A particle just clears a wall of height b at a distance a and strikes the ground at a distance c from the point of projection. The angle of projection is-

• 

• 45°

• 
• 

32.

If (2, 3, 5) is one end of a diameter of the sphere x²+ y²+ z² − 6x − 12y − 2z + 20 = 0, then the coordinates of the other end of the diameter are

• (4, 9, –3)

• (4, –3, 3)

• (4, 3, 5)

• (4, 3, 5)

• 0

• -1

• -2

• -2

Let A(h, k), B(1, 1) and C(2, 1) be the vertices of a right-angled triangle with AC as its hypotenuse. If the area of the triangle is 1, then the set of values which ‘k’ can take is given by  

• {1, 3}

• {0, 2}

• {–1, 3}

• {–1, 3}

Let P = (−1, 0), Q = (0, 0) and R = ( 3, 3 √3) be three points. The equation of the bisector of the angle PQR

If one of the lines of my²+ (1 − m²)xy − mx² = 0 is a bisector of the angle between the lines xy = 0, then m is

• −1/2

• -2

• 1

• 1

• 1/2

• 1

• 2

• 2

The solution for x of the equation 

• 2

• π

• 
• 

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

• 2/3

• 1/3

• 1/6

• 1/6