Odisha State Board CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Additional Exercise Textbook Exercise Questions and Answers.

## CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Additional Exercise

(A) Multiple Choice Questions (Mcqs) With Answers

Question 1.

Write the value of cos^{-1} cos(3π/2).

(a) π

(b) \(\frac{\pi}{2}\)

(c) \(\frac{\pi}{4}\)

(d) 2π

Solution:

(b) \(\frac{\pi}{2}\)

Question 2.

Sets A and B have respectively m and n elements. The total number of relations from A to B is 64. If m < n and m ≠ 1, write the values of m and n respectively.

(a) m = 3, n = 2

(b) m = 2, n = 2

(c) m = 2, n = 3

(d) m = 3, n = 3

Solution:

(c) m = 2, n = 3

Question 3.

Write the principal value of

sin^{-1} (\(-\frac{1}{2}\)) + cos^{-1} cos(\( -\frac{\pi}{2}\))

(a) \(\frac{\pi}{2}\)

(b) \(\frac{\pi}{3}\)

(c) \(\frac{\pi}{4}\)

(d) π

Solution:

(b) \(\frac{\pi}{3}\)

Question 4.

Write the maximum value of x + y subject to: 2x + 3y < 6, x > 0, y > 0.

(a) 3

(b) 1

(c) 2

(d) 0

Solution:

(a) 3

Question 5.

Let A has 3 elements and B has m elements. Number of relations from A to B = 4096. Find the value of m.

(a) 2

(b) 4

(c) 1

(d) 3

Solution:

(b) 4

Question 6.

Let A is any non-empty set. Number of binary operations on A is 16. Find |A|.

(a) 2

(b) 1

(c) 3

(d) 4

Solution:

(a) 2

Question 7.

Give an example of a relation which is reflexive, transitive but not symmetric.

(a) x < y on Z

(b) x = y on Z

(c) x > y on Z

(d) None of the above

Solution:

(a) x < y on Z

Question 8.

Find the least positive integer r such that – 375 ∈ [r]_{11}

(a) r = 5

(b) r = 6

(c) r = 3

(d) r = 10

Solution:

(d) r = 10

Question 9.

Find three positive integers x_{i}, i = 1, 2, 3 satisfying 3x ≡ 2 (mod 7)

(a) x = 1, 3, 9…

(b) x = 2, 4, 6…

(c) x = 3, 10, 17…

(d) x = 2, 10, 18…

Solution:

(c) x = 3, 10, 17…

Question 10.

If the inversible function f is defined as f(x) = \(\frac{3 x-4}{5}\) write f^{-1}(x)

(a) \(\frac{5 x+4}{3}\)

(b) \(\frac{4 x+5}{3}\)

(c) \(\frac{5 x-4}{3}\)

(d) \(\frac{5 x+4}{2}\)

Solution:

(a) \(\frac{5 x+4}{3}\)

Question 11.

Let f : R → R and g : R → R defined as f(x) = |x|, g(x) = |5x – 2| then find fog.

(a) |5x + 2|

(b) |5x – 2|

(c) |2x – 2|

(d) |2x – 5|

Solution:

(b) |5x – 2|

Question 12.

Let ∗ is a binary operation defined by a ∗ b = 3a + 4b – 2, find 4 ∗ 5.

(a) 20

(b) 12

(c) 30

(d) 36

Solution:

(c) 30

Question 13.

Let the binary operation on Q defined as a ∗ b = 2a + b – ab, find 3 ∗ 4.

(a) -2

(b) -1

(c) 2

(d) 1

Solution:

(a) -2

Question 14.

Let ∗ is a binary operation on Z defined as a ∗ b = a + b – 5 find the identity element for ∗ on Z.

(a) e = 1

(b) e = 5

(c) e = -5

(d) e = -1

Solution:

(b) e = 5

Question 15.

Find the number of binary, operations on the set {a, b}.

(a) 12

(b) 14

(c) 15

(d) 16

Solution:

(d) 16

Question 16.

Let ∗ is a binary operation on [0, ¥) defined as a ∗ b = \(\sqrt{a^2+b^2}\) find the identity element.

(a) e = 0

(b) e = 2

(c) e = 1

(d) e = 3

Solution:

(a) e = 0

Question 17.

Find least non-negative integer r such that 7 × 13 × 23 × 413 ≡ r (mod 11).

(a) r = 13

(b) r = 49

(c) r = 7

(d) r = 23

Solution:

(c) r = 7

Question 18.

Find least non-negative integer r such that 1237(mod 4) + 985 (mod 4) ≡ r (mod 4).

(a) r = 1

(b) r = 2

(c) r = -2

(d) r = -1

Solution:

(b) r = 2

Question 19.

