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BSc in Economics is a well-known course of Easwari School of Liberal Arts, SRM University AP. The course-wise placement stats is yet to be relaesed on the official website. Meanwhile, check out the overall highlights of Easwari School of Liberal Arts, SRM University Andhra Pradesh placements for the Class of 2024 in the table below:

Note: The above-mentioned placement stats are the cumulative stats of all the courses of SRM University AP.

Given, $f:N\to N$ defined $f(x)=\left(\begin{array}{cc}\hfill \frac{x+1}{2},if\hfill & \hfill xisodd\hfill \\ \hfill \frac{x}{2},if\hfill & \hfill xiseven\hfill \end{array}\right)$ $\cup x\in N$

Let $x_1=1$ and $x_2=2\in N,$

$f(x_1)=f(x_2)\Rightarrow f(1)=f(2)$

$\Rightarrow \frac{1+1}{2}=\frac{2}{2}$

$\Rightarrow 1=1$ but $1\ne 2$

So, $f$ is not one-one

For $x=$ odd and $x\in N$ , say $x=2C+1$ where $C\in N$

There exist $(4C+1)$ $\in N$ such that

$(4C+1)=\frac{4C+1+1}{2}=2C+1.\in N$

And for $x=$ even $\in N$ , say $x=2C$ where $C\in N$

There exist $(4C)\in N$ such that

$f(4C)=\frac{4C}{2}=2C.\in N$

So, $f$ is onto

But, $f$ is not bijective

Yes, Easwari School of Liberal Arts, SRM University Andhra Pradesh offers decent placements annually for its BSc and BA courses.

Note: The above-mentioned placement stats are the cumulative stats of all the courses of SRM University AP.

The BA course at Gokul Global University is offered to the students in offline mode. It is a clasroom programme. The duration of the course goes on for three years, which is six semesters. 

Given, $f:A\times B\to B\times A$ defined as $f(a,b)=(b,a)$

Let $(a_1,b_1),(a_2,b_2)\in A\times B$ such that

$f(a_1,b_1)=f(a_2,b_2)$

$\Rightarrow (b_1,a_1)=(b_2,a_2)$

So, $b_1=b_2$ and $a_1=a_2$

$\Rightarrow (a_1,b_1)=(a_2,b_2)$

$\therefore f$ is one-one

For $(a,b)\in B\times A$

There exist $(a,b)\in A\times B$ such that $f(a,b)=(b,a)$

$\therefore f$ is onto

Hence, $f$ is bijective

Easwari School of Liberal Arts, SRM University Andhra Pradesh has a good record of placements for its BSc and BA courses. In 2024, the the highest and average package offered stood at INR 42.82 LPA and INR 9.38 LPA, respectively. The key highlights of Easwari School of Liberal Arts, SRM University Andhra Pradesh placements for the Class of 2024 are tabulated below:

Note: The above-mentioned placement stats are the cumulative stats of all the courses of SRM University AP.

(i) $f:R\to R$ defined as $f(x)=3-4x$

For $x_1,x_2\in R$ such that $f(x_1)=f(x_2)$

$\begin{array}{l}\Rightarrow 3-4x_1=3-4x_2\\ \Rightarrow 4x_1=4x_2\\ \Rightarrow x_1=x_2\end{array}$

So, $f$ is one-one

For $y\in R$ , there exist

$f\left(\frac{.3-y}{4}\right)=3-4\left(\frac{3-y}{4}\right)=3-3+y=y$

Hence, $f$ is onto

$\therefore f$ is bijective

(ii) Given, $f:R\to R$ defined as $f(x)=1+x_{}^2$

For $x_1,x_2\in R$ such that $f(x_1)=f(x_2)$

$\begin{array}{l}\Rightarrow 1+x_1^2=1+x_2^2\\ \Rightarrow x_1^2=x_2^2\\ \Rightarrow x_1^{}=\pm x_2^{}\end{array}$

$\Rightarrow x_1^{}=x_2^{}$ or $x_1^{}=-x_2^{}$

$\therefore f$ is not one-one

The range of $f(x)$ is always a positive real number which is not equal to co-domain $R$

So, $f$ is not onto

20. (i) Domain of the given relation = {2,5,8,11,14,17}

Since every element of the domain has one and only one image, the given relation is a fxⁿ.

