JEE MAIN - Mathematics Hindi (2019 - 10th January Morning Slot - No. 19)

यदि  $${{dy} \over {dx}} + {3 \over {{{\cos }^2}x}}y = {1 \over {{{\cos }^2}x}},\,\,x \in \left( {{{ - \pi } \over 3},{\pi \over 3}} \right)$$  और  $$y\left( {{\pi \over 4}} \right) = {4 \over 3},$$  तो  $$y\left( { - {\pi \over 4}} \right)$$   के बराबर है -
$${1 \over 3} + {e^6}$$
$${1 \over 3}$$
$${1 \over 3}$$ + e3
$$-$$ $${4 \over 3}$$

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