Let ∗ is a binary operation on R – {0} defined as a ∗ b = \(\frac{a b}{5}\). If 2 ∗ (x ∗ 5) = 10, then find x:

(a) x = 25

(b) x = -5

(c) x = 5

(d) x = 1

Solution:

(a) x = 25

Question 20.

Find the principal value of cos^{-1} (\( -\frac{1}{2}\)) + 2sin^{-1} (\( \frac{1}{2}\)).

(a) \(\frac{5 \pi}{6}\)

(b) \(\frac{5 \pi}{2}\)

(c) \(\frac{5 \pi}{4}\)

(d) \(\frac{\pi}{6}\)

Solution:

(a) \(\frac{5 \pi}{6}\)

Question 21.

Evaluate sin^{-1} (\(\frac{1}{\sqrt{5}}\)) + cos^{-1} (\(\frac{3}{\sqrt{10}}\))

(a) \(\frac{\pi}{2}\)

(b) \(\frac{\pi}{5}\)

(c) \(\frac{\pi}{4}\)

(d) π

Solution:

(c) \(\frac{\pi}{4}\)

Question 22.

Evaluate cos^{-1} (\(\frac{1}{2}\)) + 2sin^{-1} (\(\frac{1}{2}\)).

(a) \(\frac{2 \pi}{5}\)

(b) \(\frac{2 \pi}{3}\)

(c) π

(d) \(\frac{\pi}{3}\)

Solution:

(b) \(\frac{2 \pi}{3}\)

Question 23.

Find the value of tan^{-1} √3 – sec^{-1} (-2).

(a) –\(\frac{\pi}{3}\)

(b) \(\frac{2 \pi}{3}\)

(c) \(\frac{\pi}{3}\)

(d) \(\frac{3 \pi}{2}\)

Solution:

(a) –\(\frac{\pi}{3}\)

Question 24.

Evaluate tan (2 tan^{-1} \(\frac{1}{3}\))

(a) 2

(b) 1

(c) 0

(d) -1

Solution:

(a) 2

Question 25.

Evaluate : sin^{-1} (sin \(\frac{3 \pi}{5}\)).

(a) \(-\frac{2 \pi}{3}\)

(b) \(\frac{\pi}{3}\)

(c) \(\frac{2 \pi}{5}\)

(d) \(\frac{\pi}{5}\)

Solution:

(c) \(\frac{2 \pi}{5}\)

Question 26.

tan^{-1} (2cos\(\frac{\pi}{3}\)) is ________.

(a) \(\frac{\pi}{2}\)

(b) \(\frac{\pi}{4}\)

(c) \(\frac{\pi}{5}\)

(d) \(\frac{\pi}{3}\)

Solution:

(b) \(\frac{\pi}{4}\)

Question 27.

Evaluate : sin^{-1} (sin \(\frac{2 \pi}{3}\)) is ________.

(a) \(\frac{\pi}{2}\)

(b) \(\frac{\pi}{4}\)

(c) \(\frac{\pi}{5}\)

(d) \(\frac{\pi}{3}\)

Solution:

(d) \(\frac{\pi}{3}\)

Question 28.

The value of sin(tan^{-1} x + tan^{-1} \( \frac{1}{x}\)), x > 0 = ________.

(a) -1

(b) 1

(c) 0

(d) 2

Solution:

(b) 1

Question 29.

2sin^{-1} \( \frac{4}{5}\) + sin^{-1} \( \frac{24}{25}\) = ________.

(a) 12

(b) 15

(c) 16

(d) 20

Solution:

(b) 15

Question 30.

Evaluate: tan^{-1} 1 = (2cos\(\frac{\pi}{3}\))

(a) \(\frac{\pi}{4}\)

(b) \(\frac{\pi}{2}\)

(c) \(\frac{\pi}{3}\)

(d) \(\frac{2 \pi}{3}\)

Solution:

(a) \(\frac{\pi}{4}\)

Question 31.

If sin^{-1} (\( \frac{\pi}{5}\)) + cosec^{-1} (\(\frac{5}{4}\)) = \(\frac{5}{2}\) then find the value of x.

(a) 1

(b) 2

(c) 3

(d) 4

Solution:

(c) 3

Question 32.

Evaluate:

tan^{-1} (\( \frac{-1}{\sqrt{3}}\)) + cot^{-1} (\( \frac{1}{\sqrt{3}}\)) + tan^{-1}(sin(\( -\frac{\pi}{2}\))).

(a) \(\frac{- \pi}{12}\)

(b) \(\frac{2 \pi}{5}\)

(c) \(\frac{\pi}{12}\)

(d) \(\frac{\pi}{6}\)

Solution:

(a) \(\frac{- \pi}{12}\)

Question 33.