So, domain = {2,5,8,11,14,17}

range = {1}

(ii) Domain of the given relation = {2,4,6,8,10,12,14}

Since every element of the domain has one and only one image, the given relation is a fxⁿ.

So, domain = {2, 4, 6, 8, 10, 12, 14}

range = {1,2,3,4,5,6,7}

(iii) Domain of the given relation = {1,2}

As element 1 has more than one image i.e., 3 and 5, the given relation is not a fxⁿ.

Given,  $f:A\to B$ and $f=\{(1,4),(2,5),(3,6)\}$

$\begin{array}{l}\therefore f(1)=4\\ f(2)=5\\ f(3)=6\end{array}$

i.e., the image elements of $A$ under the given $f_X^n$ $f$ are unique

So,  $f$ is one-one

49.

Let P(3, 9) and Q (-1, 2) be the point. Let M bisects PQ at M so,

Co-ordinate of M = $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

$=\left(\frac{3+(-1)}{2},\frac{4+2}{2}\right)$

$=\left(\frac{3-1}{2},\frac{6}{2}\right)=\left(\frac{2}{2},\frac{6}{2}\right)=(1,3)$

Now, slope of AB, m = $\frac{y_2-y_1}{x_2-x_1}=\frac{2-4}{-1-3}=\frac{-2}{-4}=\frac{1}{2}.$

As the bisects AB perpendicular it has slope $-\frac{1}{m}$ and it passes through M(1, 3) it has the equation of the form,

$\frac{-1}{m}=\frac{y-y{}_{\mathrm{\sigma }}}{x-x{}_0}$

$\frac{-1}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{y-3}{x-1}$

$\Rightarrow -2=\frac{y-3}{x-1}$

$\Rightarrow -2(x-1)=y-3$

$\Rightarrow -2x+2=y-3$

$\Rightarrow 2x+y-3-2=0$

2x + y - 5 = 0

The $f{x}_{}^n$ $f:R?R$ is given by $f(x)=\left(\begin{array}{ccc}\hfill 1_{}\hfill & \hfill if\hfill & \hfill x>0\hfill \\ \hfill 0_{}\hfill & \hfill if\hfill & \hfill x=0\hfill \\ \hfill ?1_{}\hfill & \hfill if\hfill & \hfill x<0\hfill \end{array}\right)$

For $x_1=1,x_2=2,?R$

$f(x_1)=f(1)=1$

$f(x_2)=f(2)=1$ but $1?2$

So,  $f$ is not one-one

And the range of $f(x)=\{1,0,?1\}$ hence it is not equal to the co-domain $R$

So,  $f$ is not onto

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48.

Given, slope of line 1, m₁ = 2.

Let m₂ be the slope of line 2.

If θ is the angle between the two lines then we can write,

The $f{x}_{}^n$ $f:R\to R$ is given by $f(x)=\left|x\right|$

$\Rightarrow f(x)=\left(\begin{array}{cc}\hfill x_{},if\hfill & \hfill x\ge 0\hfill \\ \hfill -x_{},if\hfill & \hfill x<0\hfill \end{array}\right)$

For $x_1=-1$ and $x_2=1$

$f(x_1)=f(-1)=\left|-1\right|=1$

$f(x_2)=f(1)=\left|1\right|=1$

So, $f(x_1)=f(x_2)$ but $x_1\ne x_2$

i.e., $f$ is not one-one

For $x=-1\in R$

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

i.e., $f(-1)=\left|-1\right|=1$

So, range of $f(x)$ is always a positive real number and is not equal to the co-domain $R$

i.e., $f$ is not onto

The $f{x}_{}^n$ $f:R\to R$ is given by $f(x)=[x]$

Let $x_1=1.5$ and $x_2=1.2\in R$ Then,

$f(x_1)=f(1.5)=[1.5]=1$

$f(x_2)=f(1.2)=[1.2]=1$

So, $f(x_1)=f(x_2)$ but $x_1\ne x_2$

i.e., $f(1.5)=f(1.2)$ but $1.5\ne 1.2$

So, $f$ is not one-one

The range of $f(x)$ is a set of all integers, $Z$ which is not a co-domain of $R$

$\therefore f$ is not onto

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