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Functions (XII)

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Question

$If f(x)= \sqrt{4 - x^{2}} + \sqrt{x^{2} - 1}, t h e n t h e m a x i m u m v a l u e o f (f {(x))}^{2} i s$

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Solution

${Let x}^{2} = 4 c o s^{2} θ + s i n^{2} θ t h e n 4 - x^{2} = 3 s i n^{2} θ a n d x^{2} - 1 = 3 c o s^{2} θ n o w f (x)= \sqrt{4 - x^{2}} + \sqrt{x^{2} - 1} ∴ f (x) = \sqrt{3} |s i n θ| + \sqrt{3} |c o s θ|$
$(f {(x))}^{2} = 3 s i n^{2} θ + 3 c o s^{2} θ + 2 (3) |s i n θ c o s θ| \Rightarrow (f {(x))}^{2} = 3 + 3 |s i n 2 θ| \Rightarrow m a x o f |s i n 2 θ| = 1 H e n c e, m a x i m u m v a l u e o f (f {(x))}^{2} i s 6$

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