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If \[F\Left( X \Right) = \Cos^2 X + \Sec^2 X\], Then

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Question

If \[f\left( x \right) = \cos^2 x + \sec^2 x\], then

Options

  • f(x) < 1

  • f(x) = 1

  • 1 < f(x) < 2

  • f(x) ≥ 2

MCQ
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Solution

\[f\left( x \right) = \cos^2 x + \sec^2 x\]
\[ = \cos^2 x + \sec^2 x - 2\cos x\sec x + 2\cos x\sec x\]
\[ = \left( \sec x - \cos x \right)^2 + 2\]
\[ \therefore f\left( x \right) \geq 2 \forall x \left[ \left( \sec x - \cos x \right)^2 \geq 0 \forall x \right]\]

Hence, the correct option is answer f(x) ≥ 2.

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Chapter 5: Trigonometric Functions - Exercise 5.5 [Page 43]

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R.D. Sharma Mathematics [English] Class 11
Chapter 5 Trigonometric Functions
Exercise 5.5 | Q 24 | Page 43

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