Evaluate sin^{-1} (cos(\( \frac{33 \pi}{5}\)))

(a) \(\frac{\pi}{10}\)

(b) \(\frac{- \pi}{10}\)

(c) \(\frac{\pi}{5}\)

(d) \(\frac{\pi}{2}\)

Solution:

(b) \(\frac{- \pi}{10}\)

Question 34.

Express the value of the following in simplest form. tan (\( \frac{\pi}{4}\) + 2cot^{-1} 3)

(a) 7

(b) 12

(c) 3

(d) 6

Solution:

(a) 7

Question 35.

Express the value of the following in simplest form sin cos^{-1} tan sec^{-1 }√2

(a) cos 0

(b) cot 0

(c) tan 0

(d) sin 0

Solution:

(d) sin 0

Question 36.

tan \(\left\{\frac{1}{2} \sin ^{-1} \frac{2 x}{1+x^2}+\frac{1}{2} \cos ^{-1} \frac{1-y^2}{1+y^2}\right\}\)

(a) \(\frac{x-y}{1+x y}\)

(b) \(\frac{x+y}{1-x y}\)

(c) \(\frac{x-y}{1+y}\)

(d) \(\frac{x+y}{x y}\)

Solution:

(b) \(\frac{x+y}{1-x y}\)

Question 37.

The relation R on the set A = [1, 2, 3] given by R = {(1, 1), (1, 2), (2, 2), (2, 3), (3, 3)} is:

(a) Reflexive

(b) Symmetric

(c) Transitive

(d) Equivalence

Solution:

(a) Reflexive

Question 38.

Let f : R → R be defined as f(x) = 3x – 2. Choose the correct answer.

(a) f is one-one onto

(b) f is many-one onto

(c) f is one-one but not onto

(d) f is neither one-one nor onto

Solution:

(a) f is one-one onto

Question 39.

Let R be a relation defined on Z as R = {(a, b) ; a^{2} + b^{2} = 25 }, the domain of R is:

(a) {3, 4, 5}

(b) {0, 3, 4, 5}

(c) {0, 3, 4, 5, -3, -4, -5}

(d) None of the above

Solution:

(c) {0, 3, 4, 5, -3, -4, -5}

Question 40.

let R be the relation in the set N given by R={(a, b) : a = b – 2, b > 6}. Choose the correct answer.

(a) (2, 4) • R

(b) (3, 8) • R

(c) (6, 8) • R

(d) (8, 10) • R

Solution:

(d) (8, 10) • R

Question 41.

Set A has 3 elements and set B has 4 elements. Then the number of injective functions that can be defined from set A to set B is

(a) 144

(b) 12

(c) 24

(d) 64

Solution:

(c) 24

Question 42.

Let R be a relation on set of lines as L_{1} R L_{2} if L_{1} is perpendicular to L_{2}. Then

(a) R is Reflexive

(b) R is transitive

(c) R is symmetric

(d) R is an equivalence relation

Solution:

(c) R is symmetric

Question 43.

A Relation from A to B is an arbitrary subset of:

(a) A × B

(b) B × B

(c) A × A

(d) B × B

Solution:

(a) A × B

Question 44.

Let T be the set of all triangles in the Euclidean plane, and let a relation R on T be defined as aRb if a is congruent to b ∀ a, b ∈ T. Then R is

(a) reflexive but not transitive

(b) transitive but not symmetric

(c) equivalence

(d) None of these

Solution:

(c) equivalence

Question 45.

The maximum number of equivalence relations on the set A = {1, 2, 3} are

(a) 1

(b) 2

(c) 3

(d) 5

Solution:

(d) 5

Question 46.

Let A = {1, 2, 3} and consider the relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}. Then R is

(a) reflexive but not symmetric

(b) reflexive but not transitive

(c) symmetric and transitive

(d) neither symmetric nor transitive

Solution:

(a) reflexive but not symmetric

Question 47.

Which of the following functions from Z into Z are bijective?

f(x) = x^{3}

f(x) = x + 2

f(x) = 2x + 1

f(x) = x^{2} + 1

Solution:

f(x) = x + 2

Question 48.

Let R be a relation on the set N of natural numbers denoted by nRm <=> n is a factor of m (i.e. n | m). Then, R is

(a) Reflexive and symmetric

(b) Transitive and symmetric

(c) Equivalence

(d) Reflexive, transitive but not symmetric

Solution:

(c) Equivalence

Question 49.

Let S = {1, 2, 3, 4, 5} and let A = S × S. Define the relation R on A as follows: (a, b) R (c, d) iff ad = cb. Then, R is

(a) reflexive only

(b) Symmetric only

(c) Transitive only

(d) Equivalence relation

Solution:

(d) Equivalence relation

Question 50.

Let X = {-1, 0, 1}, Y = {0, 2} and a function f : X → Y defined by y = 2x^{4}, is

(a) one-one onto

(b) one-one into

(c) many-one onto

(d)many-one into

Solution:

(c) many-one onto

Question 51.

