Greatest Integer Functions MCQ Quiz in मराठी - Objective Question with Answer for Greatest Integer Functions - मोफत PDF डाउनलोड करा

Calculation: 

\(\mathrm{x}=(8 \sqrt{3}+13)={ }^{13} \mathrm{C}_{0} \cdot(8 \sqrt{3})^{13}+{ }^{13} \mathrm{C}_{1}(8 \sqrt{3})^{12}(13)^{1}+\ldots \)

\(\mathrm{x}^{\prime}=(8 \sqrt{3}-13)^{13}={ }^{13} \mathrm{C}_{0}(8 \sqrt{3})^{13}-{ }^{13} \mathrm{C}_{1}(8 \sqrt{3})^{12}(13)^{1}+\ldots \)

\(\mathrm{x}-\mathrm{x}^{\prime}=2\left[{ }^{13} \mathrm{C}_{1} \cdot(8 \sqrt{3})^{12}(13)^{1}+{ }^{13} \mathrm{C}_{3}(8 \sqrt{3})^{10} \cdot(13)^{3} \ldots\right]\)

∴ , x − x' is even integer, hence [x] is even

Now, \( \mathrm{y}=(7 \sqrt{2}+9)^{9}={ }^{9} \mathrm{C}_{0}(7 \sqrt{2})^{9}+\) \({ }^{9} \mathrm{C}_{1}(7 \sqrt{2})^{8}(9)^{1} +{ }^{9} \mathrm{C}_{2}(7 \sqrt{2})^{7}(9)^{2} \ldots \ldots\)

\(\mathrm{y}^{\prime}=(7 \sqrt{2}-9)^{9}={ }^{9} \mathrm{C}_{0}(7 \sqrt{2})^{9}-\) \({ }^{9} \mathrm{C}_{1}(7 \sqrt{2})^{8}(9)^{1} +{ }^{9} \mathrm{C}_{2}(7 \sqrt{2})^{7}(9)^{2} \ldots \ldots\)

\(\rm y-y^{\prime}=2\left[{ }^{9} C_{1}(7 \sqrt{2})^{8}(9)^{1}+{ }^{9} C_{3}(7 \sqrt{2})^{6}(9)^{3}+\ldots\right]\)

y − y' = Even integer, hence [y] is even

∴ [x] + [y] is even

Hence, the correct answer is Option 1. 

Concept:

Greatest Integer Function and Definite Integral:

• The greatest integer function, denoted by [x], gives the largest integer less than or equal to x.
• To integrate a greatest integer function, divide the integral into intervals where the function is constant.
• The function inside the integral is f(x) = [1 / e^x−1] = [e^1−x].
• We need to evaluate ∫₀³ [e^1−x] dx = α − logₑ2.

Calculation:

f(x) = [e^1−x] is a decreasing function

f(0) = [e¹] = [2.718] = 2

f(1−ln2) = e^{1−(1−ln2)} = e^ln2 = 2

⇒ boundary point

f(x) = 2 for x ∈ [0, 1−ln2)

f(1) = [e⁰] = [1] = 1

f(x) = 1 for x ∈ [1−ln2, 1)

f(x) < 1 for x ≥ 1 ⇒ [f(x)] = 0

Now break the integral accordingly:

∫₀³ [e^1−x] dx = ∫₀^1−ln2 2 dx + ∫_1−ln2¹ 1 dx + ∫₁³ 0 dx

⇒ 2(1 − ln2) + (1 − (1 − ln2)) + 0

⇒ 2 − 2ln2 + ln2 = 2 − ln2

Given: ∫₀³ [e^1−x] dx = α − ln2

Comparing both sides:

α − ln2 = 2 − ln2 ⇒ α = 2

Now, α³ = 2³ = 8

∴ The value of α³ is 8.

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 :

1. Observation and representation of step bracket wherever necessary.
2. 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.

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