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S = {a, b, c}, T = {1, 2, 3}

(i) F: S → T is defined as:

F = { (a, 3), (b, 2), (c, 1)}

⇒ F (a) = 3, F (b) = 2, F (c) = 1 

Therefore, F⁻¹ : T → S is given by

F⁻¹  = { (3, a), (2, b), (1, c)}.

(ii) F: S → T is defined as:

F = { (a, 2), (b, 1), (c, 1)}

Since F (b) = F (c) = 1, F is not one-one.

Hence, F is not invertible i.e., F⁻¹  does not exist.

Thomas Jefferson University (TJU) with an acceptance rate of 86%, is much easier to secure admissions at than Florida International University (FIU).

The FIU acceptance rate of 64%, suggests the institute is fairly selective at granting admissions, & prefers highly qualified indviduals, accepting on average 6.4 applications for every 10 it receives.

However, it is worth noting though, that FIU ranks #91 as part of the country-level university rankings rolled out by Shiksha, which is considerably higher than Thomas Jefferson University that ranks #156.

Onto functions from the set {1, 2, 3, …, n} to itself is simply a permutation on n symbols 1, 2, …, n.

Thus, the total number of onto maps from {1, 2, …, n} to itself is the same as the total number of permutations on n symbols 1, 2, …, n, which is n!

28. Given, f (x)=

The given fxⁿ is valid for all x such that x – 1 ≥ 0 ⇒x≥ 1

∴ Domain of f (x)= [1,∞)

As x ≥ 1

⇒ x – 1 ≥ 1 – 1

⇒ x – 1 ≥ 0

⇒ ≥ 0

⇒ f (x) ≥ 0

So, range of f (x)= [0,∞ )

Let S be a non-empty set and P (S) be its power set. Let any two subsets A and B of S.

It is given that: $P\left(X\right)xP\left(X\right)\to P\left(X\right)$ is defined as $A.B=A\cap B\forall A,B\in P\left(X\right)$

We know that $A\cap X=A=X\cap A\forall A\in P\left(X\right)$

$\Rightarrow A.X=A=X.A\forall A\in P\left(X\right)$

Thus, X is the identity element for the given binary operation*.

Now, an element is  $A\in P\left(X\right)$ invertible if there exists $B\in P\left(X\right)$ such that

$A*B=X=B*A$  (As X is the identity element)

i.e.

$A\cap B=X=B\cap A$

This case is possible only when $A=X=B.$

Thus, X is the only invertible element in P (X) with respect to the given operation*.

Hence, the given result is proved.

Since every set is a subset of itself, ARA for all A ∈ P (X).

∴R is reflexive.

Let ARB ⇒ A ⊂ B.

This cannot be implied to B ⊂ A.

For instance, if A = {1, 2} and B = {1, 2, 3}, then it cannot be implied that B is related to A.

∴ R is not symmetric.

Further, if ARB and BRC, then A ⊂ B and B ⊂ C.

⇒ A ⊂ C

⇒ ARC 

∴ R is transitive.

Hence, R is not an equivalence relation since it is not symmetric.

Define $f:{\rm N}\to {\rm N}$ by

$f\left(x\right)=x+1$

And,  $g:{\rm N}\to {\rm N}$ by,

$g\left(x\right)\left\{\begin{array}{cc}\hfill x-1\hfill & \hfill if,x>1\hfill \\ \hfill 1\hfill & \hfill f,x=1\hfill \end{array}\right\}$

We first show that g is not onto.

For this, consider element 1 in co-domain N. it is clear that this element is not an image of any of the elements in domain.

$\therefore f$ is not onto.

Now,  $gof:{\rm N}\to {\rm N}$ is defined by,

$\begin{array}{l}gof\left(x\right)=g\left(f\left(x\right)\right)=g\left(x+1\right)=\left(x+1\right)-1\left[x,in{\rm N}=>\left(x+1\right)>1\right]\\ =x\end{array}$

Then, it is clear that for $y\in {\rm N}$ , there exists $x=y\in {\rm N}$ such that $gof\left(x\right)=gof\left(y\right)$

Hence, gof is onto.

62. 

The given eqⁿ of the lines are

 y - x = 0 _____ (1)

x + y = 0 ______ (2)

x - k = 0 ______ (3)

The point of intersection of (1) and (2) is given by

(y - x) - (x + y) = 0

⇒ y - x -x -y = 0

y = 0 and x = 0

ie, (0, 0)

The point of intersection of (2) and (3) is given by

(x + y) – (x – k) = 0

y + k = 0

y = –k and x = k

i.e, (k, –k)

The point of intersection of (3) and (1) is given by

x = k

and y = k

ie, (k, k).

