2nd PUC

Students can Download Basic Maths Exercise 11.2 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Basic Maths Question Bank Chapter 11 Linear Programming Problems Ex 11.2

Part – A

2nd PUC Basic Maths Linear Programming Problems Ex 11.2 Five Marks Questions with Answers 

Solve the following linear programming problems using graphical method

Question 1.
Maximize z = 60x + 15y
subject to x+y ≤ 50
3x + ≤ 90
x, y ≥ 0
Answer:
Consider line x + y = 50
Put x = 0, y = 50, (0,50)
Put y = 0, x = 50 (50,0)
Consider the line 3x + y =- 90
Put x = 0, y = 90 (0,90)
Put y = 0, x = 30 (30,0)
Plot the above lines on the graph we get OABC is the solution region

≥ is maximum at c (30,0)
z_max = 1800

Question 2.
Maximise z = 5x + 3y subject to
3x + 5y ≤ 15
5x + 2y ≤ 10
x ≥ 0,7 ≥ 0
Answer:
Consider 3x + 5y = 15

Plot the lines on the graph.
O ABC is the solution region.

X is maximum at B( \(\frac{20}{19}, \frac{45}{19}\) )
z_max = \(\frac{235}{19}\) = 12.37

Question 3.
Minimise z = 3x + 5y
subject to x + 3y ≥ 3
X+ y ≥ 2
x, y: ≥ 0.
Answer:
Consider x + 3y = 3

0+0 ≯ 3
x + y = 2
False

0+0 ≯  2
x + y = 2
False
∴ Shading is away from origin (0,0)

z is minimum at B(1.5, 0.5)

Question 4.
Minimise z = x – 7y + 190 subject to
x + y ≤ 8
x ≤ 5, y ≤ 5, x + y ≥ 4, x ≥ 0, y ≥ 0
Answer:


z is minimum of E (0,5) & Z_min = 155.

Question 5.
Kellogg is a new cereal formed from a mixture of bran and rice that contains atleast 88 gm of protein and atleast 36 milligram of iron per kg knowing that bran contains 80 gm of protein and 40 milligram of iron per kilogram and that rice contains 100 gm of protein and 30 milligram of iron per kilogram, find the minimum cost of producing this new cereal if bran cost ₹ 5 kilogram and rice cost ₹ 4 per kilogram.
Answer:
Let the units of Bran be x kg & Rice be y kg

Minimize Z = 5x + 4y subject to the constraints
80x + 100y ≥ 88
40x + 30 y ≥ 36
x, y ≥ 0

Plot the times on the graph.

z is minimum at B \(\left(\frac{3}{5}, \frac{2}{5}\right)\)
z_min = 4.6

Question 6.
A firm owned by Abhirami has to transport 1200 packages using large vans which can carry 200 packages each and small vans which can take 80 packages each. The cost for engaging each large van is ₹ 400 and each small van ₹ 200. Not more than ₹ 3000 is to be spent on the job and the number of large vans cannot exceed the number of small vans. Solve this L.P.P. graphically to find the minimum cost.
Answer:

Let the number of large vans be x and small vans be y minimize z = 400x + 200y subject to the constraints
200x + 80y ≥ 1,200
400x + 200y ≤ 3,000
x ≤ y


z is min at B \(\left(\frac{30}{7}, \frac{30}{7}\right)\)
Z_min = 2,571.4.

Question 7.
There are two types of fertilizers. F1 and F2 F1, consists of 10% nitrogen and 6% phosphoric acid and F, consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer Vaidhya finds that he needs atleast 14 kg of nitrogen and 14kg of phosphoric acid for the crop. If F1 costs ₹ 6 per kg and F2 costs ₹ 5 per kg, determine how much of each type of fertilizer should be used so that nutrient requirements are met at a minimum cost. Solve graphically.
Answer:

∴ z = 6x + 5y subject to the constraints


Plot the lines in graph

z is min at B_(100,80) Z_min = 1,000

Question 8.
A Television company owned by Priyanka and Bhavana operates two assembly lines, line land line II. Each of line is used to assemble the components of three types of television: colour standard and economy. The expected daily production on each line is as follows:

The daily running costs fro two lines average ₹ 6000 for line I and ₹ 4000 for line II. It is given that company must produce atleast 24 colour, 16 standard and 48 economy TV sets for which an order is pending. Formulate L.P.P. and solve graphically determining the number of days the two lines should be run to meet the requirements.
Answer:
Min Z = 6000x + 6000y subject to constraint
3x + y ≥ 24
x + y ≥ 16
2x + 6y ≥ 48
x, y ≥ 0

plot the above lines on the graph.

z is minimum at B(4, 12)
z min = 72,000

Question 9.
Old hens can be brought at ₹ 2 each and young ones at ₹ 5 each. The old hens lay 3 eggs per week and the
young ones lay 5 eggs per week, each egg being worth 30 paise. A hen cost ₹ 1 per week to feed. Deepthi has only ₹ 80 to spend for hens. How many hens of each kind should Deepthi buy to give a profit of more than ₹ 6 per week assuming that Deepthi cannot house more than 20 hens. Solve graphically.
Answer:
Max = 0.3 (3x + 5y) – (x + y) z = 0.5y – 0.1x subject to
x + y ≤ 20
2x + 5y ≤ 80
0.1 + 0.5y ≥ 6
x, y ≥ 0
Plot the lines on the graph and Solution is z is max” at (0,16) & z_max = 8

Question 10.
A company owned by Viswa Narayana concentrates on two grades of paper A and B, produced on a paper machine. Because of raw material restrictions, not more than 400 tonnes of grade A and 300 tonnes of grade B can be produced in a week. There are 160 production hours in a week. It requires 0,2 hour and 0.4 hour to produce one tone of the optimum product mix using the graphical method.
Answer:
max z = 20x + 50 y subject to
0.2x + 0.4y ≤ 160
x ≤ 400
y ≤ 300 x, y ≥ 0
Plot the lines on the graph and solution is z is maximum at (200, 300) & z_max = 19,000

Question 11.
A company owned by Navya manufactures two types of cloth, using three different colours of wool. One yard length of type A cloth required 4 oz(once) of red wool, 5 oz of green wool, 3 oz (ounce) of yellow wool. One yard length of type B cloth requires 5 oz red wool, 2 oz of green wool and 8 oz of yellow wool. The wool available for manufacture is 1000 oz of red wool 1000 oz of green wool and 1200 oz of yellow wool. The manufactures can make a profit of ₹ 5 on one yard of type A cloth and ₹ 3 on one yard of type B cloth. Find the best combination of the quantities of type A and Type B cloth which gives him maximum profit by solving the L.P.P. by graphical method.
Answer:
Maximize Z = 5x + 3y subject to
4x + 5y ≤ 1,000
5x + 2y ≤ 1,000
3x + 8y ≤ 1,200
x, y ≥ 0
Plot the lines on the graph & solution z is maximum at (0, 200) & z_max= 6,000.

Students can Download Maths Chapter 8 Application of Integrals Ex 8.2 Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Maths Question Bank Chapter 8 Application of Integrals Ex 8.2

2nd PUC Maths Application of Integrals NCERT Text Book Questions and Answers Ex 8.2

Question 1.
Find the area of the circle 4x² + 4y² = 9 which is interior to the Parobla x² = 4y.
Answer:
x² = 4y is up-ward parabola with vertex (0, 0)
Puting x² = 4y in 4x² + 4y² = 9

4 (4y) + 4y² = 9
⇒ 4y² + 16 y – 9 = 0

So required area is 2 x (shaded area is 1^st quadrant)
= 2x[(area under circle from (x = o to x = \(\sqrt{2}\) – area under parabola (from x = o to x = \(\sqrt{2}\)]

Question 2.
Find the area bounded by curves (x – 1)² + y² = 1 and x² + y² = 1.
Answer:
(1) (x – 1)² + y² =1 is a circle with centre (1,0) and radius 1
(2) x² + y² = 1 is circle with circle with centre (0,0) and radius (1)

• As shaded area is required one and is a symmetrical convex curve, so, 2 x area of half of shaded region
• It is symmentrical because radius of both are same

2x (area of curve x² + y² = 1 from x = \(\frac{1}{2}\)to x = 1
+ area curve (x -1)² + y² = 1 from x = 0 to x = \(\frac{1}{2}\))
As if we do along ‘x’ axis or ‘y’ axis, we get area of 1st quadrant, so to get from 1st and 4th quadrant for a curve is this question we need to multiply by 2.

Question 3.
Find the area of the region bounded by the curves y = x² + 2, y = x, x = 0 and x = 3.
Answer:
Curve y = x² + 2 is parabola with vertex (0, 2) (upward parabola)
y = x, line points are (1, 1) (0, 0) (-1, -1) (2, 2) etc
x = 0, line points are (0, 0) (0, 1) (0,-1), (0, 2) (0,-2) etc
x = 3, line points are (3, 0) (3, 1), (3, 2), (3, 3) (3-1) etc…..

so lines meet parabola at :-
line x = 0, y = 2 from y… y = x² + 2
line x = 3, y = 11
and line x = y, y² – y + 2 = 0 → y = not pts
so does not meet
Required area is shaded area:-
= (Area under curve y = x² + 2 from x = 0 to x = 3] – [Area under line x = y from x = 0 to x = 3]

Question 4.
Using integration find the area of region bounded by the triangle whose vertices are
(-1, 0), (1, 3) and (3, 2).
Answer:
We need line eq: for line AB, BC and AC

so line AB,


Required area is shaded area
⇒[Area under line AB from x = -1 to x = 1
+ Area under BC from x = 1 to x = 3
– Area under AC from x = -1 to x = 3]

Question 5.
Using integration find the area of the triangular region whose sides have the equations y 2x + 1, y = 3x + 1 add x = 4.
Answer:
y = 2x – 1

points are (0, 1), (1, 3) (-1, -1) (4, 9) etc y = 3x + 1
points are (0, 1), (1,4), (-1, -2) (4, 13) etc x = 4
points are (4, 0), (4, 1), (4, 2) etc….

Required area is shaded area:
= [Area under line y = 3x + 1 from (x = 0 to x = 4)]
– [Area under line y = 2x + 1 from (x = 0 to x = 4)]

Choose the correct answer in the following exercises 6 and 7.

