The domain of the function f(x)=sin−13x2+x−1(x−1)2+cos−1x−1x+1 is

f(x)=sin−1(3x2+x−1(x−1)2)+cos−1(x−1x+1) is definediff −1≤3x2+x−1(x−1)2≤1 and −1≤x−1x+1≤1case(i) −(x−1)2≤3x2+x−1≤(x−1)2let −(x−1)2≤3x2+x−1⇒ −(x2+1−2x)≤3x2+x−1⇒ −x2−1+2x≤3x2+x−1⇒ 4x2−x≥0⇒ x(4x−1)≥0⇒ x∈R−[0, 14]and 3x2+x−1≤(x−1)2⇒ 3x2+x−1≤x2−2x+1⇒ 2x2+3x−2≤0⇒ 2x2+4x−x−2≤0⇒ (2x−1)(x+2)≤0x∈[−2, 12]x∈[−2, 0]∪[14, 2]case(i): −1≤x−1x+1≤1⇒ −(x+1)≤x−1≤(x+1)let −(x+1)≤x−1,x−1≤x+1⇒ −x−1≤x−1,true for all values⇒ 2x≥0⇒ x≥0x∈[0, ∞]∴ x∈{0}∪[14, 12]

The domain of the function f(x)=sin−13x2+x−1(x−1)2+cos−1x−1x+1 is

f(x)=sin−1(3x2+x−1(x−1)2)+cos−1(x−1x+1) is definediff −1≤3x2+x−1(x−1)2≤1 and −1≤x−1x+1≤1case(i) −(x−1)2≤3x2+x−1≤(x−1)2let −(x−1)2≤3x2+x−1⇒ −(x2+1−2x)≤3x2+x−1⇒ −x2−1+2x≤3x2+x−1⇒ 4x2−x≥0⇒ x(4x−1)≥0⇒ x∈R−[0, 14]and 3x2+x−1≤(x−1)2⇒ 3x2+x−1≤x2−2x+1⇒ 2x2+3x−2≤0⇒ 2x2+4x−x−2≤0⇒ (2x−1)(x+2)≤0x∈[−2, 12]x∈[−2, 0]∪[14, 2]case(i): −1≤x−1x+1≤1⇒ −(x+1)≤x−1≤(x+1)let −(x+1)≤x−1,x−1≤x+1⇒ −x−1≤x−1,true for all values⇒ 2x≥0⇒ x≥0x∈[0, ∞]∴ x∈{0}∪[14, 12]

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

$The domain of the function f (x) = \sin^{- 1} (\frac{3 x^{2} + x - 1}{{(x - 1)}^{2}}) + \cos^{- 1} (\frac{x - 1}{x + 1}) is$

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

$[- 2, 0] \cup [\frac{1}{4}, \frac{1}{2}]$

$[\frac{1}{4}, \frac{1}{2}] \cup {0}$

$[0, \frac{1}{2}]$

$[0, \frac{1}{4}]$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} f (x) = \sin^{- 1} (\frac{3 x^{2} + x - 1}{{(x - 1)}^{2}}) + \cos^{- 1} (\frac{x - 1}{x + 1}) i s d e f i n e d \\ i f f - 1 \leq \frac{3 x^{2} + x - 1}{{(x - 1)}^{2}} \leq 1 a n d - 1 \leq \frac{x - 1}{x + 1} \leq 1 \\ c a s e (i) - {(x - 1)}^{2} \leq 3 x^{2} + x - 1 \leq {(x - 1)}^{2} \\ l e t - {(x - 1)}^{2} \leq 3 x^{2} + x - 1 \\ \Rightarrow - (x^{2} + 1 - 2 x) \leq 3 x^{2} + x - 1 \\ \Rightarrow - x^{2} - 1 + 2 x \leq 3 x^{2} + x - 1 \\ \Rightarrow 4 x^{2} - x \geq 0 \\ \Rightarrow x (4 x - 1) \geq 0 \Rightarrow x \in R - [0, \frac{1}{4}] \\ a n d 3 x^{2} + x - 1 \leq {(x - 1)}^{2} \\ \Rightarrow 3 x^{2} + x - 1 \leq x^{2} - 2 x + 1 \\ \Rightarrow 2 x^{2} + 3 x - 2 \leq 0 \\ \Rightarrow 2 x^{2} + 4 x - x - 2 \leq 0 \\ \Rightarrow (2 x - 1) (x + 2) \leq 0 x \in [- 2, \frac{1}{2}] \\ x \in [- 2, 0] \cup [\frac{1}{4}, 2] \end{array}$

$\begin{array}{l} c a s e (i) : - 1 \leq \frac{x - 1}{x + 1} \leq 1 \\ \Rightarrow - (x + 1) \leq x - 1 \leq (x + 1) \\ l e t - (x + 1) \leq x - 1, x - 1 \leq x + 1 \\ \Rightarrow - x - 1 \leq x - 1, t r u e f o r a l l v a l u e s \\ \Rightarrow 2 x \geq 0 \\ \Rightarrow x \geq 0 \\ x \in [0, \infty] \\ ∴ x \in {0} \cup [\frac{1}{4}, \frac{1}{2}] \end{array}$