Greatest Integer Functions MCQ Quiz in বাংলা - Objective Question with Answer for Greatest Integer Functions - বিনামূল্যে ডাউনলোড করুন [PDF]
Last updated on Mar 15, 2025
Latest Greatest Integer Functions MCQ Objective Questions
Top Greatest Integer Functions MCQ Objective Questions
Greatest Integer Functions Question 1:
Let [.] denote the greatest integer function. If \(\int_{0}^{3} \left[ \frac{1}{e^{x-1}} \right] dx = \alpha - \log_e 2, \text{ then } \alpha^3 \text{ is equal to } \_\_\_\_.\)
Answer (Detailed Solution Below) 0 - 8
Greatest Integer Functions Question 1 Detailed Solution
Concept:
Greatest Integer Function and Definite Integral:
The greatest integer function, denoted by [x], gives the greatest integer less than or equal to x.
When integrating a function involving the greatest integer function, it is important to break the integral over intervals where the floor value is constant.
Given integral:
\(\displaystyle \int_0^3 \left[ \frac{1}{e^{x-1}} \right] dx = \int_0^3 \left[ e^{1-x} \right] dx = \alpha - \log_e 2 \)
Calculation
\((f(x) = 2, e^{1-x} = 2 \Rightarrow x = 1 - \ln 2.\)
\((f(x) = 1, (e^{1-x} = 1 \Rightarrow x = 1\)
Also
f(0) = e1-0 = e ≈2.718 > 2
f(1) = e1 =0
f(2) = e-1 = 0.367
f(3) = e-2 = ≈ 0.1353
\[ \int_0^3 [f(x)] \, dx = \int_0^{1-\ln 2} 2 \, dx + \int_{1-\ln 2}^1 1 \, dx + \int_1^3 0 \, dx \]
\[ = 2(1-\ln 2) + (1 - (1-\ln 2)) + 0 = 2 - 2\ln 2 + \ln 2 = 2 - \ln 2 \]
Also
\[ \int_0^3 [f(x)] \, dx = \alpha - \log_e 2 \]
Comparing,
\[ 2 - \ln 2 = \alpha - \ln 2 \implies \alpha = 2 \]
Therefore,
\[ \alpha^3 = 2^3 = 8 \]
∴ The value of
Greatest Integer Functions Question 2:
Consider the function f(x) = [x + 1] - (sin\(\frac{\pi }{2}\)[x]) for x ϵ R. where [x] denotes the greatest integer less than or equal to x. Let l1 = limx→0-f(x) and l2 = limx→0+ f(x).It follows that
Answer (Detailed Solution Below)
Greatest Integer Functions Question 2 Detailed Solution
Concept :
⇒ f(x) = [x] denotes a step function whose graph is as follows :
⇒ Thus by the graph we can depict any value, for example [2.93] = 2, [-0.5] = -1, ...
Calculation :
Given the function f(x) = [x + 1] - (sin\(\frac{π }{2}\)[x]).
It is given that l1 = limx→0-f(x).
⇒ l1 = limx→0-f(x) = limx→0- {[x + 1] - (sin\(\frac{π }{2}\)[x])} = {[0-+1] - (sin\(\frac{π }{2}\)[0-])} = {[1-] - (sin\(\frac{π }{2}\)[0-])}.
⇒ l1 = {[1-] - (sin\(\frac{-π }{2}\))} = {0- (-1)} = 1.
It is given that l2 = limx→0+f(x).
⇒ l2 = limx→0+f(x) = limx→0+ {[x + 1] - (sin\(\frac{π }{2}\)[x])} = {[0++1] - (sin\(\frac{π }{2}\)[0+])} = {[1+] - (sin\(\frac{π }{2}\)[0+])}.
⇒ l2 = {[1+] - (sin(0))} = {1- 0} = 1.
Thus l1 = l2 = 1.
Mistake Points
Student often gets mistaken in two points when solving these type of problems :
- Observation and representation of step bracket wherever necessary.
- sin\(\frac{π }{2}\)(0-) = sin\(\frac{π }{2}\)(0+) = 0, but their Difference when a step function is used on them.
⇒ sin\(\frac{π }{2}\)[0-] = sin\(\frac{-π }{2}\) = -1 and sin\(\frac{π }{2}\)[0+] = sin0 = 0.
Greatest Integer Functions Question 3:
If f(x) = (x)[x] where [.] denotes greatest integer function and g(x) = x2 then find the value of g o f(5/2) ?
Answer (Detailed Solution Below)
Greatest Integer Functions Question 3 Detailed Solution
Concept:
Greatest Integer Function: (Floor function)
The function f (x) = [x] is called the greatest integer function and means greatest integer less than or equal to x i.e [x] ≤ x.
Domain of [x] is R and range is I.
If f :A → B and g : C → D. Then (fog) (x) will exist if and only if co-domain of g = domain of f i.e D = A and (gof) (x) will exist if and only if co-domain of f = domain of g i.e B = C.
Calculation:
Given: f(x) = (x)[x] where [.] denotes greatest integer function and g(x) = x2
Here, we have to find the value of g o f(5/2)
⇒ g o f(5/2) = g( f(5/2))
∵ f(x) = (x)[x] where [.] denotes greatest integer function
⇒ f(5/2) = (5/2)[5/2]
As we know that [5/2] = 2
⇒ f(5/2) = (5/2)2 = 25/4
⇒ g o f(5/2) = g(25/4)
∵ g(x) = x2 so, g(25/4) = 625/16
Hence, g o f(5/2) = 625/16