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The correct option is A f(x) 2 Since, ( x 1/x )^2≥ 0, ∀ x ϵ R, we have x^2+1/x^2≥ 2 and Hence, f(x)=cos^2 x+1/cos^2 x≥ 2 ...

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Domain and Range of Trigonometric Ratios

If f(x) = cos...

Question

$I f f (x) = c o s^{2} x + s e c^{2} x, t h e n$

f(x) <1

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f(x) = 1

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1 < f(x) < 2

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f(x) 2

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Solution

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Similar questions

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

$I f f (x) = c o s^{2} x + s e c^{2} x, t h e n$

f (x + 2) = \frac{1}{2} {f (x + 1) + \frac{4}{f (x)}}

f (x) > 0

x ϵ R

lim x \to \infty f (x)

f (0) = 1, f (1) = 2

f (x + 2) = 2 f (x + 2) + f (x + 1)

f (x + 2) = \frac{1}{2} {f (x + 1) + \frac{4}{f (x)}}

f (x) > 0

x \in R

lim x \to \infty f (x)

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