Q99 / 120 · NDA 2024 (Phase 2) · Mathematics

Paragraph: Let $2 \int \frac{x^2-1}{\sqrt{x^2+1}} d x=\mathrm{U}(x) \mathrm{V}(x)-3 \ln \{\mathrm{U}(x)+\mathrm{V}(x)\}+c$
Question: What is $\left|\mathrm{U}^2(x)-\mathrm{V}^2(x)\right|$ equal to?

  • A
    0
  • Correct answer
    B
    1
  • C
    2
  • D
    3

Solution

$\begin{aligned} & I=2 \int \frac{x^2-1}{\sqrt{x^2+1}} d x=2 \int\left(\frac{x^2+1}{\sqrt{x^2+1}}-\frac{2}{\sqrt{x^2+1}}\right) d x \\ & =2 \int\left(\sqrt{x^2+1}-\frac{2}{\sqrt{x^2+1}}\right) d x \\ & =2\left[\frac{x}{2} \sqrt{x^2+1}+\frac{1}{2} \ln \left(x+\sqrt{x^2+1}\right)-2 \ln \left(x+\sqrt{x^2+1}\right)+c\right. \\ & =x \sqrt{x^2+1}+\ln \left(x+\sqrt{x^2+1}\right)-4 \ln \left(x+\sqrt{x^2+1}\right) \\ & =x \sqrt{x^2+1}-3 \ln \left(x+\sqrt{x^2+1}\right)+c \\ & =U(x) \cdot V(x)-3 \ln \{U(x)+V(x)\}+c \\ & \therefore U(x)=x \text { and } V(x)=\sqrt{x^2+1}\end{aligned}$ $\begin{aligned} & \text { Now, }\left|\mathrm{U}^2(\mathrm{x})-\mathrm{V}^2(\mathrm{x})\right| \\ & =\left|\mathrm{x}^2-\left(\mathrm{x}^2+1\right)\right|=\left|\mathrm{x}^2-\mathrm{x}^2-1\right|=|-1|=1\end{aligned}$

Hence, the correct answer is option: 'B'.

Relevant Questions