Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Application of Integrals

Question:

Area bounded by $|x-1|≤2$ and $x^2-y^2 =1$, is

Options:

$6\sqrt{2}+\frac{1}{2}ln|3+2\sqrt{2}|$

$6\sqrt{2}+\frac{1}{2}ln|3-2\sqrt{2}|$

$6\sqrt{2}-\frac{1}{2}ln|3+2\sqrt{2}|$

none of these

Correct Answer:

none of these

Explanation:

We have,

$|x-1|≤2⇔-1≤x≤3$

∴ Required area A is the area of the shaded region and is given by

$A=2\int\limits_{1}^{3}\sqrt{x^2-1}dx$

$⇒A=\left[x\sqrt{x^2-1}-\log_e(x+\sqrt{x^2-1})\right]_1^3$

$⇒A=\left\{3\sqrt{8}-\log_e(3+\sqrt{8})\right\}=6\sqrt{2}-\log_e(3+2\sqrt{2})$