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Question Number 16089 by Tinkutara last updated on 17/Jun/17

The range of function

f (θ) = \sin^{2} θ + \frac{1}{1 + \sin^{2} θ} is

(1) [1, \infty)

(2) [2, \infty)

(3) [1, \frac{3}{2}]

(4) [\frac{3}{2}, \infty)

Commented by prakash jain last updated on 18/Jun/17

0 ⩽ \sin^{2} θ ⩽ 1

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

\frac{d}{d x} (x + \frac{1}{1 + x}) = 1 - \frac{1}{{(1 + x)}^{2}} = \frac{x (x + 2)}{{(1 + x)}^{2}} ⩾ 0

f (x) is increasing function for

x ⩾ 0

\min f (x) = f (0) = 1

m a x f (x) = f (1) = \frac{3}{2}

Commented by Tinkutara last updated on 18/Jun/17

Thanks Sir!