JEE MAIN - Mathematics Bengali (2022 - 27th July Evening Shift - No. 4)
$$p \ne q \ne 0$$ এর জন্য $$f(x) = {{\root 7 \of {p(729 + x)} - 3} \over {\root 3 \of {729 + qx} - 9}}$$ অপেক্ষকটি $$x = 0$$ বিন্দুতে সন্তত। তাহলে ঃ
$$7pq\,f(0) - 1 = 0$$
$$63q\,f(0) - {p^2} = 0$$
$$21q\,f(0) - {p^2} = 0$$
$$7pq\,f(0) - 9 = 0$$
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