Indian Institute of Technology Bombay

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1 Indian Institute of Technology Bombay MA 5 Calculus Autumn SRG/AA/VDS/KSK Solutions and Marking Scheme for the Mid-Sem Exam Date: September, Weightage: 3 % Time: 8.3 AM.3 AM Max. Marks:. (i) Consider the sequence {x n } defined by x n = n(n + ) for n N. Show that {x n } is a bounded sequence. Is {x n } convergent? Justify your answer. [ marks] (ii) Consider the sequence {y n } defined by y n = n for n N. Is {y n } convergent? Justify your answer. Solution: (i) For each n N, we have < x n = k= k(k + ) = k= ( k ) = k + n + <.... [] Hence {x n } is a bounded sequence. Moreover, lim x n = lim =, and n n + thus {x n } is convergent.... [] (ii) In view of (i) above, for each n N, we have n < y n = n (n )n = +x n = n <.... [] Hence {y n } is a bounded sequence. Moreover, since y n+ y n = (n+) > for all n N, the sequence {y n } is monotonically increasing.... [] Thus {y n } is a monotonic bounded sequence, and hence it is convergent.... [] [Note: The part marking for the last step should be given only when the first two steps are done correctly.]

2 . (i) Let f : R R be a function and let c R. Define what is meant by limit of f(x) as x tends to c. [ marks] (ii) Consider the function f : R R defined by cos if x, f(x) = x if x =. Use your definition in (i) above to show that limit of f(x) as x tends to does not exist. Solution: (i) A real number l is said to be a limit of f(x) as x tends to c if the following condition holds: {x n } sequence in R \ {c} and x n c = f(x n ) l.... [] Aliter: A real number l is said to be a limit of f(x) as x tends to c if for every ɛ >, there exists δ > such that < x c < δ = f(x) l < ɛ.... [] [Note: Deduct mark if in the sequence definition it is not specified that {x n } is sequence in R \ {c} or that x n c for all n N. Likewise, deduct mark if in the ɛ-δ definition, the inequality < x c or the stipulation x c is missing.] (ii) We will show that there is no real number l such that limit of f(x) as x tends to is l. Consider the sequences {x n } and {y n } defined by x n = nπ and y n = (n + )π for n N.... [] It is clear that {x n } and {y n } are sequences in R\{} such that x n and y n, whereas f(x n ) = and f(y n ) = for all n N. Hence there can be no real number l such that a limit of f(x) as x tends to is l. In other words, limit of f(x) as x tends to does not exist.... [] Aliter: Clearly, {/nπ} is a sequence in R \ {} that converges to.... [] But f(/nπ) = ( ) n for n N, and hence the sequence {f(/nπ)} is not convergent. So a limit of f(x) as x tends to does not exist.... [] Aliter: Suppose, if possible limit of f(x) as x tends to does exist. Then there is l R satisfying the ɛ-δ definition above. In particular, for ɛ =, there is δ > such that < x < δ = f(x) l < ɛ. Let n N be such that n > π/δ so that < δ for every m n. Now, mπ ( ) ( ) = cos nπ cos(n + )π = f f nπ (n + )π ( ) ( ) f l nπ + f l (n + )π < ɛ + ɛ =,... [] which is a contradiction. Thus, limit of f(x) as x tends to does not exist.... []

3 3. Find all points (x, y ) on the curve y = x 3 + x x 3 such that the tangent line to the curve at (x, y ) passes through the point (, 3). [5 marks] Solution: The slope of the tangent line to the curve at the point (x, y ) is and must be equal to 3x + x y 3 x.... [] From 3x + x = y 3 x and y = x 3 + x x 3, one obtains x 3 5x x + =, which is (x 5)(x ) =.... [] ( 5 Thus the required points are:, ), (, ), (, ).... [] 8. Suppose f : R R is twice differentiable and satisfies f(x) and f (x), for all x R. Show that f is a constant function. [5 marks] Solution: Since f (x) for all x R, f is increasing on R.... [] We will show that f (x) = for all x R, which implies that f is a constant function. Suppose f (q) for some q R. Then f (q) > or f (q) <. Case : f (q) >. For any r > q, there exists by Lagrange s MVT some c (q, r) such that f(r) f(q) = f (c)(r q). Since c > q, we have f (c) f (q) so that f(r) f(q) + f (q)(r q). Then f(r) > for large positive r, which is a contradiction.... [] Case : f (q) <. For any p < q, there exists by Lagrange s MVT some c (p, q) such that f(q) f(p) = f (c)(q p). Since c < q, we have f (c) f (q) so that f(q) f(p) + f (q)(q p), that is, f(p) f(q) f (q)(q p). Then f(p) > for large negative p, which is a contradiction.... []

4 5. (i) Consider the function f : (, ) R defined by f(x) = x /x. Find the intervals in (, ) where f is increasing/decreasing and find the points in (, ), if there are any, where (i) f has a local minimum, and (ii) f has a local maximum. Justify your answer. (ii) Let a, b R be such that a > b >. Consider f : (, ) R defined by f(x) = a x + b x for x >. Find the intervals in (, ) where f is convex/concave and find the points in (, ), if there are any, where f has a point of inflection. [ marks] Solution: (i) We have ( ) ln x f(x) = x /x = exp x for x >. Hence f is differentiable on (, ) and for any x (, ), ( ) ( ln x f (x) = exp x x ln x ) = x/x ( ln x).... [] x x Since exp(y) > for all y R, it follows that for any x (, ), f (x) = ln x = x = e. Moreover, f (x) > x < e and f (x) < x > e. So it follows that f is (strictly) increasing on (, e) and strictly decreasing on (e, ).... [] Moreover, from the First Derivative Test, we see that f has a local maximum at x = e and no local minimum.... [] (ii) It is clear that f is twice differentiable on (, ) and for x >, f (x) = ( b a ) x x and f (x) = ( 3a x x x b ) x = x x x (3a bx). Hence f (x) > for x (, 3a/b) and f (x) < for x (3a/b, ). Consequently, f is convex on (, 3a/b) and concave on (3a/b, ).... [] Moreover, since f changes from being convex to concave at x = 3a/b, we conclude that f has a point of inflection at x = 3a/b.... []