Question 6.
Smaller area enclosed by the circle x² + y² = 4 and the line x + y = 2 is
(A) 2 (π – 2)
(B) π- 2
(C) 2π – 1
(D)2(π+2)
Answer:
Circle x² + y² = 4 his centre (0, 0) and radius 2
line x + y = 2 pts are (1,1) (2, 0) (0, 2) etc………
circle and line intersect at pts.
It can be found by solving the equation
x² + y² = 4 put (y = 2 – x)
⇒ x² + (2 -x)² = 4 ⇒ 2x² – 4x + 0 = 0
⇒ 2x² – 4x = 0 ⇒ x² – 2x = 0
⇒ (x -2) x = 0  ⇒ x = 0 or x = 2
so y = 2 – x , x = 0 , y = 2    (0, 2)
x = 2 , y = 0 , (2,0)
so they meet at (0, 2) and (2, 0)

Required area is shaded area:­-
(Area under circle x² + y² = 4 from x = 0 to x = 2)
– (Area under line x+y = 2 from x = 0 to x = 2)

Question 7.
Area lying between the curves y² = 4x and y = 2x is
(A)\(\frac{2}{3}\)
(B)\(\frac{1}{3}\)
(C)\(\frac{1}{4}\)
(D)\(\frac{3}{4}\)
Answer:
Curve y² = 4x is right parabola with vertex (0, 0)
1. y = 2x is line with pts.
(0,0) (1, 2) (-1,-2) etc…

They meet at pts
y² = 4x but y = 2x
⇒ (2x)² = 4x ⇒ 4x² = 4x ⇒ x = 1, y = 2 (1, 2)
x =0, y = 0 (0, 0)
Required area is shaded area:-
= [Area under curve y² = 4x from x = 0 to x = 1]
– [Area under line y = 2x from x = 0 to x = 1]

Students can Download Basic Maths Exercise 11.1 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Basic Maths Question Bank Chapter 11 Linear Programming Problems Ex 11.1

Part – A

2nd PUC Basic Maths Linear Programming Problems Ex 11.1 Four Marks Questions with Answers 

Question 1.
Smaran being a manufacturer produces nuts and bolts for industrial machinery. It takes 1 hour of work on machine A and 3 hours on machine B to produce a package of nuts while it takes 3 hours on machine A and 1 hour on machine B to produce a package of bolts. He earns a profit of ₹ 2.50 per package on nuts and 1 per package on bolts. Form a linear programming problem to maximize his profit, if he operates each machine for at most 12 hours a day,
Answer:

Let the number of units of nuts x & bolts be y it takes 1 hour on machine A for a nut and 3 hours on machine 13 for a nut maximum time available for machine A is 12 hours.
∴ x + 3y ≤ 12
3x + y ≤ 12
Negative units cannot be produced: x, y ≥ 0 our objective is to maximum profit the LPP is Maximize = 2.5 x + y subject to
x + 3y ≤ 12
3x + y ≤ 12
and x ≥ 0, y ≥ 0

Question 2.
Vishal consumes two types of food, A and B everyday to obtain minimum 8 units of protein, 12 units carbohydrates and 9 units of fat which is his daily requirements. 1kg of food A contains 2, 6, 1 units of protein, carbohydrates and fat respectively. 1 kg of food B contains 1, 1 and 3 units of protein, carbohydrates and fat respectively. Food A cost ₹ 8 per kg and food B cost ₹ 5 per kg. From an L.P.P. to find how many kilogram of each food should he buy daily to minimize his cost of food and still meet minimal nutritional requirements.
Answer:

Let the units of food A be x & food B be y
Minimize z = 8x + 5y subject to the constraints
2x + y ≥ 8
6x + y ≥ 12
x + 3y ≥ 9
x, y ≥ 0 : x – y ≥ 0

Question 3.
Archana, a dietician wishes to mix two types of foods F1 and F2 in such a way that the vitamin contents of the mixture contains at least 6 units of vitamin A and 8 units of vitamin B. Food F1 contains 2 units/kg of vitamin A and 3 units/kg of vitamin B while food F2 contains 3 units/kg of vitamin A and 4 units/kg of Vitamin B. Food F1 costs ₹50 per kg and food F2 costs ₹ 75 per kg. Formulate the problem as L.P.P. to minimize the cost of the mixture.
Answer:

Let the units of food F, be x and food F, be y minimize z = 50x + 75y subject to the restrictions
2x + 3y ≥ 6
3x + 4y ≥ 8
x,y ≥ 0

Question 4.
A furniture maker Jatin has 6 units of wood and 28 hours of free time in which he will make decorative screens. Two models have sold well in the past, so he will restrict himself of those two. He estimates that | model 1 requires 2 units of wood and 7 hours of time. Model 2 requires 1 units of wood and 8 hours of time. The prices of the models are ₹ 120 and ₹ 80 respectively. Formulate LPP to determine how many screens of each model should the furniture maker assemble if the wishes to maximize his sales revenue.
Answer:
Let the number of units of model I be x and model II be y

Maximize z = 120 x + 80 y subject to the constraints
2x + y ≥ 6
7x + 8y ≥ 8
x, y ≥ 0

Question 5.
A firm owned by Sheshnag manufactures two types of products A and B and sell them at a profit of ₹ 2 on type of A and 3 on type B. Each product is processed on machines M1 and M2. Type A requires one minute of processing time on M1 and two minutes on M2. Type B requires one minute of time on M1 and one minute on M2. The machine M1 is available for not more than 6 hours 40 minutes while M2 is available for 10 hours during any working day. Formulate the L.P.P. in order to find how many products of each type, should the firm produce each day so that profit is maximum.
Answer:
Let the number of units of product A bex & product B be y

Maximize z = 2x + 3y subject to the constraints
[6 hrs + 40 minutes 360 minutes + 40 minutes = 400 minutes]
x + y ≤ 400
2x + y ≤ 600
x, y ≥ 0

Question 6.
Shreya company produces 2 types of shoes A and B. A is of superior type and B is of ordinary type. Each
shoe of the first kind requires thrice as much time as the second kind. If all shoes are the second kind only, the company can produce a total of 650 pairs a day. Only a maximum of 150 pairs of the first kind and 400 of the second kind can be sold per day. If the profit per pair of the first kind is ₹ 400 and per pair of the second kind is ₹ 105. Find the optimal products mix to be produced to maximize the profit by formulating on L.P.P. model.
Answer:
Let the number of units of shoe A be x and shoe B be y, time taken by shoe A is hours & time taken by shoe B is 1 hour company can produce 650 pairs per day
∴ 3x + y ≤ 650 Similarly maximum of 150 pairs of shoe A can be produced i.e x ≤ 150 and 400 of shoe B can be sold i.e. y ≤ 400 Maximize z = 400x + 105y subject to the constraints
3x + y ≤ 650
x ≤ 150
y ≤ 400
x ≥ 0,y ≥ 0

Question 7.
Producer Rahul has 50 and 85 units of labour and capital respectively which he can use to produce two
types of goods A and B. To produce one unit of A, 1 unit of labour and 2 units of capital are required. Similarly 3 units of labour and 2 units of capital is required to produce 1 unit of B. If A and B are priced at ₹ 100 and ₹ 150 per unit respectively, how should the producer use his resources to maximize the total revenue. Formulate the L.P.P. to maximize his total revenue?
Answer:
Let the units of product A be x and product B be y

Maximize z = 100 x + 150 y subject to the restrictions
x + 3y ≤ 50
2x + 2y ≤ 85
x,y ≥ 0

Question 8.
Nikhil pesticide company must produce 200 kg mixture consisting of chemicals A and B daily. A cost ₹ 3 per kg and B cost ₹ 8 per kg. Maximum 80 kg of chemical A and atleast 60 kg, of chemical B should be used. Formulate L.P.P. model to minimize the cost.
Answer:
Let the units of A be x and B be y maximum 80 kg of chemical A used x ≤ 80
At least 60 kg of chemical B is used y ≥ 60 Company should produce 200 kg x + y ≤ 200
Minimise z = 3x + 8y subject to the constraints
X + y ≤ 200
x ≤ 80, y ≥ 60
x, y ≥ 0.

Students can Download Basic Maths Exercise 14.2 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.2

Part – A & B

2nd PUC Basic Maths Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.2 One or Two Marks Questions and Answers

Question 1.
If sin A = \(\frac { 1 }{ 2 }\) find sin 2A
Answer:
Given sin A = \(\frac { 1 }{ 2 }\) ⇒ cos A = \(\frac{\sqrt{3}}{2}\) sin2A

Question 2.
If cos A = \(\frac{\sqrt{3}}{2}\) find cos 2A
Answer:

Question 3.
If tan A = \(\frac{1}{\sqrt{3}}\) find tan 2A
Answer:

Question 4.
If sin A = \(\frac { 3 }{ 5 }\) find sin 3A
Answer:
Given sin A = \(\frac { 3 }{ 5 }\) sin3A = 3sinA – 4sin³A = \(3 \cdot \frac{3}{5}-4\left(\frac{3}{5}\right)^{3}\)
\(=\frac{9}{5}-4 \cdot \frac{27}{125}=\frac{225-108}{125}=\frac{117}{125}\)

Question 5.
If cos A = \(\frac { 4 }{ 5 }\) 4 find cos 3A
Answer:
Given cos A = \(\frac { 4 }{ 5 }\)

Question 6.
If tan A = \(\frac { 3 }{ 4 }\) find tan 3A
Answer:

Question 7.
Find the value of 3 sin 10° – 4 sin³10°
Answer:
3sin10° – 4sin³10° is of the form 3 sin A – 4sin³A = sin3A = sin 3-10° = sin 30° = \(\frac { 1 }{ 2 }\)

Question 8.
If cot A = \(\frac { 12 }{ 5 }\) and A is acute find sin3A
Answer:
Given cot A = \(\frac { 12 }{ 5 }\) ⇒ sin A = \(\frac { 5 }{ 13 }\) cosA = \(\frac { 12 }{ 13 }\)
∴ sin 3A = 3sinA – 4 sin³A

Question 9.
Show that tan A = \(\frac{\tan (A-B)+\tan B}{1-\tan (A-B) \tan B}\)
Answer:

Question 10.
Prove that \(\frac{\cos 2 A}{1+\sin 2 A}=\frac{\cos A-\sin A}{\cos A+\sin A}\)
Answer:

Question 11.
Prove that \(\frac{1+\sin 2 \theta}{\cos 2 \theta}=\frac{1+\tan \theta}{1-\tan \theta}\)
Answer:

Question 12.
Prove that \(\frac{\sin A+\sin 2 A}{1+\cos A+\cos 2 A}=\tan A\)
Answer:

Question 13.
Prove that \(\frac{\sin 2 \theta}{1+\cos 2 \theta}=\tan \theta\)
Answer:

Question 14.
Prove that (sin A – cos A)² = 1 – sin 2A
Answer:
L.H.S (sinA – cosA)²
= sin²A + cos²A – 2sinA cosA
= 1 – sin2A = R.H.S.

Question 15.
Prove that cos⁴θ – sin⁴θ = 2 cos²θ – 1
Answer:
L.H.S. cos⁴θ – sin⁴θ
= (cos²θ)² – (sin²θ)²
= (cos²θ + sin²θ) (cos²θ – sin²θ)
= 1 . (cos²θ – (1 – cos²θ)
= cos²θ – 1 + cos²θ = 2cos²θ – 1 = R.H.S.

