The general solution of the differential equation \(\left(x-y^2\right) d x+y\left(5 x+y^2\right) d y=0\) is :
The slope of the tangent to a curve C : y=y(x) at any point [x, y) on it is \(\frac{2 e ^{2 x }-6 e ^{- x }+9}{2+9 e ^{-2 x }}\) If C passes through the points \(\left(0, \frac{1}{2}+\frac{\pi}{2 \sqrt{2}}\right) \)and \(\left(\alpha, \frac{1}{2} e ^{2 \alpha}\right)\)$ then \(e ^\alpha\) is equal to :
If two distinct points Q, R lie on the line of intersection of the planes –x + 2y – z = 0 and 3x – 5y + 2z = 0 and\(PQ = PR = \sqrt{18}\)where the point P is (1, –2, 3), then the area of the triangle PQR is equal to
Let the plane\(P : \stackrel{→}{r} . \stackrel{→}{a} = d\)contain the line of intersection of two planes\(\stackrel{→}{r} . ( \hat{i} + 3\hat{j} - \hat{k} ) = 6\)and\(\stackrel{→}{r} . ( -6\hat{i} + 5\hat{j} - \hat{k} ) = 7\). If the plane P passes through the point (2, 3, 1/2), then the value of \(\frac{| 13a→|² }{d²}\) is equal to