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Definite Integration question

2022 · 27 Jul · Shift 1 · Q30
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Definite Integration question

2022 · 27 Jul · Shift 1 · Q30

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
Let f:R→Rf: \mathbb{R} \rightarrow \mathbb{R}f:R→R be a function defined as f(x)=asin⁡(π[x]2)+[2−x],a∈Rf(x)=a \sin \left(\frac{\pi[x]}{2}\right)+[2-x], a \in \mathbb{R}f(x)=asin(2π[x]​)+[2−x],a∈R where [t][t][t] is the greatest integer less than or equal to ttt. If lim⁡x→−1f(x)\mathop {\lim }\limits_{x \to -1 } f(x)x→−1lim​f(x) exists, then the value of ∫04f(x)dx\int\limits_{0}^{4} f(x) d x0∫4​f(x)dx is equal to
  1. A
    −-− 1
  2. B
    −-− 2
  3. C
    1
  4. D
    2
View written solutionFree

Correct answer: B

  1. Given function

f(x)=asin⁡(π[x]2)+[2−x]f(x)=a\sin\left(\frac{\pi [x]}{2}\right)+[2-x]f(x)=asin(2π[x]​)+[2−x]

where [t][t][t] denotes the greatest integer function.

We are told that

lim⁡x→−1f(x)\lim_{x\to -1} f(x)limx→−1​f(x)

exists. First, we use this to determine aaa.

  1. Check left-hand and right-hand limits at x=−1x=-1x=−1

Since the floor function changes value at integers, examine intervals around −1-1−1.

For x→−1−x\to -1^-x→−1−:

Then x∈(−2,−1)x\in(-2,-1)x∈(−2,−1), so

[x]=−2[x]=-2[x]=−2

Hence,

sin⁡(π[x]2)=sin⁡(π(−2)2)=sin⁡(−π)=0.\sin\left(\frac{\pi [x]}{2}\right)=\sin\left(\frac{\pi(-2)}{2}\right)=\sin(-\pi)=0.sin(2π[x]​)=sin(2π(−2)​)=sin(−π)=0.

Also, for x∈(−2,−1)x\in(-2,-1)x∈(−2,−1),

2−x∈(3,4)  ⟹  [2−x]=3.2-x\in(3,4) \implies [2-x]=3.2−x∈(3,4)⟹[2−x]=3.

So,

f(x)=a⋅0+3=3.f(x)=a\cdot 0+3=3.f(x)=a⋅0+3=3.

Thus,

lim⁡x→−1−f(x)=3.\lim_{x\to -1^-} f(x)=3.limx→−1−​f(x)=3.

For x→−1+x\to -1^+x→−1+:

Then x∈[−1,0)x\in[-1,0)x∈[−1,0), so

[x]=−1[x]=-1[x]=−1

Hence,

sin⁡(π[x]2)=sin⁡(−π2)=−1.\sin\left(\frac{\pi [x]}{2}\right)=\sin\left(\frac{-\pi}{2}\right)=-1.sin(2π[x]​)=sin(2−π​)=−1.

Also, for x∈(−1,0)x\in(-1,0)x∈(−1,0),

2−x∈(2,3)  ⟹  [2−x]=2.2-x\in(2,3) \implies [2-x]=2.2−x∈(2,3)⟹[2−x]=2.

So,

f(x)=a(−1)+2=2−a.f(x)=a(-1)+2=2-a.f(x)=a(−1)+2=2−a.

Thus,

lim⁡x→−1+f(x)=2−a.\lim_{x\to -1^+} f(x)=2-a.limx→−1+​f(x)=2−a.

Since the limit exists,

3=2−a3=2-a3=2−a

which gives

a=−1.a=-1.a=−1.

  1. Now compute f(x)f(x)f(x) on [0,4][0,4][0,4]

Substitute a=−1a=-1a=−1:

f(x)=−sin⁡(π[x]2)+[2−x].f(x)=-\sin\left(\frac{\pi [x]}{2}\right)+[2-x].f(x)=−sin(2π[x]​)+[2−x].