5 6. (i) Determine if the following limit exists. If yes, find it. ( ) lim n n + i3 n + i. [ marks] 5 n 5 (ii) Let f : R R be continuous. For x R, let g(x) = i= f(t) sin(x t)dt. Show that the function g : R R is twice differentiable and also that g (x) = f(x) g(x) for all x R. Solution: (i) Observe that ( ) n + i3 n + i = + 5 n 5 n i= i= ( ) 3 i + n n i= ( ) i = + n n n S(P n, f)+s(p n, g),... [] where f, g : [, ] R are defined by f(x) = x 3 and g(x) = x for x [, ], and where P n = {, /n, /n,..., } denotes the equidistant partition of [, ] into n parts, and S(P n, f), S(P n, g) denote, respectively, the approximate Riemann sums for f, g with respect to P n with the functions being evaluated at the right hand endpoints of each subinterval of [, ] corresponding to P n. Riemann integrable (being continuous), it follows that S(P n, f) f(x)dx = i= and S(P n, g) Since f and g are g(x)dx = 5 as n. Also, /n as n. Hence the desired limit exists and ( ) lim n n + i3 n + i = n 5 5 = [] (ii) Expanding sin(x t) as sin x cos t cos x sin t, we see that g(x) = sin x f(t) cos t dt cos x f(t) sin t dt.... [] Since f is continuous on R, it follows from the Fundamental Theorem of Calculus and the Product Rule that g is differentiable and g (x) = cos x = cos x f(t) cos t dt + sin x [f(x) cos x] + sin x f(t) cos t dt + sin x f(t) sin t dt cos x [f(x) sin x] f(t) sin t dt.... [] Again, since f is continuous, it follows from the Fundamental Theorem of Calculus and the Product Rule that g is differentiable, i.e., g is twice differentiable and g (x) = cos x [f(x) cos x] sin x f(t) cos t dt + cos x f(t) sin t dt + sin x [f(x) sin x] = g(x) + [ cos x + sin x ] f(x)... [] = f(x) g(x). [Deduct mark if there is no mention of the Fundamental Theorem of Calculus.]

6 7. Consider the planar region D bounded by the curves y = e x and y = ln x, and the vertical lines x = and x = e. (i) Find the area of D. [ marks] (ii) Find the volume of the solid generated by revolving the planar region D about the x-axis. Solution: (i) Clearly, f, g : [, e ] R defined by f(x) = e x and g(x) = ln x are Riemann integrable and satisfy f(x) g(x) for all x [, e ]. Hence Area(D) = e (f(x) g(x)) dx = e (e x ln x) dx... [] = [e x (x ln x x)] e ( ) = e e e ( e e + ) = e e e e.... [] (ii) Using the Washer Method, we see that the volume of the solid generated by revolving D about the x-axis is given by e π ( f(x) g(x) ) e dx = π = π ( e x (ln x) ) dx... [] [ e x = π ( x(ln x) (x ln x x) ) ] e [( ) e e ( e (e e + ) )] = π [ e e e 5e + ]... [].... [] Aliter: We note that D can also be viewed as the region between the curves x = u(y) and x = v(y) for y e e, where u, v : [, e e ] R are defined by { { if y e, e y if y, u(y) = and v(y) =... [] ln y if e y e e. e if y e e. Moreover, v(y) u(y) for all y [, e ]. Hence, using the Shell Method, we see

7 that the volume of the solid generated by revolving D about the x-axis is given by e e πy [v(y) u(y)] dy { = π y (e y ) dy + = π [ye y e y y ] e { [e = π ] [ e + 5e [ ] e e = π 5e + = π [ e e 5e + ] + e y ( e ) dy + ] + [ ] y e + e + [ e e e e e e y ( e ln y ) dy [ ] [ y e e y e ] [ e e e e } ln y y e ] e e e ee e ]}... [].... []

8 8. Consider the curve C in the plane given by the parametric equations x(t) = 3 t and y(t) = t t 3 for t [ 3, 3]. Show that C is a smooth curve that intersects the x-axis in two distinct points and at one of these points, C intersects the x-axis for two distinct values of the parameter t, thus forming a loop. Find the arc length of this loop. [5 marks] Solution: Clearly, x, y : [ 3, 3] R are differentiable with x (t) = 3 t and y (t) = 3t for t [ 3, 3] so that x and y are continuous on [ 3, 3]. Thus C is a smooth curve. Moreover, intersection with the x-axis corresponds to y(t) = = t = or t = ±... [] and since x() = while x(±) = 3, it follows that C intersects the x-axis at two points, namely, the origin O = (, ) and the point P = ( 3, ). Moreover, C intersects the x-axis at the point P for two distinct values of the parameter t, viz., t = and t =.... [] The loop formed by C is given by (x(t), y(t)) as t varies from to. Hence its arc length is given by = = = + x (t) + y (t) dt... [] t + 6t + 9t dt ( + 3t ) dt... [] =.... []

MA CALCULUS. Marking Scheme

MA CALCULUS. Marking Scheme MA 05 - CALCULUS Midsemester Examination (Autumn 06-07) Marking Scheme Department of Mathematics, I.I.T. Bombay. Max. Marks: 30 Time: 5:00 to 7:00 PM Date: 06/09/06 Q. (i) Show that lim n n /n exists and