Engineering Mathematics Notes by Made Easy 2023

Engineering Mathematics Notes by Made Easy 2023

Engineering Mathematics Notes by Made Easy 2023

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Engineering Mathematics Notes by Made Easy 2023:- Today we are sharing Engineering Mathematics Notes by Made Easy 2023. This Engineering Mathematics Notes by Made Easy 2023 for GATE, NEET, JEE ADVANCE, NET, JRF, RAS, UPSC, NTPC, ALL STATE PCS EXAM. We all know that math subject is the most important subject for our all government job exams so today, we are sharing easy and important the Engineering Mathematics Notes by Made Easy 2023 in English Free for your math exams score improvement. This Engineering Mathematics books for competitive exams pdf free download Download for the preparation of upcoming government exams like SSC CGL, BANK, RAILWAYS, RRB NTPC, LIC AAO, and many other exams. Math Book pdf Download For Competitive Exams 2022-23.

 where we are sharing Engineering Mathematics Notes by Made Easy 2023 in English for free DOWNLOAD for UPSC, SSC, BANK, RAILWAY, LIC, and many other exams. Our Best Engineering Mathematics Book and Notes PDF Download for All Competitive Exams | Engineering Mathematics Notes by Made Easy 2023 is very simple and easy to understand. We also cover basic subjects like Mathematics, Geography, History, General Science, Politics, etc. We also share study material including previous year question papers, current affairs, important sources, etc. for upcoming government exams. Our PDF will help you prepare for any SARKARI EXAM.

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Engineering Mathematics Questions and Answers 

Q 1. Calculate the sum of lengths: 21 m 13 cm, 33 m 55 cm and 45 m 6 cm.

Solution:

21 m 13 cm + 33 m 55 cm + 45 m 6 cm

= (21 + 33 + 45) m (13 + 55 + 6) cm

= 99 m 74 cm

Therefore, the sum of given lengths = 99 m 74 cm.

Q 2. Beena bought 3 kg 760 grams of wool to make a carpet. How much more wool does she need to make the weight 4 kg?

Solution:

Given the weight of wool = 3 kg 760 grams

Let us convert this weight into grams.

Q 3 kg 760 grams = (3 × 1000 + 760) grams

= (3000 + 760) grams

= 3760 grams

4 kg = (4 × 1000) grams = 4000 grams

Difference = (4000 – 3760) grams = 240 grams

Therefore, 240 grams of more wool is required.

Q 3. A pile of 10 books is 10 cm high. What is the thickness of each book?

Solution:

As we know,

1 cm = 10 mm

Given that the height of a pile of 10 books = 10 cm

10 books = 10 cm

= 10 x 10 mm

= 100 mm

1 book = 100/10

= 10 mm

Therefore, the thickness of each book = 10 mm.

Q 4. A furlong is a unit of length used in horse racing; it equals one-eighth of a mile. To the nearest tenth, how many metres are equal to a furlong if 1.609 km equals a mile?

Solution:

Given,

1.609 km = 1 mile

That means 1609 m = 1 mile

Also, one furlong = one-eighth of a mile

= (1/8) × 1.609 km

= (1/8) × 1609 m

= 201.125

= 201.1 m (approx)

Q 5. The cost of 1 litre of syrup is Rs. 840.80. Find the cost of 600 ml of the syrup.

Solution:

Given,

The cost of 1 litre of syrup = Rs. 840.80

As we know,

1 litre = 1000 ml

The cost of 600 ml of the syrup = (600/1000) × Rs. 840.80 = Rs. 504.48

Q 6. One inch equals 2.54 centimetres. How many centimetres tall is an 84-inch door?

Solution:

Given,

1 inch = 2.54 cm

84 inch = ?

84 inch = 84 × 2.54 cm

= 213.36 cm

Thus, the door is 213.36 cm tall.

Q 7. Vinu and Shan together weigh 72 kg 350 g. If Vinu weighs 39 kg 185 g, what is Shan’s weight?

