MCQ
Area bounded by curves $y = {x^2}$ and $y = 2 - {x^2}$ is
- ✓$8/3$
- B$3/8$
- C$3/2$
- DNone of these
$y = 2 - {x^2}$.....$(ii)$
$\therefore $ By equation $(i)$ and $(ii),$ we get, $x = \pm 1$
$\therefore $ $y = \pm 1$
$\therefore $ Required area $ = 2\left[ {\int_0^1 {(2 - {x^2})dx - \int_0^1 {{x^2}dx} } } \right]$
$ = 2\,\left[ {2x - \frac{{2{x^3}}}{3}} \right]_0^1 = 4\left[ {x - \frac{{{x^3}}}{3}} \right]_0^1 = 4\left( {\frac{2}{3}} \right) = \frac{8}{3}$.
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| Column $I$ | Column $II$ |
| $(A)$ If $\vec{a}=\hat{j}+\sqrt{3} \hat{k}, \vec{b}=-\hat{j}+\sqrt{3} \hat{k}$ and $\vec{c}=2 \sqrt{3} \hat{k}$ form a triangle, then the internal angle of the triangle between $\vec{a}$ and $\vec{b}$ is | $(p)$ $\frac{\pi}{6}$ |
| $(B)$ If $\int_a^b(f(x)-3 x) d x=a^2-b^2$, then the value of $f\left(\frac{\pi}{6}\right)$ is | $(q)$ $\frac{2 \pi}{3}$ |
| $(C)$ The value of $\frac{\pi^2}{\ln 3} \int_{1 / 6}^{5 / 6} \sec (\pi x) d x$ is | $(r)$ $\frac{\pi}{3}$ |
| $(D)$ The maximum value of $\left|\operatorname{Arg}\left(\frac{1}{1-z}\right)\right|$ for $|z|=1, \quad z \neq 1$ is given by | $(s)$ $\pi$ |
| $(t)$ $\frac{\pi}{2}$ |