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If Cos − 1 X > Sin − 1 X , Then (A) 1 √ 2 < X ≤ 1 (B) 0 ≤ X < 1 √ 2 (C) − 1 ≤ X < 1 √ 2 (D) X > 0

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Question

If \[\cos^{- 1} x > \sin^{- 1} x\], then

Options

  • \[\frac{1}{\sqrt{2}} < x \leq 1\]

  •  \[0 \leq x < \frac{1}{\sqrt{2}}\]

  •  \[- 1 \leq x < \frac{1}{\sqrt{2}}\]

  •  x > 0

MCQ
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Solution

\[\cos^{- 1} x > \sin^{- 1} x\]
\[ \Rightarrow \cos^{- 1} x > \frac{\pi}{2} - \cos^{- 1} x\]
\[ \Rightarrow 2 \cos^{- 1} x > \frac{\pi}{2}\]
\[ \Rightarrow \cos^{- 1} x > \frac{\pi}{4}\]
\[ \Rightarrow x > \cos\frac{\pi}{4}\]
\[ \Rightarrow x > \frac{1}{\sqrt{2}}\]

We know that the maximum value of cosine fuction is 1.

\[\therefore \frac{1}{\sqrt{2}} < x \leq 1\]

Hence, the correct answer is option(a).

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Chapter 3: Inverse Trigonometric Functions - Exercise 4.16 [Page 121]

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R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 3 Inverse Trigonometric Functions
Exercise 4.16 | Q 26 | Page 121
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