Question 16.
Prove that \(\frac{\cos ^{3} A-\sin ^{3} A}{\cos A-\sin A}=1+\frac{1}{2} \sin 2 A\)
Answer:

Part – C

2nd PUC Basic Maths Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.2Five Marks Question and Answers

Prove the following

Question 1.
\(\frac{1+\cos 2 A+\sin 2 A}{1-\cos 2 A+\sin 2 A}=\cot A\)
Answer:

Question 2.
\(\frac{\cos 3 A}{2 \cos 2 A-1}=\cos A\) and hence find cos 15°
Answer:

Question 3.
\(\frac{1-\cos 2 A+\sin 2 A}{1+\cos 2 A+\sin 2 A}=\tan A\)
Answer:

Question 4.
cos⁶A+ sin⁶A = 1 – \(\frac { 3 }{ 4 }\) sin³(2A)
Answer:
cos⁶A+ sin⁶A [∵ a³ + b³ = (a + b)³ – 3ab(a+b)]
= (cos²A)³ + (sin²A)³
= (cos²A + sin²A)³ – 3cos²A – sin²A(cos³A + sin³A)
= 1³ – 3cos²A · sin²A.1

Question 5.
Provfe that \(\frac{\sin 3 \theta}{\sin \theta}-\frac{\cos 3 \theta}{\cos \theta}=2\)
Answer:

Question 6.
sec (45° + A) · sec(45° – A) = 2 sec 2A
Answer:

Question 7.
\(\frac{\cot A}{\cot A-\cot 3 A}+\frac{\tan A}{\tan A-\tan 3 A}=1\)
Answer:

Question 8.
Prove that \(\frac{\cos 2 A}{1+\sin 2 A}\) = tan(45° – A)
Answer:

Question 9.
If tan α = \(\frac { 1 }{ 3 }\), tan β = \(\frac { 1 }{ 7 }\) P.T. tan(2α + β) = 45°.
Answer:
L.H.S = tan(2α + β)

tan(2α + β) = 1 = tan 45° = R.H.S.

Question 10.
If tan²(45° + θ) = \(\frac { a }{ b }\) prove that \(\frac{b-a}{b+a}\) = – sin 2θ
Answer:

Question 11.
Prove that tan 2θ – tanθ = tan θ . sec 2θ.
Answer:
L.H.S = tan2θ – tanθ

Question 12.
Prove that cos 2α – tan α = \(\frac{\cos 3 \alpha}{\cos \alpha \cdot \sin 2 \alpha}\)
Answer:

Students can Download Basic Maths Exercise 10.1 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Basic Maths Question Bank Chapter 10 Learning Curve Ex 10.1

Part – A

2nd PUC Basic Maths Learning Curve Ex 10.1 One Mark Questions with Answers 

Question 1
Define learning curve.
Answer:
Learning curve is defined as curvilinear relation between the decrease in average labour hours’ per unit with the increase in the total output.

Question 2.
Write what does it deals.
Answer:
The theory of learning curve deals with labour efficiency.

Question 3.
Write the learning curve ratio.
Answer:
8^th line LCR = Curve ratio

Question 4.
Write the formula for learning index.
Answer:

Question 5.
Write the learning curve equation.
Answer:
Y = ax^b

Question 6.
Find the index of learning for 70% learning effect?
Answer:

Question 7.
Find the index of learning for 80% learning effect?
Answer:

Question 8.
What do you mean by 80% learning effect?
Answer:
80% learning effect means that when the cumulative output is doubled the cumulative average labour hours per unit will be 80% of the previous level.

Question 9.
Calculate the index of learning for 90% L effect.
Answer:

Question 10.
The average labour cost of 1^st150 units of production by a factory is ₹ 100, when the output was doubled
to 300 units, the ave labour cost was ₹ 50. Find the L.C.R & the learning curve percent.
Answer:

Part – B

2nd PUC Basic Maths Learning Curve Ex 10.1 Five Marks Questions with Answers 

Question 1.
An engineering company has 80% learning effect and spends 1000 hours to produce 1 lot of the product. Estimate the labour cost of producing 8 lots of the product if the labour cost is ₹ 40 per hour.
Answer:

Total time for lots = 4096 hours
Total labour cost at ₹ 40 per hour = 4096 × 40 = ₹ 163840.

Question 2.
ABC company required 1000 hours to produce first 30 engines. If the learning effect is 90% find the total labour cost at 20 per hour to produce total of ₹ 20 engines?
Answer:
Let 30 Engines = 1 unit
∴ 120 engines = units Units

∴ Total time required for 120 engines = 3240 hrs.
Total labour cost at ₹ 20 per hour = 3240 × 20 = ₹ 648000

Question 3.
The production manager of a company obtained the following equation for the learning effect. Y = 1400 X^-0.3 This function is based on the company’s experience for assembling the first 50 units of the product. The company was asked to bid a new order of 100 additional units and the labour cost for producing an additional 100 units at the rate of ₹ 20 per hour.
Answer:
Given a = 1400, 1 lot = 50 units = 2 lots = 100 units
∴ x = 2, y = 1400(2) ^-0.3
Taking logarithm both sides we get
Log Y = log(1400) – 0.3 log(2) = 3.1461 -0.3 × 0.3010
Log y = 3.1461 – 0.0903 = 3.0558
∴ y = A . L (3.0558) = 1137.0 hrs per lot
Total time required for 2 lots = 1137 × 2 = 2274.
Labour cost at for ₹ 20 per hour = 2274 × 20 = ₹ 45480.

Question 4.
An aircraft manufacturer supplies aircraft engines to different airlines. They have just completed an initial order for 30 engines involving a total of 6000 direct labour hours at ₹ 20 per hour. They have been asked to bid for a prospective contract for a supply of 90 engines. It is expected that there will be 80% learning effect. Estimate the labour cost for the new order.
Answer:
Let 30 engines = 1 unit, 120 engines = 4 units.

∴ Total hours for 90 engines = Total hours for 120 engines – Total hours for 30 engines = 15360 – 6000 = 9360.
The total labour cost for 90 engines at 20 per hour = 9360 × 20 = 1,87,000

Question 5.
A first sample batch of 50 units of product A took 80 hours to make. The company now wishes to estimate | the average time per unit will be if the total output of product A is 200 units and 80% learning rate
applies. Answer: Let 50 units = 1 lot ⇒ 200 units = 4lots

∴ Total hours for 200 units (4 lots) = 204.8 hours and average time per 4 lots
= 51.2 hours

Some Important Questions from Previous Question Papers

Part – C

2nd PUC Basic Maths Learning Curve Ex 10.1 Five Marks Questions with Answers 

Question 1.
The time required to produce the first unit of a product is 1000 hours. If the manufacturers experiences 80% learning effect calculate the average time pr unit & the time taken to produce altogether 8 units. Also find the total labour charges for the production of 8 units at the rate of ₹ 12.50 per hour.
Answer:
I. Table method

Total time to produce 8 units = 4096 hrs.
Given labour charges for 1 hour = ₹ 12.50
∴Labour charges for 4096 hours = 4096 x 12.50 = ₹ 51,200

II. Formula Method:
Given a = 1000, x = 8 units

y = ax^b
log y = loga + b log x = log1000 + (-0.3219) log 8
= 3.000 – 0.3219 x 0.9031 = 3.000 – 0.2907
log y = 2.7093
y = AL (2.7093) = 512 hrs. per unit
∴ Total time for 8 units = 512 x 8 = 4,096 hrs.
Labour charges per hour = ₹ 12.50.
∴ Labour charges for 4,096 hrs. = 4,096 x 12.50 = 51,200

Question 2.
An engineering company has won the contract for supplying aircraft engines of a new type. The prototype constructed to win the contract took 500 hours. It is expected that these will have 90% learning effect. Estimate the labour cost of producing 8 engines of new orders. If the labour cost is ₹ 40 per hour.
Answer:
Table Method

Total time to produce 8 units = 2916
Given labour charges per hour = ₹ 40
:: Labour charges for 2916 hrs. = 2916 ~ 40 = ₹ 1,16,640
Formula Method
Given a = 500 hrs. x = 8,

y = ax^b
log y = log a + b log x
= log 500 – 0.1521 log 8
= 2.6990 – 0.1521 x 0.9031
= 2.6990 – 0.1374
y = AL (2.5616)
= 364.43
∴ Ave time /unit = 364.43
Total time to produce 8 units = 364.43 × 8 = 2915.452 = 2916
Given labour charges 1hr. = ₹ 40
∴Labour charges for 2916 hrs. = 2916 × 40 = 1,16,640.

Question 3.
The manager of a company obtained the following equation for the learning effect Y= 1450 X^-0.2512. The function is based for assembling the first 50 units of the product find the total labour hours required to assemble 100 units.
Answer:
Given the total time for producing a lot of 50 units of a product.
∴ 100 units would constitute 2 lots of 50 units. Each :: 1 = 1450, x = 2, b = -0.2512.
y = ax^b
log y = log a + b log x
= log 1450 – 0.2512 log2
= 3.1614 – 0.2512 × 0.3010 = 3.0858
y = A.L (3.0858) = 1216 hrs. per lot
∴ Total time required for 2 lots = 2 × 1216 = 2432 hrs.

Question 4.
A manager of a company obtained the following equation for the learning effect y= 200 x^-0.3219. The first unit of product requires 200 labour hours. If the company is planning to assemble 100 units of the product find the total labour hours required.
Answer:
Given y = 200 x X^-0.3219 Where x = 100 units
Y = 200 × x^-0.3219
Where x = 100 units
y = 200 × (100)^-0.3219
log y = log 200 – 0.3219 log 100
= 2.3010 – 0.3219 × 2.0
= 2.3010 – 0.6438 = 1.6572
y = A.L (1.6572) = 45.41 per unit
∴ Total labour hours for 100 units = 45.41 × 100 = 4541 hours

Question 5.
The production manager of an electronic company obtained the following equation for the learning effect y = 1356 x ^-0.3219. The function is based on the company experience for assembling the first 50 units of the product. The company was asked to bid a new order of 100 additional units. Find the labour hours
required to assemble 100 units.
Answer:
Given y = 1356 x^-0.3219
Let 50 units = 1 lot
y = 1356(2)^-0.3219
∴ 100 units = 2 lots = x
log y = log 1356 -0.3219 log 2
= 3.1322 – 0.3219 (0.3010) = 3.1322 – 0.0969 = 3.0353.
y = A.L (3.0353) = 1085.00 hours /lot
∴ The total no. of labour hours required to assemble 2 lots (100 units) = 1085 2
= 2170 hours.

Question 6.
An aircraft manufacturer supplies aircraft engines to different Airlines. They have just completed an
initial trial order for 30 engines involving a total of 6000 direct labour hours at ₹ 20 per hour. They have been asked to bid prospective contract for supply of 90 engines. It is expected that these will be 80% learning effect. Estimate the labour cost for the new order.
Answer:
Given
a = 6000 hrs., 1 lot = 30 engines, x = 120 engines = 4 lots
Table method

Total time taken to produce 120 engines = 15360
Time taken to produce 30 engines = 6000
Total time taken to produce 90 engines = 15360 – 6000 = 9360
Labour cost at ₹ 20 per hour = 9360 × 20 = 1,87,200
Formula Method
a = 6000 hours, x = 4,

y = ax^b
log y = log a + b log x
= log 6000 – 0.3219 x log 4
= 3.7782 – 0.3219 × 0.6021 = 3.7782 – 0.1938 = 3.5844
y = A.L (3.5844) = 3841
Total number of hours required to produce 4 lots (120 engines) = 3841 × 4 = 15364 hours.
∴ Total number of hours required to produce 90 engines = 15364 – 6000 = 9364 hrs.
Labour cost at ₹ 20 per hour = 20 × 9364 = ₹ 1,87,280

Question 7.
A company requires 253.5 hours to produce the 1^st 30 units. If the learning effect is 80% Find the number of hours required to produce an additional number of 120 units. Also find the total labour cost at ₹ 20/hr.
Answer:
1 lot = 30 units
120 units = 4 lots

Total time required for 120 units = 648.96 hrs.
Additional time taken for 30 units = 253.5 hrs.
∴ Total time required to complete 150 units = 648.96 +253.5 = 902.46
∴ Total labour cost at 20₹ /hour = 902.46 × 20 = 18049.2

Question 8.
A company requires 2134.2 hours to produce the first 40 machines. If the learning effect is 80%. Find the number of hours required to produce next 120 machines.
Answer:
Let 40 units = 1 lot
120 units = 3 lots

Total time required for 4 lots (160 machines) = 5463.55
∴ Total time for 3 lots (120 machines) = 5463.55 – 2134.20 = 3329.35

Question 9.
The first unit of a product took 80 hours to manufacture. If the workers show 80% learning effect find the
total time taken to produce 6th, 7th & 8th units.
Answer:

Total time taken to produce 5 units = Time taken to produce
= 4 units + 1 unit = 204.8 + 80 = 284.8
∴ The total time taken to produce 6th, 7th & gth units.
= Time taken to produce 8 units – Time taken to produces 5 units
= 327.68 – 284.8 = 42.88.

