MA CALCULUS. Marking Scheme

Transcription

1 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 find it. [3] Solution: Method : For n >, n /n >. So write n /n = + h n, where h n > 0. Then for n >, (n = ( + h n ) n = + nh n + Therefore n(n ) h n < n which means 0 < h n < ( ) n h n + some positive terms. [] n By Sandwich theorem lim n h n = 0. [] Therefore lim n n /n =. [] Method : lim n n /n = lim x x /x (if any one of the two limits exist.) [] By taking logarithm, we get the required limit exists if ln x lim x x /x exists if (by H Lospital s rule) lim x exists. The last limit exists and is equal to 0. Therefore the original limit exists and is equal to e 0. Therefore lim n n /n =. [] []

2 (ii) Prove that the sequence : a =, a n+ = /(3 a n ), n is monotone and bounded. Find its limit. [3] Solution: a = a =. Therefore a > a. Inductively, assume that a n > a n+. This is equivalent to say iff 3 a n < a n (3 a n )a n > 0 Therefore in order to show that a n+ > a n+ we need to show that (3 a n+ )a n+ > 0 Now 3 a n+ )a n+ = (3 3 a n ) 3 a n = 3 an)an (3 a n) > 0 and the numerator is positive by induction hypothesis. Therefore a n > a n+ This proves that a n is decreasing. [] Therefore a n < a = for all n >. Therefore a n+ > 0 also. This proves that {a n } is bounded. [] Therefore, the limit exists. Let this limit be equal to s. Then 0 s <. Upon taking the limit in the formula it follows that a n+ = 3 a n s = 3 s. Therefore s 3s + = 0. s = 3 5. []

3 Alternatively: a =, a =. Assuming a n (0, ] we see that 0 < 3 a n <. Therefore, by induction, a 0 < a n for all n. [] Now for n, a n+ a n = a n a n (3 a n )(3 a n+ ) Since a > a > a 3 by induction {a n } is decreasing [] Note that, by sandwich theorem, the limit l has to be and can be calculated from l = 3 l Answer: l = 3 5 [] 3

4 Q. (i) Let f : R R be a function such that f(/n) = /n for all natural numbers n. (a) If f is differentiable, then find f (0). [] Solution: By continuity we have f(0) = lim n f(/n) = 0. [] Now, f (0) = lim x 0 f(x) f(0) x = lim n f(/n) /n =. [] (b) If f is twice differentiable, determine f (0)? [] Solution: By Lagrange s MVT there exists x n (/n, /n ) such that f (x n ) = f(/n ) f(/n) /n /n =. [] Therefore f (0) = lim x 0 f (x) f (0) x = lim n f (x n) x n = 0. [] (ii) Let f : R R satisfy f(x + y) = f(x) f(y) for all x, y R. If f is differentiable at 0, then show that f is differentiable at every c R and f (c) = f (0) f(c). [] Solution: f(x) = f(0)f(x) for all x. Therefore if f(0) = 0 then the function is indentically zero and the required property follows easily. So, let us assume that f(0) 0. But then f(0) = f(0 + 0) = f(0)f(0) implies that f(0) =. [] Now assume that f is differentiable at 0. Then f (c) = lim h 0 f(c+h) f(c) h = f(c) lim h 0 f(h) h = f(c)f (0). [] 4

