The true solution set of the inequality,

$\sqrt {5\,x\,\, - \,\,6\,\, - \,\,{x^2}} \,\, + \,\,\frac{\pi }{2}\,\,\int\limits_0^x {} $$dz > x \int\limits_0^\pi  {} sin^2 x \,dx$ is :

Let the function $f :[0,2] \rightarrow R$ be defined as

$f(x)=\left\{\begin{array}{cc}e^{\min \left[x^2, x-[x]\right\}}, & x \in[0,1) \\e^{\left[x-\log _e x\right]}, & x \in[1,2]\end{array}\right.$

where [t] denotes the greatest integer less than or equal to $t$. Then the value of the integral $\int \limits_0^2 x f(x) d x$ is

The number of continuous functions $f:[0,1] \rightarrow R$ that satisfy $\int \limits_0^1 x f(x) d x=\frac{1}{3}+\frac{1}{4} \int \limits_0^1(f(x))^2 d x$ is

Let for $x \in R , S_0( x )= x$,$S _{ k }( x )= C _{ k } x + k \int _0^{ x } S _{ k -1}(t) d t$, where $C _0=1, C _{ k }=1-\int_0^1 S _{ k -1}( x ) dx , k =1,2,3 \ldots$. Then $S _2(3)+6 C _3$ is equal to $...........$.

Let $I = \mathop \smallint \limits_0^1 \frac{{\sin x}}{{\sqrt x }}\;dx$ and $\;J = \mathop \smallint \limits_0^1 \frac{{\cos x}}{{\sqrt x }}\;dx$ Then which one of  the following is true?

Let $a, b$ and $c$ be positive constants. The value of $‘a’$ in terms of $‘c’$ if the value of integral $\int\limits_0^1 {(ac{x^{b + 1}} + {a^3}b{x^{3b + 5}})\,dx} $ is independent of $b$ equals