JEE MAIN - Mathematics (2019 - 11th January Morning Slot - No. 20)
Let fk(x) = $${1 \over k}\left( {{{\sin }^k}x + {{\cos }^k}x} \right)$$ for k = 1, 2, 3, ... Then for all x $$ \in $$ R, the value of f4(x) $$-$$ f6(x) is equal to
$${1 \over 4}$$
$${5 \over {12}}$$
$${{ - 1} \over {12}}$$
$${1 \over {12}}$$
Explanation
f4(x) $$-$$ f6(x)
= $${1 \over 4}$$ (sin4 x + cos4 x) $$-$$ $${1 \over 6}$$ (sin6 x + cos6 x)
= $${1 \over 4}$$ (1$$-$$ $${1 \over 2}$$ sin2 2x) $$-$$ $${1 \over 6}$$ (1 $$-$$ $${3 \over 4}$$ sin2 2x) = $${1 \over 12}$$
= $${1 \over 4}$$ (sin4 x + cos4 x) $$-$$ $${1 \over 6}$$ (sin6 x + cos6 x)
= $${1 \over 4}$$ (1$$-$$ $${1 \over 2}$$ sin2 2x) $$-$$ $${1 \over 6}$$ (1 $$-$$ $${3 \over 4}$$ sin2 2x) = $${1 \over 12}$$
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