\(lim_{x\rightarrow {\pi/2}}{2x-\pi\over cosx} \)

(a)

2

(b)

1

(c)

-2

(d)

0

\(lim_{x \rightarrow \infty}{\sqrt{x^2-1}\over 2x+1}=\)

(a)

1

(b)

0

(c)

-1

(d)

\(1\over 2\)

\(lim_{x \rightarrow 3}\left\lfloor x \right\rfloor =\)

(a)

2

(b)

3

(c)

does not exist

(d)

0

If f : \(R \rightarrow R\) is defined by f(x)=\(\left\lfloor x-3 \right\rfloor +|x-4|\) for \(x \in R\), then \(lim_{x\rightarrow 3^-}f(x)\) is equal to

(a)

-2

(b)

-1

(c)

0

(d)

1

\(lim_{n \rightarrow \infty}({1\over n^2}+{2\over n^2}+{3\over n^2}+..+{n\over n^2})\) is

(a)

\(1\over 2\)

(b)

0

(c)

1

(d)

\(\infty\)

Let f :\(R \rightarrow R\) be defined by \(f(x)= \begin{cases}x & x \text { is irrational } \\ 1-x & x \text { is rational }\end{cases}\)  then f is

(a)

discontinuous at x = \({1\over 2}\)

(b)

continuous at  x = \({1\over 2}\)

(c)

continuous everywhere

(d)

discontinuous everywhere

\(\lim _{ x\rightarrow 1 }{ \frac { { x }^{ m }-1 }{ { x }^{ n }-1 } } is\)

(a)

mn

(b)

m+n

(c)

m-n

(d)

\(\frac { m }{ n } \)

\(\lim _{ x\rightarrow \infty }{ \frac { 1+2+3+....+n }{ { 2n }^{ 2 }+6 } } \)

(a)

2

(b)

6

(c)

\(\frac { 1 }{ 4 } \)

(d)

\(\frac { 1 }{ 2 } \)

The points of discontinuity of the function \(\frac { { x }^{ 2 }+6x+8\quad }{ { x }^{ 2 }-5x+6\quad } is\)

(a)

3,2

(b)

3,-2

(c)

-3,2

(d)

-3,-2

Find the odd one of the following

(a)

x²

(b)

x⁴

(c)

cos x

(d)

sin x