The domain of the function f(x)=sin−1(log2x)+sin−1(1+x22x)+cos(sinx) is

f(x)=sin−1(logex)+sin−1(1+x22x)+cos sinx is definediff sin−1(log2x)≥0 ⇒ log2x ≥0 ⇒ x≥1and −1≤x2+12x≤1⇒ |x2+12x|≤1 ⇒ |x|2+12|x|≤1⇒ |x|2+1≤2|x|⇒ |x|2−2|x|+1≤0⇒ (|x|−1)2≤0∴ |x|−1=0⇒ |x|=1⇒ x=±1and cos(sinx)≥0 ∀x∈R∴ Domain ={1}

The domain of the function f(x)=sin−1(log2x)+sin−1(1+x22x)+cos(sinx) is

f(x)=sin−1(logex)+sin−1(1+x22x)+cos sinx is definediff sin−1(log2x)≥0 ⇒ log2x ≥0 ⇒ x≥1and −1≤x2+12x≤1⇒ |x2+12x|≤1 ⇒ |x|2+12|x|≤1⇒ |x|2+1≤2|x|⇒ |x|2−2|x|+1≤0⇒ (|x|−1)2≤0∴ |x|−1=0⇒ |x|=1⇒ x=±1and cos(sinx)≥0 ∀x∈R∴ Domain ={1}

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$T h e d o m a i n o f t h e f u n c t i o n f (x) = \sqrt{\sin^{- 1} (\log_{2} x)} + \sin^{- 1} (\frac{1 + x^{2}}{2 x}) + \sqrt{\cos (\sin x)} i s$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

${1}$

${- 1, 1}$

${x : 1 \leq x \leq 2}$

$N o t d e f i n e d f o r a n y r e a l x .$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} f (x) = \sqrt{\sin^{- 1} (\log_{e}^{x})} + \sin^{- 1} (\frac{1 + x^{2}}{2 x}) + \sqrt{\cos \sin x} i s d e f i n e d \\ i f f \sin^{- 1} (\log_{2}^{x}) \geq 0 \Rightarrow \log_{2}^{x} \geq 0 \Rightarrow x \geq 1 \\ a n d - 1 \leq \frac{x^{2} + 1}{2 x} \leq 1 \\ \Rightarrow | \frac{x^{2} + 1}{2 x} | \leq 1 \Rightarrow \frac{{| x |}^{2} + 1}{2 | x |} \leq 1 \\ \Rightarrow {| x |}^{2} + 1 \leq 2 | x | \\ \Rightarrow {| x |}^{2} - 2 | x | + 1 \leq 0 \\ \Rightarrow {(| x | - 1)}^{2} \leq 0 \\ ∴ | x | - 1 = 0 \Rightarrow | x | = 1 \Rightarrow x = \pm 1 \\ a n d \cos (\sin x) \geq 0 \forall x \in R \\ ∴ D o m a i n = {1} \end{array}$