Consider the function f (x) = cos^ - 1 ( [ 2 ^ x ]) +sin^ - 1 ([ 2 ^ x ] -1), then (where [.] represents the greatest integer part function)

Correct answer: B. Range of f (x) is singleton For f (x) to be defined [ 2 ^ x ] =0, 1 ∴ 2 ^ x ∈ [0,2) ⇒ x ∈ (– ∞,1) Range \ cos^ - 1 (0) + sin^ - 1 (-1), cos ^ - 1 (1) + sin ^ - 1 (0)\ ≡ \ 0\ ∴ f(x) =0 x ∈ (– ∞,1) So, Domain is not symmetric about the origin ∴f (x)& is neither even nor odd

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Inverse Trigonometric Functions

NTA Abhyas 2020 · Mathematics · Inverse Trigonometric Functions

Consider the function f (x) = cos^ - 1 ( [ 2 ^ x ]) +sin^ - 1 ([ 2...

Consider the function

f \left(x\right) = cos^{- 1} \left( \left[{2}^{x}\right]\right) +sin^{- 1}\left(\left[{2}^{x}\right] -1\right), then<br /> (where [.] represents the greatest integer part function)

Options

ADomain of $f \left(x\right)$ is $x \in \left(– \infty,0\right]$
BRange of $f \left(x\right) is singleton$
C $f \left(x\right) is an even function$
D $f \left(x\right) is an odd function$

Correct answer

B. Range of $f \left(x\right) is singleton$

Step-by-step solution

For

f \left(x\right)

be defined \left[{2}^{x}\right] =0, 1

∴ {2}^{x} \in \left[0,2\right)

⇒ x \in \left(– \infty,1\right)

<br />Range \left\{cos^{- 1}\left(0\right) + sin^{- 1}\left(-1\right),\text{ cos }^{- 1}\left(1\right) +\text{ sin }^{- 1} \left(0\right)\right\}

≡ \left\{0\right\}

∴

f\left(x\right) =0 \forall x \in \left(– \infty,1\right)
So,

Domain is not symmetric about the origin<br />

∴f \left(x\right)& is neither even nor odd

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