Functions: JEE Advanced

Algebra Level 3

Find the range of the function f ( x ) = 1 4 cos ⁡ 2 x + 6 sin ⁡ x cos ⁡ x + 8 sin ⁡ 2 x f(x) = \dfrac1{4\cos^2 x + 6\sin x \cos x + 8\sin^2 x} .

[ 0 , 2 ] [0,2] [ 6 + 13 , 6 − 13 ] [6+\sqrt{13}, 6-\sqrt{13} ] [ 0 , 1 6 − 13 ] \left [ 0 , \frac1{6-\sqrt{13}} \right ] [ 1 6 + 13 , 1 6 − 13 ] \left [ \frac1{6+\sqrt{13}} , \frac1{6-\sqrt{13}} \right ]

Rajdeep Bharati by Rajdeep Bharati , 21, India

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1 solution

Vignesh Suresh
Jan 27, 2016

Hence it follows min(f(x))= 1/(6+sqrt(13)) and max(f(x))=1/(6-sqrt(13)) Hence the 4th option is the answer.

3 Helpful 1 Interesting 0 Brilliant 0 Confused

Right! Try out this: f(x)=sin^2 x+cos^4 x Find domain and range

Rajdeep Bharati - 5 years, 4 months ago

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Domain:All real numbers. Range : 0.75 to 1 , both inclusive

Vignesh Suresh - 5 years, 4 months ago

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Absolutely correct

Rajdeep Bharati - 5 years, 4 months ago

You can also try this: https://brilliant.org/problems/minimum-maximum/?group=seR3vnZFJhnc

Rajdeep Bharati - 5 years, 4 months ago

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