Let f(x)=(sin−1x−cos−1x),g(x)=(tan−1x−cot−1x) and h(x)=(sec−1x−cosec−1x) then correct statement(s) is (are) :

Domain of f ( x ) + g ( x ) is { 1 } , Domain of g ( x ) + h ( x ) is [ 2 , ∞ ) , Domain of f ( x ) + g ( x ) + h ( x ) is ϕ

Let f(x)=(sin−1x−cos−1x),g(x)=(tan−1x−cot−1x) and h(x)=(sec−1x−cosec−1x) then correct statement(s) is (are) :

f ( x ) = sin − 1 x − cos − 1 x sin − 1 x − cos − 1 x ≥ 0 ⇒ sin − 1 x − π 2 + sin − 1 x ≥ 0 ⇒ sin − 1 x ≥ π 4 ⇒ x ∈ [ 1 2 , 1 ] g ( x ) = tan − 1 x − cot − 1 x tan − 1 x − cot − 1 x ≥ 0 tan − 1 x − π 2 + tan − 1 a ≥ 0 ⇒ tan − 1 x ≥ π 4 → x ∈ [ 1 , ∞ ) h ( x ) = sec − 1 x − cos e c − 1 x sec − 1 x − cos e c − 1 x ≥ 0 ⇒ sec − 1 x − π 2 + sec − 1 x ≥ 0 ⇒ sec − 1 x ≥ π 4 x ∈ ( − ∞ , − 1 ] ∪ [ 2 , ∞ ) = domain of h(x) and D g x ∈ [ 1 , ∞ ) Domain x ∈ [ 1 ] Domain of g ( x ) + h ( x ) is x ∈ [ 2 , ∞ ) Domain of h ( x ) + f ( x ) is x ∈ ∞ Domain of h ( x ) + f ( x ) + g ( x ) is x ∈ ∞ Hence, (A), (B) and (D) are correct.

Domain of g ( x ) + h ( x ) is [ 2 , ∞ )

Domain of h ( x ) + f ( x ) is [ 1 2 , 1 ]

Book Online Demo

Check Your IQ

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

Let $f (x) = \sqrt{(\sin^{- 1} x - \cos^{- 1} x)}, g (x) = \sqrt{(\tan^{- 1} x - \cot^{- 1} x)}$ and $h (x) = \sqrt{(\sec^{- 1} x - \cos e c^{- 1} x)}$ then correct statement(s) is (are) :

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

Domain of $f (x) + g (x)$ is ${1}$

Domain of $g (x) + h (x)$ is $[\sqrt{2}, \infty)$

Domain of $h (x) + f (x)$ is $[\frac{1}{\sqrt{2}}, 1]$

Domain of $f (x) + g (x) + h (x)$ is $ϕ$

answer is A, B, D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$f (x) = \sqrt{\sin^{- 1} x - \cos^{- 1} x} \sin^{- 1} x - \cos^{- 1} x \geq 0 \Rightarrow {sin}^{- 1} x - \frac{π}{2} + \sin^{- 1} x \geq 0 \Rightarrow \sin^{- 1} x \geq \frac{π}{4} \Rightarrow x \in [\frac{1}{\sqrt{2}}, 1] g (x) = \sqrt{\tan^{- 1} x - \cot^{- 1} x} \tan^{- 1} x - \cot^{- 1} x \geq 0 \tan^{- 1} x - \frac{π}{2} + \tan^{- 1} a \geq 0 \Rightarrow \tan^{- 1} x \geq \frac{π}{4} \to x \in [1, \infty) h (x) = \sqrt{\sec^{- 1} x - \cos e c^{- 1} x} \sec^{- 1} x - \cos e c^{- 1} x \geq 0 \Rightarrow \sec^{- 1} x - \frac{π}{2} + \sec^{- 1} x \geq 0 \Rightarrow \sec^{- 1} x \geq \frac{π}{4}$

$x \in (- \infty, - 1] \cup [\sqrt{2}, \infty) =$ domain of h(x)

and $D_{g} x \in [1, \infty)$
Domain $x \in [1]$
Domain of $g (x) + h (x)$ is $x \in [\sqrt{2}, \infty)$
Domain of $h (x) + f (x)$ is $x \in \infty$
Domain of $h (x) + f (x) + g (x)$ is $x \in \infty$
Hence, (A), (B) and (D) are correct.