Indian Institute of Information Technology, Allahabad. End Semester Examination - Tentative Marking Scheme

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1 Idia Istitute of Iformatio Techology, Allahabad Ed Semester Examiatio - Tetative Markig Scheme Course Name: Mathematics-I Course Code: SMAT3C MM: 75 Program: B.Tech st year (IT+ECE) ate of Exam:..7 ( st Sessio) Attempt all parts of questios, 6 & together. Numbers idicated o the right i [ ] are full marks of that particular questio. All otatios are stadard ad same as used i lectures. Be precise i your aswer. o ot write o the questio paper ad cover pages of your aswer booklet except your details. Use of calculator is ot allowed. This questio paper has two pages.. Give a short aswer to the followig questios with justificatio. [ 6] (a) Fid the supremum ad ifimum of the set {x + : x > }. [+] x Solutio. The set is ot bouded above. if. (b) For ay ratioal umber r, show that there exists a sequece of irratioal umbers covergig to r. Solutio. Let x r + R \ Q x r. (c) Fid the values of x for which the power series x coverges. Solutio. Sice x /, by the root test the series coverges for all x R. (d) oes the poit (, π/) lies o the curve r cos θ. Solutio. Yes. The poit (, π/) (, π/) ad (, π/) satisfies the equatio r cos θ. (e) What does the equatio θ π/4 represets i spherical coordiates. Solutio. The equatio θ π/4 represets a half-plae, higed alog the z-axis, makig a agle of π/4 radias with the positive x-axis. [] (f) Let f : R R be defied by f(x, y) xy. oes the partial derivative of f with respect to first variable at the poit (, ) exists. Solutio. Yes. f f(h,) f(,) x (,) lim h h lim. h h. Let J { : N}. Show that every fuctio f : J R is cotiuous o J. [5] Solutio. Let a be fixed, ad ɛ >. Choose δ > such that δ < k k k+ [] The for ay x J such that x a < δ x a []

2 f(x) f(a) < ɛ. Therefor, f is cotiuous at a, ad hece o J. 3. Let f : (, ] R be differetiable with f (x) <. efie a sequece a : f( ). Show that (a ) coverges. [4] Solutio. ( ) is coverget ad hece Cauchy. By Mea Value Theorem, c such that f( m f (c)( ) m < as, m. m Therefore, (a ) is Cauchy. Therefore, (a ) is coverget. 4. Usig Taylor s theorem, show that si x Solutio. By Taylor s theorem, c betwee ad x such that ( ) ( + )! x+, x R. [5] si x x x3 3! + x5 5! + + ( ) x + + f (+) (c) ( + )! ( + )! x + [] Sice f () (x) for all N ad x R, E (x) f (+) (c) x + x + ( + )! ( + )! (By ratio test) Therefore, E (x), ad hece si x 5. Prove that ( ) ( + )! x+. si x dx coverges but ot absolutely for < p. [8] Solutio. Proof of covergece of Let for < p. The Moreover, si x is cotiuous, ad By irichlet s Test, the itegral Proof of divergece of Sice si x si x, si x By irichlet s Test, But si x si x dx: is decreasig ad lim. x x si t dt for x >. dx: si x si x dx coverges. cos x. cos x dx coverges for all p >. dx diverges for p.

3 Hece, cos x dx diverges. Therefore, by compariso test si x dx diverges. 6. Cosider the curve C defied by x(t) cos 3 (t), y(t) si 3 (t), t π. [3+3+] (a) Fid the legth of the curve. Solutio. The legth L π/ 3 cos t si t dt π/ x (t) + y (t) dt 3. (b) Fid the area of the surface geerated by revolvig C about the x-axis. Solutio. The surface area S 6π π/ si 4 t cos t dt π/ πy(t) x (t) + y (t) dt 6π 5. (c) If (x, y) is the cetroid of C, the fid y. Solutio. By Pappus theorem, S πyl, 6π πy 3 or y Fid the uit taget vector, pricipal ormal ad curvature for the curve R(t) ( cos t, si t, si t), t R [6] Solutio. The uit taget vector T (t) The pricipal ormal vector N(t) The curvature is κ(t) R (t) R (t) ( si t, cos t, cos t). [] T (t) T (t) ( cos t, si t, si t). [] T (t) R (t). [] 8. Assume that amog all rectagular boxes with fixed surface area of square meters, there is a box of largest possible volume. Fid its dimesios. [7] Solutio. Let the box have sides of legth x, y, z >. The the volume V (x, y, z) xyz ad xy + yz + xz. Let g(x, y, z) xy + yz + xz By Lagrage multipliers method, V (x, y, z) λ g(x, y, z) (yz, xz, xy) λ(y + z, x + z, y + x). yz λ(y + z), xz λ(x + z) ad xy λ(x + y). [3] Solvig we get x y z. Therefore, x 3 Alterate Solutio. Let the box have sides of legth x, y, z >. The the volume V (x, y, z) xyz 3

4 ad xy + yz + xz. Let g(x, y, z) xy + yz + xz z(x + y) xy or z xy x+y. Let f(x, y) xy( xy) x+y f x (x, y) y ( xy x ) (x+y), f y (x, y) x ( xy y ) (x+y). For critical poits, f x (x, y) xy y. ad f y (x, y) xy x Solvig above two equatios, we get x y or x y. Puttig x y i xy x we get x y. 3 z xy x+y Usig chage of variables, evaluate y x + y (y x) dx dy. [9] Solutio. The regio R of the itegratio is bouded by the lies x, y ad x + y. Let u x + y ad v y x. [+] Solvig above equatios i x, y we get x u v ad y u+v. [/+/] 3 3 The Jacobia J(u, v) is X Y /3 /3 J(u, v) u X v u Y v Now, for the regio T i the uv-plae we have /3 /3 /3. x v u, [/] y v u, [/] ad x + y u [/] So, T is bouded by the lies u, u v ad v u. y Therefore, x + y (y x) u dx dy u v dv du [+] 3 9. (a) Write the itegral 4 y u [/] dz dy dx i the order dx dy dz. [6] Solutio. 4 x y x dz dy dx z y dx dy dz. [6] 4

5 (b) Let be the solid that lies iside the cylider x + y, below the coe z 4(x + y ) ad above the plae z. Evaluate x dx dy dz. [5] Solutio. Usig Cylidrical coordiates: The projectio of the solid o the xy-plae is the circular disk {(x, y) R : x + y }. We use cylidrical coordiates. x r cos θ, y r si θ, z z. The Jacobia J(r, θ, z) r. The solid is bouded by z ad z r. π r Therefore, x dx dy dz r cos θ r dz dr dθ. [3] π/5. Usig Spherical coordiates: ρ varies from to cosec φ φ varies from ta (/) to π/. θ varies from to π. The Jacobia J(ρ, θ, φ) ρ si φ. The solid is bouded by z ad z r. π π/ cosecφ Therefore, x dx dy dz ρ si φ cos θ ρ si φ dρ dφ dθ ta (/) π/5. 5

Assignment 1 : Real Numbers, Sequences. for n 1. Show that (x n ) converges. Further, by observing that x n+2 + x n+1

Assignment 1 : Real Numbers, Sequences. for n 1. Show that (x n ) converges. Further, by observing that x n+2 + x n+1 Assigmet : Real Numbers, Sequeces. Let A be a o-empty subset of R ad α R. Show that α = supa if ad oly if α is ot a upper boud of A but α + is a upper boud of A for every N. 2. Let y (, ) ad x (, ). Evaluate