Marking Scheme for the end semester examination of MTH101, (I) for n N. Show that (x n ) converges and find its limit. [5]
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1 Marking Scheme for the end semester examination of MTH, 3-4 (I). (a) Let x =, x = and x n+ = xn+x for n N. Show that (x n ) converges and find its limit. [5] Observe that x n+ x = x x n [] The sequence satisfies the Cauchy criterion and hence it converges. Note that x n+ + x = x + xn. [] If x n l then l + l = and hence l = 3. (b) Suppose that f : [, ] R is three times differentiable s.t. f( ) =, f() = and f () =. Show that there exists c in (, ) such that f (c) 3. [5] By Taylor s Theorem f() = f() + f () + f () + f (c ) for some c (, ) [] and f( ) = f() f () + f () f (c ) for some c (, ) Therefore f (c )+f (c ) =. Hence either f (c ) or f (c ) 3. (c) Let f : [, ] R be continuous and a n = f(sin n ) f(sin i. Show that n= a n converges and find its sum/limit. ) for n =,,... ii. If f is differentiable and f (x) < for all x [, ], discuss the convergence/divergence of n= a n. [] i. The partial sum S n = f() f(sin Since f(sin ) which converges. [] ) f(), the sum of the series is f() f() =. ii. By MVT, f(sin n ) f(sin ) sin n sin n. [] Since a n n(), by comparison test, a n converges.. (a) Let a n = n n 3 +, n N. Find the largest term of the sequence (a n). [5] If f(x) = x x 3 +, x > then f (x) = x(4 x3 ). [] (x 3 +) The point of maximum for f is 4 3. Since 7 < 4 3 < 8 and a 7 = > a 8 = 8 89, a 7 is the largest term []
2 (b) Find lim n n 8 n k= k. [4] [ n Note that n n 8 k= k = n n k= ( k n )] [] and n n k= ( k n ) x dx. Therefore lim n n n 8 k= k =. (c) Let f : [, ] R be twice differentiable such that f(x)dx < f( ). Show that there exists x [, ] such that f (x ). [7] Suppose f (x) > for all x [, ]. [] Then by Taylor s Theorem, f(x) f( ) + f ( )(x ) for all x [, ]. [3] This implies that f(x)dx f( ) + f ( ) f ( ) = f( ). [] which is a contradiction. 3. (a) Using the Riemann criterion show that every increasing function on [, ] is integrable. [5] Let f[, ] R be increasing. For n N, consider the partition P n = {x, x, x,..., x n }, where x i = i n. Then M i = sup{f(x) : x [x i, x i ]} = f(x i ) and m i = inf{f(x) : x [x i, x i ]} = f(x i ). [] Then U(P n, f) L(P n, f) = n n i= [f(x i) f(x i )] = n [f() f()]. [] By the Riemann Criterion the function is integrable. (b) Derive an equation for the surface generated by revolving the curve 4x + 9y = 3, z = around the y-axis. [5] Let P (x, y, z) be any point on the surface. Consider a point Q = (x, y, ) on the curve for some x. [] Note that the distance from Q to the y-axis and the distance from P to the y-axis are the same. Therefore x = x + z. An equation of the surface is 4(x + z ) + 9y = 3.
