Let $$f$$ and $$g$$ be real valued functions defined on interval $$(-1, 1)$$ such that $$g''(x)$$ is continuous, $$g\left( 0 \right) \ne 0.$$ $$g'\left( 0 \right) = 0$$, $$g''\left( 0 \right) \ne 0$$, and $$f\left( x \right) = g\left( x \right)\sin x$$

STATEMENT - 1: $$\mathop {\lim }\limits_{x \to 0} \,\,\left[ {g\left( x \right)\cot x - g\left( 0 \right)\cos ec\,x} \right] = f''\left( 0 \right)$$ and

STATEMENT - 2: $$f'\left( 0 \right) = g\left( 0 \right)$$

If $$0 < x < 1$$, then

$$\sqrt {1 + {x^2}} {\left[ {{{\left\{ {x\cos \left( {{{\cot }^{ - 1}}x} \right) + \sin \left( {{{\cot }^{ - 1}}x} \right)} \right\}}^2} - 1} \right]^{1/2}} = $$

Consider the two curves $${C_1}:{y^2} = 4x,\,{C_2}:{x^2} + {y^2} - 6x + 1 = 0$$. Then,

Let $$f(x)$$ be a non-constant twice differentiable function defined on $$\left( { - \infty ,\infty } \right)$$


such that $$f\left( x \right) = f\left( {1 - x} \right)$$ and $$f'\left( {{1 \over 4}} \right) = 0.$$ Then,