Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.7

Students can download 12th Business Maths Chapter 2 Integral Calculus I Ex 2.7 Questions and Answers, Samacheer Kalvi 12th Business Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.7

Integrate the following with respect to x.

Question 1.
\(\frac{1}{9-16 x^{2}}\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.7 Q1

Question 2.
\(\frac{1}{9-8 x-x^{2}}\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.7 Q2

Question 3.
\(\frac{1}{2 x^{2}-9}\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.7 Q3

Question 4.
\(\frac{1}{x^{2}-x-2}\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.7 Q4

Question 5.
\(\frac{1}{x^{2}+3 x+2}\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.7 Q5
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.7 Q5.1

Question 6.
\(\frac{1}{2 x^{2}+6 x-8}\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.7 Q6

Question 7.
\(\frac{e^{x}}{e^{2 x}-9}\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.7 Q7

Question 8.
\(\frac{1}{\sqrt{9 x^{2}-7}}\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.7 Q8
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.7 Q8.1

Question 9.
\(\left(\frac{1}{\sqrt{x^{2}+6 x+13}}\right)\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.7 Q9

Question 10.
\(\left(\frac{1}{\sqrt{x^{2}-3 x+2}}\right)\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.7 Q10

Question 11.
\(\frac{x^{3}}{\sqrt{x^{8}-1}}\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.7 Q11

Question 12.
\(\sqrt{1+x+x^{2}}\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.7 Q12

Question 13.
\(\sqrt{x^{2}-2}\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.7 Q13

Question 14.
\(\sqrt{4 x^{2}-5}\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.7 Q14
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.7 Q14.1

Question 15.
\(\sqrt{2 x^{2}+4 x+1}\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.7 Q15

Question 16.
\(\frac{1}{x+\sqrt{x^{2}-1}}\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.7 Q16

Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.6

Students can download 12th Business Maths Chapter 2 Integral Calculus I Ex 2.6 Questions and Answers, Samacheer Kalvi 12th Business Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.6

Integrate the following with respect to x.

Question 1.
\(\frac{2 x+5}{x^{2}+5 x-7}\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.6 Q1

Question 2.
\(\frac{e^{3 \log x}}{x^{4}+1}\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.6 Q2

Question 3.
\(\frac{e^{2 x}}{e^{2 x}-2}\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.6 Q3

Question 4.
\(\frac{(\log x)^{3}}{x}\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.6 Q4

Question 5.
\(\frac{6 x+7}{\sqrt{3 x^{2}+7 x-1}}\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.6 Q5

Question 6.
\((4 x+2) \sqrt{x^{2}+x+1}\)
Solution:
\((4 x+2) \sqrt{x^{2}+x+1}\)
Let f(x) = x2 + x + 1
then f'(x) = 2x + 1
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.6 Q6

Question 7.
x8 (1 + x9)5
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.6 Q7

Question 8.
\(\frac{x^{e-1}+e^{x-1}}{x^{e}+e^{x}}\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.6 Q8

Question 9.
\(\frac{1}{x \log x}\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.6 Q9

Question 10.
\(\frac{x}{2 x^{4}-3 x^{2}-2}\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.6 Q10
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.6 Q10.1

Question 11.
ex (1 + x) log(x ex)
Solution:
ex (1 + x) log(x ex) = (ex + x ex) log (x ex)
Let z = x ex, Then dz = d(x ex)
dz = (x ex + ex) dx (Using product rule)
So ∫ ex (1 + x) log (x ex) dx
= ∫ log (x ex) (ex + x ex) dx
= ∫ log z dz
= z (log z – 1) + c
= x ex [log (x ex) – 1] + c

Question 12.
\(\frac{1}{x^{2}\left(x^{2}+1\right)}\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.6 Q12

Question 13.
\(e^{x}\left[\frac{1}{x^{2}}-\frac{2}{x^{3}}\right]\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.6 Q13

Question 14.
\(e^{x}\left[\frac{x-1}{(x+1)^{3}}\right]\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.6 Q14

Question 15.
\(e^{3 x}\left[\frac{3 x-1}{9 x^{2}}\right]\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.6 Q15

Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.5

Students can download 12th Business Maths Chapter 2 Integral Calculus I Ex 2.5 Questions and Answers, Samacheer Kalvi 12th Business Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.5

Integrate the following with respect to x.

Question 1.
x e-x
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.5 Q1

Question 2.
x3 e3x
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.5 Q2

Question 3.
log x
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.5 Q3

Question 4.
x log x
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.5 Q4

Question 5.
xn log x
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.5 Q5

Question 6.
\(\boldsymbol{x}^{\boldsymbol{5}} \boldsymbol{e}^{\boldsymbol{x}^{2}}\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.5 Q6
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.5 Q6.1
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.5 Q6.2
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.5 Q6.3

Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.4

Students can download 12th Business Maths Chapter 2 Integral Calculus I Ex 2.4 Questions and Answers, Samacheer Kalvi 12th Business Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.4

Integrate the following with respect to x.

