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Yes, you can take admission through both quotas including General and OBC, however, the courses may vary in both. Considering the IIT Kanpur JEE Advanced cutoff 2023 for the General AI category, the rank stood between 238 and 11824. Similarly, for OBC AI quota, the rank varied from 146 to 5256. At 7k under General AI category, there are options available for the candidates. However, for OBC AI category, at 6k, you cannot get anything. Further, check out courses available under 7k for General AI category candidates.

25. Given, f(x)= $\{\begin{array}{ll}{x}^2,\hfill & 0\le x\le 3\hfill \\ 3x,\hfill & 3\le x\le 10\hfill \end{array}$

f(x)={(0,0),(1,1),(2,4),(3,9),(4,12),(5,15),(6,18),(7,21),(8,24),(9,27),(10,30)}

So, the elements in domain of f  has one and only one image.

 ? f(x) is a function.

Given, g(x)= $\{\begin{array}{ll}{x}^2,\hfill & 0\le x\le 2\hfill \\ 3x,\hfill & 2\le x\le 10\hfill \end{array}$ .

g(x)={(0,0),(1,1),(2,4),(2,6),(3,9),(4,12),(5,15),(6,18),(7,21),(8,24),(9,27),(10,30)}

So, the element 2 of the domain has more than one image i.e., 4 and 6.  

? g(x) is not a function.

It is given that $f:R\to R$ is defined as $f\left(x\right)=10x+7.$

One-one:

$\begin{array}{l}Let,f\left(x\right)=f\left(y\right),where,x,y\in R\\ \Rightarrow 10x+7=10y+7\\ \Rightarrow x=y\end{array}$

$\therefore f$ is a one-one function.

Onto:

$\begin{array}{l}For,y\in R,let,y=10x+7\\ \Rightarrow x=\frac{y-7}{10}\in R\end{array}$

Therefore, for any $y\in R$ ,there exists $x=\frac{y-7}{10}\in R$

Such that

$f\left(x\right)=f\left(\frac{y-7}{10}\right)=10\left(\frac{y-7}{10}\right)+7=y-7+7=y$

$\therefore f$ is onto.

Therefore, f is one-one and onto.

Thus, f is an invertible function.

Let us define $g:R\to R$ as $g\left(y\right)=\frac{y-7}{10}$

Now, we have

$\begin{array}{l}gof\left(x\right)=g\left(f\left(x\right)\right)=g\left(10x+7\right)=\frac{\left(10x+7\right)-7}{10}=10\frac{x}{10}=10\\ And\\ fog\left(y\right)=f\left(g\left(y\right)\right)=f\left(\frac{y-7}{10}\right)=10\left(\frac{y-7}{10}\right)+7=y-7+7=y\end{array}$

Hence, the required function $g:R\to R$ is defined as $g\left(y\right)=\frac{y-7}{10}$

Karnavati University offers a full-time BCom (Hons) in Economics & Accounts. The cost of pursuing the four-year programme at the university is INR 9.5 lakh. The total tuition fee is distributed over the course duration. Candidates may have to submit additional one-time charges during admission. It must be noted that the mentioned fee is sourced from the official website of the university/ sanctioning body. It is still subject to change, and hence, is indicative.

On N, the operation * is defined as a * b = a³ + b³.

For, a, b, ∈ N, we have:

a * b =  a³ + b³ = b³ + a³  = b * a [Addition is commutative in N]

Therefore, the operation * is commutative.

It can be observed that:

  (1*2)*3 = (1³+2³)*3 = 9 * 3 = 93 + 33 = 729 + 27 = 756

Also, 1* (2*3) = 1* (2³ +3³) = 1* (8 +27) = 1 × 35

= 1³ +35³ = 1 + (35)³ = 1 + 42875 = 42876.

∴ (1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ N

Therefore, the operation * is not associative.

Hence, the operation * is commutative, but not associative. Thus, the correct answer is B.

(i) Define an operation * on N as:

a * b = a + b  a, b ∈ N

Then, in particular, for b = a = 3, we have:

3 * 3 = 3 + 3 = 6 ≠ 3

Therefore, statement (i) is false.

(ii) R.H.S. = (c * b) * a

= (b * c) * a [* is commutative]

= a * (b * c) [Again, as * is commutative]

= L.H.S.

∴ a * (b * c) = (c * b) * a

Therefore, statement (ii) is true.

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• PGCET (Karnataka Post Graduate Common Entrance Test)
• MAT (Management Aptitude Test)
• CAT (Common Admission Test)
• XAT (Xavier Aptitude Test)
• CMAT (Common Management Admission Test)
• ATMA (AIMS Test for Management Admissions)

24. (i) f(x)=2 – 3x, x $\in$ R, x>0.

Given, x>0

3x>3 × 0

3x>0

(–1) × 3x<(–1) × 0.

–3x<0

2 – 3x<0+2

2 – 3x<2

i.e., f(x) < 2

Hene, range of f(x) = (– ?, 2)

(ii) Given, f(x) = x²+2, x is a real number.

Since, x is a real number,

x² ≥ 0 (x²=0 for x=0)

x²+2 ≥ 0+2

x²+2 ≥ 2

f(x) ≥ 2

?Range of f(x) = [2, ?) 

(iii) Given, f(x) = x, x is a real number.

As, f(x) = x, the range of f(x) is also real.

i.e., Range of f(x) = R.

Let the identity be I.

An element $e\in Q$ will be the identity element for the operation* if

$a*e=a=e*a,\forall a\in Q.$

We are given

$\begin{array}{l}a*b=ab4\\ \Rightarrow a*e=a\Rightarrow ae4=a\Rightarrow e=4\end{array}$

Similarly, it can be checked for $e*a=a$ , we get e=4. Thus, e=4 is the identity

The students who are looking forward to get admission in the IIIT Kalyani must appear for the JEE Mains exam and clear the required cut off. They are adviced to prepare well and get the required percentile. Therefore, no the students can not take direct admission in the BTech course at IIIT Kalyani.

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