Break the interval [0,4][0,4][0,4] into subintervals where floor values remain constant:

[0,1), [1,2), [2,3), [3,4).[0,1),\ [1,2),\ [2,3),\ [3,4).[0,1), [1,2), [2,3), [3,4).

  1. Evaluate on each interval

On x∈[0,1)x\in[0,1)x∈[0,1):

[x]=0  ⟹  −sin⁡(π⋅02)=0.[x]=0 \implies -\sin\left(\frac{\pi\cdot 0}{2}\right)=0.[x]=0⟹−sin(2π⋅0​)=0.

Also, 2−x∈(1,2]  ⟹  [2−x]=1 for x∈(0,1),2-x\in(1,2] \implies [2-x]=1 \text{ for } x\in(0,1),2−x∈(1,2]⟹[2−x]=1 for x∈(0,1), and endpoint values do not affect the integral.

So, f(x)=1.f(x)=1.f(x)=1.

Contribution: ∫01f(x) dx=∫011 dx=1.\int_0^1 f(x)\,dx=\int_0^1 1\,dx=1.∫01​f(x)dx=∫01​1dx=1.

On x∈[1,2)x\in[1,2)x∈[1,2):

[x]=1  ⟹  −sin⁡(π2)=−1.[x]=1 \implies -\sin\left(\frac{\pi}{2}\right)=-1.[x]=1⟹−sin(2π​)=−1.

Also, 2−x∈(0,1]  ⟹  [2−x]=0 for x∈(1,2).2-x\in(0,1] \implies [2-x]=0 \text{ for } x\in(1,2).2−x∈(0,1]⟹[2−x]=0 for x∈(1,2).

So, f(x)=−1.f(x)=-1.f(x)=−1.

Contribution: ∫12f(x) dx=∫12(−1) dx=−1.\int_1^2 f(x)\,dx=\int_1^2 (-1)\,dx=-1.∫12​f(x)dx=∫12​(−1)dx=−1.

On x∈[2,3)x\in[2,3)x∈[2,3):

[x]=2  ⟹  −sin⁡(π)=0.[x]=2 \implies -\sin(\pi)=0.[x]=2⟹−sin(π)=0.

Also, 2−x∈(−1,0]  ⟹  [2−x]=−1 for x∈(2,3).2-x\in(-1,0] \implies [2-x]=-1 \text{ for } x\in(2,3).2−x∈(−1,0]⟹[2−x]=−1 for x∈(2,3).

So, f(x)=−1.f(x)=-1.f(x)=−1.

Contribution: ∫23f(x) dx=∫23(−1) dx=−1.\int_2^3 f(x)\,dx=\int_2^3 (-1)\,dx=-1.∫23​f(x)dx=∫23​(−1)dx=−1.

On x∈[3,4)x\in[3,4)x∈[3,4):

[x]=3  ⟹  −sin⁡(3π2)=1.[x]=3 \implies -\sin\left(\frac{3\pi}{2}\right)=1.[x]=3⟹−sin(23π​)=1.

Also, 2−x∈(−2,−1]  ⟹  [2−x]=−2 for x∈(3,4).2-x\in(-2,-1] \implies [2-x]=-2 \text{ for } x\in(3,4).2−x∈(−2,−1]⟹[2−x]=−2 for x∈(3,4).

So, f(x)=1+(−2)=−1.f(x)=1+(-2)=-1.f(x)=1+(−2)=−1.

Contribution: ∫34f(x) dx=∫34(−1) dx=−1.\int_3^4 f(x)\,dx=\int_3^4 (-1)\,dx=-1.∫34​f(x)dx=∫34​(−1)dx=−1.

  1. Add all contributions

∫04f(x) dx=1+(−1)+(−1)+(−1)=−2.\int_0^4 f(x)\,dx = 1+(-1)+(-1)+(-1)=-2.∫04​f(x)dx=1+(−1)+(−1)+(−1)=−2.

  1. Check options

The value is

−2\boxed{-2}−2​

So the correct option is B.

  1. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

Hence, they agree.

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