The inverse of f(x) = \(\begin{cases} x\quad if\quad x<1 \\ { x }^{ 2 }\quad if\quad 1\le x\le 4 \\ 8\sqrt { x } \quad if\quad x>4 \end{cases}\) is

(a)

\({ f }^{ -1 }(x)=\begin{cases} x\quad if\quad x<1 \\ \sqrt { x } \quad if\quad 1\le x\le 16 \\ \frac { { x }^{ 2 } }{ 64 } \quad if\quad x>16 \end{cases}\)

(b)

\({ f }^{ -1 }(x)=\begin{cases} -x\quad if\quad x<1 \\ \sqrt { x } \quad if\quad 1\le x\le 16 \\ \frac { { x }^{ 2 } }{ 64 } \quad if\quad x>16 \end{cases}\)

(c)

\({ f }^{ -1 }(x)=\begin{cases} { x }^{ 2 }\quad if\quad x<1 \\ \sqrt { x } \quad if\quad 1\le x\le 16 \\ \frac { { x }^{ 2 } }{ 64 } \quad if\quad x>16 \end{cases}\)

(d)

\({ f }^{ -1 }(x)=\begin{cases} { 2x }\quad if\quad x<1 \\ \sqrt { x } \quad if\quad 1\le x\le 16 \\ \frac { { x }^{ 2 } }{ 8 } \quad if\quad x>16 \end{cases}\)

If 8 and 2 are the roots of x²+ ax + c = 0 and 3, 3 are the roots of x² + dx + b = 0; then the roots of the equation x²+ ax + b = 0 are

(a)

1, 2

(b)

-1, 1

(c)

9, 1

(d)

-1, 2

The product of r consecutive positive integers is divisible by

(a)

r!

(b)

(r-1)!

(c)

(r+1)!

(d)

r^r

The sum up to n terms of the series \(\frac { 1 }{ \sqrt { 1 } +\sqrt { 3 } } +\frac { 1 }{ \sqrt { 3 } +\sqrt { 5 } } +\frac { 1 }{ \sqrt { 5 } +\sqrt { 7 } } +\)....is 

(a)

\(\sqrt { 2n+1 } \)

(b)

\(\frac { \sqrt { 2n+1 } }{ 2 } \)

(c)

\(\sqrt { 2n+1 } -1\)

(d)

\(\frac { \sqrt { 2n+1 } -1 }{ 2 } \)

If a is the arithmetic mean and g is the geometric mean of two numbers, then

(a)

a \(\le \) g

(b)

a \(\ge\) g

(c)

a = g

(d)

a > g

The equation of the line with slope 2 and the length of the perpendicular from the origin equal to \(\sqrt5\) is

(a)

x - 2y = \(\sqrt5\)

(b)

2x - y =\(\sqrt5\)

(c)

2x - y = 5

(d)

x - 2y - 5 = 0

\(\theta\) is acute angle between the lines x²- xy - 6y² = 0, then \(\frac{2\cos\theta+3\sin\theta}{4\sin\theta+5\cos\theta}\) is

(a)

1

(b)

\(-\frac{1}{9}\)

(c)

\(\frac{5}{9}\)

(d)

\(\frac{1}{9}\)

If 7x² - 8xy +A = 0 represents a pair of perpendicular lines, the A is ______________

(a)

7

(b)

-7

(c)

-8

(d)

8

If \(\triangle\) = \(\begin{vmatrix} a&b &c \\ x & y & z \\ p &q &r \end{vmatrix}\), then \(\begin{vmatrix} ka&kb &kc \\ kx & ky & kz \\k p &kq &kr \end{vmatrix}\) is

(a)

\(\triangle\)

(b)

k\(\triangle\)

(c)

3k\(\triangle\)

(d)

k³\(\triangle\)

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

(a)

1

(b)

0

(c)

-1

(d)

\(1\over 2\)