Question 10.
A company requires 100 hrs. to produce the fst ten units at 15/per hour. If the learning effect is 80%. Find the total number of hours required to produce 160 units & also the total cost to produce 160 units.
Answer:
Given
a = 100 hrs. 10 units = 1 lot
∴ 160 units = 16 lots

y = ax^b
log y = log a + b log x
= log 100 – 0.3219 log 16
= 2 – 0.3219 1.2041 = 2 – 0.3875 = 1.625
y = A(1.6125) = 40.98 hours/unit
∴ Total time required for 160 units or 16 lots = 40.98 16 = 655.68 hrs.
∴ Total cost to produce 16 units at ₹ 15/hr. = 655.68 × 15 = ₹ 9835.20
Table method

Total time for 16 lots = 655.36 hrs.
∴ Total cost to produce 16 lots at ₹ 15/hr. = 655.36 × 15 = ₹ 19830.4.

Question 11.
ABC company Ltd., has observed that a 90% learning effect applies to all labour related costs whenever a new product is taken up for production. The anticipated production is 320 units for the coming year. The production is done in lots of 10 units each & each lot requires 1000 labour hours at ₹16 per hour. You are required to compute the total labour hours & labour costs to manufacture 320 units.
Answer:
10 units = 1lot 320 units = 32 lots

Total labour hours for 32 lots = 18,895.9 hours
Labour cost at 16₹ per hour = 18,895.9 × 16 = ₹ 3,02,334.

Question 12.
A company requires 200 hrs. to produce the first ten units at ₹ 10 per hour. If the learning effect is expected is 90%. Find the total no. of hours required to produce 320 units & also the total cost to produce 320 units.
Answer:
10 units = 1 lot
320 units = 32 lots

Total number of hours to produce 320 units = 3778.88
∴ Total cost at ₹ 10 per hour = 37762.88 × 10 = 37788.8

Students can Download Maths Chapter 8 Application of Integrals Ex 8.1 Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Maths Question Bank Chapter 8 Application of Integrals Ex 8.1

2nd PUC Maths Application of Integrals NCERT Text Book Questions and Answers Ex 8.1

Question 1.
Find the area of the region bounded by the curve y² = x and the lines x = 1, x = 4 and the x-axis.
Answer:
We know that y² = x represents right hand parabola with vertex (0,0)
Required area is shaded area:-

If not mentioned above/below/any quadrant, so we should take above and below
\(\Rightarrow 2 \times \frac{14}{3}=\frac{28}{3} \text { sq. units. }\)

Question 2.
Find the area of the region bounded by y² = 9x, x = 2, x = 4 and the x-axis in the first quadrant.
Answer:
We know y² = 9x is right handed parabola with vertex (0,0)

Question 3.
Find the area of the region bounded by x² = 4y, y = 2, y = 4 and the y – axis in the first quadrant.
Answer:
We know x² = 4y is upward parabola with vertex (0,0).
Required area is shaded area:-

Question 4.
Find the area of the region bounded by the ellipse
\(\frac{x^{2}}{16}+\frac{y^{2}}{9}=1\)
Answer:
We know curve is ellipse vertex (0,0) major axis is ‘x’ and major ‘y’
Area of given area is = 4 x (Area of shaded)

Question 5.
Find the area of the region bounded by the ellipse
\(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\)
Answer:
We know curve is ellipse vertex (0,0) major axis is ‘y’ and major ‘x’
Area of given area is = 4 x (Area of shaded)
\(=4 \times \int_{0}^{2}|y| d x\)

Question 6.
Find the area of the region in the first quadrant enclosed by x-axis, line \(x=\sqrt{3}\) y and the circle x² + y² = 4.
Answer:

Question 7.
Find the area of the smaller part of the circle x² + y² = a² cut off by the line \(x=\frac{a}{\sqrt{2}} \)
Answer:
Circle x² + y² = a² is circle with centre (0,0) and radius

Question 8.
The area between x = y² and x = 4 is divided into two equal parts by the line x = a, find the value of a.
Answer:
(1) x = y² is right hand parabola with vertex (0, 0) ⇒ y² = x
As area between y² = x, x = 4 is divided into 2 parts by x = a,

so,(area under y² = x, x = 0 to x = a) =(area under y² = x, x = a, x = 4)
⇒area (A) = area (B)

Question 9.
Find the area of the region bounded by the parabola y = x² and y = |x|
Answer:
(1) As we know x² = y is a upward parabola with vertex (0,0)
(2) line ⇒ y = |x|
So if y = 1,x = 1 (1,1)
y = -1,x = 1 (-1, 1)
y = 2, x = 2 (2, 2)
y = 2, x = 2 (-2,2)

Area required is the shaded area
= 2 x (area of one half (lets take it of 1^st quadrant))
= 2 x [(area under line y = |x|) – area under parabola y = x]

Question 10.
Find the area bounded by the curve x² = 4y and the line x = 4y – 2.
Answer:
(1) Curve x² = 4y is upward parabola vertex (0, 0)
(2) Line x = 4y – 2 (put 4y = x² in it)
⇒ x = x² – 2 ⇒ x²– x – 2 = 0
⇒ x = 2, -1

So when x = 2 , y = 1 (2,1)
\(x=-1, \quad y=\frac{1}{4} \quad\left(-1, \frac{1}{4}\right)\)
So, Area required is shaded area:-
= [Area under line x = 4y – 2 from (-1 to 2) (x axis) – [Area under x² = 4y from
-1 to 2 (x axis)]
\(\int_{-1}^{2}\left(\frac{x+2}{4}\right) d x-\int_{-1}^{2} \cfrac{x^{2}}{4} d x\)
As solving along ‘x’ axis ,so we take values in terms of x to solve

Question 11.
Find the area of the region bounded by the curve y² = 4x and the line x = 3.
Answer:
(1) Curve y² = 4x is right handed parabola with vertex (0,0) y
(2) Line x = 3, so points are (3,0) (3, 1) (3, 2)
\(\Rightarrow y^{2}=12 \Rightarrow y=\pm 2 \sqrt{3}\)

(3) But as we solve with respect to ‘x’ axis, so need only ‘x’ values:-
⇒Required area is shaded area = 2 X area of upper half of shaded region

Choose the correct answer in the following Exercise 12 and 13

Question 12.
Area lying in the first quadrant and bounded by the circle x² + y² = 4 and the lines x = 0 and x = 2 is
(A) π
(B)\(\frac{\pi}{2}\)
(C)\(\frac{\pi}{3}\)
(D)\(\frac{\pi}{4}\)
Answer:
(1) Circle is x² + y² = 4 with centre (0, 0) and radius 2.
se required area is shaded area of circle from


so answer is (A) π unit ……
OR
As circle of radius 2. so area of circle = πr²
π x 2² = 4π
As we need area of only 1 quadrant
\(\Rightarrow \frac{1}{4} \times \text { area of circle }=\frac{1}{4} \times 4 \pi=\pi \mathrm{sq.} \text { units. }\)
option (A)

Question 13.
Area of the region bounded by the curve y² = 4x, y – axis and the line y = 3 is
(A) 2
(B) \(\frac{9}{4}\)
(C) \(\frac{9}{3}\)
(D) \(\frac{9}{2}\)
Answer:
(1) Curve y² = 4x is right handed parabola
(2) line y = 3 ⇒ points in line are (0,3), (1,3), (2,3),(3, 3), (-1,3) (-2, 3) etc

(3) Pt where line meets y² = 4x is solved by . solving the equation
as we know y = 3 \(\text { so } x=\frac{3^{2}}{4} \Rightarrow \frac{9}{4}\)
\(\text { so pt is }\left(\frac{9}{4}, 3\right)\)
(If we solve along ‘y’ axis)
[Here to solve along ‘x’ axis, it is tough as to get equation of line in terms of ‘x’ is tough.
(4) So area is area under parabola from y = 0 to y = 3.

Students can Download Basic Maths Exercise 9.1 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Basic Maths Question Bank Chapter 9 Stocks and Shares Ex 9.1

Part – A

2nd PUC Basic Maths Stocks and Shares Ex 9.1 One Mark Questions with Answers ( 1 ×  10 = 10 )

Question 1.
Find the income obtained by investing 3600 in 5% stock at 90.
Answer:

Question 2.
How much stock at ₹ 75 can be bought for ₹ 3375.
Answer:
Amount 75 – Stock 100
Amount 3375 → \(\frac{3,375 \times 100}{75}\) = 4,500
∴ Stock = ₹ 4,500

Question 3.
6% stock is being sold at 15 discount. How much money is required to buy ₹ 6000 stock.
Answer:
Amount 85 – Stock 100
Amount? Stock 6,000
∴ Amount = \(\frac{6,000 \times 85}{100}\) = 5,100

Question 4.
Define yield.
Answer:
Yield is defined as nominal interest divided by Amount invested

Question 5.
What is the yield obtained when ₹ 5000, 3% stock is purchased at 125.
Answer:

Question 6.
What income can be obtained from an investment of ₹ 10,725 in 6.5% stock at 143. What is the amount of stock purchased?
Answer:

Question 7.
10% stock is quoted at ₹ 120. Find the rate of interest.
[Hint: here rate of interest means yield]
Answer:
Here rate of interest means yield]
Interest = \(=\frac{D}{A}\) X 100
= \(=\frac{10}{120}\) x 100 = 8.33%

Question 8.
How much of 8% stock at 96 can be purchased for ₹ 4800? Also find the income obtained?
Answer:
Income = \(=\frac{D}{A}\) X Cash
Income = \(=\frac{8}{96}\) X 4800
Income = ₹ 400

Question 9.
How much money will a man get by selling ₹ 6000 at 5% stock at 20 premium?
Answer:
Stock 100 – Amount 120
Stock 6000 – ?
Amount = \(\frac{6,000 \times 120}{100}\) = ₹ 7,200

Question 10.
Abhi purchased 850 shares of a company the nominal value being 10 each. Company declared an annual
dividend of 12%. How much dividend did abhi receive?
Answer:

Part – B

2nd PUC Basic Maths Stocks and Shares Ex 9.1 Three or Four Marks Questions with Answers

Question 1.
Rakshitha invests ₹ 15000 in a company paying 7% per arum, when a share of value ₹ 100 is selling for ₹ 150. What is her annual income and what percentage does she get on her money?
Answer:
Income = \(\frac{\text { Dividend }}{\text { Amount }} \times Cash\)
= \(\frac{7}{150}\)x 1500 = ₹ 700

