Functions
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How can MBA students at PIBM Guwahati develop a better understanding of different business functions?
At PIBM Guwahati, students can explore different management domains through academic learning, practical projects, industry interactions, and exposure to business professionals. This helps them understand how functions such as Marketing, Finance, Human Resources, and Business Analytics work individu
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For the Class of 2024 the Full time MBA graduates of the Rady School of Management of UC San Diego, the top functions for employment stood to be the following-
- Finance/Accounting- 27.8%
- Other-27.8%
- Consulting- 22.2%
- Marketing/Sales- 16.7%
- Operations/Logistics- 5.6%
Upvote Number of integers in the domain of the function, f(x) =
Number of integers in the domain of the function, f(x) =
Let n(A) = 4 and n(B) = 6, then the number of one- one functions from A to B is
Let n(A) = 4 and n(B) = 6, then the number of one- one functions from A to B is
Let f : R → R be a function defined by f(x) = 2e²ˣ / (e²ˣ + e). Then f(1/100) + f(2/100) + f(3/100) + . + f(99/100) is equal to.
Let f : R → R be a function defined by f(x) = 2e²ˣ / (e²ˣ + e). Then f(1/100) + f(2/100) + f(3/100) + . + f(99/100) is equal to.
f (x) = 2e²? / (e²? +e? ) and f (1-x) = 2e²? / (e²? +e¹? )
∴ f (x) + f (1-x) = 1/2
i.e. f (x) + f (1-x) = 2
∴ f (1/100) + f (2/100) + . + f (99/100)
Σ? f (x/100) + f (1-x/100) + f (1/2)
= 49 x 2 + 1 = 99
Upvote Let a be an integer such that lim (x→a) [18 - [1-x]] / [x-3a] exists, where [t] is greatest integer ≤ t. Then a is equal to:
Let a be an integer such that lim (x→a) [18 - [1-x]] / [x-3a] exists, where [t] is greatest integer ≤ t. Then a is equal to:
lim? (x→7) (18- [1-x])/ ( [x-3a])
exist & a∈I.
= lim? (x→7) (17- [-x])/ ( [x]-3a)
exist
RHL = lim? (x→7? ) (17- [-x])/ ( [x]-3a) = 25/ (7-3a) [a ≠ 7/3]
LHL = lim? (x→7? ) (17- [-x])/ ( [x]-3a) = 24/ (6-3a) [a ≠ 2]
LHL = RHL
25/ (7-3a) = 8/ (2-a)
∴ a = -6
Upvote If f(x) = { 1/|x| ; |x| ≥ 1, ax² + b ; |x| < 1 } is differentiable at every point of the domain, then the values of a and b are respectively:
If f(x) = { 1/|x| ; |x| ≥ 1, ax² + b ; |x| < 1 } is differentiable at every point of the domain, then the values of a and b are respectively:
A function f (x) is continuous at x=1, so lim (x→1? ) f (x) = lim (x→1? ) f (x) = f (1).
Assuming a piecewise function like f (x) = { -x, x<1; ax+b, x≥1 } (structure inferred from derivative).
Continuity at x=1: f (1) = 1. a (1)+b = 1 => a+b=1.
The function is differentiable at x=1. The derivative of f (x) at x=1 from the left is -1. The derivative from
Upvote The real function f(x) = arccosec(x) / √(x - [x]), where [x] denotes the greatest integer less than or equal to x, is defined for all x belonging to:
The real function f(x) = arccosec(x) / √(x - [x]), where [x] denotes the greatest integer less than or equal to x, is defined for all x belonging to:
Given the function f (x) = cosec? ¹ (x) / √ {x - [x]} where [x] is the greatest integer function.
The domain of cosec? ¹ (x) is (-∞, -1] U [1, ∞).
For the denominator to be defined, x - [x] ≠ 0, which means {x} ≠ 0 (the fractional part of x is not zero). This implies that x cannot be an integer (x ∉ I
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Let A=(a,b,c) and B=(1,2,3,4). Then the number of elements in the set C={f: A→B | 2∈f(A) and f is not one-one} is
Let A=(a,b,c) and B=(1,2,3,4). Then the number of elements in the set C={f: A→B | 2∈f(A) and f is not one-one} is
19.00
Only '2' in range → 1 function
one element out of 1, 3, 4 is in range with '2'
number of ways = ³C? × (3!/(2!1!)) × 2! = 18
(Select one from 1, 3, 4 and distribute among a, b, c)
Total function = 1 + 18 = 19
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