If the general solution of the differential equation $(y-x+1) dy - (y+x+2) dx = 0$ is $f(x, y, c) = 0$, then the value of $c$ such that $f(1, 1, c) = 0$ is

Let $f$ be a twice differentiable function such that $f(x) = \int_{0}^{x} \tan(t-x) dt - \int_{0}^{x} f(t) \tan t dt$, where $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Then $f''\left(\frac{\pi}{6}\right) + f\left(\frac{\pi}{6}\right)$ is equal to . . . . . . .

The solution of $(1 + xy)y\,dx + (1 - xy)x\,dy = 0$ is

The slope of the normal at any point $(x, y), x > 0, y > 0$ on the curve $y=y(x)$ is given by $\frac{x^{2}}{x y-x^{2} y^{2}-1}$. If the curve passes through the point $(1, 1)$,then $e \cdot y(e)$ is equal to

Let $f:(-1,1) \rightarrow R$ be a differentiable function satisfying $(f^{\prime}(x))^4 = 16(f(x))^2$ for all $x \in (-1,1)$ and $f(0)=0$. The number of such functions is:

Let $S = (0, 2 \pi) - \left\{\frac{\pi}{2}, \frac{3 \pi}{4}, \frac{3 \pi}{2}, \frac{7 \pi}{4}\right\}$. Let $y = y(x)$,$x \in S$,be the solution curve of the differential equation $\frac{dy}{dx} = \frac{1}{1 + \sin 2x}$ with $y\left(\frac{\pi}{4}\right) = \frac{1}{2}$. If the sum of abscissas of all the points of intersection of the curve $y = y(x)$ with the curve $y = \sqrt{2} \sin x$ is $\frac{k \pi}{12}$,then $k$ is equal to: