1.Relation and Function
easy

વિધેય $f(x) = \frac{{{{\sin }^{ - 1}}(x - 3)}}{{\sqrt {9 - {x^2}} }}$ નો પ્રદેશ મેળવો.

A

$[1, 2)$

B

$[2, 3)$

C

$[1, 2]$

D

$[2, 3]$

(AIEEE-2004)

Solution

(b) To define $f(x)$,

$9 – {x^2} > 3 \Rightarrow – 3 < x < 3…..(i)$

$ – 1 \le (x – 3) \le 1 \Rightarrow 2 \le x \le 4…..(ii)$

From $(i)$ and $(ii)$, $2 \le x < 3$   $i.e., \,\, [2, 3).$

Standard 12
Mathematics

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