Question 1.
2 cos x – 3 sin x + 4 sec2 x – 5 cosec2 x
Solution:
∫2 cos x – 3 sin x + 4 sec2 x – 5 cosec2 x
= 2 ∫ cos x dx – 3 ∫ sin x dx + 4 ∫ sec2 x – 5 ∫ cosec2 x dx
= 2 sin x + 3 cos x + 4 tan x + 5 cot x + c

Question 2.
∫ sin3 x dx
Solution:
We know that, sin 3x = 3 sin x – 4 sin3 x
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.4 Q2

Question 3.
\(\frac{\cos 2 x+2 \sin ^{2} x}{\cos ^{2} x}\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.4 Q3

Question 4.
\(\frac{1}{\sin ^{2} x \cos ^{2} x}\)
[Hint: sin2 x + cos2 x = 1]
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.4 Q4

Question 5.
\(\sqrt{1-\sin 2} x\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.4 Q5
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.4 Q5.1

Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Miscellaneous Problems

Students can download 12th Business Maths Chapter 4 Differential Equations Miscellaneous Problems and Answers, Samacheer Kalvi 12th Business Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Miscellaneous Problems

Question 1.
Suppose that Qd = 30 – 5P + 2 \(\frac{d p}{d t}+\frac{d^{2} p}{d t^{2}}\) and Qs = 6 + 3P. Find the equilibrium price for market clearance.
Solution:
For equilibrium price Qd = Qs
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Miscellaneous Problems Q1
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Miscellaneous Problems Q1.1

Question 2.
Form the differential equation having for its general solution y = ax2 + bx
Solution:
Given y = ax2 + bx
Since there are 2 constants a, b we have to differentiate twice to eliminate them
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Miscellaneous Problems Q2
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Miscellaneous Problems Q2.1

Question 3.
Solve yx2 dx + e-x dy = 0
Solution:
The given equation can be written as
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Miscellaneous Problems Q3

Question 4.
Solve: (x2 + y2) dx + 2xy dy = 0
Solution:
The given equation can be written as \(\frac{d y}{d x}=-\frac{\left(x^{2}+y^{2}\right)}{2 x y}\)
It is a homogeneous differential equation
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Miscellaneous Problems Q4
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Miscellaneous Problems Q4.1
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Miscellaneous Problems Q4.2

Question 5.
Solve: x \(\frac{d y}{d x}\) + 2y = x4
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Miscellaneous Problems Q5

Question 6.
A manufacturing company has found that the cost C of operating and maintaining the equipment is related to the length ‘m’ of intervals between overhauls by the equation \(m^{2} \frac{d C}{d m}\) + 2mC = 2 and c = 4 and when m = 2. Find the relationship between C and m.
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Miscellaneous Problems Q6
given that c = 4 when m = 2
4(4) = 2(2) + k
k = 12
So the relation ship between C and m is Cm2 = 2m + 12 = 2(m + 6)

Question 7.
Solve (D2 – 3D + 2)y = e4x given y = 0 when x = 0 and x = 1.
Solution:
(D2 – 3D + 2)y = e4x
The auxiliary equations is m2 – 3m + 2 = 0
(m – 2) (m – 1) = 0
m = 2, 1
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Miscellaneous Problems Q7
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Miscellaneous Problems Q7.1
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Miscellaneous Problems Q7.2

Question 8.
Solve: \(\frac{d y}{d x}\) + y cos x = 2cos x
Solution:
The given equation can be written as \(\frac{d y}{d x}\) + (cos x)y = 2 cos x
It is of the form \(\frac{d y}{d x}\) + Py = Q
Where P = cos x, Q = 2 cos x
Now ∫Pdx = ∫cos x dx = sin x
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Miscellaneous Problems Q8

Question 9.
Solve: x2y dx – (x3 + y3) dy = 0
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Miscellaneous Problems Q9
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Miscellaneous Problems Q9.1

Question 10.
Solve: \(\frac{d y}{d x}\) = xy + x + y + 1
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Miscellaneous Problems Q10

Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Ex 4.6

Students can download 12th Business Maths Chapter 4 Differential Equations Ex 4.6 Questions and Answers, Samacheer Kalvi 12th Business Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Ex 4.6

Choose the correct answer.

Question 1.
The degree of the differential equation \(\frac{d^{4} y}{d x^{4}}-\left(\frac{d^{2} y}{d x^{2}}\right)^{4}+\frac{d y}{d x}=3\) is _________
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(a) 1
Hint:
Since the power of \(\frac{d^{4} y}{d x^{4}}\) is 1

Question 2.
The order and degree of the differential equation \(\sqrt{\frac{d^{2} y}{d x^{2}}}=\sqrt{\frac{d y}{d x}+5}\) are respectively
(a) 2 and 3
(b) 3 and 2
(c) 2 and 1
(d) 2 and 2
Answer:
(c) 2 and 1
Hint:
Squaring both sides, we get \(\frac{d^{2} y}{d x^{2}}=\frac{d y}{d x}+5\)
So order = 2, degree = 1

Question 3.
The order and degree of the differential equation \(\left(\frac{d^{2} y}{d x^{2}}\right)^{\frac{3}{2}}-\sqrt{\left(\frac{d y}{d x}\right)}-4=0\) are respectively.
(a) 2 and 6
(b) 3 and 6
(c) 1 and 4
(d) 2 and 4
Answer:
(a) 2 and 6
Hint:
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Ex 4.6 Q3

Question 4.
The differential equation \(\left(\frac{d x}{d y}\right)^{3}+2 y^{\frac{1}{2}}=x\) is _________
(a) of order 2 and degree 1
(b) of order 1 and degree 3
(c) of order 1 and degree 6
(d) of order 1 and degree 2
Answer:
(b) of order 1 and degree 3