Solution:

Given,

Vinu’s weight = 39 kg 185 g

Let x be Shan’s weight.

According to the given,

39 kg 185 g + x = 72 kg 350 g

x = 72 kg 350 g – 39 kg 185 g

= (72 – 39) kg (350 – 185) g

= 33 kg 165 g

Therefore, Shan’s weight is 33 kg 165 g.

Q 8. A tabletop measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the tabletop?

Solution:

We know that,

1 m = 100 cm

1 cm = 0.01 m

Length of tabletop = 2 m 25 cm = 2.25 m

Breadth of tabletop = 1 m 50 cm = 1.50 m

Perimeter of tabletop = 2 (Length + Breadth)

= 2 (2.25 + 1.50)

= 2 (3.75)

= 2 × 3.75

= 7.5 m

Therefore, the perimeter of the tabletop is 7.5 m.

Q 9. Two sides of a triangle are 11 cm and 15 cm. If the perimeter of the triangle is 36 cm, find its third side.

Solution:

Let x cm be the third side of the triangle.

Given,

Two sides: 11 cm and 15 cm

Perimeter of triangle = 36 cm

11 + 15 + x = 36

26 + x = 36

x = 36 – 26

x = 10 cm

Hence, the third side of the triangle is 10 cm.

Q 10. The weight of a rice packet is 40 kg 360 g. What is the total weight of 6 rice packets of the same weight?

Solution:

Given,

Weight of a rice packet = 40 kg 360 g

Number of rice packets = 6

Total weight = 6 × weight of one rice packet

= 6 × (40 kg 360 g)

= (6 × 40) kg (6 × 360) g

= 240 kg 2160 g

= 240 kg (2000 + 160) g

= (240 + 2) kg 160 g

= 242 kg 160 g

Thus, the total weight of 6 rice packets = 242 kg 160 g.

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Engineering Mathematics Questions and Answers 

Q 1. The pth derivative of a qth degree monic polynomial, where p, q are positive integers and 2p4 + 3pq3⁄2 = 3q3⁄2 + 2qp3 is given by?
a) Cannot be generally determined
b) (q – 1)!
c) (q)!
d) (q – 1)! * pq
Answer: c
Explanation: First consider the equation
2p4 + 3pq3⁄2 = 3q3⁄2 + 2qp3
After simplification, we get
(2p3 + 3q1⁄2) (p – q) = 0
This gives us two possibilities
2p3 = – 3q1⁄2
OR
p = q
The first possibility can’t be true as we are dealing with positive integers
Hence, we get
p = q
Thus the pth derivative of any monic polynomial of degree p(p = q) is
p! =q!.
Q 2. The first and second derivatives of a quadratic Polynomial at x = 1 are 1 and 2 respectively. Then the value of f(1) – f(0) Is given by?
a) 3⁄2
b) 1⁄2
c) 1
d) 0
Answer: d
Explanation: Let the quadratic polynomial be
f(x) = ax2 + bx + c
The first derivative at x = 1 is given by
2a + b = 1
Now consider the second derivative at x = 1 which is given by
2a = 2
Solving for the coefficients using equations, we get the values as a = 1 and b = -1
Putting these values back in the polynomial yields
f(x) = x2 – x + c
Now the required value can be computed as
f(1) – f(0) = (12 – 1 + c) – (02 – 0 + c)
= (0 + c) – (0 + c) = 0.
Q 3. Let f(x) = sin(x)x–54, then the value of f(100)(54) is given by?
a) Undefined
b) 100
c) 10
d) 0
Answer: a
Explanation: The key here is to expand the numerator into a taylor series centered at 54
Doing this gives us the following
Sin(x)=Sin(54)+(x−54)×Cos(54)1!−(x−54)2×Sin(54)2!…∞
Now the function transforms into f(x)=Sin(54)(x−54)+Cos(54)1!−(x−54)×Sin(54)2!…∞
Q 4. Let f(x) = ln(x2 + 5x + 6) then the value of f(30)(1) is given by?
a) (29!)(1330+1430)
b) (-29!)(1330+1430)
c) (30!)(1330+1430)
d) (-30!)(1330+1430)
Answer: b
Explanation: Given function
f(x) = ln(x2 + 5x + 6)
Factorising the inner polynomial we get
f(x) = ln((x + 3) (x + 2))
Now using the rule of logarithms ln (m * n) = ln(m) + ln(n)
we get
f(x) = ln(x + 3) + ln(x + 2)
Now using the nth derivative of logarithmic function dn(ln(x+a))dxn=(−1)n+1×(n−1)!(x+a)n
We have
dn(f(x))dxn=dn(ln(x+3))dxn+dn(ln(x+2))dxn
Which simplifies to dn(f(x))dxn=(−1)n+1×(n−1)!(x+3)n+(−1)n+1×(n−1)!(x+2)n
Substituting x=1 and n=30
Gives us the answer as (-29!)(1330+1430)

