The solution for x of the equation 

• 2

• π

• 
• 

22.

• 
• 
• 
• 

The area enclosed between the curves y2 = x and y = |x| is

• 2/3

• 1/3

• 1/6

• 1/6

The area enclosed between the curve y = log_e (x + e) and the coordinate axes is

• 1

• 2

• 3

• 3

The parabolas y² = 4x and x² = 4y divide the square region bounded by the lines x = 4, y = 4 and the coordinate axes. If S₁, S₂, S₃ are respectively the areas of these parts numbered from top to bottom; then S₁ : S₂: S₃ is

• 1 : 2 : 1

• 1 : 2 : 3

• 2 : 1 : 2

• 2 : 1 : 2

• e

• e+1

• e-1
• e-1

The area of the region bounded by the curves y = |x – 2|, x = 1, x = 3 and the x-axis is

• 1

• 2

• 3

• 3

Let g(x) = cos x², f(x) = $\sqrt{x}$ and $\alpha , \beta (\alpha <\hspace{0.17em}\beta )$ be the roots of the quadrtic equation 18x² - 9πx + π² = 0. Then the area (in sq. units) bounded by the curve y = (gof)(x) and the lines x = α, x = β and y = 0 is

• $\frac{1}{2}\left(\sqrt{2-1}\right)$

• $\frac{1}{2}\left(\sqrt{3-1}\right)$

• $\frac{1}{2}\left(\sqrt{3+1}\right)$

• $\frac{1}{2}(\sqrt{3}-\sqrt{2})$

If sum of all the solutions of the equation $8 \cos x. \left(\cos \left(\frac{\pi }{6}+x\right).\cos \left(\frac{\pi }{6}-x\right)-\frac{1}{2}\right) = 1$ in [0,π] is kπ, then k is equal to:

• 20/9

• 2/3

• 13/9

• 8/9

The area enclosed by y = $\sqrt{5 - {\mathrm{x}}^2}$ and y = $\left|\mathrm{x} - 1\right|$ is

• $\left(\frac{5\mathrm{\pi }}{4} - 2\right)$ sq. units

• $\left(\frac{5\mathrm{\pi } - 2}{2}\right)$ sq. units

• $\left(\frac{5\mathrm{\pi }}{4} - \frac{1}{2}\right)$ sq. units

• $\left(\frac{\mathrm{\pi }}{2} - 5\right)$ sq. units