Question 5.
The differential equation formed by eliminating a and b from y = aex + be-x is _______
(a) \(\frac{d^{2} y}{d x^{2}}-y=0\)
(b) \(\frac{d^{2} y}{d x^{2}}-\frac{d y}{d x}=0\)
(c) \(\frac{d^{2} y}{d x^{2}}=0\)
(d) \(\frac{d^{2} y}{d x^{2}}-x=0\)
Answer:
(a) \(\frac{d^{2} y}{d x^{2}}-y=0\)
Hint:
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Ex 4.6 Q5

Question 6.
If y = cx + c – c3 then its differential equation is ______
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Ex 4.6 Q6
Answer:
(a) \(y=x \frac{d y}{d x}+\frac{d y}{d x}-\left(\frac{d y}{d x}\right)^{3}\)

Question 7.
The integrating factor of the differential equation \(\frac{d x}{d y}\) + Px = Q is _____
(a) e∫Pdx
(b) ∫Pdx
(c) ∫Pdy
(d) e∫Pdy
Answer:
(d) e∫Pdy

Question 8.
The complementary function of (D2 + 4) y = e2x is _______
(a) (Ax + B) e2x
(b) (Ax + B) e-2x
(c) A cos 2x + B sin 2x
(d) Ax-2x + Be2x
Answer:
(c) A cos 2x + B sin 2x
Hint:
A.E = m2 + 4 = 0 ⇒ m = ±2i
C.F = e0x (A cos 2x + B sin 2x)

Question 9.
The differential equation of y = mx + c is (m and c are arbitrary constants).
(a) \(\frac{d^{2} y}{d x^{2}}=0\)
(b) y = x \(\frac{d y}{d x}\)
(c) x dy + y dx = 0
(d) y dx – x dy = 0
Answer:
(a) \(\frac{d^{2} y}{d x^{2}}=0\)

Question 10.
The particular integral of the differential equation \(\frac{d^{2} y}{d x^{2}}-8 \frac{d y}{d x}+16 y=2 e^{4 x}\) is ________
(a) \(\frac{x^{2} e^{4 x}}{2 !}\)
(b) \(\frac{e^{4 x}}{2 !}\)
(c) x2 e4x
(d) xe4x
Answer:
(c) x2 e4x
Hint:
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Ex 4.6 Q10

Question 11.
Solution of \(\frac{d x}{d y}\) + px = 0
(a) x = cepy
(b) x = ce-py
(c) x = py + c
(d) x = cy
Answer:
(b) x = ce-py
Hint:
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Ex 4.6 Q11

Question 12.
If sec2 x is an integrating factor of the differential equation \(\frac{d y}{d x}\) + Py = Q then P = _____
(a) 2 tan x
(b) sec x
(c) cos2 x
(d) tan2 x
Answer:
(a) 2 tan x
Hint:
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Ex 4.6 Q12

Question 13.
The integrating factor of x\(\frac{d y}{d x}\) – y = x2 is _____
(a) \(\frac{-1}{x}\)
(b) \(\frac{1}{x}\)
(c) log x
(d) x
Answer:
(b) \(\frac{1}{x}\)
Hint:
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Ex 4.6 Q13

Question 14.
The solution of the differential equation \(\frac{d y}{d x}\) + Py = Q where P and Q are the function of x is ______
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Ex 4.6 Q14
Answer:
(c) \(y e^{\int p d x}=\int \mathrm{Q} e^{\int p d x} d x+c\)

Question 15.
The differential equation formed by eliminating A and B from y = e-2x (A cos x + B sin x) is _______
(a) y2 – 4y1 + 5y = 0
(b) y2 + 4y1 – 5y = 0
(c) y2 – 4y1 – 5y = 0
(d) y2 + 4y1 + 5y = 0
Answer:
(d) y2 + 4y1 + 5y = 0
Hint:
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Ex 4.6 Q15

Question 16.
The particular integral of the differential equation f(D) y = eax where f(D) = (D – a)2 ________
(a) \(\frac{x^{2}}{2} e^{2 x}\)
(b) xeax
(c) \(\frac{x}{2} e^{2 x}\)
(d) x2 e2x
Answer:
(a) \(\frac{x^{2}}{2} e^{2 x}\)

Question 17.
The differential equation of x2 + y2 = a2 is _____
(a) x dy + y dx = 0
(b) y dx – x dy = 0
(c) x dx – y dx = 0
(d) x dx + y dy = 0
Answer:
(d) x dx + y dy = 0
Hint:
x2 + y2 = a2
⇒ 2x + 2y \(\frac{d y}{d x}\) = 0
⇒ x dx + y dy = 0

Question 18.
The complementary function of \(\frac{d^{2} y}{d x^{2}}-\frac{d y}{d x}=0\) is ______
(a) A + B ex
(b) (A + B) ex
(c) (Ax + B) ex
(d) (Aex + B)
Answer:
(a) A + B ex
Hint:
A.E is m2 – m = 0
⇒ m(m – 1) = 0
⇒ m = 0, 1
CF is Ae0x + Bex = A + Bex

Question 19.
The P.I of (3D2 + D – 14)y = 13e2x is _______
(a) \(\frac{x}{2}\) e2x
(b) x e2x
(c) \(\frac{x^{2}}{2}\) e2x
(d) 13xe2x
Answer:
(b) xe2x
Hint:
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Ex 4.6 Q19