Let A = R – {3}, B = R – {1}. Let f : A → B be defined by f(x) = (x-2)/(x-3). Then,

(a) f is bijective

(b) f is one-one but not onto

(c) f is onto but not one-one

(d) None of these

Solution:

(a) f is bijective

Question 52.

The function f : R → R given by f(x) = x^{3} – 1 is

(a) a one-one function

(b) an onto function

(c) a bijection

(d) neither one-one nor onto

Solution:

(c) a bijection

Question 53.

Let f : [0, ∞) → [0, 2] be defined by f(x) = 2x/1+x, then f is

(a) one-one but not onto

(b) onto but not one-one

(c) both one-one and onto

(d) neither one-one nor onto

Solution:

(a) one-one but not onto

Question 54.

If N be the set of all natural numbers, consider f : N → N such that f(x) = 2x, ∀ × ∈ N, then f is

(a) one-one onto

(b) one-one into

(c) many-one onto

(d) None of these

Solution:

(b) one-one into

Question 55.

Let f : R → R be a function defined by f(x) = x^{3} + 4, then f is

(a) injective

(b) surjective

(c) bijective

(d) none of these

Solution:

(c) bijective

Question 56.

Given set A = {a, b, c}. An identity relation in set A is

(a) R = {(a, b), (a, c)}

(b) R = {(a, a), (b, b), (c, c)}

(c) R = {(a, a), (b, b), (c, c), (a, c)}

(d) R= {(c, a), (b, a), (a, a)}

Solution:

(b) R = {(a, a), (b, b), (c, c)}

Question 57.

Set A has 3 elements and the set B has 4 elements. Then the number of injective functions that can be defined from set A to set B is

(a) 144

(b) 12

(c) 24

(d) 64

Solution:

(c) 24

Question 58.

sin(sec^{-1} x + cosec^{-1} x) =

(a) 1

(b) -1

(c) π/2

(d) π/3

Solution:

(a) 1

Question 59.

The principle value of sin^{-1} (√3/2) is:

(a) 2π/3

(b) π/6

(C) π/4

(d) π/3

Solution:

(d) π/3

Question 60.

Simplified form of cos^{-1} (4x^{3} – 3x)

(a) 3 sin^{-1} x

(b) 3 cos^{-1} x

(c) π – 3 sin^{-1} x

(d) None of these

Solution:

(b) 3 cos^{-1} x

Question 61.

tan^{-1} √3 – sec^{-1} (-2) is equal to

(a) π

(b) -π/3

(c) π/3,

(d) 2π/3

Solution:

(b) -π/3

Question 62.

If y = sec^{-1} x then

(a) 0 ≤ y ≤ π

(b) 0 ≤ y ≤ π/2

(c) -π/2 ≤ y ≤ π/2

(d) None of these

Solution:

(d) None of these

Question 63.

If x + (1/x) = 2 then the principal value of sin^{-1} x is

(a) π/4

(b) π/2

(c) π

(d) 3π/2

Solution:

(b) π/2

Question 64.

The principle value of sin^{-1} (sin 2π/3) is :

(a) 2π/3

(b) π/3

(c) -π/6

(d) π/6

Solution:

(b) π/3

Question 65.

The value of cos^{-1} (1/2) + 2sin^{-1} (1/2) is equal to

(a) π/4

(b) π/6

(c) 2π/3

(d) 5π/6

Solution:

(b) π/6

Question 66.

Principal value of tan^{-1} (-1) is

(a) π/4

(b) -π/2

(c) 5π/4

(d) -π/4

Solution:

(d) -π/4

Question 67.

A Linear function, which is minimized or maximized is called

(a) an objective function

(b) an optimal function

(c) A feasible function

(d) None of these

Solution:

(a) an objective function

Question 68.

The maximum value of Z = 3x + 4y subject to the constraints : x+ y ≤ 4, x ≥ 0, y ≥ 0 is:

(a) 0

(b) 12

(c) 16

(d) 18

Solution:

(c) 16

Question 69.

The maximum value of Z = 2x +3y subject to the constraints : x + y ≤ 1, 3x + y ≤ 4, x, y ≥ 0 is

(a) 2

(b) 4

(c) 5

(d) 3

Solution:

(c) 5

Question 70.

The point in the half plane 2x + 3y – 12 ≥ 0 is:

(a) (-7,8)

(c) (-7,-8)

(b) (7, -8)

(d) (7, 8)

Solution:

(d) (7, 8)

Question 71.

Any feasible solution which maximizes or minimizes the objective function is Called:

(a) A regional feasible solution

(b) An optimal feasible solution

(c) An objective feasible solution

(d) None of these

Solution:

(b) An optimal feasible solution

Question 72.