Question 2.
Which is a better investment?
Answer:
(a) 8% stock at 110 = \(\frac{8}{110}\) × 100 = 7.272%
9% stock at 98 = \(\frac{9}{98}\) × 100 = 9.18%
98 9.18% > 7.272
∴ 9% stock at 98 is a better investment

(b) 6 \(\frac{1}{2}\) % stock at 84 \(\frac{\frac{13}{2}}{\frac{84}{84}}\) × 100 = 5.9%
4% stock at 102 = \(\frac{4}{102}\) × 100 = 3.92%
∴ 6\(\frac{1}{2}\) % stock at 84 is a better investment

(C) 7% stock at 115 = \(\frac{7}{115}\) × 100 = 6.08%
5% stock at 88 = \(\frac{5}{88}\) × 100 = 5.68%
Here 7% stock at 115 is better investment

(d) 4% stock at 118 =
\(\frac{4}{118}\) × 100 = 3.38%
6% stock at 124 = \(\frac{6}{124}\) × 100 = 6.83%
∴ 6% stock at 124 is better investment

Question 3.
A man invests equal sums of money in 4%, 5% and 6% stock, each stock being at par. If the total income of the man is 23,600. Find his total investment.
Answer:
Let the amount invested be x

Question 4.
What is the market value of 9.5% stock when an investment of ₹ 12,400 produces an income of ₹1472.5?
Answer:
Income = \(\frac{\text { Dividend }}{\text { Amount pershare }} \times cash\)
1472.5 = \(\frac{9.5}{\mathrm{A}}\) × 12,400
⇒ A = \(\frac{9.5 \times 12,400}{1972.5}\) = ₹ 80
Market value per share = ₹ 80

Question 5.
What is the quoted value of 12% stock if it earns an interest of 8% after deducting the income tax of 8%.
Answer:
Net income 92 – open income is 100
Net income 8 – ? \(\frac{100 \times 8}{92}\) , Dividend = 12
Income = \(\frac{D}{A}\) x cash

Question 6.
Venu invested ₹ 5,02,950 in Infosys when price of ₹5 shares was 3353. He sold shares worth ₹ 500 when the price went upto 3583 and the remaining he sold when the price was ₹ 3253. How much did Venu gain.
Answer:
Number of shares = \(\frac{50,2950}{3,353}\) = 150
∴ Stock = 5 x 150 = 750
Stock work 500 means 100 shares
Price of 1 share is 3,585
Price of 100 shares = 3,58,300
Remaining shares = 50
Price of 1 share is 3,253
Price of 50 shares = 1,62,650
∴Total amount received after selling the share’s is 3,58,300 + 1,62,650 = ₹ 5,20,950
Hence venu gaind 5,20,950 – 5,02,950 = ₹ 18,000

Question 7.
Ramesh has invested ₹ 4,300 partly in 4.5% stock at ₹ 72 and partly in 5% stock at ₹ 95. If the total income
from both is ₹ 250, find the investment in both the types of stock.
Answer:
Let the amount invested in 4.5% stock be x
Let the amount invested in 5% stock be 4,300 – x
45 Income obtained from 4.5% stock is \(\frac{4.5}{72}\) × x
Income obtained from 5% stock is \(\frac{5}{95}\) × (4,300 – x)
Total interest income from both = 250

19x + 16 (4,300 – x) = 250 R 16 × 19
3x = 16 (4,750 – 4,300) ⇒ 3x = 16 × 450 ⇒ x = 2,400
The amount invested in 4.5% stock is ₹ 2,400
The amount invested in 5% stock is ₹ 1,900

Question 8.
Ramesh holds ₹ 2,100 of 3% stock. He sells at
₹ 121 and invests the proceeds in 5% stock. Thereby his income increases by 14. Find the market price of 5% stock.
Answer:
Stock 100 – amount 121
Stock 2100 – ?
Amount = \(\frac{2,100 \times 121}{100}\) = 2,541
Income = \(\frac{D}{A}\) × 100
77 = \(\frac{8}{A}\) x 2,541
Amount = \(\frac{5 \times 2,541}{77}\) = ₹ 165
Market price of 5% stock = ₹ 165

Question 9.
How much must be invested in 14.25% stock at 98 to produce the same income as would be obtained by investing ₹ 9,975 in 15% stock at 105.
Answer:
Let the cash invested be x
Income from 14.25% stock at 98 = \(\frac{14.25}{98}\) x x
Income from 15% stock at 105 by investing
₹ 9,975 = \(\frac{15}{105}\) × 9,975
Given both the incomes are equal

⇒ x = 9,800
∴ Amount to be invested = ₹ 9800

Question 10.
A stock yields 5% to an investor. A fall of 75 in its price causes it to yield 5 42%. What was its market value if the two income are equal.
Answer:
Let dividend I = D₁ = 5,& D₂ = 5 \(\frac{1}{2}\)
& Amount per share ₹ x & ₹ x-5
Incomes are equal i.e. \(\frac{D_{i}}{A_{1}}=\frac{D_{2}}{A_{2}}\)
\(\frac{5}{x-5}=\frac{11}{2 x}\)
10x = 11x – 55; 55 = 11x – 10x = x in
∴ its market value is 55.

Question 11.
Sanjana invests ₹ 3240 in a stock at 108 and sells when the price falls to 104. How much stock at 130 can Sanjana now buy.
Answer:

Question 12.
A person invested 4200 partly in 5% stock at 125 and the remaining in 7.5% stock at 75. If income derived
from the two stocks is the same. Find the respective investments in each stock. Also find the total income
Answer:
Let ₹ x be invested in 5 % stock at 125 & ₹(4,200 – x) will be invested in 7.5% at 75
I₁ = \(\frac{5}{125}\) × x
I₂ = \(\frac{7.5}{75}\) × (4,200 – x)
Both incomes are equal

50x = 5,25,000 – 125x;
175x = 5,25,000 ⇒ x= 3,000
Amount invested in 5% stock is ₹ 3,000 & 7.5% is ₹ 1,200
Income from I stock = \(\frac{5}{125}\) ⇒ x 3,000 =120 & Total income ⇒ 120 + 120 = 240

Question 13.
If a person wishes to obtain 18% yield from his investment at what price should he buy 13.5% stock.
Answer:
Given yield = 18%, D = 13.5, A = ?
Yield = \(\frac{\mathrm{D}}{\mathrm{A}}\) × 100
18 = \(\frac{13.5}{A}\) x 100 ⇒ A = \(\frac{13.5 \times 100}{18}\) ₹ 75

Question 14.
(a) A invests a sum of money in 5.5% stock at 90 and B and equal sum in 3.5% stock. If A’s income is
10% more than B’s find the price of 3.5% stock.
(b) How much money has to be invested in 11.5% stock at 73 to obtain an income of ₹ 150 after a tax deduction at source of 20%.
Answer:
(a) A⁹ income = \(\frac{55}{90}\) × x
B⁹ income = \(\frac{3.5}{y}\) × x
Where x = investment
y = price of 3.5% stock
A⁹ income: B¹⁹ income :: 110 : 100

(b)
net income is 150
tax deduction at 20%
Net income 80 – gross income 100
If net income is 150 – ?

Question 15.
(a) A person invests ₹15000 partly in 3% stock at 75 and partly in 6% stock at 125. If the income from
both is ₹ 675 find his investment in the 2 types of stock.
(b) At what price a person has bought 25% shares with face value 100 which has give him 20% returns.
Answer:
(a) Let x be invested in I stock & (15,000 – x) will be invested in II stock.
I₁ = Income obtained from I^st stock is \(\frac{3 x}{75}\)
I₂ = Income obtained from 2nd stock = \(\frac{6}{125}\) (15,000-x)
Given Total income = I₁ + I₂ = 675
\(\frac{3 x}{75}\) + \(\frac{6}{125}\) (15,000-x) = 675
0.008 x = 45 ⇒ x= 5,625
∴ Amount invested in 1^st stock is ₹ 5,625 and in 2nd stock is ₹ 9,375

(b) Yield = 20%, dividend = 25%
Yield = \(\frac{\mathrm{D}}{\mathrm{A}}\) = x 100
20 = \(\frac{25} {\mathrm{A}}\) × 100
Α = \(\frac{25 \times 100}{20}\) = 125

Question 16.
Rohan invested ₹ 55000 partly in 8% stock at 80 and partly in 12% stock at 150 in such a way as to get a
return of 9% for his money. How much did Rohan invest in each.
Answer:
Let ₹ x be invested in 8% stock at 80 & ₹ (55,000 – x) be invested in 12% stock at – 150.
Total income = \(\frac{9}{100}\) × 55,000 = 4,950
Income in I stock = \(\frac{8}{800}\) x
Income in II stock = \(\frac{12}{150}\) (55,000 – x)
Total income = 4,950

12.5 x + 5,50,000 – 10x = 4,950 × 10 × 12.5
2.5x = 61,87,500 – 5,50,000
2.5x = 68,750 ⇒ x= 27,500
Rohan invests 27,500 in each

Question 17.
Prathik sells out ₹ 6000 of 7.5% stock at 108 and reinvests the proceeds in 9% stock. If Prathik’s income
increases by 270. At what price did Prathik buy 9% stock.
Answer:

Question 18.
Jane sells her ₹ 12500, 4.5% stock at 94. How much of 9% stock at 125 can Jane purchase from the sale
proceeds of the former stock. What is the change in Jane’s income.
Answer:
Amount obtained by selling 1,25,000 stock at 94 is \(\frac{12,500 \times 94}{100}\) = ₹ 11.750
100 Jane 13 income from 4.5% stock is
= \(\frac{45}{100}\) × 12,500 562.5
11,750 x 100 9% stock purchased at 125 is = \(\frac{11,750 \times 100}{125}\) = 9,400
125 Income from 9% stock is , \(\frac{9}{100}\) × 9400 = 846
100 Change in income = 846 – 562.5 = ₹ 283.5

Question 19.
Aman sells ₹ 25000, 13.5% stock when the shares were selling at a premium of 20. He invests the proceeds
partly in 15% stock at 75 and partly in 16% stock at 128. Find how much he has invested in each stock if
his income increases by ₹ 1875.
Answer:
Amount obtained by Selling 25,000
Stock at ₹ 120 is \(\frac{25,000 \times 120}{100}\) = 30.000
100 Let ₹ x be invested in 15% stock at 75
& ₹ (30,000 – x) invested in 16% stock at 128
Income from 25,000 stock is \(\frac{13.5}{100}\) × 25,000 = 3,375
100 New income = 3,375 + 1,875 = 5,250
Income from 15% stock at 75 = \(\frac{15 x}{75}\)
Income from 16% stock at 128 = \(\frac{16}{128}\) (30,000 – x)
Total income = 5,250

8x + 1,50,000 – 5x = 5,250 × 40
3x = 60,000 ⇒ x = 20,000
∴ He invested 20,000 at 15% & 10,000 at 16%

