Let $$f:R\to R$$ be a function defined by : f(x) = \left\{ {\matrix{ {\max \,\{ {t^3} - 3t\} \,t \le x} & ; & {x \le 2} \cr {{x^2} + 2x - 6} & ; & {2 5} \cr } } \right. where [t] is the greatest integer less than or equal to t. Let m be the number of points where f is not differentiable and $$I=\underset{-2}{\overset{2}{\int }}f(x)dx$$. Then the ordered pair (m, I) is equal to :

\int_0^5 {\cos \left( {\pi \left( {x - \left[ {{x \over 2}} \right]} \right)} \right)dx}, where [t] denotes greatest integer less than or equal to t, is equal to:

Let f be a real valued continuous function on [0, 1] and $$f(x)=x+\underset{0}{\overset{1}{\int }}(x-t)f(t)dt$$. Then, which of the following points (x, y) lies on the curve y = f(x) ?

If \int\limits_0^2 {\left( {\sqrt {2x} - \sqrt {2x - {x^2}} } \right)dx = \int\limits_0^1 {\left( {1 - \sqrt {1 - {y^2}} - {{{y^2}} \over 2}} \right)dy + \int\limits_1^2 {\left( {2 - {{{y^2}} \over 2}} \right)dy + I} } }, then I equals

Let f : R $$\to$$ R be a differentiable function such that f\left( {{\pi \over 4}} \right) = \sqrt 2 ,\,f\left( {{\pi \over 2}} \right) = 0 and f'\left( {{\pi \over 2}} \right) = 1 and let $$g(x)={\int }_x^{\pi /4}(f^{\prime }(t)\sec t+\tan t\sec tf(t))dt$$ for x \in \left[ {{\pi \over 4},{\pi \over 2}} \right). Then \mathop {\lim }\limits_{x \to {{\left( {{\pi \over 2}} \right)}^ - }} g(x) is equal to :

Let f : R $$\to$$ R be a continuous function satisfying f(x) + f(x + k) = n, for all x $$\in$$ R where k > 0 and n is a positive integer. If $$I_1=\underset{0}{\overset{4nk}{\int }}f(x)dx$$ and $$I_2=\underset{-k}{\overset{3k}{\int }}f(x)dx$$, then :

Let [t] denote the greatest integer less than or equal to t. Then, the value of the integral $$\underset{0}{\overset{1}{\int }}[-8x^2+6x-1]dx$$ is equal to :

If m and n respectively are the number of local maximum and local minimum points of the function f(x) = \int\limits_0^{{x^2}} {{{{t^2} - 5t + 4} \over {2 + {e^t}}}dt}, then the ordered pair (m, n) is equal to

Let f be a differentiable function in \left( {0,{\pi \over 2}} \right). If $$\underset{\cos x}{\overset{1}{\int }}{t}^{2}f(t)dt={\sin }^{3}x+\cos x$$, then {1 \over {\sqrt 3 }}f'\left( {{1 \over {\sqrt 3 }}} \right) is equal to

The integral \int\limits_0^1 {{1 \over {{7^{\left[ {{1 \over x}} \right]}}}}dx}, where [ . ] denotes the greatest integer function, is equal to

The value of the integral \int\limits_{ - 2}^2 {{{|{x^3} + x|} \over {({e^{x|x|}} + 1)}}dx} is equal to :

If {b_n} = \int_0^{{\pi \over 2}} {{{{{\cos }^2}nx} \over {\sin x}}dx,\,n \in N}, then

The value of \int\limits_0^\pi {{{{e^{\cos x}}\sin x} \over {(1 + {{\cos }^2}x)({e^{\cos x}} + {e^{ - \cos x}})}}dx} is equal to:

The value of the integral \int\limits_{ - \pi /2}^{\pi /2} {{{dx} \over {(1 + {e^x})({{\sin }^6}x + {{\cos }^6}x)}}} is equal to

\mathop {\lim }\limits_{n \to \infty } \left( {{{{n^2}} \over {({n^2} + 1)(n + 1)}} + {{{n^2}} \over {({n^2} + 4)(n + 2)}} + {{{n^2}} \over {({n^2} + 9)(n + 3)}} + \,\,....\,\, + \,\,{{{n^2}} \over {({n^2} + {n^2})(n + n)}}} \right) is equal to :

\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{r \over {2{r^2} - 7rn + 6{n^2}}}} is equal to :

For any real number $$x$$, let $$[x]$$ denote the largest integer less than equal to $$x$$. Let $$f$$ be a real valued function defined on the interval $$[-10,10]$$ by $$f(x)=\{\begin{array}{l}x-[x],\text{ if }[x]\text{ is odd }\\ 1+[x]-x,\text{ if }[x]\text{ is even }.\end{array}$$ Then the value of $$\frac{{\pi }^2}{10}{\int }_{-10}^{10}f(x)\cos \pi xdx$$ is :

\mathop {\lim }\limits_{n \to \infty } {1 \over {{2^n}}}\left( {{1 \over {\sqrt {1 - {1 \over {{2^n}}}} }} + {1 \over {\sqrt {1 - {2 \over {{2^n}}}} }} + {1 \over {\sqrt {1 - {3 \over {{2^n}}}} }} + \,\,...\,\, + \,\,{1 \over {\sqrt {1 - {{{2^n} - 1} \over {{2^n}}}} }}} \right) is equal to

Let $$[t]$$ denote the greatest integer less than or equal to $$t$$. Then the value of the integral $${\int }_{-3}^{101}\left([\sin (\pi x)]+e^{[\cos (2\pi x)]}\right)dx$$ is equal to

If a = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {{{2n} \over {{n^2} + {k^2}}}} and f(x) = \sqrt {{{1 - \cos x} \over {1 + \cos x}}}, $$x\in (0,1)$$, then :