Objective function of a LPP is

(a) a constraint

(b) a function to be optimized

(c) a relation between the variables

(d) none of these

Solution:

(b) a function to be optimized

(B) Very Short Type Questions With Answers

Question 1.

If R is a relation on A such that R = R^{-1}, then write the type of the relation R.

Solution:

We know that (a, b) ∈ R ⇒ (b, a) ∈ R^{-1}

As R = R^{-1}, so R is symmetric. [2019(A)

Question 2.

Write the value of cos^{-1} cos (\(\frac{3 \pi}{2}\)). [2019(A)

Solution:

cos^{-1} cos (\(\frac{3 \pi}{2}\)) = cos^{-1} (0) = \(\frac{\pi}{2}\)

Question 3.

Sets A and B have respectively m and n elements. The total number of relations from A to B is 64. If m < n and m ≠ 1, write the values of m and n respectively. [2018(A)

Solution:

|A| = m and |B| = n

Number of relations from A to B = 2^{mn}.

A.T.Q. 2^{mn} = 64 = 2^{6}.

⇒ mn = 6, m < n with m ≠ 1.

∴m = 2, n = 3

Question 4.

Write the principal value of sin^{-1} (\(\frac{- 1}{2}\)) + cos^{-1} cos(\(\frac{- \pi}{2}\)) [2018(A)

Solution:

sin^{-1} (\(\frac{- 1}{2}\)) + cos^{-1} (cos\(\frac{- \pi}{2}\)) = \(\frac{- \pi}{6}\) + \(\frac{\pi}{2}\) = \(-\frac{\pi}{3}\)

Question 5.

Write the maximum value of x + y subject to: 2x + 3y ≤ 6, x ≥ 0, y ≥ 0. [2011(A)

Solution:

2x + 3y = 6 intersects the axes at (3, 0) and (0, 2)

∴ The maximum value of x + y = 3.

Question 6.

Define ‘feasible’ solution of an LPP. [2009(A)

Solution:

The solutions of LPP which satisfy all the constraints and non-negative restrictions are called feasible solutions.

Question 7.

Mention the quadrant in which the solution of an LPP with two decision variables lies when the graphical method is adopted. [2008(A)

Solution:

The solution lies in XOY or 1st quadrant.

Question 8.

Write the smallest equivalence relation on A = {1, 2, 3}.

Solution:

The relation R = {(1, 1), (2, 2), (3, 3)} is the smallest equivalence relation on set A.

Question 9.

Congruence modulo 3 relation partitions the set Z into how many equivalence classes?

Solution:

The relation congruence modulo 3 on the set Z partitions Z into three equivalence classes.

Question 10.

Give an example of a relation which is reflexive, symmetric but not transitive.

Solution:

The relation R = {(a, b), (b, a), (a, c), (c, a), (a, a), (b, b), (c, c)} defined on the set {a, b, c} is reflexive, symmetric but not transitive.

Question 11.

Give an example of a relation which is reflexive, transitive but not symmetric.

Solution:

‘‘The relation x ≤ y on Z” is reflexive, transitive but not symmetric.

Question 12.

Give an example of a relation which is reflexive but neither symmetric nor transitive.

Solution:

The relation R = {(a, a), (b, b), (c, c), (a, b), (b, c)} defined on the set A = {a, b, c} is reflexive but neither symmetric nor transitive.

Question 13.

Find three positive integers x_{i}, i =1, 2, 3 satisfying 3x ≡ 2 (mod 7)

Solution:

3x ≡ 2 mod 7

Least positive value of x = 3

Each member of [3] is a solution

∴ x = 3, 10, 17…

Question 14.

State the reason for relation R on {1, 2, 3} defined as {(1, 2), (2, 1)} is not transitive.

Solution:

(1, 2), (2, 1) ∈ R but (1, 1) ∉ R

∴ R is not transitive.

Question 15.

Give an example of a function which is injective but not surjective.

Solution:

f(x) = \(\frac{x}{2}\) from Z → R is injective but not surjective.

Question 16.

Let X = {1, 2, 3, 4}. Determine whether

f : X → X defined as given below have inverses. Find f^{-1} if it exists:

f = {(1, 2), (2, 2), (3, 2), (4, 2)}

Solution:

f is not injective hence not invertible.

Question 17.

Let ∗ is a binary operation defined by a ∗ b = 3a + 4b – 2, find 4 ∗ 5.

Solution:

4 ∗ 5 = 3 × 4 + 4 × 5 – 2

= 12 + 20 – 2

= 30

Question 18.

Let the binary operation on Q defined as a ∗ b = 2a + b – ab, find 3 ∗ 4.

Solution:

3 ∗ 4 = 6 + 4 – 12 = -2

Question 19.

Let ∗ is a binary operation on Z defined as a ∗ b = a + b – 5 find the identity element for ∗ on Z.

Solution:

Let e is the identity element.

