Q109 / 155 · WBJEE 2021 · Mathematics

Whichever of the following is/are correct?

  • A
    To evaluate $I_{1}=\int_{-2}^{2} \frac{d x}{4+x^{2}}$, it is possible to put $x=\frac{1}{t}$
  • B
    To evaluate $I_{2}=\int_{0}^{1} \sqrt{\left(x^{2}+1\right)} d x$, it is possible to put $x=\sec t$
  • Correct answer
    C
    To evaluate $I_{2}=\int_{0}^{1} \sqrt{\left(x^{2}+1\right)} \mathrm{dx}$, it is not possible to put $x=\operatorname{cosec} \theta$
  • Correct answer
    D
    To evaluate $\mathrm{I}_{1}$, it is not possible to put $\mathrm{x}=\frac{1}{\mathrm{t}}$

Solution

$\mathrm{I}_{1}=\int_{-2}^{2} \frac{\mathrm{dx}}{4+\mathrm{x}^{2}}=2 \int_{0}^{2} \frac{\mathrm{dx}}{4+\mathrm{x}^{2}}>0$ By $\mathrm{x}=1 / \mathrm{t}, \quad \mathrm{I}_{1}=\int_{-1 / 2}^{1 / 2} \frac{-\mathrm{dt}}{4 \mathrm{t}^{2}+1} < 0$ So not possible $\mathrm{I}_{2}=\int_{0}^{1} \sqrt{\mathrm{x}^{2}+1} \mathrm{dx}, \mathrm{x}=\operatorname{cosec} \theta$ Since $x \in(0,1)$ and $\operatorname{cosec} \theta \in(-\infty,-1] \cup[1, \infty)$ So not possible

Hence, the correct answer is options: 'C', 'D'.

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