- A$2$
- ✓$3$
- C$4$
- D$6$
$\overrightarrow {AC} = \overrightarrow {AD} - \overrightarrow {CD} $
Therefore, $\overrightarrow {AB} + \overrightarrow {AC} + \overrightarrow {AD} + \overrightarrow {AE} + \overrightarrow {AF} $
= $3\overrightarrow {AD} + (\overrightarrow {AE} - \overrightarrow {BD} ) + (\overrightarrow {AF} - \overrightarrow {CD} ) = 3\overrightarrow {AD} $
Hence $\lambda = 3$, [Since $\overrightarrow {AE} = \overrightarrow {BD,} \,\overrightarrow {AF} \, = \overrightarrow {CD} ]$.
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$\mathrm{F}(\mathrm{x})=\int\limits_{1}^{\mathrm{x}} \mathrm{t}^{2} \mathrm{g}(\mathrm{t}) \mathrm{dt},$ where $\mathrm{g}(\mathrm{t})=\int\limits_{1}^{\mathrm{t}} \mathrm{f}(\mathrm{u}) \mathrm{du}$
Then for the function $\mathrm{F}$, the point $\mathrm{x}=1$ is