Let fx=sin^(-1)x√(1-x)-√(x1-x^2), ∀0≤x≤1, then fx is

Correct answer: (D) non-positive. fx=sin^(-1)x√(1-x^2)-√x√(1-x^2) ∵ sin^(-1)a-sin^(-1)b=sin^(-1)a√(1-b^2)-b√(1-a^2) ⇒fx=sin^(-1)x-sin^(-1)√x≤0 as x≤√x

Q35 / 71 · NTA Abhyas (JEE Main) 2020 · Mathematics · Inverse Trigonometric Functions

Let $f (x) = s i n^{- 1} \{x \sqrt{1 - x} - \sqrt{x (1 - x^{2})}\}, \forall 0 \leq x \leq 1$ , then $f (x)$ is

A
negative
B
positive
C
non-negative
Correct answer
D
non-positive

Solution

f (x) {= s i n}^{- 1} (x \sqrt{1 - {(\sqrt{x})}^{2}} - \sqrt{x} \sqrt{1 - x^{2}})

∵ s i n^{- 1} (a) - s i n^{- 1} (b) = s i n^{- 1} (a \sqrt{1 - b^{2}} - b \sqrt{1 - a^{2}})

\Rightarrow f (x) = \sin^{- 1} x - \sin^{- 1} \sqrt{x} \leq 0 (as x \leq \sqrt{x})

Hence, the correct answer is option: 'D'.

Let fx=sin-1x1-x-x1-x2, ∀0≤x≤1, then fx is

Solution

More Inverse Trigonometric Functions Questions

Let $f (x) = s i n^{- 1} \{x \sqrt{1 - x} - \sqrt{x (1 - x^{2})}\}, \forall 0 \leq x \leq 1$ , then $f (x)$ is