- Home
- Standard 11
- Mathematics
Basic of Logarithms
hard
${\log _{1/2}}({x^2} - 6x + 12) \ge - 2$ નું સમાધાન કરે તેવી $x$ ની વાસ્તવિક કિમતોનો ગણ મેળવો.
A
$\left( { - \infty ,\,2} \right]$
B
$[2,\,4]$
C
$\left[ {4, + \infty } \right)$
D
એકપણ નહી.
Solution
(b) ${\log _{1/2}}({x^2} – 6x + 12) \ge – 2$…..$(i)$
For log to be defined, ${x^2} – 6x + 12 > 0$
$ \Rightarrow $${(x – 3)^2} + 3 > 0$, which is true $\forall x \in R$.
From $(i),$ ${x^2} – 6x + 12 \le {\left( {{1 \over 2}} \right)^{ – 2}}$
$ \Rightarrow $${x^2} – 6x + 12 \le 4$ $ \Rightarrow $ ${x^2} – 6x + 8 \le 0$
$ \Rightarrow $ $(x – 2)(x – 4) \le 0$ $ \Rightarrow $ $2 \le x \le 4$;
$\therefore x \in [2,\,4]$.
Standard 11
Mathematics