Question 20.
Nihal has ₹ 5 Maruthi shares worth ₹ 1000. He sells 120 shares when the shares are selling at ₹ 1400 and
the remaining shares when the price goes upto ₹ 1600. He invests the proceeds in ₹ 10 Maruthi shares selling at ₹ 1184 find the number of Maruthi shares purchased by Nihal.
Answer:
No. of Shares = \(\frac{1,000}{5}\) = 200
Amount received by selling 120 shares at 1400 = 120 × 1,400 = 1,68,000
Amount received by selling 80 shares art 1,600 = 80 × 1,600 = 1,28,000
∴ Total amount = 1,68,000 + 1,28,000 = 2,96,000
∴ Number of shares purchased = \(\frac{29,6000}{1,184}\) = 250

Question 21.
A man owns 50 SBI shares which are now selling at the rate of 1800. He needs 50,000 for his daughter’s
education. He decides to sell 25 SBI shares. The brokerage charged is 25%. How much more money does
he need to arrange after selling the share.
Answer:
Amount received after selling 25 shares at ₹ 1800 = 45,000
Brokerage = \(\frac{25}{100}\) × 45,000 = 112.5
Money obtained = 45,000 – 112.5 = 4,4887.5
He has to arrange for 50,000 – 4,4887.5 = 5112.5

Question 22.
Rakshith decides to invest in TCS shares which are selling at 2020 per share. How much money is required
to purchase 10 shares if the brokerage is 0.5%.
Answer:
Selling price of 10 shares at 2020 per share = 20,200
0.5 Brokerage = \(\frac{0.5}{100}\) × 20,200 = ₹ 101
100 Amount required to purchase = 20,200 + 101 = 20,301

Question 23.
Ritu purchased 200 HDFC shares when the price was 625 and then sold all the shares when the price
went upto 715. If the brokerage for each transaction was 1%. How much did Ritu gain.
Answer:
Selling price of 200 shares at 625 = 1,25,000
Brokerage is \(\frac{1}{100}\) × 1,25,000 = 1,250
Amount paid = 1,25,000 + 1,250 = 1,26,250.
Selling price of 200 shares at 715 = 1,43,000
Brokerage is \(\frac{1}{100}\) × 1,43,000 = 1,430
Amount received = 1,43,000 – 1,430 = 1,41,570
Ritu gained = 1,41,570 – 1,26,250 = ₹ 15,320

Question 24.
If a client buys shares worth ₹90,000 and sells shares worth ₹ 1,10,000 through a stock broker calculate
the brokerage payable to the stock broker if the brokerage rate is 0.5% each side.
Answer:
Brokerage is \(\frac{0.5}{100}\) × (1,10,000 – 90,000) = \(\frac{0.5}{100}\) × 20,000 = ₹ 100

Question 25.
Veena buys 100 shares of Karnataka bank at ₹ 101 per share. She pays ₹ 10,130.3 to her broker. What is
the total brokerage she paid and calculate the percentage rate of brokerage.
Answer:
Selling price of 100 shares at 101 = 10,100
Amount paid is 10130.3
Brokerage is 10130.3 – 10,100 = 30.3
100 x 30.3 Rate of brokerage is \(\frac{100 \times 30.3}{10,100}\) = 0.3%

Question 26.
Mr. Ravi sold ₹ 2,250 stock at 75 and bought stock at 88.5 with the proceeds. How much stock does he buy
if the brokerage is 2% for selling and 1.5% for buying.
Answer:
Mr. Ravi selling price of 2,250 stock at 75 is = \(\frac{2,250 \times 75}{100}\) = 1,687.5
Brokerage is \(\frac{2}{100}\) × 1,687.5 = 33.75
100 Amount received = 1,687.5 – 33.75 = 1,653.75
Brokerage for buying is 1.5%
Brokerage is on cash = \(\frac{1,653.75 \times 1.5}{100}\) = 24.81
100 Cash available for purchasing shares = 1,653.75 – 24.81 = 1,628.94
Stock purchased at 88.5 = \(\frac{1,628.94 \times 100}{88.5}\) = 1,840.61

Students can Download Basic Maths Exercise 14.1 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.1

Part – A

2nd PUC Basic Maths Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.1 Two Marks Questions and Answers

Question 1.
Obtain the values of the trigonometric functions of 75°, 15° and 105° and prove that
(i) tan 75° + cot 75° = 4
(ii) sin 105° + cos 105° = \(\frac{1}{\sqrt{2}}\)
(iii) Sec 15° + cosec 15° = \(2 \sqrt{6}\)
Answer:
sin(45° + 30°) = sin45°. cos30° + cos45° · sin 30° = \(\frac{1}{\sqrt{2}} \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}} \cdot \frac{1}{2}\)
∴ sin 75° = \(\frac{\sqrt{3}+1}{2 \sqrt{2}}\)
sin15° = sin(45° – 30°) = \(\frac{\sqrt{3}-1}{2 \sqrt{2}}\)
sin(105°) = sin(60° + 45°) = sin60°. sin45° + cos60°.cos45°
= \(\frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}}+\frac{1}{2} \frac{1}{\sqrt{2}}\) ; sin 105° = \(\frac{\sqrt{3}+1}{2 \sqrt{2}}\)
cos(75°) = cos(45° + 30°) = cos 45° – cos30° + sin45° . sin30° = \(\frac{\sqrt{3}+1}{2 \sqrt{2}}\)
cos105° = cos(60° + 45°) = cos60°. cos45° – sin60° . sin45° = \(\frac{+1-\sqrt{3}}{2 \sqrt{2}}\)

Question 2.
If sin A = \(\frac { 3 }{ 5 }\), cos B = \(\frac { 4 }{ 5 }\) find sin(A + B) and cos(A – B). Where A and B are acute angles.
Answer:
Given sin A = \(\frac { 3 }{ 5 }\) , cos B = \(\frac { 4 }{ 5 }\)
⇒ cos A = \(\frac { 4 }{ 5 }\) & sin B = \(\frac { 3 }{ 5 }\) ∵ sin²A + cos²A = 1
(i) sin(A + B) = sinA cosB + cosA sinB = \(\frac{3}{5} \cdot \frac{4}{5}+\frac{4}{5} \cdot \frac{3}{5}=\frac{24}{25}\)
(ii) cos(A – B) = COSA COSB + sinA sinB = \(\frac{4}{5} \cdot \frac{4}{5}+\frac{3}{5} \frac{3}{5}=\frac{16}{25}+\frac{9}{25}=\frac{25}{25}=1\)

Question 3.
If cos A = \(\frac { 5 }{ 13 }\), cos B = \(\frac { 24 }{ 25 }\) find cos(A + B) and sin (A – B). A and B are acute angles.
Answer:
Given
cos A = \(\frac { 5 }{ 13 }\) cosB = \(\frac { 24 }{ 25 }\)

cos (A + B) = cosA cosB – sinA sinB = \(\frac{5}{13} \cdot \frac{24}{25}-\frac{12}{13} \cdot \frac{7}{25}=\frac{120-84}{325}=\frac{36}{325}\)
sin(A – B) = sinA . cosB – cosA sinB = \(\frac{12}{13} \cdot \frac{24}{25}-\frac{5}{13} \cdot \frac{7}{25}=\frac{288-35}{325}=\frac{253}{325}\)

Question 4.
If sec A = \(\frac { 17 }{ 8 }\) , cosec B = \(\frac { 5 }{ 4 }\) find set (A + B). cosec(A + B), A and B are acute angles
Answer:
Given sec = \(\frac { 17 }{ 8 }\) ⇒ cos A = \(\frac { 8 }{ 17 }\) ∴ sin A = \(\frac { 15 }{ 17 }\) cosec B = \(\frac { 5 }{ 4 }\), sinB = \(\frac { 4 }{ 5 }\), cosB = \(\frac { 3 }{ 5 }\)

sin(A – B) = sinA cosB – cosA sinB
\(=\frac{15}{17} \cdot \frac{3}{5}-\frac{8}{17} \cdot \frac{4}{5}=\frac{45}{85}-\frac{32}{85}=\frac{13}{85}\) ; cosec (A – B) = \(\frac { 85 }{ 13 }\)

Question 5.
If sin A = \(\frac { 3 }{ 5 }\) cos B = \(\frac { -8 }{ 17 }\) < A < π and \(\frac{\pi}{2}\) < B < π. Find the values of sin(A + B) and cos(A – B).
Answer:
Given sin A = \(\frac { 3 }{ 5 }\) . cos A = \(\frac { -4 }{ 5 }\)
cos B = \(\frac { -8 }{ 17 }\), sin B = \(\frac { 15 }{ 17 }\)
sin(A + B) = sinA cosB + Cos A sinB = \(\frac{3}{5}\left(\frac{-8}{17}\right)+\left(\frac{-4}{5}\right) \frac{15}{17}=\frac{-24-60}{85}=\frac{-84}{85}\)

Question 6.
If sin A = \(\frac { 7 }{ 25 }\), cos B = \(\frac { -12 }{ 13 }\) where \(\frac{\pi}{2}\) < A < π and π < B <  \(\frac{3 \pi}{2}\) find the values of:
(i) sin(A + B)
(ii) cos(A + B)
(iii) sin(A – B)
(ii) cos(A – B)
(v) tan(A + B)
(vi) tan(A – B)
Answer:

Question 7.
If tan A = \(\frac { 1 }{ 2 }\), tan B = \(\frac { 1 }{ 3 }\) Find tan(A + B), tan(A – B)
Answer:

Question 8.
tan(A – B) = \(\frac { 1 }{ 7 }\), tan A = \(\frac { 1 }{ 2 }\) show that A + B = 45°.
Answer:

Question 9.
If tan A = \(\frac { 1 }{ 3 }\) tan (A + B) = \(\frac { 1 }{ 7 }\) find tan B.
Answer:

Question 10.
tan A = \(\frac { 3 }{ 4 }\) and tan B = \(\frac { 1 }{ 7 }\) . show that tan (A + B) = 1.
Answer:
Given tan A = \(\frac { 3 }{ 4 }\), tanB = \(\frac { 1 }{ 7 }\)

Question 11.
If tan A = \(\frac { 5 }{ 6 }\) and tan(A + B) = 1 Show that tanB = \(\frac { 1 }{ 11 }\)
Answer:
Given

Question 12.
If tan α = \(\frac{\mathrm{n}}{\mathrm{n}+1}\) and tan β = \(\frac{1}{2 n+1}\) Show that α + β = \(\frac{\pi}{4}\)
Answer:

Question 13.
Prove that \(\frac{\cos 2 A}{\sec A}+\frac{\sin 2 A}{\csc A}=\cos A\)
Answer:
L.H.S. = cos2A . \(\frac{1}{\sec A}\) + sin 2A . \(\frac{1}{\csc A}\)
= cos2A . cosA + sin2A . sinA
= cos(2A – A) = cosA = R.H.S.

