Iff(x) = \left\{ {\matrix{ {{{\sin (p + 1)x + \sin x} \over x}} & {,x 0} \cr } } \right. is continuous at x = 0, then the ordered pair (p, q) is equal to

If \mathop {\lim }\limits_{x \to 1} {{{x^2} - ax + b} \over {x - 1}} = 5, then a + b is equal to :

If $$\alpha$$ and $$\beta$$ are the roots of the equation 375x2 – 25x – 2 = 0, then $$\underset{n\to \infty }{\lim }\sum _{r=1}^n{\alpha }^r+\underset{n\to \infty }{\lim }\sum _{r=1}^n{\beta }^r$$ is equal to :

\mathop {\lim }\limits_{x \to 0} {{x + 2\sin x} \over {\sqrt {{x^2} + 2\sin x + 1} - \sqrt {{{\sin }^2}x - x + 1} }} is :

Let f(x) = 5 – |x – 2| and g(x) = |x + 1|, x $$\in$$ R. If f(x) attains maximum value at $$\alpha$$ and g(x) attains minimum value at $$\beta$$, then \mathop {\lim }\limits_{x \to -\alpha \beta } {{\left( {x - 1} \right)\left( {{x^2} - 5x + 6} \right)} \over {{x^2} - 6x + 8}} is equal to :

\mathop {\lim }\limits_{x \to {1^ - }} {{\sqrt \pi - \sqrt {2{{\sin }^{ - 1}}x} } \over {\sqrt {1 - x} }} is equal to :

Let f be a differentiable function such that f(1) = 2 and f '(x) = f(x) for all x $$\in$$ R R. If h(x) = f(f(x)), then h'(1) is equal to :

Let f : R $$\to$$ R be a function defined as f(x) = \left\{ {\matrix{ 5 & ; & {x \le 1} \cr {a + bx} & ; & {1 < x < 3} \cr {b + 5x} & ; & {3 \le x < 5} \cr {30} & ; & {x \ge 5} \cr } } \right. Then, f is

For each x$$\in$$R, let [x] be the greatest integer less than or equal to x. Then \mathop {\lim }\limits_{x \to {0^ - }} \,\,{{x\left( {\left[ x \right] + \left| x \right|} \right)\sin \left[ x \right]} \over {\left| x \right|}} is equal to :

For each t $$\in$$ R , let [t] be the greatest integer less than or equal to t Then \mathop {\lim }\limits_{x \to 1^ + } {{\left( {1 - \left| x \right| + \sin \left| {1 - x} \right|} \right)\sin \left( {{\pi \over 2}\left[ {1 - x} \right]} \right)} \over {\left| {1 - x} \right|.\left[ {1 - x} \right]}}

Let f\left( x \right) = \left\{ {\matrix{ {\max \left\{ {\left| x \right|,{x^2}} \right\}} & {\left| x \right| \le 2} \cr {8 - 2\left| x \right|} & {2 < \left| x \right| \le 4} \cr } } \right. Let S be the set of points in the interval (– 4, 4) at which f is not differentiable. Then S

Let f : ($$-$$1, 1) $$\to$$ R be a function defined by f(x) = max $$\left\{-\left|x\right|,-\sqrt{1-x^2}\right\}.$$ If K be the set of all points at which f is not differentiable, then K has exactly -

Let [x] denote the greatest integer less than or equal to x. Then \mathop {\lim }\limits_{x \to 0} {{\tan \left( {\pi {{\sin }^2}x} \right) + {{\left( {\left| x \right| - \sin \left( {x\left[ x \right]} \right)} \right)}^2}} \over {{x^2}}}

Let K be the set of all real values of x where the function f(x) = sin |x| – |x| + 2(x – $$\pi$$) cos |x| is not differentiable. Then the set K is equal to :

\mathop {\lim }\limits_{x \to 0} {{x\cot \left( {4x} \right)} \over {{{\sin }^2}x{{\cot }^2}\left( {2x} \right)}} is equal to :

\mathop {\lim }\limits_{x \to \pi /4} {{{{\cot }^3}x - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}} is :

Let S be the set of all points in (–$$\pi$$, $$\pi$$) at which the function, f(x) = min{sin x, cos x} is not differentiable. Then S is a subset of which of the following ?

Let f\left( x \right) = \left\{ {\matrix{ { - 1} & { - 2 \le x < 0} \cr {{x^2} - 1,} & {0 \le x \le 2} \cr } } \right. and $$g(x)=\left|f\left(x\right)\right|+f\left(\left|x\right|\right).$$ Then, in the interval (–2, 2), g is :

If \alpha = {\cos ^{ - 1}}\left( {{3 \over 5}} \right), \beta = {\tan ^{ - 1}}\left( {{1 \over 3}} \right) where 0 < \alpha ,\beta < {\pi \over 2} , then $$\alpha$$ - $$\beta$$ is equal to :