular to the vector (4, 7, –4).

Sol:

 Let a = 3i – 2j + k and n = 4i + 7j – 4k

 The equation of the plane passing through point A(a) and perpendicular to the vector n is (r – a). n = 0

 ⟹ [r – (3i – 2j + k)]. (4i + 7j – 4k) = 0

 ⟹ r. (4i + 7j – 4k)– [(3i – 2j + k). (4i + 7j – 4k)] = 0

r. (4i + 7j – 4k)– (12 – 14 – 4) = 0

r. (4i + 7j – 4k)– 6 = 0

r. (4i + 7j – 4k) = 6

Question 11

Find the unit vector parallel to the XOY-plane and perpendicular to the vector 4i – 3j + k

Sol:

The vector which is parallel to the XOY-plane is of the form xi + yj

The vector which is parallel to the XOY-plane and perpendicular to 4i – 3j + k

 is 3i + 4j

 its magnitude =     = 5

∴ The unit vector parallel to the XOY-plane and perpendicular to the vector 4i – 3j + k =

Question 12

If a + b + c = 0, = 3,  = 5 and    = 7, then find the angle between a and b

Sol:

Given, a + b + c = 0,  = 3,  = 5 and    = 7

 a + b = – c

 (a + b)² =

 

 3² + 5² + 2  cos θ = 7²

  9 + 25 + 2.3.5 cos θ = 49

 34 + 30cos θ = 49

 30cos θ = 49 – 34

 30cos θ = 15

cos θ = 15/30 = 1/2

∴ θ = π/3

Question 13

If a = 2i – 3j + 5k, b = – i + 4j + 2k, then find a × b and unit vector perpendicular to both a and b

Sol:

 Given, a = 2i – 3j + 5k, b = – i + 4j + 2k             

 a × b =

         = i (–6 – 20) – j (4 + 5) + k (8 – 3)

         = –26i – 9j + 5k

The unit vector perpendicular to both a and b    =  

  =

=

Question 14

If a = i + j + 2k and b = 3i + 5j – k are two sides of a triangle, then find its area.

Sol:

Given, a = i + 2j + 3k and b = 3i + 5j – k

If a, b are two sides of a triangle, then area of the triangle =

a × b =

         = i (–2 – 15) – j (–1 – 9) + k (5 – 6)

         = –17i + 10 j – k

 =

Area of triangle = =

                              =

Question 15

Find the area of the parallelogram for which the vectors a = 2i – 3j and b = 3i – k are adjacent sides.

Sol:

Given, a = 2i – 3j and b = 3i – k are adjacent sides of a parallelogram

The area of the parallelogram whose vectors a , b are adjacent sides =  

  a × b=

              = i (3 – 0) – j (–2 – 0) + k (0 + 9)

             =3 i +2 j +9 k

=

                =

∴ The area of the parallelogram =

Question 16

Let a, b be two non-collinear unit vectors. If α = a – (a . b) b and β = a × b, then show that

Sol:

=

          =

              = 1 – cos² θ

             = sin² θ

 =

        = 1 + cos² θ – 2cos² θ

       = 1– cos² θ

       = sin² θ

∴

Question 17

For any vector a, show that     

Sol:

Let a = xi + yj + zk

       a × i =

                = i ( 0 – 0) – j (0 – z) + k (0 – y)

           = – yk + zj

   = y² + z²        

Similarly,  = x² + z²      and      = y² + x²        

+ +  = y² + z² + x² + z² + y² + x²        

                                                        = 2(x² + y² +z²)

                                                        =2

Question 18

If = 2,  = 3 and (p, q) = , then find

Sol:

Given = 2,  = 3 and (p, q) =

  =  sin (p, q)

                = 2 × 3 sin

                = 2 × 3×1/2

               = 3

 = 3² = 9

Question 19

If 4i + j + pk is parallel to the vector i + 2j + 3k, then find p.

Sol:

Given 4i +  j + pk is parallel to the vector i + 2j + 3k

   =  =

 = 4

⇒ p = 12

Question 20

Compute a× (b + c) + b× (c + a) + c× (a + b)

Sol:

a× (b + c) + b× (c + a) + c× (a + b)

= a× b + a× c + b× c + b× a + c × a + c × b

= a× b + a× c + b× c –a × b – a × c – b ×c

= 0

Question 21

Compute 2j× (3i – 4k) + (i + 2j) × k

Sol:

2j× (3i – 4k) + (i + 2j) × k

= 6(j × i) – 8(j × k) + (i × k) + 2(j × k)

 = – 6k – 8i – j + 2i

 = – 6i –j –6 j

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These solutions designed by the ‘Basics in Maths’ team. These notes to do help the intermediate First-year Maths students.