Question 14.
Prove that sin(45° + A) + cos(45° + A) = \(\sqrt{2}\) cosA ,
Answer:
L.H.S. = sin(45° + A) + cos(45° + A)

Question 15.
Prove that cos \(\left(A+\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}\) (cos A – sin A).
Answer:

Question 16.
Prove that cos \(\left(\frac{\pi}{6}+\mathbf{A}\right)\) . cos \(\left(\frac{\pi}{6}-\mathbf{A}\right)\) – sin\(\left(\frac{\pi}{6}+\mathbf{A}\right)\) . sin \(\left(\frac{\pi}{6}+\mathbf{A}\right)\) = \(\frac { 1 }{ 2 }\).
Answer:

Part – B

2nd PUC Basic Maths Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.1 Five Marks Questions and Answers

Question 1.
Prove the following \(\frac{\sin (A+B)}{\cos A \cdot \cos B}\) =tan A + tanB.
Answer:

Question 2.
\(\frac{\sin (A-B)}{\sin A \sin B}=\cot B-\cot A\)
Answer:

Question 3.
Show that cos(A + B) cos(A – B) = cos²A – sin²B = cos²B – sin²A.
Answer:
L.H.S = cos(A + B).cos(A – B)
= (cos A . cosB – sin A sinB) (cos A cosB + sinA sinB)
= cos2A . cos2B – sin2A · sin 2B = cos2A(1 – sin2B) – (1 – cos2A) · sin2B

Similarly cos2A . cos2B – sin2A · sin2B = (1 – sin2A) cos2B – sin2A(1 – cos2B)

Question 4.
sin(A + B) sin(A – B) = cos²B – Cos²A
Answer:
L.H.S = sin(A + B) sin(A – B) = (sinA cosB + cosA sinB) (sin A cosB – CosA sinB)
= sin²A cos²B – cos²A sin²B = (1 – Cos²A) cos²B – cos²A (1 – cos²B)

Question 5.
\(\cos \left(\frac{\pi}{4}-A\right)-\sin \left(\frac{\pi}{4}+A\right)=0\)
Answer:

Question 6.
\(\sin \left(\frac{\pi}{3}-A\right) \cdot \cos \left(\frac{\pi}{6}+A\right)+\cos \left(\frac{\pi}{3}-A\right) \cdot \sin \left(\frac{\pi}{6}+A\right)=1\)
Answer:

Question 7.
tan 15° + cot 15° = 4.
Answer:

Question 8.
sin 105° + cos 105° = cos 45°
Answer:
L.H.S = sin105° + cos(105°) = sin(60° + 45°) + cos(60° + 459)

Question 9.
cot 2A + tan A = cosec 2A.
Answer:
L.H.S. = cot2A + tanA

Question 10.
P.T tan A tan 3A, tan 4A = tan 4A – tan3A – tanA.
Answer:
Consider tan4A = tan(3A + A)
\(\frac{\tan 4 A}{1}=\frac{\tan 3 A+\tan A}{1-\tan 3 A \cdot \tan A}\) ; tan4A(1 – tan3A · tanA) = tan3A + tanA
tan4A – tan4A tan3A tanA = tan3A + tanA
tan4A – tan3A – tanA = tan4A · tan3A tanA

Question 11.
Prove that cos(45° – A) · cos(45° – B) – sin(45° – A) sin(45° – B) = sin(A + B).
Answer:
L.H.S = cos(45° – A) · cos(45° – B) – sin(45° – A) · sin(45° – B)
= cos(45° – A + 45° – B) ∵ cosA . cosB – sinA . sinB
= cos(90° – (A + B)) = cos (A + B) = sin(A + B) = R.H.S.

Question 12.
Show that \(\sum \frac{\sin (A-B)}{\cos A \cdot \cos B}=0\)
Answer:

Question 13.
cos(120° + A) + cos(120° – A) = -cos A
Answer:
L.H.S = cos(120° + A) + cos(120° – A)

Question 14.
\(\frac{\sin (A+B)}{\sin (A-B)} \quad \frac{\tan A+\tan B}{\tan A-\tan B}\)
Answer:

Students can Download Maths Chapter 7 Integrals Miscellaneous Exercise Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Maths Question Bank Chapter 7 Integrals Miscellaneous Exercise

Question 1.
\(\frac{1}{x-x^{3}}\)
Answer:

Question 2.
\(\frac{1}{\sqrt{x+a}+\sqrt{x+b}}\)
Answer:

Question 3.
\(\frac{1}{x \sqrt{a x-x^{2}}}\left[\text { Hint: Put } x=\frac{a}{t}\right]\)
Answer:

Question 4.
\(\frac{1}{x^{2}\left(x^{4}+1\right)^{\frac{3}{4}}}\)
Answer:

Question 5.
\(\frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}\)
\(\left[\text { Hint : } \frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}=\frac{1}{\frac{1}{x^{3}}\left(1+x^{\frac{1}{6}}\right)}, \text { put } x=t^{6}\right]\)
Answer:

Question 6.
\(\frac{5 x}{(x+1)\left(x^{2}+9\right)}\)
Answer:

Question 7.
\(\frac{\sin x}{\sin (x-a)} d x\)
Answer:

Question 8.
\(\frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}} d x\)
Answer:

Question 9.
\(\frac{\cos x}{\sqrt{4-\sin ^{2} x}}\)
Answer:

Question 10.
\(\frac{\sin ^{8}-\cos ^{8} x}{1-2 \sin ^{2} x \cos ^{2} x}\)
Answer:

Question 11.
\(\frac{1}{\cos (x+a) \cos (x+b)}\)
Answer:

Question 12.
\(\frac{x^{3}}{\sqrt{1-x^{8}}}\)
Answer:

Question 13.
\(\frac{\mathbf{e}^{\mathbf{x}}}{\left(\mathbf{1}+\mathbf{e}^{\mathbf{x}}\right)\left(2+\mathbf{e}^{\mathbf{x}}\right)}\)
Answer:

Question 14.
\(\frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}\)
Answer:

Question 15.
cos³x e^{log sinx}
Answer:

Question 16.
e^{3 log x} (x⁴+1)¹
Answer:

Question 17.
f'(ax+b)[f(ax+b)ⁿ
Answer:

Question 18.
\(\frac{1}{\sqrt{\sin ^{3} x \sin (x+\alpha)}}\)
Answer:

Question 19.
\(\frac{\sin ^{-1} \sqrt{x}-\cos ^{-1} \sqrt{x}}{\sin ^{-1} \sqrt{x}+\cos ^{-1} \sqrt{x}}, x \in[0,1]\)
Answer:

Question 20.
\(\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\)
Answer:

Question 21.
\(\frac{2+\sin 2 x}{1+\cos 2 x} e^{x}\)
Answer:

Question 22.
\(\frac{x^{2}+x+1}{(x+1)^{2}(x+2)}\)
Answer:

Question 23.
\(\tan ^{-1} \sqrt{\frac{1-x}{1+x}}\)
Answer:

Question 24.
\(\frac{\sqrt{x^{2}+1}\left[\log \left(x^{2}+1\right)-2 \log x\right]}{x^{4}}\)
Answer:

Evaluate the definite integrals in Exercises 25 to 33

Question 25.
\(\int_{\frac{\pi}{2}}^{\pi} \mathrm{e}^{x}\left(\frac{1-\sin x}{1+\cos x}\right) d x\)
Answer:

Question 26.
\(\int_{0}^{\frac{\pi}{4}} \frac{\sin x \cos x}{\cos ^{4} x+\sin ^{4} x} d x\)
Answer:

Question 27.
\(\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x d x}{\cos ^{2} x+4 \sin ^{2} x}\)
Answer:

Question 28.
\(\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x\)
Answer:

Question 29.
\(\int_{0}^{1} \frac{d x}{\sqrt{1+x}-\sqrt{x}}\)
Answer:

Question 30.
\(\int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x\)
Answer:

Question 31.
\(\int_{0}^{\frac{\pi}{2}} \sin 2 x \tan ^{-1}(\sin x) d x\)
Answer:

Question 32.
\(\int_{0}^{\pi} \frac{x \tan x}{\sec x+\tan x} d x\)
Answer:

Question 33.
\(\int_{1}^{4}[|x-1|+|x-2|+|x-3|] d x\)
Answer:

Prove the following (Exercise 34 to 39)

Question 34.
\(\int_{1}^{3} \frac{d x}{x^{2}(x+1)}=\frac{2}{3}+\log \frac{2}{3}\)
Answer:

Question 35.
\(\int_{0}^{1} x e^{x} d x=1\)
Answer:

Question 36.
\(\int_{-1}^{1} x^{17} \cos ^{4} x d x=0\)
Answer:

Question 37.
\(\int_{0}^{\frac{\pi}{2}} \sin ^{3} x d x=\frac{2}{3}\)
Answer:

Question 38.
\(\int_{0}^{\frac{\pi}{4}} 2 \tan ^{3} x d x=1-\log 2\)
Answer:

Question 39.
\(\int_{0}^{1} \sin ^{-1} x d x=\frac{\pi}{2}-1\)
Answer:

Question 40.
Evaluate \(\int_{0}^{1} e^{2-3 x} d x\) as a a= limit of a sum.
Answer:

Choose the correct answers in Exercises 41 and 44.

Question 41.
\(\int \frac{d x}{e^{x}+e^{-x}} \text { is equal to }\)
(A) tan-1 (e^x) + C
(B) tan¹ (e ^x) + C
(C) log (e^x – e ^-x) + C
(D) log (e^x + e^-x) + C
Answer:

Question 42.
\(\int \frac{\cos 2 x}{(\sin x+\cos x)^{2}} d x \text { is equal to }\)

Answer:

Question 43.
If f (a+b-x) = f (x), then \(\int_{a}^{b} x f(x) d x \)is equal to

Answer:

Question 44.
The value of \(\int_{0}^{1} \tan ^{-1}\left(\frac{2 x-1}{1+x-x^{2}}\right) d x \text { is }\)
(A) 1
(B) 0
(C) -1
(D) \(\frac{\pi}{4}\)
Answer:

2nd PUC Maths Integrals Miscellaneous Exercise Extra Questions and Answers

Question 1.
\(\int \frac{x}{1+x^{4}} d x=\)
Answer:

Question 2.
\(\int \frac{x^{3}}{1+x^{4}} d x\)
Answer:

Question 3.
\(\int \frac{x^{2}}{1+x^{4}} d x\)
Answer:

Question 4.
\(\int \frac{1}{1+x^{4}} d x\)
Answer:

Question 5.
\(\int \sqrt{\cot x} d x\)
Answer:

Question 6.
\(\int \sqrt{\tan x} d x\) (CBSE 2009)
Answer:

Question 7.
\(\int \frac{x^{2}+1}{1+x^{4}} d x\)
Answer:

Question 8.
\(\text { If } \int \frac{1}{f(x)} d x=\log (f(x))^{2}+C, \text { find } f(x)\)
Answer:

Question 9.
\(\int e^{\tan ^{-1} x}\left(1+\frac{x}{1+x^{2}}\right) d x\)(CBSE 2006)
Answer:

Question 10.
\(\int \sin ^{-1} \sqrt{\frac{x}{1+x}} d x\) (CBSE 2010)
Answer:

Question 11.
∫|x| dx
Answer:

Question 12.
\(\int\left(\log x+\frac{1}{(\log x)^{2}}\right) d x\)
Answer:

Question 13.
∫e^ax (af(x) + f'(x))dx = f (x). e^ax
Answer:

Question 14.
∫e ^3-x(cosx – 3sinx) dx   (Kerala CET)
Answer:
∫ e^-3x (-3 sin x + cos x)
f(x) = sinx,f'(x) =cosx
∫ e^-3x (-3 sin x + cos x) = e^-3x sin x + C

Question 15.
∫ sec² (7-4x) dx (CBSE – 2010)
Answer:

Question 16.
\(\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin 5 x \cdot d x\) (CBSE – 2010)
Answer:

Question 17.
\(\int \frac{x+2}{\sqrt{(x-2)(x-3)}} d x\) (CBSE – 2010)
Answer:

