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

NTA Abhyas 2020 · Mathematics · Inverse Trigonometric Functions

Let f (x) = sin^ - 1 \ x 1 - x - x (1 - x^ 2 ) \ , 0 ≤ x ≤ 1 , then...

Let ff\left(x\right) = sin^{- 1} \left\{x\sqrt{1 - x} - \sqrt{x \left(1 - x^{2}\right)}\right\}, \forall 0 ≤ x ≤ 1,thenf(x), then f \left(x\right) is

Options

  1. Anegative
  2. Bpositive
  3. Cnon-negative
  4. Dnon-positive

Correct answer

D. non-positive

Step-by-step solution

f(x)=sin−1$lleft(x1−(x)2f \left(x\right){= sin}^{- 1} \$ lleft(x\sqrt{1 - {\left(\sqrt{x}\right)}^{2}} - \sqrt{x}\sqrt{1 - x^{2}})<br/><br />∵ sin^{- 1} \left(a\right)−sin−1(b)=- sin^{- 1} \left(b\right) = sin^{- 1}\lleft(a1−b2−left(a\sqrt{1 - b^{2}} --b\sqrt{1 - a^{2}})<br/><br />⇒ f \left(x\right) = \sin^{-1}x - \sin^{-1}\sqrt{x} \le0(asx≤x)\left(asx \le \sqrt{x}\right)
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