JEE Main 2022 Mathematics question
The slope of normal at any point (x, y), x > 0, y > 0 on the curve y = y(x) is given by . If the curve passes through the point (1, 1), then e·y(e) is equal to
Answer: (4)
Solution
The slope of the normal is −dx/dy = , so dx/dy... equivalently the slope of the tangent satisfies dy/dx = ( − xy + . Put xy = v: y + x dy/dx = dv/dx, which gives dv/dx =
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