Let M and m respectively be the maximum and minimum values of the function f(x) = tan$$-$$1 (sin x + cos x) in $$\left[ {0,{\pi \over 2}} \right]$$, then the value of tan(M $$-$$ m) is equal to :
JEE · Math · previous-year question
- A.$$2 + \sqrt 3 $$
- B.$$2 - \sqrt 3 $$
- C.$$3 + 2\sqrt 2 $$
- D.$$3 - 2\sqrt 2 $$correct
Answer
D. $$3 - 2\sqrt 2 $$
Explanation
Let g(x) = sin x + cos x = $$\sqrt 2 $$ sin$$\left( {x + {\pi \over 4}} \right)$$ g(x)$$\in$$ $$\left[ {1,\sqrt 2 } \right]$$ for x$$\in$$ [0, $$\pi$$/2] f(x) = tan$$-$$1 (sin x + cos x) $$\in$$ $$\left[ {{\pi \over 4},{{\tan }^{ - 1}}\sqrt 2 } \right]$$ tan$$({\tan ^{ - 1}}\sqrt 2 - {\pi \over 4}) = {{\sqrt 2 - 1} \over {1 + \sqrt 2 }} \times {{\sqrt 2 - 1} \over {\sqrt 2 - 1}} = 3 - 2\sqrt 2 $$
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