The equation sin -1 x - x - a - will have atleast one solution if a -A- -1-1-B- -2- -2-C- -1-1-2-D- 1-2- 1-2

The correct option is D (1 π/2, 1 + π/2)sin^ 1x = |x a| The domain of sin^ 1x is [ π/2,π/2] ⇒ π/2≤ x ≤π/2 ...(1) 1 ≤ sin^ 1x ≤ 1 Given that sin^ 1x=|x a| ≥ ...

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Standard XII

Mathematics

Properties Derived from Trigonometric Identities

The equation ...

Question

The equation ${sin}^{- 1} x = | x - a |$ will have atleast one solution if $a \in$

[- 1, 1]

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[- \frac{π}{2}, \frac{π}{2}]

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[1, 1 + \frac{π}{2}]

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(1 - \frac{π}{2}, 1 + \frac{π}{2})

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Solution

The correct option is D $(1 - \frac{π}{2}, 1 + \frac{π}{2})$
${sin}^{- 1} x = | x - a |$
The domain of $s i n^{- 1} x$ is $[- \frac{π}{2}, \frac{π}{2}]$

$\Rightarrow - \frac{π}{2} \leq x \leq \frac{π}{2} . . . (1)$

$- 1 \leq s i n^{- 1} x \leq 1$
Given that $s i n^{- 1} x = | x - a | \geq 0 \Rightarrow 0 \leq | x - a | \leq 1 \Rightarrow - 1 \leq x - a \leq 1 . . . (2)$

$(1) - (2) \Rightarrow 1 - \frac{π}{2} \leq a \leq \frac{π}{2} - 1$

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