JEE MAIN - Mathematics Hindi (2020 - 4th September Morning Slot - No. 6)
यदि $$\left( {a + \sqrt 2 b\cos x} \right)\left( {a - \sqrt 2 b\cos y} \right) = {a^2} - {b^2}$$
जहां a > b > 0, तब $${{dx} \over {dy}}\,\,पर\left( {{\pi \over 4},{\pi \over 4}} \right)$$ है :
जहां a > b > 0, तब $${{dx} \over {dy}}\,\,पर\left( {{\pi \over 4},{\pi \over 4}} \right)$$ है :
$${{a - 2b} \over {a + 2b}}$$
$${{a - b} \over {a + b}}$$
$${{a + b} \over {a - b}}$$
$${{2a + b} \over {2a - b}}$$
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