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The binary operation? on the set {1, 2, 3, 4, 5} is defined as a? b = min {a, b} &mn For E; a, b? {1, 2, 3, 4, 5}.

Thus, the operation table for the given operation? can be given as:

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Disclaimer: The information provided above is taken from external sources and may vary.

BA fee structure at NIPS Hotel Management Bhubaneswar includes tuition fees as well as other components such as application charges, refundable security deposits, hostel accommodation, and mess fees. Students should confirm the latest fee details from official university sources, as the structure is subject to change.

To be eligible for admission into the BA in Hotel Management, candidates must have passed Class 12 or an equivalent examination from a recognised board, such as CBSE, ISC, or Odisha CHSE. Meeting the eligibility criteria is essential for consideration during the admission process.

55. We have (k - 3) x - (4 - k²) y + k² - 7 y + 6 = 0.

(i) When the line is parall to x-axis, all x coefficient = 0. then,

(k - 3)x - (4 -k²)y + k² - 7y + 6 = 0 x.x - a x y       where a = constant

Equating the co-efficient,

K – 3 = 0

=> k = 3

(ii) When the line is parallel to y-axis all y co-efficient = 0 then

- (4 -k)² = 0

=> – 4 + x² = 0

k² = 4

k = ± 2.

(iii) When the line pares through origin, (0, 0) need satisfy the given eqn then,

k² - 7k + 6 = 0

k² - k – 6k + 6 = 0

k (k- 1) - 6 (k - 1) = 0

(k = 1) (k - 6) = 0

k = 1 and k = 6

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It is given that $f:R-\left\{-\frac{4}{3}\right\}\to R$ is defined as $f\left(x\right)=\frac{4x}{3x-4}$

Let y be an arbitrary element of Range f.

Then, there exists $x\in R-\left\{-\frac{4}{3}\right\}$ such that $y=f\left(x\right)$

$\Rightarrow y=\frac{4x}{3x+4}$

$\Rightarrow 3xy+4y=4x$ let us define g: Range $f\to R-\left\{-\frac{4}{3}\right\}as,g\left(y\right)=\frac{4y}{4-3y}$

$\begin{array}{l}\Rightarrow x\left(4-3y\right)=4y\\ Now,\left(gof\right)\left(x\right)=g\left(f\left(x\right)\right)=g\left(\frac{4x}{3x+4}\right)\\ \Rightarrow x=\frac{4y}{4-3y}\\ =\frac{4\left(4\frac{x}{3x+4}\right)}{4-3\left(\frac{4x}{3x+4}\right)}=\frac{16x}{12x+16-12x}=\frac{16x}{16}=x\\ And,\left(fog\right)\left(y\right)=f\left(g\left(y\right)\right)=f\left(\frac{4y}{4-3y}\right)\\ =\frac{4\left(\frac{4y}{4-3y}\right)}{3\left(\frac{4y}{4-3y}\right)+4}=\frac{16y}{12y+16-12y}=\frac{16y}{16}=y\\ \therefore gof=I_{R-\left(-\frac{4}{3}\right)}and,fog=I\_"Range,f"\end{array}$

Thus, g is the inverse of f i.e., $f^{-1}=g.$

Hence, the inverse of f is the map g: Range $f\to R-\left\{-\frac{4}{3}\right\}$ which is given by

$g\left(y\right)=\frac{4y}{4-3y}$

The correct answer is B.

NIPS Hotel Management Bhubaneswar offers a single, focused specialisation under its BA programme, Hotel Management. This specialisation is designed to prepare students for various roles within the hospitality industry, including hotels, resorts, catering, and tourism services.

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$f:R\to R$ is given as $f\left(x\right)={\left(3-x^{\wedge }3\right)}^{\wedge }\left(1/3\right)$

$\begin{array}{l}f\left(x\right)={\left(3-x^3\right)}^{\frac{1}{3}}\\ \therefore fof\left(x\right)=f\left(f\left(x\right)\right)=f\left({\left(3-x^3\right)}^{\frac{1}{3}}\right)={\left[3-{\left({\left(3-x^3\right)}^{\frac{1}{3}}\right)}^3\right]}^{\frac{1}{3}}\\ ={\left[3-\left(3-x^3\right)\right]}^{\frac{1}{3}}={\left(x^3\right)}^{\frac{1}{3}}=x\\ \therefore fof\left(x\right)=x\end{array}$

The correct answer is C.

