Question
If $\text{F}(\text{x})=\cos^2\text{x}+\sec^2\text{x},$ then
  1. $\text{F}(\text{x})<1$
  2. $\text{F}(\text{x})=1$
  3. $2<\text{F}(\text{x})<1$
  4. $\text{F}(\text{x})\geq2$

Answer

  1. $\text{F}(\text{x})\geq2$

Solution:

$\text{f}(\text{x}) = \cos^2\text{x} + \sec^2\text{x}$

$=\cos^2\text{x} +\sec^2\text{x}-2\cos\text{x}\sec\text{x}+2\cos\text{x}\sec\text{x}$

$=(\sec\text{x}-\cos\text{x})^2 +2$

$\therefore\text{f}\text({x})\geq 2 \ \forall \text{ x }$$\Big[(\sec\text{x}-\cos\text{x})^2\geq 0 \ \forall \text{ x}\Big]$

Hence, the correct option is answer D.

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