JEE Advance - Mathematics (1980 - No. 9)
Given $$A = \left\{ {x:{\pi \over 6} \le x \le {\pi \over 3}} \right\}$$ and
$$f\left( x \right) = \cos x - x\left( {1 + x} \right);$$ find $$f\left( A \right).$$
$$f\left( x \right) = \cos x - x\left( {1 + x} \right);$$ find $$f\left( A \right).$$
$$\left[ {{{\sqrt 3 } \over 2} - {\pi \over 6}\left( {1 + {\pi \over 6}} \right), {1 \over 2} - {\pi \over 3}\left( {1 + {\pi \over 3}} \right)} \right]$$
$$\left[ {{{\sqrt 3 } \over 2} - {\pi \over 6}\left( {1 - {\pi \over 6}} \right), {1 \over 2} - {\pi \over 3}\left( {1 - {\pi \over 3}} \right)} \right]$$
$$\left[ {{1 \over 2} - {\pi \over 3}\left( {1 + {\pi \over 3}} \right),,{{\sqrt 3 } \over 2} - {\pi \over 6}\left( {1 + {\pi \over 6}} \right)} \right]$$
$$\left[ {{{\sqrt 3 } \over 2} + {\pi \over 6}\left( {1 + {\pi \over 6}} \right), {1 \over 2} + {\pi \over 3}\left( {1 + {\pi \over 3}} \right)} \right]$$
$$\left[ {\frac{\sqrt{2}}{2} - {\pi \over 4}(1 + {\pi \over 4}) } \right]$$
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