MT1710: Test: 13 January, 2016: Solutions 1. Considering y =: ?3 -2x2+5x x4 + 1 . Vertical asymptotes occur when x3- 2x2+5x=x(x2-2x+5)=x((x-1)2 +4) = 0. Since (x - 1)2 + 4 > 4 for all x, the only vertical asymptote occurs when x = 0 (or use the fact that the discriminant is negative to prove there are no other vertical asymptotes), for which we see that as x -> 0 then y -> too or, alternatively, lim y = +00. Also using long division of x4 + 1 by x3 - 2x2 + 5x, we get @4+1=(x+2) (x3-2x2+5x) +1-x2-10x, giving, x4 + 1 x3-2x2+5x 1 =x+2+ x3-2x2+5x . As lim x100 x3-2x2+5x ?4 +1 - (x + 2) = lim x+100 x3 - 2x2 + 5x 1-x2-10x = 0, there is a slant asymptote of equation y = x + 2. Hence for this function, there is a slant asymptote y = x + 2, together with the vertical asymptote x = 0. 2 (a). Take y = arcsinhx; by its definition we have that sinh y = x. Differentiating implicitly this equation gives, dy dx 1 cosh y . As cosh2y - sinh2y = 1, and as cosh y > 0 for all real arguments, we write coshy= v1+x2 and replacing in our expression gives, dx dy 1 V1 + x2' as required. 2 (b). By definition, dx d(arcsinh x)2 = d e x ln(arcsinh x) dx e x ln(arcsinh x) . Differentiating using the chain rule, dx d(arcsinh x)ª = d [x ln (arcsinh x)] dx = ln (arcsinh x) + x arcsinh x Vx2+1 1 (arcsinh x)a where we used the product rule and the result from question 2 (a). Alternatively you could apply ln to the equation y = (arcsinh x)" to get ln y = x ln (arcsinh x)
and then differentiate implicitly to get 1 dy 1 , y dx = ln (arcsinh x) + arcsinh x x Vx2 +1' from which follows the same final result as above. 2x 1-x2 dv COS V 3. Using the chain rule twice with u = and v = arctan u, dy dx dv du dx COS U = du dx 1 du - COS U = 1 + u2 dx 1 = COS U , using either the quotient rule on u = 2x 2 2(1+x2) 2 1- 22, or the product rule on u = (2x) (1 - x2)-1, to evaluate dx du . Then, substituting back for v and u, = cos arctan( 2x dx dy 1 - 22)] 1+x2 2 cos arctan( 2x (1+ x2)2 (1-x2)2 (1-x2)2 2(1+x2) 4. Note that = 1 - 22)] . (+2 -+-2) = (+-2) - 3 2 2 . Then using the substitution t - > = 2 cosh u, we get arccosh [](x- 1)] 1. 1 (+2 - t - 2)1/2 3 dt = 2 Z [(coshu)2 - (3)271/2 sinh u du or, 2 1 (t2 - t - 2)1/2 1 dt = Z du,