The domain of the function f(x)=^{-1}≤ft({3x²+x-1}{(x-1)²})+^{-1}≤ft(x-1/x+1) is [31 Aug 2021 Shift 2]

Correct Answer is : [14,12]∪{0}≤ft[1/4, 1/2] \{0\}[41,21]∪{0}

Correct Answer is : [14,12]∪{0}≤ft[1/4, 1/2] \{0\}[41,21]∪{0}

Test Index

Inverse Trigonometric Functions

Section: Mathematics

Question : 63 of 99

Marks: +1, -0

f(x)=\sin^{-1}\left(\frac{3x^{2}+x-1}{(x-1)^{2}}\right)+\cos^{-1}\left(\frac{x-1}{x+1}\right)

Solution:

f(x)=\sin^{-1}\left(\frac{3x^{2}+x-1}{(x-1)^{2}}\right)+\cos^{-1}\left(\frac{x-1}{x+1}\right)

-1 \leq \frac{x-1}{x+1} \leq 1

\Rightarrow -1-1 \leq \frac{x-1}{x+1}-1 \leq 1-1

\Rightarrow -2 \leq \frac{-2}{x+1} \leq 0

\Rightarrow 0 \leq \frac{1}{x+1} \leq 1

\Rightarrow 1 \leq x+1 < \infty

\Rightarrow 0 \leq x < \infty

\Rightarrow x \in [0, \infty)

...(i)

\text{ and } -1 \leq \frac{3x^{2}+x-1}{(x-1)^{2}} \leq 1

\Rightarrow -(x-1)^{2} \leq 3x^{2}+x-1 \leq (x-1)^{2}, x \neq 1

\Rightarrow -(x^{2}-2x+1) \leq 3x^{2}+x-1

\text{ and } 3x^{2}+x-1 \leq x^{2}-2x+1

\Rightarrow 4x^{2}-x \geq 0

\text{ and } 2x^{2}+3x-2 \leq 0

\Rightarrow x(4x-1) \geq 0

\text{ and } (x+2)(2x-1) \leq 0

\Rightarrow x \in \left(-\infty, 0] \cup [\frac{1}{4}, \infty\right)

and

x \in \left[-2, \frac{1}{2}\right]

\Rightarrow x \in (-2,0] \cup \left[\frac{1}{4}, \frac{1}{2}\right]

...(ii)

Domain of f in Eq. (i) ∩ Eq. (ii)

\therefore x \in \{0\} \cup \left[\frac{1}{4}, \frac{1}{2}\right]

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