⇒ a ∗ e = e ∗ a = a

⇒ a + e – 5 = a

⇒ e = 5

Question 20.

Find the number of binary operations on the set {a, b}.

Solution:

Number of binary operations on

{a, b} = 2^{22} = e^{4} =16.

Question 21.

Let * is a binary operation on [0, ¥) defined as a * b = \(\sqrt{\mathbf{a}^2+\mathbf{b}^2}\) find the identity element.

Solution:

Let e is the identity element

⇒ a * e = \(\sqrt{\mathbf{a}^2+\mathbf{e}^2}\) = a

⇒ a^{2} + e^{2} = a^{2}

⇒ e = 0

Question 22.

Evaluate cos^{-1} (\(\frac{1}{2}\)) + 2 sin^{-1} (\(\frac{1}{2}\)).

Solution:

cos^{-1} (\(\frac{1}{2}\)) + 2 sin^{-1} (\(\frac{1}{2}\))

= \(\frac{\pi}{3}\) + 2 × \(\frac{\pi}{6}\) = \(\frac{2 \pi}{3}\)

Question 23.

Find the value of tan^{-1} √3 – sec^{-1} (-2)

Solution:

tan^{-1} √3 – sec^{-1} (-2)

= \(\frac{\pi}{3}\) – \(\frac{2 \pi}{3}\) = – \(\frac{\pi}{3}\).

Question 24.

Evaluate tan (2 tan^{-1} \(\frac{1}{3}\))

Solution:

tan (2 tan^{-1} \(\frac{1}{3}\)) = tan tan^{-1} \(\left(\frac{\frac{2}{3}}{1-\frac{2}{3}}\right)\)

= tan tan^{-1} (2) = 2

Question 25.

Evaluate: sin^{-1} (sin \(\frac{3 \pi}{5}\)).

Solution:

sin^{-1} (sin \(\frac{3 \pi}{5}\)) = sin^{-1} sin (\(\pi \frac{-2 \pi}{5}\))

= sin^{-1} sin \(\frac{2 \pi}{5}\) = \(\frac{2 \pi}{5}\)

Question 26.

Evaluate tan^{-1} 1 = (2 cos \(\frac{\pi}{3}\))

Solution:

tan^{-1} (2 cos \(\frac{\pi}{3}\))

= tan^{-1} (2 × 1/2) = tan^{-1} 1 = \(\frac{\pi}{4}\)

Question 27.

Define the objective function.

Solution:

If C_{1}, C_{2}, C_{3} …. Cn are constants and x_{1}, x_{2}, …… x_{n} are variables then the linear function z = C_{1}x_{1} + C_{2}x_{2} +…… C_{n}x_{n} which is to be optimized is called an objective function.

Question 28.

Define feasible solution.

Solution:

A set of values of the variables x_{1}, x_{2}, …… x_{n} is called a feasible solution of LPP if it satisfies the constraints and non-negative restrictions of the problem.

Question 29.

Define a convex set.

Solution:

A set is convex if every point on the line segment joining any two points lies on it.

Question 30.

State extreme point theorem.

Solution:

Let S is a convex polygon bounded by the straight lines. The linear function z = Ax + By attains its optimum value at the vertices of S.

(C) Short Type Questions With Answers

Question 1.

Construct the multiplication table X_{7} on the set {1, 2, 3, 4, 5, 6}. Also find the inverse element of 4 if it exists. [2019(A)

Solution:

Given set A = { 1, 2, 3, 4, 5, 6} Binary operation ∗ defined on A is X_{7}.

i.e. a ∗ b = a × b mod 7

= The remainder on dividing a × b by 7

The composition table for this operation is:

∗ | 1 | 2 | 3 | 4 | 5 | 6 |

1 | 1 | 2 | 3 | 4 | 5 | 6 |

2 | 2 | 4 | 6 | 1 | 3 | 5 |

3 | 3 | 6 | 2 | 5 | 1 | 4 |

4 | 4 | 1 | 5 | 2 | 6 | 3 |

5 | 5 | 3 | 1 | 6 | 4 | 2 |

6 | 6 | 5 | 4 | 3 | 2 | 1 |

As the second row is identical to first row, we have ‘1’ as the identity element.

As 4 ∗ 2 = 2 ∗ 4 = 1 we have 4^{-1} = 2

Question 2.

Let R be a relation on the set R of real numbers such that aRb iff a – b is an integer. Test whether R is an equivalence relation. If so, find the equivalence class of 1 and \(\frac{1}{2}\). [2019(A)

Solution:

The relation R on the set of real numbers is defined as

R = { (a, b) : a – b ∈ Z}

Reflexive:

∀ a ∈ R (set of real numbers)

a – a = 0 ∈ Z

⇒ (a, a) ∈ R

⇒ R is reflexive

Symmetric:

Let (a, b) ∈ R

⇒ a – b ∈ Z

⇒ b – a ∈ Z

⇒ (b, a) ∈ R

⇒ R is symmetric.

