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1.Relation and Function
normal
If $x = {\log _2}\left( {\sqrt {56 + \sqrt {56 + \sqrt {56 + .... + \infty } } } } \right)$ then
A
$x < 0$
B
$0 < x < 2$
C
$2 < x < 4$
D
$3 < x < 4$
Solution
$x=\log _2(\sqrt{56+\sqrt{56+\ldots \infty}})$
het $t=\sqrt{56+t}$
$t^2-t-56=0$
$(t+7)(t-8)=0$
$t=8,-7 \quad$ [log can't be negative]
$\therefore x=\log _2 8$
$\therefore x=3$ (c)
Standard 12
Mathematics