Inequalities

Step 1: Kailash starts by facing north. After turning to his right, he will face east.
- He walks 25 metres in the east direction.
Step 2: Kailash then turns to his left, which will make him face north.
- He walks 30 metres in the north direction.
Step 3: Kailash then turns to his right again, which will make him face east.
- He walks 25 metres in the east direction.
Step 4: Kailash turns to his right once more, which will make him face south.
- He walks 55 metres in the south direction.
Step 5: Finally, Kailash turns to his right again, which will make him face west.
- He walks 40 metres in the west direction.
Step 6: The net displacement is:
- He has moved 50 metres east (25 + 25) and 40 metres west, so his net movement in the east-west direction is metres east.
- He has moved 30 metres north and 55 metres south, so his net movement in the north-south direction is metres south. Thus, Kailash is 10 metres east and 25 metres south from his starting point, which puts him in the South-East direction.

The vertex of a parabola lies halfway between the focus and the directrix.
The focus is at the origin and the directrix is the line .
The midpoint between these points is at . Thus, the vertex of the parabola is at .

The general equation of the ellipse whose axes are along the coordinate axes is given by: where and are the lengths of the semi-major and semi-minor axes, respectively. The eccentricity of the ellipse is related to and by the formula: We are given that , so: Thus, the relation between and is: The equation of the ellipse now becomes: Next, we substitute the point (-3, 1) into this equation to find the value of . Substituting and : Now, substitute into the equation of the ellipse: Multiply the entire equation by 32: Thus, the equation of the ellipse is: .

Step 1: The function consists of a constant 3 and the absolute value function , which is always non-negative.
Step 2: The absolute value function reaches its minimum value of 0 at . Therefore, the minimum value of occurs when : Thus, the local minimum value of is 3.

Step 1: For the function to be continuous at , the left-hand limit and the right-hand limit must be equal to the value of the function at .
The left-hand limit is the value of for , i.e., .
The right-hand limit is the value of for , i.e., .
Step 2: For continuity at , set the two expressions equal:
Step 3: Solve for in terms of : Thus, the relationship is .

Step 1: The standard equation of a hyperbola with vertical transverse axis is: where is the distance from the center to the vertices and is related to the asymptotes. The eccentricity is given by: We are given that and the vertices are at , so .
Step 2: Using the formula for eccentricity: Multiplying both sides by 6: Squaring both sides: Thus, the equation of the hyperbola is:

Step 1: Expressing the integral. We need to evaluate the integral: We will split the integral into two parts:
Step 2: Evaluating the first integral. Consider the integral . By the substitution , we have and . This gives: Thus, both integrals and are equal.
Step 3: Simplifying the result. Therefore, we can write: Using known integral results, we know that: Thus: