Let the definite integral be defined by the formula
$$\int\limits_a^b {f\left( x \right)dx = {{b - a} \over 2}\left( {f\left( a \right) + f\left( b \right)} \right).} $$ For more accurate result for
$$c \in \left( {a,b} \right),$$ we can use $$\int\limits_a^b {f\left( x \right)dx = \int\limits_a^c {f\left( x \right)dx + \int\limits_c^b {f\left( x \right)dx = F\left( c \right)} } } $$ so
that for $$c = {{a + b} \over 2},$$ we get $$\int\limits_a^b {f\left( x \right)dx = {{b - a} \over 4}\left( {f\left( a \right) + f\left( b \right) + 2f\left( c \right)} \right).} $$

If $$f''\left( x \right) < 0\,\forall x \in \left( {a,b} \right)$$ and $$c$$ is a point such that $$a < c < b,$$ and
$$\left( {c,f\left( c \right)} \right)$$ is the point lying on the curve for which $$F(c)$$ is
maximum, then $$f'(c)$$ is equal to

A curve $$y=f(x)$$ passes through $$(1,1)$$ and at $$P(x,y),$$ tangent cuts the $$x$$-axis and $$y$$-axis at $$A$$ and $$B$$ respectively such that $$BP:AP=3:1,$$ then

There are $$n$$ urns, each of these contain $$n+1$$ balls. The ith urn contains $$i$$ white balls and $$(n+1-i)$$ red balls. Let $${u_i}$$ be the event of selecting ith urn, $$i=1,2,3........,n$$ and $$w$$ the event of getting a white ball.

If $$P\left( {{u_i}} \right) \propto i,\,$$ where $$i=1,2,3,.......,n,$$ then $$\mathop {\lim }\limits_{n \to \infty } P\left( w \right) = $$

There are $$n$$ urns, each of these contain $$n+1$$ balls. The ith urn contains $$i$$ white balls and $$(n+1-i)$$ red balls. Let $${u_i}$$ be the event of selecting ith urn, $$i=1,2,3........,n$$ and $$w$$ the event of getting a white ball.

Let $$P\left( {{u_i}} \right) = {1 \over n},$$ if $$n$$ is even and $$E$$ denotes the event of choosing even numbered urn, then the value of $$P\left( {w/E} \right)$$ is