Question 20.
The general solution of the differential equation \(\frac{d y}{d x}\) = cos x is _______
(a) y = sin x + 1
(b) y = sin x – 2
(c) y = cos x + c, c is an arbitrary constant
(d) y = sin x + c, c is an arbitrary constant
Answer:
(d) y = sin x + c, c is an arbitrary constant

Question 21.
A homogeneous differential equation of the form \(\frac{d y}{d x}=f\left(\frac{y}{x}\right)\) can be solved by making substitution
(a) y = vx
(b) v = yx
(c) x = vy
(d) x = v
Answer:
(a) y = vx

Question 22.
A homogeneous differential equation of the form \(\frac{d x}{d y}=f\left(\frac{x}{y}\right)\) can be solved by making substitution,
(a) x = vy
(b) y = vx
(c) y = v
(d) x = v
Answer:
(a) x = vy

Question 23.
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Ex 4.6 Q23
Answer:
(d) \(\frac{1+v}{2 v^{2}} d v=-\frac{d x}{x}\)
Hint:
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Ex 4.6 Q23.1

Question 24.
Which of the following is the homogeneous differential equation?
(a) (3x – 5) dx = (4y – 1) dy
(b) xy dx – (x3 + y3) dy = 0
(c) y2 dx + (x2 – xy – y2) dy = 0
(d) (x2 + y) dx = (y2 + x) dy
Answer:
(c) y2 dx + (x2 – xy – y2) dy = 0

Question 25.
The solution of the differential equation \(\frac{d y}{d x}=\frac{y}{x}+\frac{f\left(\frac{y}{x}\right)}{f^{\prime}\left(\frac{y}{x}\right)}\) is ______
(a) f(\(\frac{y}{x}\)) = kx
(b) x f(\(\frac{y}{x}\)) = k
(c) f(\(\frac{y}{x}\)) = ky
(d) y f(\(\frac{y}{x}\)) = k
Answer:
(a) f(\(\frac{y}{x}\)) = kx
Hint:
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Ex 4.6 Q25

Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Ex 4.5

Students can download 12th Business Maths Chapter 4 Differential Equations Ex 4.5 Questions and Answers, Samacheer Kalvi 12th Business Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Ex 4.5

Solve the following differential equations:

Question 1.
\(\frac{d^{2} y}{d x^{2}}-6 \frac{d y}{d x}+8 y=0\)
Solution:
Given (D2 – 6D + 8) y = 0, D = \(\frac{d}{d x}\)
The auxiliary equations is
m2 – 6m + 8 = 0
(m – 4)(m – 2) = 0
m = 4, 2
Roots are real and different
The complementary function (C.F) is (Ae4x + Be2x)
The general solution is y = Ae4x + Be2x

Question 2.
\(\frac{d^{2} y}{d x^{2}}-4 \frac{d y}{d x}+4 y=0\)
Solution:
The auxiliary equations A.E is m2 – 4m + 4 = 0
(m – 2)2 = 0
m = 2, 2
Roots are real and equal
The complementary function (C.F) is (Ax + B) e2x
The general solution is y = (Ax + B) e2x

Question 3.
(D2 + 2D + 3) y = 0
Solution:
The auxiliary equations A.E is m2 + 2m + 3 = 0
⇒ m2 + 2m + 1 + 2 = 0
⇒ (m + 1)2 = -2
⇒ m + 1 = ± √2i
⇒ m = – 1 ± √2i
It is of the form α ± iβ
The complementary function (C.F) = e-x [A cos √2 x + B sin √2 x]
The general solution is y = e-x [A cos √2 x + B sin √2 x]

Question 4.
\(\frac{d^{2} y}{d x^{2}}-2 k \frac{d y}{d x}+k^{2} y=0\)
Solution:
Given (D2 – 2kD + k2)y = 0, D = \(\frac{d}{d x}\)
The auxiliary equations is m2 – 2km + k = 0
⇒ (m – k)2 = 0
⇒ m = k, k
Roots are real and equal
The complementary function (C.F) is (Ax + B) ekx
The general solution is y = (Ax + B) ekx

Question 5.
(D2 – 2D – 15) y = 0 given that \(\frac{d y}{d x}\) = 0 and \(\frac{d^{2} y}{d x^{2}}\) = 2 when x = 0
Solution:
A.E is m2 – 2m – 15 = 0
(m – 5)(m + 3) = 0
m = 5, -3
C.F = Ae5x + Be-3x
The general solution is y = Ae5x + Be-3x …….. (1)
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Ex 4.5 Q5

Question 6.
(4D2 + 4D – 3) y = e2x
Solution:
The auxiliary equations is 4m2 + 4m – 3 = 0
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Ex 4.5 Q6

Question 7.
\(\frac{d^{2} y}{d x^{2}}\) + 16y = 0
Solution:
Given (D2 + 16) y =0
The auxiliary equation is m2 + 16 = 0
⇒ m2 = -16
⇒ m = ± 4i
It is of the form α ± iβ, α = 0, β = 4
The complementary function (C.F) is e0x [A cos 4x + B sin 4x]
The general solution is y = [A cos 4x + B sin 4x]

Question 8.
(D2 – 3D + 2)y = e3x which shall vanish for x = 0 and for x = log 2
Solution:
A.E is m2 – 3m + 2 = 0
⇒ (m – 2) (m – 1) = 0
⇒ m = 2, 1
CF = Ae2x + Bex
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Ex 4.5 Q8