Engineering Mathematics Questions and Answers 

Q 5. f(x) = ∫π/20sin(ax)da then the value of f(100)(0) is?
a) a(100) sin(a)
b) -a(100) sin(a)
c) a(100) cos(a)
d) 0
Answer: d
Explanation: First solve the integral f(x)=∫π/20sin(ax)da
Which gives =[−cos(ax)x]π/20=1−cos(πx2)x
Now expanding into Taylor series yields
=((πx2)2×2!−(πx2)4×4!….∞)
Observe that every term in the expansion is odd powered
Hence even derivative at x = 0 has to be 0.
Check this: Numerical Methods MCQ | Engineering Mathematics Books
Q 6. Let f(x) = x9 ex then the ninth derivative of f(x) at x = 0 is given by?
a) 9!
b) 9! * e9
c) 10!
d) 21!
Answer: a
Explanation: The key here is to expand ex as a mclaurin series and then multiply it by x9
We have ex=x00!+x11!+x22!…∞
Now multiplying it by x9 we get
f(x)=x90!+x101!+x112!…∞
This simplifies into polynomial differentiation and the ninth derivative is as follows
f(9)(x)=9!+10!×x1!+p119×x22!…∞
Substituting x = 0 gives us the result
f(9)(0) = 9!.
Q 7. The following moves are performed on g(x).
(i) Pick (x0, y0) on g(x) and travel toward the left/right to reach the y = x line. Now travel above/below to reach g(x). Call this point on g(x) as (x1, y1)
(ii) Let the new position of (x0, y0)be (x0, y1)
This is performed for all points on g(x) and a new function is got. Again these steps may be repeated on new function and another function is obtained. It is observed that, of all the functions got, at a certain point (i.e. after finite number of moves) the nth derivatives of the intermediate function are constant, and the curve passes through the origin. Then which of the following functions could be g(x)
a) y = 1–x2−−−−√
b) xy3⁄2 + y = constant
c) x9 y3⁄2 + y6 x3⁄2 = constant
d) x7 y 8 + 4y = constant
Answer: d
Explanation: The phrasing is a bit convoluted. It is asking us to convert a given function f(x) into its own composition f(f(x))
Now the set of finite number of moves asks us to find the composition of composition i.e. f(f(x)) and f(f(x)) and so on.
It also says that the intermediate composition function has the property of all nth derivatives being constant. This is possible only if the intermediate function is of the form k(x) = ax + b (Linear Function)
Given another condition that it has to pass through origin leads to the conclusion that b = 0
So the intermediate function has the form
k(x) = ax
The first possibility gives rise to
k(x) = x
Now one has to select from the options as to which curve would give k(x) = f(f(….(x)…)) = x when composed with itself a finite number of times.
This can be done by simply interchanging the position of y and x
in the options and check whether it preserves its structure.
For y = 1–x2−−−−√ we have after changing the position
This has the same structure.

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