JEE MAIN - Mathematics Hindi (2022 - 24th June Evening Shift - No. 14)
यदि $$y = {\tan ^{ - 1}}\left( {\sec {x^3} - \tan {x^3}} \right),\,{\pi \over 2} < {x^3} < {{3\pi } \over 2}$$, है, तो
$$x y^{\prime \prime}+2 y^{\prime}=0$$
$$x^{2} y^{\prime \prime}-6 y+\frac{3 \pi}{2}=0$$
$$x^{2} y^{\prime \prime}-6 y+3 \pi=0$$
$$x y^{\prime \prime}-4 y^{\prime}=0$$
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