Let f be a non-negative function in [0, 1] and twice differentiable in (0, 1). If $$\int_0^x {\sqrt {1 - {{(f'(t))}^2}} dt = \int_0^x {f(t)dt} } $$, $$0 \le x \le 1$$ and f(0) = 0, then $$\mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}}}\int_0^x {f(t)dt} $$ :

Let $$\overrightarrow a $$ and $$\overrightarrow b $$ be two vectors
such that $$\left| {2\overrightarrow a + 3\overrightarrow b } \right| = \left| {3\overrightarrow a + \overrightarrow b } \right|$$ and the angle between $$\overrightarrow a $$ and $$\overrightarrow b $$ is 60$$^\circ$$. If $${1 \over 8}\overrightarrow a $$ is a unit vector, then $$\left| {\overrightarrow b } \right|$$ is equal to :

The function

$$f(x) = \left| {{x^2} - 2x - 3} \right|\,.\,{e^{\left| {9{x^2} - 12x + 4} \right|}}$$ is not differentiable at exactly :

Three numbers are in an increasing geometric progression with common ratio r. If the middle number is doubled, then the new numbers are in an arithmetic progression with common difference d. If the fourth term of GP is 3 r², then r² $$-$$ d is equal to :