Inter Maths – 1A two marks questions and solutions are very useful in IPE examinations.

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Addition of Vectors

QUESTION 1

Find the unit vector in the direction of

Sol: Given vector is

 The unit vector in the direction of a vector  is given by 

QUESTION 2

Find a vector in the direction of a where that has a magnitude of 7 units.

Sol: Given vector is

 The unit vector in the direction of a vector is

 The vector having the magnitude 7 and in the direction of  is

 

QUESTION 3

Find the unit vector in the direction of the sum of the vectors, a = 2i + 2j – 5k and b = 2i + j + 3k

Sol Given vectors are a = 2i + 2j – 5k and b = 2i + j + 3k

a + b = (2i + 2j – 5k) + (2i + j + 3k) = 4i + 3j – 2k

      

 QUESTION 4

Write the direction cosines of the vector

Sol: Given vector is

    ∴ Direction cosines are

QUESTION 5

Show that the points whose position vectors are – 2a + 3b + 5c, a + 2b + 3c, 7 a – c are collinear when a, b, c are non-collinear vectors

Sol: Let OA = – 2a + 3b + 5c, OB = a + 2b + 3c, OC = 7 a – cA

B = OB – OA = a + 2b + 3c – (– 2a + 3b + 5c)

         AB = 3a – b – 2c

 AC = OC – OA = 7 a – c – (– 2a + 3b + 5c)

 AC = 9a – 3b – 6c = 3(3a – b – 2c)

          AC = 3 AB

   A, B and C are collinear

 QUESTION 6

ABCD is a parallelogram if L and M are middle points of BC and CD. Then find (i) AL and AM in terms of AB and AD (ii) 𝛌, if AM = 𝛌 AD – LM

Sol: Given, ABCD is a parallelogram and L and M are middle points of BC and CD

(i) Take A as the origin

 M is the midpoint of CD

  AM =

          = AD + ½ AB (∵ AB = DC)

 L is the midpoint of BC

  AL =

      = AB + ½ AD ((∵ BC = AD)

(ii) AM = 𝛌 AD – LM

AM + LM= 𝛌 AD 

AD + ½ AB + AD + ½ AB – (AB + ½ AD) = 𝛌 AD 

AD + ½ AB + AD + ½ AB –  AB – ½ AD = 𝛌 AD 

3/2 AD = 𝛌 AD 

∴𝛌 = 3/2

QUESTION 7

If G is the centroid of the triangle ABC, then show that OG =  when,  are the position vectors of the vertices of triangle ABC.

Sol: OA = a, OB = b, OC = c and OD = d

 D is the midpoint of BC

OD =

G divides median AD in the ratio 2: 1

OG =

∴ OG =

QUESTION 8

If = , =    are collinear vectors, then find m and n.

Sol: Given  , are collinear vectors

⟹  = λ

 Equating like vectors

 2 = 4 λ; 5 = m λ; 1 = n λ

 λ =

 5 = m ⟹ m =10

1 = n ⟹ n = 2

∴ m = 10, n = 2

QUESTION 9

Let If  , . Find the unit vector in the direction of a + b.

Sol: Given vectors are and 

     a + b =

    The unit vector in the direction of a + b =   

       =

        = 

QUESTION 10

If the vectors – 3i + 4j + λk and μi + 8j + 6k. are collinear vectors, then find λ and μ.

Sol: let a = – 3i + 4j + λk, b = μi + 8j + 6k

     ⟹   a = tb

 – 3i + 4j + λk = t (μi + 8j + 6k)

 – 3i + 4j + λk = μt i + 8t j + 6t k

Equating like vectors

– 3 = μt; 4 = 8t, λ = 6t

4 = 8t

 t =

– 3 = μ ⟹μ=– 6

λ =  6 ⟹ λ = 3

∴ μ=– 6, λ = 3

QUESTION 11

ABCD is a pentagon. If the sum of the vectors AB, AE, BC, DC, ED and AC is 𝛌 AC then find the value of 𝛌

Sol: Given, ABCD is a pentagon

        AB + AE + BC + + DC + ED + AC = 𝛌 AC

         (AB + BC) + (AE + DC + ED) + AC = 𝛌 AC

         AC + AC + AC = 𝛌 AC

         3 AC = 𝛌 AC

         𝛌 = 3

QUESTION 12

If the position vectors of the points A, B and C are – 2i + j – k and –4i + 2j + 2k and 6i – 3j – 13k respectively and AB = 𝛌 AC, then find the value of 𝛌

Sol: Given, OA = – 2i + j – k , OB = –4i + 2j + 2k and OC  = 6i – 3j – 13k

  AB = OB – OA = –4i + 2j + 2k – (– 2i + j – k)

