- A$6$
- B$-1$
- ✓$6$ or $-1$
- DNone of these
When $x < 2,|x - 2| = - (x - 2),|x - 3| = - (x - 3)$
$\therefore $ The given equation reduces to $2 - x + 3 - x = 7$
==> $x = - 1 < 2$
$\therefore $ $x = - 1$ is a solution.
When $2 \le x < 3,\,\,|x - 2| = x - 2,|x - 3| = - (x - 3)$
$\therefore $ The equation reduces to $x - 2 + 3 - x = 7$==> $1=7$
$\therefore $ No solution in this case.
When $x \ge 3$, the equation reduces to
$x - 2 + x - 3 = 7$ ==> $x = 6 > 3$
Hence we get, $x = 6$or $-1$
Trick : By inspection, we have that both the values $x = 6, - 1$ satisfy the given equation.
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($1$) Let $p_i$ be the probability that a randomly chosen point has $i$ many friends, $i=0,1,2,3,4$. Let $X$ be a random variable such that for $i=0,1,2,3,4$, the probability $P(X=i)=p_i$. Then the value of $7 E(X)$ is
($2$) Two distinct points are chosen randomly out of the points $A_1, A_2, \ldots, A_{4 g}$. Let $p$ be the probability that they are friends. Then the value of $7 p$ is
$\left| {\begin{array}{*{20}{c}}
{{{\log }_e}\,a_1^ra_2^k}&{{{\log }_e}\,a_2^ra_3^k}&{{{\log }_e}\,a_3^ra_4^k} \\
{{{\log }_e}\,a_4^ra_5^k}&{{{\log }_e}\,a_5^ra_6^k}&{{{\log }_e}\,a_6^ra_7^k} \\
{{{\log }_e}\,a_7^ra_8^k}&{{{\log }_e}\,a_8^ra_9^k}&{{{\log }_e}\,a_9^ra_{10}^k}
\end{array}} \right| = 0$
Then the number of elements in $S$, is