If the domain of the function f(x) = $\frac{c o s^{- 1} \sqrt[]{x^{2} - x + 1}}{\sqrt[]{s i n^{- 1} (\frac{2 x - 1}{2})}}$ is the interval (a,b], then a + b is equal to:

Option 1 -

Option 2 -

$\frac{1}{2}$

Option 3 -

Option 4 -

$\frac{3}{2}$

3 Views | Posted a month ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

a month ago

Correct Option - 4

Detailed Solution:

$f (x) = \frac{c o s^{- 1} \sqrt[]{x^{2} - x + 1}}{\sqrt[]{s i n^{- 1} (\frac{2 x - 1}{2})}}$

For domain $0 \leq x^{2} - x + 1 \leq 1$

$& x^{2} - x + 1 \geq 0 \forall x \in R$

$x^{2} - x \leq 0 & x (x - 1) \leq 0 \Rightarrow x \in [0, 1] . . . . . . . . . . . (i)$ .

$\frac{1}{2} \leq x \leq \frac{3}{2} . . . . . . . . . . . . . . . . (i i)$

$(i) \cap (i i) x \in (\frac{1}{2}, 1] \equiv (α, β]$

then α + β = $\frac{3}{2}$

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If the domain of the function f(x) = $\frac{c o s^{- 1} \sqrt[]{x^{2} - x + 1}}{\sqrt[]{s i n^{- 1} (\frac{2 x - 1}{2})}}$ is the interval (a,b], then a + b is equal to:

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If the domain of the function f(x) = c o s − 1 x 2 − x + 1 s i n − 1 ( 2 x − 1 2 ) is the interval (a,b], then a + b is equal to:

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If the domain of the function f(x) = $\frac{c o s^{- 1} \sqrt[]{x^{2} - x + 1}}{\sqrt[]{s i n^{- 1} (\frac{2 x - 1}{2})}}$ is the interval (a,b], then a + b is equal to: