Application of ...

For the parabola $$y^{2}=8x$$, tangent and normal are drawn at $$P(2, 4)$$ which meet the axis of the parabola in $$A$$ and $$B$$, then the length of the diameter of the circle through $$A, P, B$$ is

I. lf $$f'(\mathrm{a})>0$$ then $$\mathrm{f}$$ is increasing at $$\mathrm{x}=\mathrm{a}$$
II:  If f is increasing at $$\mathrm{x}=\mathrm{a}$$ then $$f'(\mathrm{a})$$ need not to be positive

The stationary point of $$\mathrm{f}(\mathrm{x})=\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}}$$ is

$$\mathrm{f}(\mathrm{x})=(\sin^{-1}\mathrm{x})^{2}+(\cos^{-1}\mathrm{x})^{2}$$ is stationary at

The arrangment of the slopes of the normals to the curve  $$y=e^{\log(cosx)}$$ in the ascending order at the points given below.
$$A) \displaystyle x=\frac{\pi}{6},  B) \displaystyle x=\frac{7\pi}{4},  C)x=\frac{11\pi}{6},  D)x=\frac{\pi}{3}$$

Assertion(A): The tangent to the curve $$y=x^{3}-x^{2}-x+2$$ at (1, 1) is parallel to the x axis.
Reason(R): The slope of the tangent to the above curve at (1, 1) is zero.

Assertion A: The curves $$x^{2}=y,\ x^{2}=-y$$  touch each other at (0, 0).
Reason R: The slopes of the tangents at (0, 0) for both the curves are equal.

Match the points on the curve  $$2y^{2}=x+1$$ with the slope of normals at those points and choose 

P(1, 1) is a point on the parabola  $$y=x^{2}$$ whose vertex is A. The point on the curve at which the tangent drawn is parallel to the chord  $$\overline{AP}$$   is

lf the parametric equation of a curve given by $$x=e^{t}\cos t,\ y=e^{t}\sin t$$, then the tangent to the curve at the point $$t=\dfrac{\pi}{4}$$ makes with axis of $$x$$ the angle.