- A$27$
- ✓$25$
- C$30$
- D$21$
$ \overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{a}}=3 $
$ \overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{a}}=\mathrm{k}(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}) $
$ 3=\mathrm{k}(2+6-15+3-2+3 \lambda) $
$ 3=\mathrm{k}(-6+3 \lambda) $ ...............($1$)
$ \overrightarrow{\mathrm{r}}=\mathrm{k}(5 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-(5-\lambda) \hat{\mathrm{k}}) $
$ |\overrightarrow{\mathrm{r}}|=\mathrm{k} \sqrt{25+4+25+\lambda^2-10 \lambda}=1 . $ ...............($2$)
$ \mathrm{k}=\frac{3}{-6+3 \lambda}=\frac{1}{-2+\lambda} \quad \text { put in }(2) $
$ 4+\lambda^2-4 \lambda=54+\lambda^2-10 \lambda $
$ 6 \lambda=50 $
$ 3 \lambda=25$
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($1$) The probability that, on the examination day, the student $S_1$ gets the previously allotted seat $R_1$, and $NONE$ of the remaining students gets the seat previously allotted to him/her is
$(A)$ $\frac{3}{40}$ $(B)$ $\frac{1}{8}$ $(C)$ $\frac{7}{40}$ $(D)$ $\frac{1}{5}$
($2$) For $i =1,2,3,4$, let $T _{ i }$ denote the event that the students $S _{ i }$ and $S _{ i +1}$ do NOT sit adjacent to each other on the day of the examination. Then, the probability of the event $T _1 \cap T _2 \cap T _3 \cap T _4$ is
$(A)$ $\frac{1}{15}$ $(B)$ $\frac{1}{10}$ $(C)$ $\frac{7}{60}$ $(D)$ $\frac{1}{5}$
Give the answer or quetion ($1$) and ($2$)
$f(x) \rightarrow \frac{\lambda\left|x^{2}-5 x+6\right|}{\mu\left(5 x-x^{2}-6\right)}, x<2$
$\quad\quad\quad\quad e^{\frac{\tan (x-2)}{x-[x]}}, \quad x>2$
$\quad\quad\quad\quad \mu \quad\quad\quad\quad x=2$
Where $[x]$ is the greatest integer less than or equal to $x$. If $f$ is continuous at $x=2$, then $\lambda+\mu$ is equal to:
$(A)$ $| FE |=| I - FE || FGE |$
$(B)$ $| I - FE |( I + FGE )= I$
$(C)$ $EFG = GEF$
$(D)$ $( I - FE )( I - FGE )= I$