5 Q.3 (i) Let f be continuous on [a, b] and differentiable on (a, b). If f(b) f(a) = b a, then prove that for every natural number n, there exists distinct points c,, c n (a, b) such that n j= f (c j ) = n. [] Solution: Cut [a, b] into n equal parts: put x j = a + j b a. Apply Lagrange s n MVT to the function on each interval [x i, x i ]. We get points c j (x i, x i ) such that f (c j ) = f(x i) f(x i (b a)/n. [] Clearly then c j are distinct. Also, j f (c j ) = n b a j (f(x i) f(x i )) = n (f(b) f(a)) = n. [] b a (ii) Let f : [0, ] R be such that its second derivative f exists for all x [0, ] with f. If f(0) = f(), then show that f (x) <. [] Solution: By MVT (Rolle s or Lagrange s) there exists c (0, ) such that f (c) = 0. Now let x (0, ) be point, x c. Again by Lagrange s theorem applied to f, there exists y between c and x such that f (y) = f (x) f (c) x c Therefore = f (x) x c. [] f (y) = f (x) x c > f (x) Therefore for all x (0, ), f (x) <. [] (iii) Find a curve through (, ) whose length between the lines x = and x = 4 is 4 given by the integral + dx. [] 4x Solution: Consider the function y := f(x) := x, x 4. [] Let the curve C be the graph of this function. Then clearly it passes through (, ). The arc-length is given by L(C) = 4 + (f (x)) dx = 4 + dx. [] 4x 5

6 Q.4 (i) Compute lim n n n cos jπ n. [] j= Solution: Cut the interval [0, π] into n equal parts and choose points c j = jπ/n. Since the norm of this partition tends to 0, by a theorem on Riemann sequences, we have π 0 cos t dt = lim n n cos jπ n. [] (ii) Sketch the graph of the function f = 3x x, x (, 5) by locating the interval of increase/decrease, interval of concave up/ down, points of local maxima/minima. [5] Solution: f(x) = 3x x, x (, 5) Since there is a term involving x, we cut the interval into two parts (, 0] and [0, 5) and discuss them separately. On [, 0], we have f(x) = 3x + x. f() = 7, f(0) = ; f( ) = 0. f (x) = 6x + ; f (x) = 0 iff x = /3. f( /3) = 4/3. f (x) = 6 > 0; In particular f ( /3) > 0. Therefore /3 is a local minimum point. f is strictly increasing and hence f is strictly convex upwards. On [0, 5], we have f(x) = 3x x. f(0) =, f(5) = 64, f() = 0. f (x) = 6x which is strictly increasing; f (x) = 0 only at x = /3. f (x) = 6 > 0 thereofore, f is strictly convex. minimum. Also x = /3 is a point of local f (/3) = 4/3. 6

7 Putting the informations from the two intervals together, 4/3 is the global minimum attained at the point x = ±/3 and f has no maximum value in (, 5). Clearly 0 is the only point at which the function is not differentiable.. Marking Scheme f is decreasing in (, /3) (0, /3) and increasing in ( /3, 0) (/3, 5) [] Local minima at x = ±/3 and local maximum at x = 0. [] (If this is shown in the sketch clearly, then award this mark) f is concave up (some people may write convex up) in (, 0) (0, 5) [] Reasonably good sketch [] 7

8 Q.5 (i) Let S be the solid obtained by revolving the region lying in the first quadrant and bounded by the parabolas x = y and x = y around the y-axis. Compute the volume of S using Washer and Shell methods. [4] Solution The point of intersection of the two curves in the first quadrant is given by y = y, i.e., (, ) The other points on the boundary are (0, 0) and (, 0). Washer Method: The area of the washer is: A(y) = π[( y ) y 4 ] = π(4 4y ). [] Therefore the volume is given by V = π 0 (4 4y ) dy = 8π 3. [] Shell Method: Therefore the volume is given by Area of the cylindrical shell is πx 3/ 0 x A(x) = πxy = πx( x) x. V = π[ 0 x3/ dx + x( x) dx] = π[ 0 x3/ dx + 0 ( t)t/ dt] = 4π 0 t/ dt = 8π 3. [] (ii) Find the surface area of the surface of revolution obtained by revolving the curve : x = t, y = 4 3 (t + )3/, 0 t around the y-axis. [] Solution: The area of the surface of revolution is A = 0 πx(t) (ẋ(t)) + (ẏ(t)) dt [] = π 0 t [t + 4(t + )] dt = π 0 [t (t + ) dt = π. [] [] 8

9 PAPER ENDS 9