Question 9.
(D2 + D – 6)y = e3x + e-3x
Solution:
A.E is m2 + m – 6 = 0
(m + 3) (m – 2) = 0
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Ex 4.5 Q9

Question 10.
(D2 – 10D + 25)y = 4e5x + 5
Solution:
A.E is m2 – 10m + 25 = 0
⇒ (m – 5)2 = 0
⇒ m = 5, 5
C.F = (Ax + B) e5x
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Ex 4.5 Q10

Question 11.
(4D2 + 16D + 15) y = 4\(e^{\frac{-3}{2} x}\)
Solution:
A.E is 4m2 + 16m + 15 = 0
(2m + 3) (2m + 5) = 0
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Ex 4.5 Q11

Question 12.
(3D2 + D – 14)y = 13e2x
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Ex 4.5 Q12

Question 13.
Suppose that the quantity demanded Qd = 13 – 6p + 2\(\frac{d p}{d t}+\frac{d_{2} p}{d t^{2}}\) and quantity supplied Qs = -3 + 2p where p is the price. Find the equilibrium price for market clearance.
Solution:
For market clearance, the required condition is Qd = Qs
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Ex 4.5 Q13

Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.8

Students can download 12th Business Maths Chapter 2 Integral Calculus I Ex 2.8 Questions and Answers, Samacheer Kalvi 12th Business Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.8

I. Using second fundamental theorem, evaluate the following:

Question 1.
\(\int_{0}^{1} e^{2 x} d x\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.8 I Q1

Question 2.
\(\int_{0}^{\frac{1}{4}} \sqrt{1-4 x} d x\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.8 I Q2

Question 3.
\(\int_{1}^{2} \frac{x d x}{x^{2}+1}\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.8 I Q3

Question 4.
\(\int_{0}^{3} \frac{e^{x} d x}{1+e^{x}}\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.8 I Q4

Question 5.
\(\int_{0}^{1} x e^{x^{2}} d x\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.8 I Q5

Question 6.
\(\int_{1}^{e} \frac{d x}{x(1+\log x)^{3}}\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.8 I Q6
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.8 I Q6.1

Question 7.
\(\int_{-1}^{1} \frac{2 x+3}{x^{2}+3 x+7} d x\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.8 I Q7

Question 8.
\(\int_{0}^{\frac{\pi}{2}} \sqrt{1+\cos x} d x\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.8 I Q8

Question 9.
\(\int_{1}^{2} \frac{x-1}{x^{2}} d x\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.8 I Q9
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.8 I Q9.1

II. Evaluate the following:

Question 1.
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.8 II Q1
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.8 II Q1.1

Question 2.
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.8 II Q2
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.8 II Q2.1

Question 3.
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.8 II Q3
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.8 II Q3.1

Question 4.
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.8 II Q4
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.8 II Q4.1

Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems

Students can download 12th Business Maths Chapter 4 Differential Equations Additional Problems and Answers, Samacheer Kalvi 12th Business Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems

I. One Mark Questions

Choose the correct answer.

Question 1.
The differential equation of straight lines passing through the origin is ______
(a) \(\frac{x d y}{d x}=y\)
(b) \(\frac{d y}{d x}=\frac{x}{y}\)
(c) \(\frac{d y}{d x}=0\)
(d) \(\frac{x d y}{d x}=\frac{1}{y}\)
Answer:
(a) \(\frac{x d y}{d x}=y\)
Hint:
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems I Q1

Question 2.
The solution of x dy + y dx = 0 is _______
(a) x + y = c
(b) x2 + y2 = c
(c) xy = c
(d) y = cx
Answer:
(c) xy = c
Hint:
x dy + y dx = 0
d(xy) = 0
xy = c

Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems

Question 3.
The solution of x dx + y dy = 0 is _____
(a) x2 + y2 = c
(b) \(\frac{x}{y}\) = c
(c) x2 – y2 = c
(d) xy = c
Answer:
(a) x2 + y2 = c
Hint:
x dx = -y dy
\(\frac{x^{2}}{2}=\frac{-y^{2}}{2}+c_{1}\)
x2 + y2 = c

Question 4.
The solution of \(\frac{d y}{d x}\) = ex-y is ______
(a) ey ex = c
(b) y = log cex
(c) y = log(ex + c)
(d) ex+y = c
Answer:
(c) y = log(ex + c)
Hint:
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems I Q4

Question 5.
The solution of \(\frac{d p}{d t}\) = ke-t (k is a constant) is ________
(a) \(c-\frac{k}{e^{t}}=p\)
(b) p = ket + c
(c) t = log\(\frac{c-p}{k}\)
(d) t = logc p
Answer:
(a) \(c-\frac{k}{e^{t}}=p\)
Hint:
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems I Q5

Question 6.
The integrating factor of (1 + x2) \(\frac{d y}{d x}\) + xy = (1 + x2)3 is _______
(a) \(\sqrt{1+x^{2}}\)
(b) log(1 + x2)
(c) \(e^{\tan ^{-1} x}\)
(d) \(\log ^{\left(\tan ^{-1} x\right)}\)
Answer:
(a) \(\sqrt{1+x^{2}}\)
Hint:
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems I Q6

