MCQ
$\mathop \smallint \limits_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} \frac{{dx}}{{1 + \cos x}} = $ . . . .
- A$-1$
- B$-2$
- ✓$2$
- D$4$
$I = \int\limits_{\frac{{3\pi }}{4}}^{\frac{{3\pi }}{4}} {\frac{{{\rm{dx}}}}{{1 - \cos x}}} $
Using$\int\limits_a^b {f\left( x \right)dx = \int\limits_a^b {f\left( {a + b - x} \right)} } dx$
Adding $(i)$ and $(ii)$
$2I = \int\limits_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} {\frac{2}{{{{\sin }^2}x}}dx} $
$I = \int\limits_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} {{{\csc }^2}xdx} $
${\rm{I}} = - (\cot x)_{\pi /4}^{3\pi /4}$
$ = - \left[ {\cot \frac{{3\pi }}{4} - \cot \frac{\pi }{4}} \right] = 2$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
($A$) There exists an $\mathrm{h} \in\left[\frac{1}{4}, \frac{2}{3}\right]$ such that the area of the green region above the line $\mathrm{L}_{\mathrm{h}}$ equals the area of the green region below the line $\mathrm{L}_{\mathrm{h}}$
($B$) There exists an $\mathrm{h} \in\left[\frac{1}{4}, \frac{2}{3}\right]$ such that the area of the red region above the line $\mathrm{L}_{\mathrm{h}}$ equals the area of the red region below the line $\mathrm{L}_h$
($C$) There exists an $\mathrm{h} \in\left[\frac{1}{4}, \frac{2}{3}\right]$ such that the area of the green region above the line $\mathrm{L}_{\mathrm{h}}$ equals the area of the red region below the line $L_h$
($D$) There exists an $\mathrm{h} \in\left[\frac{1}{4}, \frac{2}{3}\right]$ such that the area of the red region above the line $\mathrm{L}_{\mathrm{h}}$ equals the area of the green region below the line $\mathrm{L}_{\mathrm{h}}$
Let $F(x)=\int_0^{x^2} f(\sqrt{t}) d t$ for $x \in[0,2]$. If $F^{\prime}(x)=f^{\prime}(x)$ for all $x \in(0,2)$, then $F(2)$ equals