Mathematics I 2019

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

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

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

So, we need to find derivatives of $f\left(x\right)=\sin \left(x\right)$.

Thus, the Maclaurin series for sin(x) is

Simplifying this yields the following:

Given

Applying,

Hence proved.

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

Accordingly, the vectors  the plane PQR.

Hence,  plane PQR.

Finally, the required unit vector will be

Finally, the desired unit vector is

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

Given,

Now,

Multiplying eqn. (i) & (ii)

Hence proved.

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,

 9x² + 4y² - 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:

Given,

T(x₁, x₂) = (x₁ + x₂, x₂, x₁)

We have,

T(e₁) = T(1, 0) = (1+0, 0, 1) = (1, 0, 1)

T(e₂) = T(0, 1) = (0+1, 1, 0) = (1, 1, 0)

Hence

A = |T(e₁), T(e₂)|

   = 

We confirm

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

Let us assume on the contrary that $\sqrt{2}$ 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 $1.$

$On\; squaring\; both\; sides,\; we\; get$

$p²=\; 2q²\; ......\; (1)$

$Clearly,\; 2\; is\; a\; factor\; of\; 2q²\Rightarrow \; 2\; is\; a\; factor\; of\; p²\; [since,2q²=p²]\Rightarrow \; 2\; is\; a\; factor\; of\; p$

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

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

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

Given,

f(x) = 2x + 1 

g(x) = x² - 2

Now,

fog(x) = f(g(x)) = 2(x² - 2)+1

          = 2x²-4+1

          = 2x²-3

gof(x) = g(f(x)) = (2x+1)²-2

          = 4x²+4x-1

Hence,

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

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 = ¹¹C₅ = 

The case where all committee has boys = ⁶C₅ = 6

The case where all committee has girls = ⁵C₅ =1

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