Question 7.
The complementary function of the differential equation (D2 – D) y = ex is _____
(a) A + B ex
(b) (Ax + B) ex
(c) A + B e-x
(d) (A + Bx) e-x
Answer:
(a) A + B ex
Hint:
m2 – m = 0
m(m – 1) = 0
CF = Ae0x + Bex = A + B ex

Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems

Question 8.
Match the following
(a) \(\frac{d^{4} y}{d x^{4}}\) + sin y = 0 – (i) order 1, degree 1
(b) y’ + y = ex – (ii) order 3, degree 2
(c) y'” + 2y” + y’ = 0 – (iii) order 4, degree 1
(d) (y”’)2 + y’ + y5 = 0 – (iv) order 3, degree 1
Answer:
(a) – (iii)
(b) – (i)
(c) – (iv)
(d) – (ii)

Question 9.
Fill in the blanks
(a) The general solution of the equation \(\frac{d y}{d x}+\frac{y}{x}=1\) is ______
(b) Integrating factor of \(\frac{x d y}{d x}\) – y = sin x is ______
(c) The differential equation of y = A sin x + B cos x is _______
(d) The D.E \(\frac{d y}{d x}+\frac{y}{x \log x}=\frac{1}{x}\) is a ________ differential equation.
Answer:
(a) \(y=\frac{x}{2}+\frac{c}{x}\)
(b) \(\frac{1}{x}\)
(c) \(\frac{d^{2} y}{d x^{2}}+y=0\)
(d) linear

Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems

Question 10.
State true or false
(a) y = 3 sin x + 4 cos x is a particular solution of the differential equation \(\frac{d^{2} y}{d x^{2}}\) + y = 0
(b) The solution of \(\frac{d y}{d x}=\frac{x+2 y}{x}\) is x + y = kx2
(c) y = 13ex + 4e-x is a solution of \(\frac{d^{2} y}{d x^{2}}\) – y = 0
Answer:
(a) True
(b) True
(c) True

II. 2 Marks Questions

Question 1.
Form the D.E of the family of curves y = ae3x + bex where a, b are parameters
Solution:
y = ae3x + bex
\(\frac{d y}{d x}\) = 3 ae3x + bex
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems II Q1

Question 2.
Find the D.E of a family of curves y = a cos (mx + b), a and b are constants.
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems II Q2

Question 3.
Find the D.E by eliminating the constants a and b from y = a tan x + b sec x
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems II Q3

Question 4.
Solve: \(\frac{d y}{d x}=e^{7 x+y}\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems II Q4

Question 5.
Solve: (x2 – ay) dx = (ax – y2) dy
Solution:
Writing the equation as
x2 dx + y2 dy = a (x dy + y dx)
x2 dx + y2 dy = a d(xy)
∫x2 dx + ∫y2 dy = a ∫d(xy) + c
\(\frac{x^{3}}{3}+\frac{y^{3}}{3}\) = axy + c
Hence the general solution is x3 + y3 = 3axy + c

Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems

Question 6.
Solve (sin x + cos x) dy + (cos x – sin x) dx = 0
Solution:
The given equation can be written as
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems II Q6

III. 3 and 5 Marks Questions

Question 1.
Solve \(\frac{x d y}{d x}\) + cos y = 0, given y = \(\frac{\pi}{4}\) when x = √2
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems III Q1

Question 2.
The slope of a curve at any point is the reciprocal of twice the ordinate of the point. The curve also passes through the point (4, 3). Find the equation of the curve.
Solution:
Slope at any point (x, y) is the slope of the tangent at (x, y)
\(\frac{d y}{d x}=\frac{1}{2 y}\)
⇒ 2y dy = dx
⇒ ∫2y dy = ∫dx + c
⇒ y2 = x + c
Since the curve passes through (4, 3)
we have 9 = 4 + c ⇒ c = 5
Equation of the curve is y2 = x + 5

Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems

Question 3.
The net profit P and quantity x satisfy the differential equation \(\frac{d p}{d x}=\frac{2 p^{3}-x^{3}}{3 x p^{2}}\). Find the relationship between the net profit and demand given that p = 20 when x = 10
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems III Q3
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems III Q3.1

Question 4.
Solve: \(\frac{d y}{d x}\) + ay = ex (a ≠ -1)
Solution:
The given equation is of the form \(\frac{d y}{d x}\) + Py = Q
Here P = a, Q = ex
The general solution is y (I.F) = ∫Q (I.F) dx + c
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems III Q4

Question 5.
Solve: \(\frac{d y}{d x}\) + y cos x = \(\frac{1}{2}\) sin 2x
Solution:
Here P = cos x, Q = \(\frac{1}{2}\) sin 2x
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems III Q5

Question 6.
Solve (D2 – 6D + 25) y = 0
Solution:
The auxiliary equations is m2 – 6m + 25 = 0
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems III Q6
The Roots are complex and of the form,
α ± β with α = 3 and β = 4
The complementary function = e3x (A cos 4x + B sin 4x)
The general solution is y = e3x (A cos 4x + B sin 4x)

Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems

Question 7.
Solve (D2 + 10D + 25) y = \(\frac{5}{2}\) + e-5x
Solution:
The auxiliary equations is m2 + 10m + 25 = 0
(m + 5)2 = 0
m = -5, -5
The complementary function = (Ax + B) e-5x
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems III Q7

