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Question
The function f : R → R is defined by f(x) = cos2 x + sin4 x. Then, f(R) =
Options
(a) [3/4, 1)
(b) (3/4, 1]
(c) [3/4, 1]
(d) (3/4, 1)
Solution
(c) [3/4, 1]
Given:
f(x) = cos2x + sin4x
\[\Rightarrow f\left( x \right) = 1 - \sin^2 x + \sin^4 x\]
\[\Rightarrow f\left( x \right) = \left( \sin^2 x - \frac{1}{2} \right)^2 + \frac{3}{4}\] The minimum value of \[f\left( x \right)\] is \[\frac{3}{4}\]
Also,
\[\sin^2 x \leq 1\]
\[ \Rightarrow \sin^2 x - \frac{1}{2} \leq \frac{1}{2}\]
\[ \Rightarrow \left( \sin^2 x - \frac{1}{2} \right)^2 \leq \frac{1}{4}\]
\[ \Rightarrow \left( \sin^2 x - \frac{1}{2} \right)^2 + \frac{3}{4} \leq \frac{1}{4} + \frac{3}{4}\]
\[ \Rightarrow f\left( x \right) \leq 1\]
The maximum value of
\[f\left( x \right)\] is 1.
∴ f(R) = (3/4, 1)
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