Transitive:

Let (a, b), (b, c) ∈ R

⇒ a – b and b – c ∈ Z

⇒ a – b + b – c ∈ Z

⇒ a – c ∈ Z

⇒ (a, c) ∈ R

⇒ R is transitive

Thus R is an equivalence relation

[1] = { x ∈ R : x – 1 ∈ Z} = Z

\(\frac{1}{2}\) = { x ∈ R : x – \(\frac{1}{2}\) ∈ Z}

= {x ∈ R : x = \(\frac{2 k+1}{2}\), k ∈ Z}

Question 3.

Two types of food X and Y are mixed to prepare a mixture in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. These vitamins are available in 1 kg of food as per the table given below: [2019(A)

Vitamin | |||

Food | ^{A} |
^{B} |
^{C} |

^{X} |
^{1} |
^{2} |
^{3} |

^{Y} |
^{2} |
^{2} |
^{1} |

1 kg of food X costs ₹16 and 1 kg of food Y costs ₹20. Formulate the LPP so as to determine the least cost of the mixture containing the required amount of vitamins.

Solution:

Let x kg of food X and Y kg of food y are to be mixed to prepare the mixture.

Total cost = 16x + 20y to be minimum.

According to the question

Total vitamin A = x + 2y ≥ 10 units

Total vitamin B = 2x + 2y ≥ 12 units

Total vitamin C = 3x + y ≥ 8 units.

∴ The required LPP is minimize

Z= 16x + 20y

Subject to : x + 2y ≥ 10

x + y ≥ 6

3x + y ≥ 8

x, y ≥ 0

Question 4.

Let ~ be defined by (m, n) ~ (p, q) if mq = np, where m, n, p, q e Z – {0}. Show that it is an equivalence relation. [2018(A)

Solution:

Let A = z – {0}

~ is a relation on A x A defined as (m, n) ~ (p, q) ⇔ mq = np

Reflexive : For all (m, n) ∈ A × A

We have mn = nm

⇒ (m, n) ~ (m, n)

∴ ~ is reflexive.

Symmetric: Let {m, n), (p, q) ∈ A × A and (m, n) ~ (p, q)

⇒ mq = np

⇒ pn = qm

(p, q) ~ (m, n)

∴ ~ is symmetric.

Transitive: Let (m, n), (p, q), (x, y) ∈ A × A

and (m, n) ~ (p, q), (p, q) ~ (x, y)

⇒ mq = np and py = qx

⇒ mqpy = npqx

⇒ my = nx

⇒ (m, n) ~ (x, y)

∴ ~ is transitive.

Thus ~ is an equivalence relation.

Question 5.

Solve the following LPP graphically:

Minimize Z = 4x + 3y

subject to 2x + 5y ≥ 10

x, y ≥ 0. [2018(A)

Solution:

Given LPP is:

Minimize: Z = 4x + 3y

Subject to: 2x + 5y ≥ 10

x, y ≥ 0

Step – 1 Considering the constraints as equations we have 2x + 5y = 10

x | 5 | 0 |

y | 0 | 2 |

Let us draw the graph.

Step – 2 As 0(0, 0) does not satisfy 2x + 5y > 0 and x, y > 0 is the first quadrant, we have the shaded region is the feasible region whose vertices are A(5, 0) B(0, 2).

Step – 3 Z (5, 0) = 20

Z (0, 2) = 6 … Minimum

As the feasible region is unbounded. Let us draw the half plane.

4x + 3y < 6

x | 0 | \(\frac{3}{2}\) |

y | 2 | 0 |

Step – 4 As there is no point common to the feasible region and the half plane 4x + 3y < 6, we have Z is minimum for x = 0, y = 2 and Z(min) = 6

Question 6.

Find the feasible region of the system 2y – x > 0, 6y – 3x < 21, x > 0, y > 0. [2017 (A)

Solution:

Step – 1: Treating the constraints as equations we have

2y – x = 0

6y – 3x = 21

⇒ 2y – x = 0

2y – x = 7

Step – 2: Let us draw the lines.

Table – 1

x | 0 | 2 |

y | 0 | 1 |

x | 1 | 37 |

y | 4 | 5 |

Step – 3: Clearly A(1, 3) Satisfies both the constraints, x > 0, y > 0 is the first quadrant. Thus the shaded region is the feasible region.

Question 7.

Solve the following LPP graphically:

Maximize: Z = 20x + 30y

Subject to: 3x + 5y ≤ 15

x, y ≥ 0. [2014 (A), 2016 (A), 2017 (A)

Solution:

Given LPP is

Maximize: Z = 20x + 30y

Subject to: 3x + 5y ≤ 15

x, y ≥ 0

Step – 1 Treating the constraints as equations we get 3x + 5y = 15.