Question 8.
Suppose that the quantity demanded Qd = 40 – 4p – 4\(\frac{d p}{d t}+\frac{d^{2} p}{d t^{2}}\) and quantity supplied Qs = -6 + 8p where p is the price. Find the equilibrium price for market clearance.
Solution:
For market clearance, the required condition is Qd = Qs
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems III Q8

Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems

Students can download 12th Business Maths Chapter 10 Operations Research Miscellaneous Problems and Answers, Samacheer Kalvi 12th Business Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems

Question 1.
The following table summarizes the supply, demand and cost information for four factors S1, S2, S3, S4 shipping goods to three warehouses D1, D2, D3.
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 1
Find an initial solution by using the north-west corner rule. What is the total cost of this solution?
Solution:
Let ‘ai‘ denote the supply and ‘bj‘ denote the demand.
Then total supply = 5 + 8 + 7 + 14 = 34 and Total demand = 7 + 9 + 18 = 34
Σai = Σbj. So the problem is a balanced transportation problem and we can find a basic feasible solution, by North-west comer rule.
First allocation:
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 2
Second allocation:
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 3
Third allocation:
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 4
Fourth allocation:
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 5
Fifth allocation:
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 6
We first allot 4 units to cell (S3, D3) and then the balance 14 units to cell (S4, D3).
Thus we get the following allocations:
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 7
The transportation schedule:
S1 → D1, S2 → D1, S2 → D2, S3 → D2, S3 → D3, S4 → D3
(i.e) x11 = 5, x21 = 2, x22 = 6, x32 = 3, x33 = 4, x43 = 14
Total cost = (5 × 2) + (2 × 3) + (6 × 3) + (3 × 4) + (4 × 7) + (14 × 2)
= 10 + 6+ 18 + 12 + 28 + 28
= 102
Thus the initial basic solution is got by NWC method and minimum cost is Rs. 102.

Question 2.
Consider the following transportation problem
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 8
Determine an initial basic feasible solution using
(a) Least cost method
(b) Vogel’s approximation method
Solution:
Let ‘ai‘ denote the availability and ‘bj‘ denote the requirement. Then,
Σai = 30 + 50 + 20 = 100 and Σbj = 30 + 40 + 20 + 10 = 100
Since Σai = Σbj. the given problem is a balanced transportation problem and we can get an initial basic feasible solution.
(a) Least Cost Method (LCM)
First allocation:
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 9
Second allocation:
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 10
Third allocation:
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 11
Fourth allocation:
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 12
Fifth allocation:
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 13
We first allocate 10 units to cell (O2, D2). Since it has a minimum cost. Then we allocate the balance 10 units to cell (O1, D2). Thus we get the final allocation table as given below.
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 14
Transportation schedule:
O1 → D2, O1 → D3, O2 → D1, O2 → D2, O2 → D4, O3 → D2
(i.e) x12 = 10, x13 = 20, x21 = 30, x22 = 10, x24 = 10, x32 = 20
The total cost is = (10 × 8) + (20 × 3) + (30 × 4) + (10 × 5) + (10 × 4) + (20 × 2)
= 80 + 60+ 120 + 50 + 40 + 40
= 390
Thus the LCM, we get the minimum cost for the transportation problem as Rs. 390.
(b) Vogel’s approximation method:
First allocation:
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 15
The largest penalty = 3. So allocate 20 units to the cell (O3, D2) which has the least cost in column D2
Second allocation:
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 16
Largest penalty = 4. So allocate min (20, 30) units to the cell (O1, D3) which has the least cost in column D3
Third allocation:
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 17
The largest penalty is 3. So allocate min (20, 50) to the cell (O2, D2) which has the least cost in column D2.
Fourth allocation:
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 18
Largest penalty = 2. So allocate min (10, 30) to cell (O2, D4) which has the least cost in column D4
Fifth allocation:
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 19
Largest penalty = 1. Allocate min (30, 20) to cell (O2, D1) which has the least cost in column D1. Finally allot the balance 10 units to cell (O1, D1)
We get the final allocation table as follows.
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 20
Transportation schedule:
O1 → D1, O1 → D3, O2 → D1, O2 → D2, O2 → D4, O3 → D2
(i.e) x11 = 10, x13 = 20, x21 = 20, x22 = 20, x24 = 10, x32 = 20
Total cost is given by = (10 × 5) + (20 × 3) + (20 × 4) + (20 × 5) + (10 × 4) + (20 × 2)
= 50 + 60 + 80+ 100 + 40 + 40
= 370
Hence the minimum cost by YAM is Rs. 370.

Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems

Question 3.
Determine an initial basic feasible solution to the following transportation problem by using (i) North-West Corner rule (ii) least-cost method.
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 21
Solution:
Total supply = 25 + 35 + 40 = 100 = Σai
Total requirement = 30 + 25 + 45 = 100 = Σbj
Since Σai = Σbj the given transportation problem is balanced and we can find an initial basic feasible solution.
(i) North West Corner Rule
First allocation:
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 22
Second allocation:
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 23
Third allocation:
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 24
Fourth allocation:
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 25
We first allot 5 units to cell (S2, D3). Then balance 40 units we allot to cell (S3, D3).
The final allotment is given as follows
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 27
Transportation schedule:
S1 → D1, S2 → D1, S2 → D2, S2 → D3, S3 → D3
(i.e) x11 = 25, x21 = 5, x22 = 25, x23 = 5, x33 = 40
Total cost is = (25 × 9) + (5 × 6) + (25 × 8) + (5 × 4) + (40 × 9)
= 225 + 30 + 200 + 20 + 360
= 835
Hence the minimum cost is Rs. 835 by NWC method.