BA in Hotel Management at NIPS Hotel Management Bhubaneswar is a three-year undergraduate programme, divided into six semesters. This structure allows students to gain a thorough understanding of hospitality operations through a combination of academic learning and practical exposure.

Let f: X → Y be an invertible function. 

Then, there exists a function g: Y → X such that gof = IX and fog = IY. 

Here, f⁻¹ = g. 

Now, gof = IX and fog = IY

⇒ f⁻¹ of = IX and fof⁻¹ = IY

Hence, f⁻¹ : Y → X is invertible and f is the inverse of f⁻¹ 

i.e., (f−1)⁻¹  = f.

$f\left(1\right)=a,f\left(2\right)b,and,f\left(3\right)=c$

If we define $g:\left\{a,b,c\right\}\to \left\{1,2,3\right\}as,g\left(a\right)=1,g\left(b\right)=2,g\left(c\right)=3,$ then we have:

  $\begin{array}{l}\left(fog\right)\left(a\right)=f\left(g\left(a\right)\right)=f\left(1\right)=a\\ \left(fog\right)\left(b\right)=f\left(g\left(b\right)\right)=f\left(2\right)=b\\ \left(fog\right)\left(c\right)=f\left(g\left(c\right)\right)=f\left(3\right)=c\\ And\\ \left(gof\right)\left(1\right)=g\left(f\left(1\right)\right)=f\left(a\right)=1\\ \left(gof\right)\left(2\right)=g\left(f\left(2\right)\right)=f\left(b\right)=2\\ \left(gof\right)\left(3\right)=g\left(f\left(3\right)\right)=f\left(c\right)=3\\ \therefore gof=I_{X}and,fog=I_Y\\ Where,X=\left\{1,2,3\right\},and,Y=\left\{a,b,c\right\}.\end{array}$

Thus, the inverse of f exists and $f^{-1}=g.$

$\therefore f^{-1}:\left\{a,b,c\right\}\to \left\{1,2,3\right\}$ is given by,

$\begin{array}{l}{f}^{-1}\left(a\right)=1,f^{-1}\left(b\right)=2,f^{-1}\left(c\right)=3\\ \end{array}$

Let us now find the inverse of $f^{-1}$ i.e., find the inverse of g.

If we define $h:\left\{1,2,3\right\}\to \left\{a,b,c\right\}as$

$h\left(1\right)=a,h\left(2\right)=b,h\left(3\right)=c$ , then we have

$\begin{array}{l}\left(goh\right)\left(1\right)=g\left(h\left(1\right)\right)=g\left(a\right)=1\\ \left(goh\right)\left(2\right)=g\left(h\left(2\right)\right)=g\left(b\right)=2\\ \left(goh\right)\left(3\right)=g\left(h\left(3\right)\right)=g\left(c\right)=3\\ And\\ \left(hog\right)\left(a\right)=h\left(g\left(a\right)\right)=h\left(1\right)=a\\ \left(hog\right)\left(b\right)=h\left(g\left(b\right)\right)=h\left(2\right)=b\\ \left(hog\right)\left(c\right)=h\left(g\left(c\right)\right)=h\left(3\right)=c\\ \therefore goh=I_{X}and,hog=I_Y\\ Where,X=\left\{1,2,3\right\},and,Y=\left\{a,b,c\right\}.\end{array}$

Thus, the inverse of g exists and $g^{-1}=h\Rightarrow {\left(f^{-1}\right)}^{-1}=h.$

It can be noted that h=f.

Hence, ${\left(f^{-1}\right)}^{-1}=f.$

54. The equation of line whose intercept on axes are a and b is given by,

$\frac{x}{a}+\frac{y}{b}=1.$

Multiplying both sides by ab we get,

$\frac{abx}{a}+\frac{aby}{b}=ab$

$\Rightarrow bx+ay-ab=0.$