Step- 2 Let us draw the graph

x | 5 | 0 |

y | 0 | 3 |

Step – 3

As 0(0, 0) satisfies 3x + 5y ≤ 15 the shaded region is the feasible region.

Step – 4

The vertices ofthe feasible region are 0(0, 0), A(5, 0) and B(0, 3).

Z(0) = 0, Z(A) = 100, Z(B) = 90

Z attains maximum at A for x = 5 and y = 0.

The given LPP has a solution, x = 5, y = 0 and Z(max) = 100.

Question 8.

Find the feasible region of the following system:

2x + y ≥ 6, x – y ≤ 3, x ≥ 0, y ≥ 0. [2016 (A)

Solution:

Given system of inequations are

2x + y ≥ 6, x – y ≤ 3, x ≥ 0, y ≥ 0

Step- 1: Consider 2x + y = 6

x – y = 3

Step – 2: Let us draw the graph

Table- 1

x | 3 | 0 |

y | 0 | 6 |

Table- 2

x | 3 | 0 |

y | 0 | -3 |

Step – 3: 0(0, 0) satisfies x – y < 3, does not satisfy 2x + y > 6 and x > 0, y > 0 is the first quadrant. Thus the shaded region is the feasible region.

Question 9.

Solve the following LPP graphically:

Minimize: Z = 6x_{1} + 7x_{2}

Subject to: x_{1} + 2x_{2} ≥ 1

x_{1}, x_{2} ≥ 0. [2015 (A)

Solution:

Given LPP is

Minimize: Z = 6x_{1}+ 7x_{2}

Subject to: x_{1} + 2x_{2} ≥ 2

x_{1}, x_{2} ≥ 0

Let us draw the line x_{1} + 2x_{2} = 2

x_{1} |
0 | 2 |

x_{2} |
1 | 0 |

Clearly (0, 0) does not satisfy x_{1} + 2x_{2} ≥ 2 and x_{1}, x_{2} ≥ 0 is the first quadrant.

The shaded region is the feasible region.

The coordinates of vertices are A(2, 0) and B(0, 1).

Point | Z = 6x_{1} + 7x_{2} |

A (2, 0) | 12 |

B (0,1) | 7 → Minimum |

As there is no point common to the half plane 6x_{1} + 7x_{2} < 7 and the feasible region.

Z is minimum when x_{1} = 0, y_{1} =1 and the minimum value of z = 7

Question 10.

Find the feasible region of the following system:

2y – x ≥ 0, 6y – 3x ≤ 21, x ≥ 0, y ≥ 0. [2015 (A)

Solution:

Given system is

2y – x ≥ 0

6y – 3x ≤ 21

x, y ≥ 0.

Considering the constraints as equations we have

2y – x = 0

and 6y – 3x = 21

x | 0 | 2 |

x | 0 | 1 |

⇒ 3y – x = 7

x | -7 | 2 |

x | 0 | 3 |

Let us draw the lines

Clearly (2, 0) does not satisfy 2y – x ≥ 0 and satisfies 6y – 3x ≤ 21

∴ The shaded region is the feasible region.

Question 11.

Find the maximum value of z = 50x_{1} + 60x_{2}

subject to 2x_{1} + 3x_{2} ≤ 6

x_{1}, x_{2} ≥ 0. [2013 (A)

Solution:

Let us consider the constraints as equations.

2x_{1} + 3x_{2} = 6 … (1)

The table of some points on (1) is

x_{1} |
0 | 3 |

x_{2} |
2 | 0 |

Let us draw the line 2x_{1} + 3x_{2} = 6

As (0, 0) satisfies the inequality 2x_{1} + 3x_{2} ≤ 6 and x_{1}, x_{2} ≥ 0 is the first quadrant, the shaded region is the feasible region with corner points O(0, 0), A(3, 0) and B(0, 2).

Corner point | z = 50x_{1} + 60x_{2} |

O(0, 0) | 0 |

A(3, 0) | 150 |

B(0, 2) | 120 |

Thus Z_{(max)} = 150 for x_{1} = 3, x_{2} = 0

Question 12.

Shade the feasible region satisfying the inequations 2x + 3y ≤ 6, x ≥ 0, y ≥ 0 in a rough sketch. [2011(A)

Solution:

Let us consider the line 2x + 3y = 6

x_{1} |
0 | 3 |

x_{2} |
2 | 0 |

Let us draw the line on the graph

The feasible region is shaded in the figure.

Question 13.

Show the feasible region for the following constraints in a graph:

2x + y ≤ 4, x ≥ 0, y ≥ 0. [2010(A)

Solution:

Let us draw the graph of 2x + y = 4.

The shaded region shows the feasible region.