(ii) Least cost method (LCM)
First allocation:
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 28
Second allocation:
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 29
Third allocation:
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 30
Fourth allocation:
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 31
We first allot 15 units to cell (S3, D1) since it has the least cost. Then we allot the balance 15 units to cell (S1, D1).
The final allotment is given as follows
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 32
Transportation schedule:
S1 → D1, S1 → D3, S2 → D3, S3 → D1, S3 → D2
(i.e) x11 = 15, x13 = 10, x23 = 35, x31 = 15, x32 = 25
Total cost is = (15 × 9) + (10 × 5) + (35 × 4) + (15 × 7) + (25 × 6)
= 136 + 50 + 140 + 105 + 150
= 580
The optimal cost by LCM is Rs. 580.

Question 4.
Explain Vogel’s approximation method by obtaining an initial basic feasible solution to the following transportation problem.
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 33
Solution:
Total supply (ai) = 6+1 + 10=17
Total demand (bj) = 7 + 5 + 3 + 2 = 17
Σai = Σbj. So given transportation problem is a balanced problem and hence we can find a basic feasible solution by VAM.
First allocation:
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 34
The largest penalty is 6. So allot min (2, 1) to the cell (O2, D4) which has the least cost in column D4.
Second allocation:
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 35
The largest penalty is 5. So allot min (5, 6) to the cell (O1, D2) which has the least cost in column D2.
Third allocation:
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 36
The largest penalty is 5. So allot min (7, 1) to the cell (O1, D1) which has the least cost in row O1.
Fourth allocation:
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 37
Largest penalty = 4. So allocate min (6, 10) to the cell (O3, D1) which has the least cost in row O3
Fifth allocation:
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 38
The largest penalty is 6. So allot min (1, 4) to the cell (O3, D4) which has the least cost in row O3. The balance 3 units is allotted to the cell (O3, D3). We get the final allocation as given below.
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 39
Transportation schedule:
O1 → D1, O1 → D2, O2 → D4, O3 → D1, O3 → D3, O3 → D4
(i.e) x11 = 1, x12 = 5, x24 = 1, x31 = 6, x33 = 3, x34 = 1
Total cost is given by = (1 × 2) + (5 × 3) + (1 × 1) + (6 × 5) + (3 × 15) + (1 × 9)
= 2 + 15 + 1 + 30 + 45 + 9
= 102
Thus the optimal (minimal) cost of the given transportation problem by Vogel’s approximation method is Rs. 102.

Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems

Question 5.
A car hire company has one car at each of five depots a, b, c, d and e. A customer in each of the fine towers A, B, C, D and E requires a car. The distance (in miles) between the depots (origins) and the towers (destinations) where the customers are given in the following distance matrix.
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 40
How should the cars be assigned to the customers so as to minimize the distance travelled?
Solution:
In the given assignment problem, the number of rows equals the number of columns. So the problem is balanced and we can get an optimal solution.
Step 1:
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 41
We have subtracted the minimum distance of each row from all elements of that row.
Step 2:
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 42
We have subtracted the minimum distance of each column from all elements of that column.
Step 3: (Assignment)
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 43
We are not able to assign depots for C and E. So we proceed as below.
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 44
We have drawn a minimum number of lines to cover all zeros. Now 15 is the smallest element from all uncovered elements. Subtract this from all uncovered elements and add to the elements which lie at the intersection of two lines. Thus we obtain the following reduced matrix for fresh assignment.
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 45
Assignment Schedule
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 46
Thus the minimum distance is 570 miles.

Question 6.
A natural truck-rental service has a surplus of one truck in each of the cities 1, 2, 3, 4, 5 and 6 and a deficit of one truck in each of the cities 7, 8, 9, 10, 11 and 12. The distance (in kilometres) between the cities with a surplus and the cities with a deficit are displayed below:
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 47
How should the truck be dispersed so as to minimize the total distance travelled?
Solution:
The number of rows equals the number of columns. So the problem is a balanced assignment problem and we can get an optimal solution.
Step 1: Subtract the minimum element of each row from all elements of that row.
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 48
Step 2: Subtract the least element of each column from all elements of that column.
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 49
Step 3: (Assignment)
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 50
We are not able to assign city for starting place 4. We proceed as follows.
Draw a minimum number of lines to cover all the zeros of the matrix. Subtract the smallest element from all uncovered elements and add to the elements which lie at the intersection of two lines.
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 51
Then we obtain a new reduced matrix for fresh assignment.
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 52
Final dispersal
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 54
Thus the minimum total distance the trucks should travel is 125 km.

Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems

Question 7.
A person wants to invest in one of three alternative investment plans: Stock, Bonds and Debentures. It is assumed that the person wishes to invest all of the funds in a plan. The pay-off matrix based on three potential economic conditions is given in the following table:
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 55
Determine the best investment plan using each of the following criteria (i) Maximin (ii) Minimax
Solution:
(i) Maximin rule
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 56
Max (3000, 1000, 6000) = 6000. Since the maximum value is 6000, he must invest in debentures by the maximin criteria.
(ii) Minimax rule
Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems 57
Min (10000, 8000, 6000) = 6000. Since the minimum value is 6000, he must invest in debentures by the minimax criteria.