Title Apr 2025 PB1 - MSTC Kippap Solutions.pdf ✅

Problem 5. Parabolas A decorative beam is to be put along the latus rectum of a parabolic arch. Find the length of the beam if the arch is 15 m wide at the base and 20 m high. a. 0.70 m c. 11.25 m b. 2.81 m d. 5.63 m Solution: If the vertex of the parabola is set at the origin, then the equation of the parabola is 4a(y − 0) = (x − 0) 2 4ay = x 2 At the base of the arch, y is the height of the parabola and x is the distance from the vertex which is half the width. 4a(20) = ( 15 2 ) 2 4a = 2.813 m The value of 4a is the length of the latus rectum. Thus, the beam is 2.813 m long. Problem 6. Binomial Expansion What is the coefficient of the x 2 term in the expansion of (x 3 + 4 x ) 6 ? Hint: the rth term in a binomial expansion (a +b) n is ( n r −1 ) a n−r+1b r−1 a. 4830 c. 3840 b. 4380 d. 3480 Solution: The rth term in a binomial expansion (a + b) n is ( n r − 1 ) a n−r+1b r−1 Substituting for the binomial expansion of (x 3 + 4 x ) 6 ( n r −1 ) a n−r+1b r−1 = ( 6 r − 1 ) (x 3 ) 6−r+1 ( 4 x ) r−1 = ( 6 r − 1 ) (x 21−3r ) ( 4 r−1 x r−1 ) = (4 r−1 ) ( 6 r − 1 ) ( x 21−3r x r−1 ) = (4 r−1 ) ( 6 r − 1 ) (x 21−3r−(r−1) ) = (4 r−1 ) ( 6 r − 1 ) (x 22−4r ) Therefore, the coefficient is (4 r−1 ) ( 6 r ). If x 22−4r = x 2 for the required term, then 22− 4r = 2 r = 5 Thus, coefficient = (4 5−1 ) ( 6 5 −1 ) = 3840 Alternatively, the terms can be expressed as ( n! a! b! ) (x 3 ) a ( 4 x ) b
Where: a + b = 6 3a − b = 2 Solving these equations simultaneously, a = 2, b = 4 coefficient = ( 6! 2! 4! ) (4 4 ) = 3840 Problem 7. Trigonometric Levelling A surveyor sights the top of a water tank from a hill (el. 400 m) through an instrument with an internal focusing telescope. The inclined distance to the top of the tank is 600 m with a vertical angle of −12°25′. If the height of the instrument is 1.45 m and the height of the tank is 32 m, what is the elevation of the base of the tank? a. 498.4 m c. 272.4 m b. 530.4 m d. 240.4 m Solution: Thus, the vertical distance is V = d sin θ = 600 sin(−12°25′ ) = −129.012 m For the elevation of the tank, Elevation at B = 400 + 1.45 −129.012 − 32 = 240. 438 m Problem 8. Ellipse An ellipse is formed by tying a 10-inch string on two pins “x” inches apart on a paper. A pencil pulls the string and traces a smooth arc. If the ellipse should be 6 inches wide, how far should the pins be? a. 4 inches c. 8 inches b. 6 inches d. 10 inches Solution: The sum of the distances of any point on the ellipse to the pins is equal to 2a since the pins act as the foci. 2a = 10 a = 5