# Mathematics I 2019

**Tribhuwan University**

**Faculty of Humanities and Social Sciences**

**OFFICE OF THE DEAN**

**2019**

**Bachelor of Computer Applications**

**Course Title: Mathematics I**

**Code No: CACS104**

**Semester: First Semester**

Group A

Attempt any SIX questions. [6x5 = 30]

1. In class of 100 students 40 students failed in Mathematics, 70 failed in English and 20 failed in both subjects. Find

a) How many students passed in both subjects?

b) How many students passed in Mathematics only?

c) How many students failed in mathematics only?

Let M and E be the set of students who failed in Mathematics and English respectively.

Then,

n(U) = 100

n(M) = 40

n(E) = 70

n(M∩E) = 20

**i)** We have,

Therefore, number of students passed in both subjects = 10.

**ii) **

Number of students passed in Mathematics only =

**iii)**

Number of students failed in mathematics only = n(M) - n(M∩E) = 40 - 20 = 20

2. Find the domain and range of the function

Given,

For domain; when (3-x)=0, f(x) doesn't exist, it means f(x) is exist for all real number except x = 3. Therefore domain is

For range,

Let,

i.e.

Here, x is exist for all real number except y = -2. Therefore range is

3. Find the Maclurin series of the function f(x) = sinx.

Maclaurin series for a function *f*(*x*) is found by:

So, we need to find derivatives of

Thus, the Maclaurin series for sin(x) is

Simplifying this yields the following:

4. Prove that = (x - y) (y - z) (z - x).

Given

Applying,

Hence proved.

5. Find a unit vector perpendicular to the plane containing points P(1, -1, 0), Q(2, 1, -1) and R(-1, 1, 2).

The given points P(1, -1, 0), Q(2, 1, -1) and R(-1, 1, 2) lies in the plane PQR.

Accordingly, the vectors

Hence, plane PQR.

Finally, the required unit vector will be

Finally, the desired unit vector is

6. In how many ways can be letter of words “Sunday” be arranged? How many of these arrangement begin with S? How many begin with S and don’t end with y?

The word SUNDAY be arranged in 6!=6*5*4*3*2=720 ways

When a word begins with S. Its position is fixed, i.e. the first position. Now rest 5 letters are to be arranged in 5 places. So,

No. of arrangement begin with S = 5! = 120

The are 5! ways of ordering the word SUNDAY if the first letter is restricted to S, because only the 5 remaining ones are allowed to change.

There are 4! ways of ordering the word SUNDAY if the first letter is restricted to S "and" the last letter is restricted to Y

Therefore, there are 5!-4! ways that the word SUNDAY can be arranged such that the first letter is S and the last letter "is not" Y

5!-4! = 120-24 =96

7. If then show that x^{2} + y^{2} = 1.

Given,

Now,

Multiplying eqn. (i) & (ii)

Hence proved.

Group C

Attempt any TWO questions. [2x10 = 20]

8. a) Define conic section. Find the coordinates of vertices, eccentricity and foci of the ellipse

9x^{2} + 4y^{2} - 18x - 16y - 11 = 0. (**1+5)**

A conic section (or simply conic) is a curve obtained as the intersection of the surface of a cone with a plane. The three types of conic sections are the hyperbola, the parabola, and the ellipse.

Given ellipse,

9x^{2} + 4y^{2} - 18x - 16y - 11 = 0

Converting to standard form of ellipse:

Here, h=1, k=2, a=2, b=3

Now,

*Coordinate of Vertices:*

i.e. (1, 5) & (1, -1)

*Eccentricity:*

*Foci:*

b) If T(x_{1}, x_{2}) = (x_{1} + x_{2}, x_{2}, x_{1})defined by be the linear transformation, then find matrix associated with linear map T. **(4)**

Given,

T(x_{1}, x_{2}) = (x_{1} + x_{2}, x_{2}, x_{1})

We have,

T(e_{1}) = T(1, 0) = (1+0, 0, 1) = (1, 0, 1)

T(e_{2}) = T(0, 1) = (0+1, 1, 0) = (1, 1, 0)

Hence

A = |T(e_{1}), T(e_{2})|

=

We confirm

9. a) Define irrational number. Prove that √2 is an irrational number. (**1+4)**

An** irrational number **is a number that cannot be expressed as a fraction for any integers and .

## Let us assume on the contrary that $2 $ is a rational number. Then, there exist positive integers p and q such that

## , where, p and q, are co-prime i.e. their $HCF$ is $1.$

## $On squaring both sides, we get$

## $p²= 2q² ...... (1)$

## $Clearly, 2 is a factor of 2q²⇒ 2 is a factor of p² [since,2q²=p²]⇒ 2 is a factor of p$

## $Let p =2m for all m ( where m is a positive integer)Squaring both sides, we getp²= 4 m² .....(2)$

## $From (1) and (2), we get2q² = 4m² ⇒ q²= 2m²Clearly, 2 is a factor of 2m²⇒2 is a factor of q² [since, q² = 2m²]⇒ 2 is a factor of q$

## $Thus, we see that both p and q have common factor 2 which is a contradiction that H.C.F. of (p,q)= 1Therefore, Our supposition is wrongHence √2 is not a rational number i.e., irrational number.$

b) If functions defined by f(x) = 2x + 1 and defined by g(x) = x^{2} - 2. Find the formulae for composite functions and also verify that (4+1)

Given,

f(x) = 2x + 1

g(x) = x^{2} - 2

Now,

fog(x) = f(g(x)) = 2(x^{2} - 2)+1

= 2x^{2}-4+1

= 2x^{2}-3

gof(x) = g(f(x)) = (2x+1)^{2}-2

= 4x^{2}+4x-1

Hence,

10. a) If arithmetic mean, geometric mean and harmonic mean between two unequal positive numbers are A, G, H respectively. Then prove that A > G > H.** (4)**

Let $a$ and $b$ be two real positive and unequal numbers and let $A,G,H$ be arithmetic, geometric and harmonic mean respectively then,

A =

G =

H =

Now,

**A - G** =

**A - G > 0** i.e. A>G

Again,

**G - H** =

**G - H > 0** i.e. G>H

Hence proved that

A > G > H

b) What is the relation between permutation and combination of n objects taken r at a time? A committee of 5 is to be constituted from 6 boys and 5 girls. In how many ways can this be done so as to include at least a boy and a girl? (1+5)

The relation between permutation and combination of n objects taken r at a time is derived as:

Now,

Total no. of students required to be chosen = **5 **

No. of boys = **6**

No. of girls = **5**

Total no. of ways to choose 5 students out of 11
students = ^{11}C_{5} =

The case where all committee has boys = ^{6}C_{5}
= 6

The case where all committee has girls = ^{5}C_{5}
=1

Therefore, The number of ways when at least1 boy and 1 girls are in
committee = 462-(6+1) = 455 ways