          = –4i + 2j + 2k +2i – j + k

        = –2i + j + 3k

 AC = OC – OA = 6i – 3j – 13k – (– 2i + j – k)

       = 6i – 3j – 13k +2i – j + k

     = 8i –4 j –12k

     = – 4 (2i + j + 3k)

AC = – 4 AB

 Given AB = 𝛌 AC

    𝛌 = – 1/4

QUESTION 13

If OA = i + j +k, AB = 3i – 2j + k, BC = i + 2j – 2k, CD = 2i + j +3k then find the vector OD

Sol: Given OA = i + j +k, AB = 3i – 2j + k, BC = i + 2j – 2k, CD = 2i + j +3k

         OD = OA + AB + BC + CD

                = i + j +k + 3i – 2j + k + i + 2j – 2k + 2i + j +3k

                = 7i + 2j +4k

QUESTION 14

Let a = 2i +4 j –5 k, b = i + j+ k, c = j +2 k, then find the unit vector in the opposite direction of a + b + c

Sol:  Given, a = 2i +4 j –5 k, b = i + j+ k, c = j +2 k

         a + b + c = 2i +4 j –5 k + i + j+ k + j +2 k

                           = 3i +6j –2k

 The unit vector in the opposite direction of a + b + c is

 ⟹

      =

QUESTION 15

Is the triangle formed by the vectors 3i +5j +2k, 2i –3j –5k, –5i – 2j +3k

Sol: Let a =3i +5j +2k, b = 2i –3j –5k, c = –5i – 2j +3k

 

 ∴ Given vectors form an equilateral triangle.

QUESTION 16

Using the vector equation of the straight line passing through two points, prove that the points whose position vectors are a, b,  (3a – 2b) are collinear.

Sol: the vector equation of the straight line passing through two points a, b is

         r = (1 – t) a+ t b

         3a – 2b = (1 – t) a+ t b

          Equating like vectors

          1 – t = 3 and t = – 2

∴ Given points are collinear.

QUESTION 17

OABC is a parallelogram If OA = a and OC = c, then find the vector equation of the side BC

Sol: Given, OABC is a parallelogram and OA = a, OC = c

 The vector equation of BC is a line which is passing through C(c) and parallel to OA

 ⟹ the vector equation of BC is r = c + t a

QUESTION 18

If a, b, c are the position vectors of the vertices A, B, and C respectively of triangle ABC, then find the vector equation of the median through the vertex A

Sol: Given OA = a, OB = b, OC = c

 D is mid of BC

  OD = =

 The equation of AD is

  r = (1 – t) a + t (  )

QUESTION 19

Find the vector equation of the line passing through the point 2i +3j +k and parallel to the vector 4i – 2j + 3k

Sol: Let a =2i +3j +k, b = 4i – 2j + 3k  

         The vector equation of the line passing through a and parallel to the vector b is r = a + tb

       r = 2i +3j +k + t (4i – 2j + 3k)

          = (2 + 4t) i + (3 – 2t) j + (1 + 3t) k

QUESTION 20

Find the vector equation of the plane passing through the points i – 2j + 5k, – 2j –k and – 3i + 5j

Sol: The vector equation of the line passing through a, b and cis r = (1 – t – s) a + tb + sc

      ⟹ r = (1 – t – s) (i – 2j + 5k) + t (– 2j –k) + s (– 3i + 5j)

              = (1 – t – 4s) i + (– 2 – 3t + 7s) j + (5 – 6t – 5s) k

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This note is designed by the ‘Basics in Maths’ team. These notes to do help the intermediate First-year Maths students.

Inter Maths – 1A two marks questions and solutions are very useful in IPE examinations.

These notes cover all the topics covered in the intermediate First-year Maths syllabus and include plenty of solutions to help you solve all the major types of Math problems asked in the

IPE examinations.  

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Matrices

QUESTION 1

If A = , then show that A² = –I

Sol: Given A =

  ∴  A² = –I

QUESTION 2

If A = , and A² = 0, then find the value of k.

Sol: Given A =  and

 A² = 0

⟹ A. A =0 ⟹     = 0

⟹  = 0

     8 + 4k = 0, – 2 – k = 0 and –4 + k² = 0

    4k = –8; k = –2; k² = 4

       k = –2; k = –2; k = ± 2

   ∴ k =– 2

QUESTION 3

Find the Trace of A, If A =

Sol: Given A =

       Trace of A = 1 – 1 + 1 = 1

QUESTION 4

If A = , B = and 2X + A = B, then find X.

Sol: Given A = , B =  and 2X + A = B

        2X = B – A

        2X =  –

              =

              =

           X =  

         ∴ X =  

QUESTION 5

Find the additive inverse of A, If A =