MT1710: Test: January 10, 2018: Summary of solutions First of all apologies: in question 5, last line, it should have been written y(T) = 1, instead of y (T/2) = 1. Students who atempted that question were given full marks for that part of the question. 1. y = x2+2x-1 ?3 +1 Vertical asymptotes are given by the roots of the equation x2 + 2 x - 1 = 0, which are x =- 1+ v2, giving two vertical asymptotes of equations, x1 = - 1+1/2 and x2 = - 1 - v2. This is justified because, lim x x3 + 1 = 00 . x->12 Also there is a slant asymptote, as x2 +2x - 1 x3 x+ 2x- x + 1 2 -2x2+x+ 1 2x2- 4x+ 2 5x - 1 and we write x3 +1= (x2 +2x-1) (x-2) + 5x - 1, giving then x2+2x-1 x3 +1 == (x-2) + x2+2x-1 5x -1 . lim x100 - (x - 2) ) = ?3 +1 x2+2x-1 lim 5x-1 x2+2x-1 = lim x 100 1+2 - 1 5 x ? 1 ) =( c as n > 0, as x -> 00, for n E N and c E R. Hence, x y > x - 2 as x> showing that y = x - 2 is a slant asymptote. 2. Using the definition of sinh,we get 2 sinh x = 3 (1 + e2) eª - e -* = 3+ 3e-ª e" - 4e-ª - 3 = 0. Multiplying this equation by e", e2 ª - 3 eª - 4 = 0. Now let y = eª to get the quadratic equation, y2 - 3y - 4 = 0,
which factorises trivially to give, (y-4)(y+1)= 0, with 2 solutions, y = 4 and y = - 1. y = - 1 does not yield a solution for finite values of x as y = eª > 0. y = 4, or e" = 4 gives the only possible solution x = In 4. 3. (a) Let y = arcsin x. By definition of arcsin we write x = sin y, for - "< < 2 . Using the chain rule and differentiating both sides with respect to x, dx dx d (sin y) dx dy 1 = cos y dx 1 cos y dx dy . But cosy = +11 - sin2y for 2 TI < y < - 2, giving, as required, dy dx 1 V1-x2 = (1-x2)-1/2 . 1 v1-1+x2 dx d(1-x2)1/2 (b) Using the chain rule, dy dx x = ? (1 - x2) - 1/2 Vx2 (You can't simplify this further because of the symmetry of the function for x > 0 and x < 0. 4. Considering the complementary equation, day +2 dx dy -3y = 0 dx2 and letting y = ema, with m E Z gives the equation, (m2 + 2m - 3) ema = 0. As ema # 0, we write the auxiliary equation, which factorises trivially, as follows, m2 +2m