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The corresponding values of y (x) are −24, 36 and 72. Since y(±∞) = ∞, the lowest value of y is that corresponding to the lowest minimum, which can only be at x = 1 or x = −2, as y must be positive at a minimum. The values of y(x) at these two points are y(1) = 1 and y(−2) = −26, and so the lowest value taken is −26. 15 Show that y(x) = xa2x exp x2 has no stationary points other than x = 0, if √ √ exp(− 2) < a < exp( 2). Since the logarithm of a variable varies monotonically with the variable, the stationary points of the logarithm of a function of x occur at the same values of x as the stationary points of the function.

Using d(u/v)/dx = (vu − uv )/v 2 , and obtain 1 + sin x , f(x) = cos x cos x(cos x) − (1 + sin x)(− sin x) f (x) = cos2 x 1 + sin x = cos2 x f(x) . = cos x Thus, since sec x = f (x)/f(x), it follows that sec x dx = ln[ f(x) ] + c = ln 1 + sin x cos x + c = ln(sec x + tan x) + c. 37 By making the substitution x = a cos2 θ + b sin2 θ, evaluate the deﬁnite integrals J between limits a and b (> a) of the following functions: (a) [(x − a)(b − x)]−1/2 ; (b) [(x − a)(b − x)]1/2 ; (c) [(x − a)/(b − x)]1/2 .

Write it in such a form that x appears only in a term that is the square of a linear function of x. We then examine the overall sign of the terms that do not contain x; this determines the form of the integral. In case (iii) there is no such term. We write b2 − 4ac as ∆2 > 0, or 4ac − b2 as ∆ 2 > 0, as needed. (i) For ∆2 = b2 − 4ac > 0, dx J= a = 1 a x+ 2 b 2a b2 4a2 − − c a − c a dx x+ b 2 2a b 2a b 2a − − ∆2 4a2 ∆ 2a ∆ 2a = 1 a x+ ln a∆ x+ = 1 2ax + b − ∆ ln + k. ∆ 2ax + b + ∆ + +k (ii) For −∆ 2 = b2 − 4ac < 0, dx J= a = 1 a x+ b 2 2a b2 4a2 − dx x+ = 1 2a tan−1 a∆ = 2 tan−1 ∆ b 2 2a + x+ ∆2 4a2 b 2a ∆ 2a 2ax + b ∆ 30 +k + k.