Q107 / 155 · WBJEE 2016 · Mathematics

If $\phi(t)=\left\{\begin{array}{l}1, \text { for } 0 \leq t < 1 \\ 0, \text { otherwise }\end{array},\right.$ then $\int_{-3000}^{3000}\left(\sum_{r^{\prime}=2014}^{2016} \phi(t-r') \phi(t-2016)\right) d t$ is

  • Correct answer
    A
    a real number
  • Correct answer
    B
    1
  • C
    0
  • D
    does not exist

Solution

$\int_{-3000}^{3000}\left(\sum_{r'=2014}^{2016} \phi\left(t-r^{\prime}\right) \phi(t-2016)\right) d t$ $=\int_{-3000}^{3000}[\phi(t-2014) \phi(t-2016)+\phi(t-2015) \phi(t-2016) +\phi(t-2016) \phi(t-2016)] \mathrm{dt}$ $=\int_{-3000}^{3000} \phi(t-2016)[\phi(t-2014)+\phi(t-2015)$ $+\phi(t-2016)] d t$ $=\int_{-3000}^{2016} 0 d t+\int_{2016}^{2017} 1 \cdot(0+0+1) dt+\int_{2017}^{3000} 0 d t$ $=0+1+0=1$ which is a real number.

Hence, the correct answer is options: 'A', 'B'.

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