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Definite Integration question

2019 · 10 Jan · Shift 2 · Q34
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Definite Integration question

2019 · 10 Jan · Shift 2 · Q34

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
The value of ∫−π/2π/2dx[x]+[sin⁡x]+4,\int\limits_{ - \pi /2}^{\pi /2} {{{dx} \over {\left[ x \right] + \left[ {\sin x} \right] + 4}}} ,−π/2∫π/2​[x]+[sinx]+4dx​, where [t] denotes the greatest integer less than or equal to t, is
  1. A
    112(7π−5){1 \over {12}}\left( {7\pi - 5} \right)121​(7π−5)
  2. B
    112(7π+5){1 \over {12}}\left( {7\pi + 5} \right)121​(7π+5)
  3. C
    310(4π−3){3 \over {10}}\left( {4\pi - 3} \right)103​(4π−3)
  4. D
    320(4π−3){3 \over {20}}\left( {4\pi - 3} \right)203​(4π−3)
View written solutionFree

Correct answer: D

  1. We need to evaluate I=∫−π/2π/2dx[x]+[sin⁡x]+4.I=\int_{-\pi/2}^{\pi/2} \frac{dx}{[x]+[\sin x]+4}.I=∫−π/2π/2​[x]+[sinx]+4dx​. Here [t][t][t] is the greatest integer function.

  2. First, determine [sin⁡x][\sin x][sinx] on the interval [−π2,π2]\left[-\frac\pi2,\frac\pi2\right][−2π​,2π​].

Since sin⁡x∈[−1,1]\sin x\in[-1,1]sinx∈[−1,1]:

  • for x∈[−π2,0)x\in\left[-\frac\pi2,0\right)x∈[−2π​,0), we have sin⁡x∈[−1,0)\sin x\in[-1,0)sinx∈[−1,0), so [sin⁡x]=−1;[\sin x]=-1;[sinx]=−1;
  • at x=0x=0x=0, [sin⁡0]=0[\sin 0]=0[sin0]=0;
  • for x∈(0,π2)x\in(0,\frac\pi2)x∈(0,2π​), we have sin⁡x∈(0,1)\sin x\in(0,1)sinx∈(0,1), so [sin⁡x]=0;[\sin x]=0;[sinx]=0;
  • at x=π2x=\frac\pi2x=2π​, [sin⁡x]=1[\sin x]=1[sinx]=1, but a single point does not affect the integral.

So effectively,

\end{cases}$$ 3. Next, determine $[x]$ on $\left[-\frac\pi2,\frac\pi2\right]$. Since $-\frac\pi2\approx -1.57$ and $\frac\pi2\approx1.57$, - for $x\in\left[-\frac\pi2,-1\right)$, $[x]=-2$, - for $x\in[-1,0)$, $[x]=-1$, - for $x\in[0,1)$, $[x]=0$, - for $x\in[1,\frac\pi2]$, $[x]=1$. 4. Break the integral into these intervals. ### (i) On $\left[-\frac\pi2,-1\right)$ Here $[x]=-2$ and $[\sin x]=-1$, so denominator is $$-2-1+4=1.$$ Hence, $$\int_{-\pi/2}^{-1} \frac{dx}{1}=\frac\pi2-1.$$ ### (ii) On $[-1,0)$ Here $[x]=-1$ and $[\sin x]=-1$, so denominator is $$-1-1+4=2.$$ Hence, $$\int_{-1}^{0} \frac{dx}{2}=\frac12.$$ ### (iii) On $[0,1)$ Here $[x]=0$ and $[\sin x]=0$, so denominator is $$0+0+4=4.$$ Hence, $$\int_0^1 \frac{dx}{4}=\frac14.$$ ### (iv) On $[1,\pi/2]$ Here $[x]=1$ and $[\sin x]=0$, so denominator is $$1+0+4=5.$$ Hence, $$\int_1^{\pi/2} \frac{dx}{5}=\frac15\left(\frac\pi2-1\right).$$ 5. Add all parts: $$I=\left(\frac\pi2-1\right)+\frac12+\frac14+\frac15\left(\frac\pi2-1\right).$$ Combine terms: $$I=\frac\pi2+\frac\pi{10}-1-\frac15+\frac12+\frac14.$$ Now, $$\frac\pi2+\frac\pi{10}=\frac{3\pi}{5},$$ and $$-1-\frac15+\frac12+\frac14= -\frac{20}{20}-\frac{4}{20}+\frac{10}{20}+\frac{5}{20}=-\frac{9}{20}.$$ Therefore, $$I=\frac{3\pi}{5}-\frac{9}{20}= rac{12\pi-9}{20}= rac{3}{20}(4\pi-3).$$ 6. So the correct option is $$\boxed{\text{D }\frac{3}{20}(4\pi-3)}.$$
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