Question 18.
\(\int \frac{5 x+3}{\sqrt{x^{2}+4 x+10}} d x\) (CBSE – 2010)
Answer:

Question 19.
∫(ax + b)³ dx (CBSE – 2011)
Answer:

Question 20.
\(\int_{0}^{\frac{\pi}{2}} \frac{x+\sin x}{1+\cos x} d x\) (CBSE – 2011)
Answer:

Question 21.
\(\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{d x}{1+\sqrt{\tan x}} d x\)(CBSE – 2011)
Answer:

Question 22.
\(\int \frac{(\log x)^{2}}{x} d x\) (CBSE – 2011)
Answer:

Question 23.
\(\int_{0}^{\frac{\pi}{4}} \log (1+\tan x) d x\) (CBSE – 2011)
Answer:
\(\frac{\pi}{8} \log 2\)

Question 24.
\(\int_{-6}^{0}|x+3| d x\) (CBSE – 2011)
Answer:

Question 25.
\(\int \frac{6 x+7}{(x-5)(x-4)} d x\) (CBSE – 2011)
Answer:

Students can Download Basic Maths Exercise 8.1 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Basic Maths Question Bank Chapter 8 Bill Discounting Ex 8.1

Part – A and B

2nd PUC Basic Maths Bill Discounting Ex 8.1 One or Two Marks Questions with Answers

Question 1.
A bill was drawn on 14-3-2013 for 3 months. When does the bill fall legally due?
Answer:
L.D.D = DD + BP + 3 days

∴L.D.D is 17^th June 2013

Question 2.
A bill drawn for 3 months was legally due on 18-8-2012. Find the date of drawing of the bill.
Answer:
Drawn date = L.D.D – Bill period – 3days grace = 14 – 3 – 2013
= 18 – 8 – 2012

∴ Drawn date is 15^th May 2012

Question 3.
Find the present value of ₹ 750 due 4 months hence at 15% p.a.
Answer:
Given
F = 750, T = 4 months = \(\frac{4}{12}\), r= 15% = 0.15.

Question 4.
Find the banker’s discount on a bill of ₹ 415 due 9 months hence at 15% p.a.
Answer:
Given
F = 415, t = \(\frac{9}{12}\) year,
r = 15% = 0.15
B.D = Ftr = 415 x \(\frac{9}{12}\) × 0.15= 46.68.

Question 5.
Define Banker’s Gain.
Answer:
Banker gain is difference between BD & TD i.e., BG = BD – TD

Question 6.
Define Banker’s Discount.
Answer:
Banker’s Discount is the Simple interest calculated by the banker on the face value of the bill i.e., BD = Ftr

Question 7.
Find the present value of 2320, due 2 years hence, at 8% per annum.
Answer:
Given F= 2320,t = 2years, r = 0.08, P = ?

Question 8.
Find the BD on ₹ 1015, payable after 3 months at 6% p.a.
Answer:
Given
F = 1015,
F=1015, t = \(\frac{3}{12}\) year, r = \(\frac{6}{100}\) = 0.06, BD = ?
BD = Ftr = 1015 x \(\frac{3}{12}\) × 0.06 = 15.22

Question 9.
Find the true discount on ₹ 1380, due 1\(\frac{1}{2}\) years after, at 10% p.a.
Answer:
Given
F = ₹ 1380, t = \(\frac{3}{2}\) year , r = 10% = 0.1.

Question 10.
TD on a bill was ₹ 100 and BG was ₹ 10. What is the face value of the bill?
Answer:
Given TD = 100, BG = 10, F = ?
BD = TD + BG = 100 + 10 = 110

Question 11.
A banker pays ₹ 2380 on a bill of ₹ 2500, 73 days before the legal due date. Find the rate of discount charged by the banker.
Answer:
Given Discounted value = 2380, F = 2500
t = \(\frac{73}{365}=\frac{1}{5}\) year , r = ?
D.V. = F(1 – tr)
2380 = 2500 \(\left(1-\frac{1}{5} r\right)\)
\(\frac{2380}{2500}=\left(1-\frac{r}{5}\right)\)
0.952 = 1 – \(\frac{r}{5}\)
\(\frac{r}{5}\) = 1 – 0.952 = 0.048
∴ r = 5 × 0.048 = 0.24 × 100 = 24%

Question 12.
The Banker’s discount and true discount on a sum of money due 3 months hence are ₹ 154.50 and ₹ 150 respectively. Find the sum of money and the rate of interest.
Answer:
Given BD = 154.5, t = \(\frac{3}{12}\) , TD = 150, t = ? r = ?

Question 13.
BD and BG on a certain bill due after sometime are 1250 and 350 respectively. Find the face value of the bill.
Answer:
BD and BG on a
Given
BD = 1250, BG = 50, F = ?
BG = BD – TD
TD = BD – BG = 1250 – 50 = 1200

Part – C

2nd PUC Basic Maths Bill Discounting Ex 8.1 Three or Four Mark Questions with Answers

Question 1.
The difference between banker’s discount and true discount on a bill due after 6 months at 4% interest p.a. is ₹ 20. Find the true discount, banker’s discount and face value of the bill.
Answer:
Given
BG = ₹ 20, r = 4% = 0.04, t = 6 Months = \(\frac{6}{12}=\frac{1}{2}\)
TD = ?, BD = ?, F = ?
W.K.T. = BG = TD. tr
20 = TD \(\frac{1}{2}\) x 0.04
40 = 0.04 TD ⇒ TD = \(\frac{40}{0.04}\) = ₹ 1000
BD = BG + TD; = 20 + 1000 = 1020

Question 2.
The BG on a certain bill due 6 months hence is ₹ 10, the rate of interest being 10% p.a. find the face value of the bill and the true present value.
Answer:
BG = 10, ⇒ BD – TD = 10,t = \(\frac{6}{12}=\frac{1}{2}\)year, r= 10% = 0.1, F = ? P = ?
BG = TD.tr
10 = TD × \(\frac{1}{2}\) × 0.1
TD = \(\frac{20}{0.1}\) = 200 ;
BD = TD + BG = 200 + 10 = 210.

Question 3.
A banker discounts a bill for a certain amount having 32 days to run before it matures at 15% p.a. The discounted value of the bill is ₹ 995.90. What is the face value of the bill, banker’s discount, true discount and banker’s gain?
Answer:
Given t = \(\frac{32}{365}\) year, r = 0.15, D.V = ₹995.90, F = ?, BD = ?, TD = ?, BG = ?
Discounted Value = F(1 – tr), 995.90 = F ( 1 – \( \frac{32}{365}\) × 0.05)

D= F – Discounted Value;
= 1009.017 – 995.90 = 13.117

BG = BD – TD = 13.117 – 12.948 = 0.169

Question 4.
A bill for ₹ 14,600 drawn at 3 months after date was discounted on 11-11-99 for ₹ 14,320. If the discount rate is 20% p.a. on what date was the bill drawn?
Answer:
Given DV = 14320, F = 14,600, r=0.2,
DD = ?
W.K. Discounted value = F(1 – tr)
14320 14,320 = 14,600 (1 – 0.2t);
\(\frac{14320}{14600}\) = 1-0.2t
t = 0.1 year = 0.1 ~ 365 = 36 days
L.D.D = 36 days + 11-11-99 = Dec. 17
Drawn date = L.D.D – BP – 3 grace days

∴Bill was drawn on October 14th 3

Question 5.
A bill for ₹ 2920 was drawn on September 11 for 3 months after date and was discounted at 16% p.a. for ₹ 2875.20. On what date was the bill discounted?
Answer:
Given drawn date = 11^th Sep., F = 2920, DV = 2875.20, r = 0.16
BD = F – Discounted value
= 2920 – 2875.20
BD = 44.8
& BD = Ftr.
44.8 = 2920 × 0.16 × t

t = \(\frac{44.8}{2920 \times 0.16}\) = 0.0958 years = 0.0958 × 365 = 35 days
Discounted date = L.D.D – Discount period = Dec. 14^th – 35 days = 9^th November

Question 6.
A bill for ₹ 3500 due for 3 months was drawn on 27 march 2012 and was discounted on 18 April 2012, at the rate of 7% p.a. Find the banker’s discount and discounted value of the bill.
Answer:
Given
F = 3500, r = 0.07, B.P = 3 Months
DD = 27th March 2012, Discounted on 18^th April 2012; BD = ?, DV = ?
L.D.D = DD + BD + 3 grace days

t = No. of days from April 18th to 30th June
12 + 31 + 30 = 73 days =\(\frac{73}{365}\) Year
DV = F(1 – tr) = 3500 1 – \(\frac{73}{365}\) x 0.07)
DV = 3451.00;
BD = F – DV = 3500 – 3451 = ₹ 49

Question 7.
A bill for 12900 was drawn on 3 Feb 2004 at 6 months and discounted on 13 March 2004 at 8% p.a. For what sum was the bill discounted and how much did the banker gain in this transaction?
Answer:
Given
F = 12900, r = 0.08, BG = ?
L.D.D = DD + BP + 3days

t = No. of days from discounted date till LDD = 13 – 3 – 200 to 6 – 8 – 2004
Mar. + Apr. + May + June + July + Aug
= 18 + 30 + 31 + 30 + 31 + 6 = 146 days = \(\frac{146}{865}\)
∴OV = F(1 – tr)
= 12900 (1 – \(\frac{146}{365}\) x 0.08
= 12900 (1 -0.032)
DV = 12487.2
BD = F – Discounted value
= 12900 – 12487.2 = 412.8
TD = \(\frac{B D}{1+t r}=\frac{412.8}{1+0.032}\) = 400
BG = BD – TD = 412.8 – 400 = ₹ 12.8

Question 8.
The banker’s gain on a bill is \(\frac{1}{9}\) th of the banker’s discount, rate of interest being 10% p.a. Find the unexpired period of the bill.
Answer:
Given BG = \(\frac{1}{9}\)BD & r= 0.1
9(BD – TD) = BD
8BD = 9TD
BD 8BD=9. \(\frac{\mathrm{BD}}{1+\mathrm{tr} \mathrm{r}}\)
1+ tr = \(\frac{9}{8}\) = 1.125
tr = 1.125 – 1
t(0.1) = 0.125
t = \(\frac{0.125}{0.1}\) = 1.25years
t = 1.25 × 12 = 15 months

Question 9.
A bill for ₹ 2725.25 was drawn on 03-06-2010 and made payable 3 months after due date. It was discounted on 15-6-2010 at 16% per annum. What is the discounted value of the bill and how much did the banker gain?
Answer:
Given F = 2725.25, r = 0.16, DV = ?, BG = ?, DD = 03-6-2010, BP = 3 months
∴ L.D.D = DD + BD + 3days
= 6-9-2010.
t = No. of days from 15^th June to 6^th Sept = June + July + August + September
= 15 +31 +31 + 6 = 83 days = \(\frac{83}{365}\) years
Discounted value = F (1 – tr)
= 2725.25 (1 – \(\frac{83}{365}\) × 0.16 = 2725.25 (1 – 0.036)
DV = 2627.14
BD = F – Discounted value = 2725.25 – 2627.14 = 98.10
TD = \(\frac{B D}{1+t r}=\frac{98.10}{1.036}\) = 94.69
BG = BD – TD = 98.10 – 94.69 = 3.41