- A$20$
- B$25$
- ✓$22$
- D$65$
$={ }_{12} \int_0^3\left|\left( x -\frac{3}{2}\right)^2-\frac{1}{4}\right| dx$
If $x-\frac{3}{2}=t$
$dx = dt$
$=24 \int \limits_0^{3 / 2}\left| t ^2-\frac{1}{4}\right| dt$
$=24\left[-\int^{1 / 2}\left(t^2-\frac{1}{4}\right) d t+\int \limits_0^{3 / 2}\left(t^2-\frac{1}{4}\right) d t\right]=22$
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If $\overrightarrow{ c } \cdot(\hat{ i }+\hat{ j }+3 \hat{ k })=8$ then the value of $\overrightarrow{ c } \cdot(\overrightarrow{ a } \times \overrightarrow{ b })$ is equal to ...... .
($A$) 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 $\mathrm{L}_{\mathrm{h}}$