Hence area of triangle whose vertex are (0, 0), (k, –k)

and (k, k) is

27. Given, f (x)= $\frac{x^2+2x+1}{x^2-8x+12}$

The given function is valid if denominator is not zero.

So, if x² – 8x+12=0.

⇒ x² – 2x – 6x+12=0

⇒ x (x – 2) –6 (x – 2)=0

⇒ (x – 2) (x – 6)=0

⇒ x=2 and x=6.

So,  f (x) will be valid for all real number x except x=2,6.

∴ Domain of f (x)=R – {2,6}

Getting direct admission to the BCom programme of  Karnavati University may not be possible. This is because the university grants BCom admission to applicants who qualify in the selection rounds conducted by the university. The first round is the KUAT entrance test. Alternatively, candidates can also submit their CUET UG scores. The next round is to appear for a personal interview conducted by the university. Further, applicants are required to meet the course-specific eligibility criteria.

Define $f:N\to Z$ as $f\left(x\right)$ and $g:Z\to Z$ as $g\left(x\right)=\left|x\right|$

We first show that g is not injective.

It can be observed that:

$\begin{array}{l}g\left(-1\right)=\left|-1\right|=1\\ g\left(1\right)=\left|1\right|=1\\ \therefore g\left(-1\right)=g\left(1\right),but,-1\ne 1\end{array}$

$\therefore g$ is not injective.

Now, $gof:N\to Z$ is defined as

$gof\left(x\right)=y\left(f\left(x\right)\right)=y\left(x\right)=\left|x\right|$

Let $x,y\in N$ such that $gof\left(x\right)-gof\left(y\right)$

$\Rightarrow \left|x\right|=\left|y\right|$

Since $x,y\in N$ , both are positive.

$\therefore \left|x\right|=\left|y\right|\Rightarrow x=y$

Hence, gof is injective

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61. The given Eqⁿ of the line is $\frac{x}{4}+\frac{y}{6}$ = 1 ______ (1)

so, Slope of line = -$\frac{6}{4}=-\frac{3}{2}.$

The line ⊥ to line (1) say l₂ has

Slope of l₂ = $\frac{-1}{(-3/2)}=\frac{2}{3}.$

Let P (0, y) be the point of on y-axis where it is cut by the line (1)

Then,  $\frac{0}{4}+\frac{y}{6}=1$

y = 6

i.e, the point P has co-ordinate (0, 6)

Eqⁿ of line ⊥ to $\frac{x}{4}+\frac{y}{6}=1$ and cuts y-axis at P (0,6) is

y – 6 = $\frac{2}{3}$ (x – 0)

3y – 18 = 2x

2x – 3y + 18 = 0

f: R → R is given as f (x) = x³.

Suppose f (x) = f (y), where x, y ∈ R.

⇒ x³ = y³  … (1)

Now, we need to show that x = y.

Suppose x ≠ y, their cubes will also not be equal.

⇒ x³ ≠ y³

However, this will be a contradiction to (1).

∴ x = y

Hence, f is injective.

60. The given eqn of lines are

x - 7y + 5 = 0 ______ (1) ⇒ x = 7y - 5

and 3x + y = 0 _________ (2)

Solution (1) and (2) we get,

3 [7y – 5] + y = 0 .

⇒ 21y - 15 + y = 0

⇒ 22y = 15

26. Given, f(x)=x².

$\frac{f(1.1)-f(1)}{1.1-1}=\frac{{(1.1)}^2-1^2}{1.1-1}=\frac{1.21-1}{0.1}=\frac{0.21}{0.1}=2.1$

It is given that $f:R\to R$ is defined as $f\left(x\right)=x^2-3x+2.$

$\begin{array}{l}f\left(f\left(x\right)\right)=f\left(x^2-3x+2\right)\\ ={\left(x^2+3x+2\right)}^2-3\left(x^2-3x+2\right)+2\\ =x^4+9x^2+4-6x^2-12x+4x^2-3x^2+9x-6+2\\ =x^4-6x^2+10x^2-3x\end{array}$

Thomas Jefferson & Montana State University are both part of the most prominent higher educational institutes in the United States. 

Both perform similarly when it comes to Shiksha Popularity Rankings, securing ranks #156 & #158 respectively.

In terms of acceptance rate, Thomas Jefferson is comparitively easier to secure admisisions at, with its acceptance rate of 86%, comfortably higher than that of Montana State University's 64%.

For a more detailed comparison, check out the table below: