Consider :
f $$\left( x \right) = {\tan ^{ - 1}}\left( {\sqrt {{{1 + \sin x} \over {1 - \sin x}}} } \right),x \in \left( {0,{\pi \over 2}} \right).$$

A normal to $$y = $$ f$$\left( x \right)$$ at $$x = {\pi \over 6}$$ also passes through the point:

Let $$f(x)$$ be a polynomial of degree four having extreme values
at $$x=1$$ and $$x=2$$. If $$\mathop {\lim }\limits_{x \to 0} \left[ {1 + {{f\left( x \right)} \over {{x^2}}}} \right] = 3$$, then f$$(2)$$ is equal to :

The normal to the curve, $${x^2} + 2xy - 3{y^2} = 0$$, at $$(1,1)$$

If $$f$$ and $$g$$ are differentiable functions in $$\left[ {0,1} \right]$$ satisfying
$$f\left( 0 \right) = 2 = g\left( 1 \right),g\left( 0 \right) = 0$$ and $$f\left( 1 \right) = 6,$$ then for some $$c \in \left] {0,1} \right[$$