Full Grade 12 test

Find an antiderivative by integrating each term. Evaluate it at both integration bounds and subtract the lower-bound value from the upper-bound value. The integration constant cancels.

A rectangular prism has three pairs of equal rectangular faces: S = 2(ab + ah + bh). Surface area uses cm², not cm³.

A cone axial section is an isosceles triangle with base 2r. A triangular pyramid section has area base × perpendicular section height / 2. Use the section dimensions rather than those of a different face.

A cylinder axial section is a rectangle with sides 2r and h. Its area is 2rh. Use the diameter, not just the radius.

Velocity is the position derivative v(t) = s′(t), and acceleration is a(t) = v′(t). For s(t) = At² + Bt, v(t) = 2At + B. Negative velocity means the position coordinate decreases, not necessarily negative acceleration.

Find the domain and critical points, then the derivative signs on intervals. f′ > 0 means increasing and f′ < 0 decreasing. A change from − to + gives a minimum; + to − a maximum. f′ = 0 alone does not guarantee an extremum. For f′ = 2(x − r), the minimum is at r.

Determine the sign of the difference between x and the approach point on the specified side. Remove the absolute-value signs accordingly, cancel and finish the calculation.

Rewrite the fraction as a negative power of x, apply the power rule and differentiate to check. The logarithm rule applies to exponent −1, not to every fraction.

The oblique segment, its orthogonal projection and the perpendicular to the plane form a right triangle. First find the perpendicular length using the Pythagorean theorem. The sine of the angle between the segment and the plane is the perpendicular length divided by the oblique length.

A closed cylinder surface includes two bases and the curved side: S = 2πr² + 2πrh. Use π = 3.14 and cm².

The vertices of the cross-section are the midpoints of the three cube edges meeting at one vertex. Each side of the cross-section is a midline in a triangular half of a cube face, so its length is a/√2. The cross-section is an equilateral triangle. Calculate its area using S = s²√3/4.

The derivative f′(a) is the instantaneous rate of change and tangent slope. Tangent equation: y − f(a) = f′(a)(x − a). Find f(a) and f′(a) separately. If f′(a) = 0, the tangent is horizontal, but an extremum is not guaranteed.

Velocity is the position derivative v(t) = s′(t), and acceleration is a(t) = v′(t). For s(t) = At² + Bt, v(t) = 2At + B. Negative velocity means the position coordinate decreases, not necessarily negative acceleration.

The distance from a point to a plane is the length of the perpendicular. The perpendicular, the oblique segment and its projection form a right triangle, with the oblique segment as the hypotenuse. Apply the Pythagorean theorem: if the oblique length is l and the projection is p, then h = √(l² − p²).