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BA graduates from NIPS Bhubaneswar generally receive starting salaries ranging between INR 15,000 to INR 20,000 per month in domestic roles. For those securing international offers, the salary can go up to INR 1.5 Lacs per month. The package largely depends on the role, recruiter, and whether the placement is national or overseas.

Yes, Karnavati University grants entrance-based admissions to its BCom (Hons) course. To shortlist prospective candidates, the university conducts the Karnavati University Admission Test (KUAT). It also accepts the CUET UG scores of applicants. However, merely qualifying in the entrance test is not enough to secure BCom (Hons) admission at the university. Candidates must also qualify in the Personal Interview (PI) conducted by the university in addition to securing a valid score in KUAT/ CUET UG. 

UTET stands for Uttarakhand Teachers Eligibility Test. It is conducted every year by the Uttarakhand Board of School Education (UBSE). It is a state-level teacher's recruitment test. Candidates who are interested to build a career in the teaching field must clear this examination.

31. Given, f(x) = x+1. and g(x) = 2x – 3.

So, (f +g)(x) = f(x)+g(x) = (x+1)+(2x – 3) = x+1+2x – 3 = 3x – 2

(f – g)(x) = f(x) –g(x) = (x+1)–(2x–3) = x+1 – 2x+3 = 4 – x

$\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{x+1}{2x-3} such that x\ne \frac{3}{2}$

68. The given eqⁿ of line is

l1: x + y = 4

Let R divides the line joining two points P (?1,1) and Q (5,7) in ratio k:1. Then,

Co-ordinate of R = ($\left(\frac{5k-1}{k+1},\frac{7k+1}{k+1}\right)$)

As l₁ divides line joining PQ, then R lies on l₁

i e,  $\left(\frac{5k-1}{k+1},\frac{7k+1}{k+1}\right)$=4

5k ?1 + 7k + 1= 4 (k + 1)

12k = 4k + 4

8k = 4

k = $\frac{1}{2}$

The ratio in which x + y = 4 divides line joining (?1,1) ad (5,7) is $\frac{1}{2}$ :1 i.e., 1: 2.

Karnavati University BCom (Hons) application process is conducted online. Candidates willing to take admission to the university's BCom (Hons) programme must apply though the university's official website. The steps included in the application process are listed below:

1. Visit the official website of Karnavati University.
2. Click on the 'Admission' tab on the Homepage.
3. Register by filling in the required details, including your name, email ID, and mobile number.
4. Verify you email address.
5. Fill out the application form.
6. Pay the application fee and submit the form.

Once selected, the final step is to complete the fee payment within the deadline provided by the institute. This includes tuition and other applicable charges. Admission is only confirmed after successful payment, so make sure to act promptly to secure your spot.

Hi, the top specialisations offered by top private Science colleges in India:

Hope this helps!

67. The given eqⁿ of line is.

l₁ : y = mx + c.

Slope of l₁ = m

Let m? be the slope of line passing through origin (0, 0) and making angle θ with l₁

Thus, (y 0) = m? (x 0)

y = m? x

m? =  $\mathrm{y}$
$\frac{\mathrm{}\mathrm{x}}{}$______ (1)

And tanθ = $\left|\frac{m^{\prime }-m}{1+m^{\prime }·m}\right|$ = $\frac{m^{\prime }-m}{1+m^{\prime }m}$

When, tanθ = $\frac{m^{\prime }-m}{1+m^{\prime }m}.$

tanθ + m? m tanθ = m’ - m

m + tanθ = m? - m?m tanθ

m' = $\frac{m+tan\theta }{1-mtan\theta }.$

When tan θ = $-\left(\frac{m^{\prime }-m}{1+m^{\prime }m}\right)$

tan θ + m? m tanθ = -m? + m

m' = $\frac{m-tan\theta }{1+mtan\theta }.$

Hence combining the two we get,

$m^{\prime }=\frac{m\pm tan\theta }{1?mtan\theta }.$

$\Rightarrow \frac{y}{x}=\frac{m\pm tan\theta }{1?mtan\theta }.$ {-: eqⁿ (1) }

30. Given, f (x)= $\left\{\left(x,\frac{x^2}{1+x^2}\right):x\in R\right\}$

We know that, for x $\in$ R.

So,  x²≥ 0 ⇒ $\frac{x^2}{x^2+1}\ge \frac{0}{x^2+1}\Rightarrow \frac{x^2}{x^2+1}\ge 0\Rightarrow f(x)\ge 0$

and x²+1>x²

⇒ $1>\frac{x^2}{x^2+1}$

⇒1 > f (x).

So, 0 ≤ f (x) < 1

∴ Range of f (x) = [0,1).

66. The given eqn of the line is.

4x + 7y – 3 = 0 _____ (1)

2x – 3y + 1 = 0 _______ (2)

Solving (1) and (2) using eqn (1) 2 x eqn (2) we get,

(4x + 7y – 3) 2 [ (2x – 3y + 1)] = 0

4x + 7y – 3 – 4x + 6y – 2 = 0

13y = 5

y = $\frac{5}{13}$

And 2x – 3 $\left(\frac{5}{13}\right)$ + 1 = 0

2x = $\frac{15}{13}$ – 1 = $\frac{2}{13}$

$\Rightarrow x=\frac{1}{13}$

Point of intersection of (1) and (2) is $\left(\frac{1}{13},\frac{5}{13}\right)$

Since, the line passing through $\left(\frac{1}{13},\frac{5}{13}\right)$ has equal intercept say c then it is of the form

$\frac{x}{c}+\frac{y}{c}=1.$

x + y = c

$\Rightarrow \frac{1}{13}+\frac{5}{13}=c$

c = $\frac{6}{13}$

the read eqn of line is x + y = $\frac{6}{13}$

13x + 13y – 6 = 0

To be eligible for JEE Advanced candidates must fulfill certain criteria such as meeting the cutoff marks for JEE Advanced and satisfying the academic eligibility criteria. As per the IIT Kanpur cutoff 2023 for the JEE Advanced exam, the OBC NCL cutoff stood between 146 and 5256 under All India quota. As you can see the cutoff for OBC AI candidates, 7000 rank is comparatively very less for the selection under OBC category. Candidates need to aim under 5256 to get admission at IIT Kanpur.

Hi, admission to most top private Science colleges in India are done on the basis of merit or you can go for entrance exams like CUET, IUET, etc. 

65. x – 2y = 3

y = $\frac{\mathrm{x}}{2}$ - $\frac{3}{2}$______ (1)

Slope of line (1) is $\frac{1}{2}\cdot$

Let the line through P (3, 2) have slope m

Then, angle between the line = $\left|\frac{m-\frac{1}{2}}{1+m\frac{1}{2}}\right|$

$\Rightarrow tan45^{°}=\left|\frac{2m-1}{2+m}\right|$

$\Rightarrow 1=\left|\frac{2m-1}{2+m}\right|$

$\Rightarrow \frac{2m-1}{2+m}=\pm 1.$

When,  $\frac{2m-1}{2+m}=1$ =>2m – 1 = 2 + m=> m = 3.

The eqn of line through (3, 2) is

y – 2 = 3 (x – 3) 3x – y – 7 = 0.

When $\frac{2m-1}{2+m}$ = – 1=> 2m – 1 = – 2 – m =>3m = – 1 m = $-\frac{1}{3}$

The equation of line through (3,2) is,

y – 2 = $-\frac{1}{3}$ (x – 3) => 3y – 6 = – X + 3

x + 3y – 9 = 0

The highest package offered at IIIT Kalyani for the BTech course in 2025 and previous two years is presented below:

Hi, please refer to the table below for the top private Science colleges in India according to the Week for the year 2024:

Click below for NCERT Solutions:

The cutoff for BA in Ancient Indian History, Culture, and Archaeology at Fergusson College typically ranges from 87-88% for Maharashtra State candidates and 90% for non-Maharashtra candidates in the first merit list. The cutoff may decrease to around 80-81% in the second merit list if seats remain vacant. These are general estimations, and the exact cutoff can vary slightly from year to year.