3 (c) Find a point on the curve y = e x at which the curvature is maximum. [] Observe that κ(x) = f (x) [+(f (x)) ] 3 = e x (+e x ) 3 and κ (x) = ex (+e x ) ( e x ). (+e x ) 3 [] The curvature is maximum at ( ln, ). [] [] 4. Consider the function f(x, y) = 3x y y 3 x +y for (x, y) (, ) and f(, ) =. (a) Verify whether f is continuous at (, ). [3] f(x, y) f(, ) y 3x y x +y y 3x +3y x +y 3 y as (x, y) (, ). [3] (b) Find the directional derivatives of f at (, ) in the directions (, ), (, ) and (, ). [4] f(t,) f x (, ) = lim t t = f(,t) f y (, ) = lim t t = Df (,) ( f( t (,)) (, )) = lim t t Therefore Df (,) ( (, )) = (c) Using (b) (NOT the definition of differentiability) verify whether f is differentiable at (, ). [4] If f is differentiable at (, ), then Df (,) ( (, )) = (f x, f y ) (,) (, ) [3] But Df (,) ( (, )) = and (f x, f y ) (,) (, ) = (d) Evaluate f y (x, ) for x. [3] f(x,t) f(x,) Since f y (x, ) = lim t t [] f y (x, ) = 3. (e) Verify whether f y is continuous at (, ). [] Note that f y (x, ) = 3 f y (, ) = as x. []
4 5. (a) Let S be the sphere x + y + z =. Evaluate the surface integral I = S (x y + z + 3e z x e x y + z cos y)dσ. [] Note that I = [ ] S x + 3e z, y e x, z + cos y (x, y, z)dσ [] [ ] n = (x, y, z) is the unit normal to S and let F (x, y, z) = x + 3e z, y e x, z + cos y. Hence I = S F ndσ. [] By divergence Theorem I = D divf dv where D is the solid sphere. Therefore I = 4. (b) Evaluate the line integral I = C zdx + (x + ey )dy + (y + e z )dz where C is the curve which is the intersection of the plane y + z = and the cylinder x + y =. Orient C counterclockwise as viewed from above. [] By Stokes Theorem I = S (curlf ) ndσ [] where S is the portion of the plane lying inside the cylinder, F (x, y, z) = (z, x + e y, y + e z ) and n is the unit outward normal to the said portion. Note that curlf = i + j + k and n = (,, ) Hence I = R + fx + fy dxdy where R := x + y and f(x, y) = y. Therefore I = (c) Consider the circle in the yz-plane with center y = 5, z = and radius 3. Let S be the surface obtained by rotating this circle about the z-axis. [] i. Find a parametric equation/representation to describe this surface with one parameter θ, where θ is described below. If (x, y, z) is any point on the surface then θ is the angle between the x-axis and the line joining (,, ) and (x, y, ). x = (5 + 3 cos φ) cos θ, y = (5 + 3 cos φ) sin θ, z = 3 sin φ [] where θ, φ and φ is the angle between the line joining (x, y, z) and the center of the moving circle (which contains (x, y, z)) with the xy-plane ii. Set up a single integral (with one variable) to find S zdσ. S zdσ = (3 sin φ) EG F dθdφ. Since EG F = 3(5 + 3 cos φ) S zdσ = 8 (sin φ)(5 + 3 cos φ)dφ.
5 . (a) Find the points of absolute maximum and absolute minimum of the function f(x, y) = x + y x + on the region {(x, y) : x + y 4 with y }. [] f x = x = and f y = y = There is no critical point in the interior of the region. On the curve x + y = 4, y, the function is x + 4 x x + = x + The candidates for the points of maxima/minima are (, ) and (, ). On the line segment joining (, ) and (, ), the function is x x + and the critical point is (, ). Since f(, ) = and f(, ) = and f(, ) =, the point of maximum is (, ) and the point of minimum is (, ). (b) Find the volume of the solid that lies above the cone z = x + y and below the sphere x + y + z =. [] Solving z = x + y and x + y + z = we get x + y =. The required volume is I = R ( x y x + y )dxdy [] where R = {(x, y) : x + y }. Use polar co-ordinates to get I = ( r r)rdθdr [] The value of I = 3 [ ] (c) Let D be the solid cone bounded below by z = x + y and above by z =. Convert the integral D zdv as iterated integrals of the form b d f a c e g(ρ, θ, φ)dρdφdθ for some a, b, c, d, e, f, g where ρ, φ and θ are spherical co-ordinates. [] 4 cos φ ρ cos φ(ρ sin φ)dρdφdθ The required integral is a =, c = and e = b = and d = 4 [] f = cos φ [] g(ρ, θ, φ) = ρ 3 cos φ sin φ
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