The number of points, where f(x) = [sin(x) + cos(x)] (where [] denotes the greatest integerfunction) and x $\in$ (0, 2) is not continuous, is

• 3

• 4

• 5

• 6

$\underset{\mathrm{x}\to 2}{\lim }\frac{2 - \sqrt{2 + \mathrm{x}}}{2^{1/3} - (4 - \mathrm{x}{)}^{1/3} } \mathrm{is} \mathrm{equal} \mathrm{to}$

• $2 . 3^{-1/2}$

• $3 . 2^{-4/3}$

• $- 3 . 2^{4/3}$

• None of the above

If 2a + 3b + 6c = 0, then the equation ax² + bx + c = 0 has atleast one real root in

• (0, 1)

• $\left(0, \frac{1}{2}\right)$

• $\left(\frac{1}{4}, \frac{1}{2}\right)$

• None of the above

994.

If f(x) = $$\begin{gathered} \left\{\sin \left(\frac{\mathrm{\pi x}}{2}\right), \mathrm{if} \mathrm{x} < 1 \\ 3 - 2\mathrm{x}, \mathrm{if} \mathrm{x} \ge 1\right\} \end{gathered}$$, then f(x) has

• local minimum at x = 1

• local maximum at x = 1

• Both local maximum and local minimum at x = 1

• None of the above

Let f : R $\to$ R be a differentiable function and f(1) = 4. Then, the value of $\underset{\mathrm{x}\to 1}{\lim }{\int }_4^{\mathrm{f}\left(\mathrm{x}\right)}\frac{2\mathrm{t}}{\mathrm{x} - 1}\mathrm{dt}$ is

• 8f'(1)

• 4f'(1)

• 2f'(1)

• f'(1)

If f''(x) = - f(x) and g(x) = f'(x) anf F(x) = ${\left[\mathrm{f}\left(\frac{\mathrm{x}}{2}\right)\right]}^2 + {\left[\left(\frac{\mathrm{x}}{2}\right)\right]}^2$ and given that F(5) = 5, then the value of F(10) is

• 15

• 0

• 5

• 10

A function g defined for all real x > 0 satisfies g(1) = 1, g'(x²) = x³ for all x > 0, then value of g(4) is

• $\frac{13}{3}$

• 3

• $\frac{67}{5}$

• None of these

A particle moving on a curve has the position at a time t is given by x = f'(t)sin(t) + f'(t)cos(t), y = f'(t)cos(t) - f'(t)sin(t), where f is a twice differentiable function. Then, the velocity of the particle at time t is

• f'(t) + f''(t)

• f'(t) - f''(t)

• f'(t) + f'''(t)

• f'(t) - f''(t)

f(x) and g(x) are differentiable in the interval [0, 1] such that f(0) = 2, g(0) = 0, f(1) = 6, g(1) = 2, then Rolle's theorem is applicable for which of the following in [0, 1] ?

• f(x) - g(x)

• f(x) - 2g(x)

• f(x) + 3g(x)

• None of the above

The positive root of equation x² - 2x - 5 = 0 lies in the interval

• {0, 1}

• (1, 2)

• (2, 3)

• (3, 4)