Real_Analysis_DPP_32_Of_Lec_42_Saakaar_2.0_Batch_for_IIT_JAM_2024_Mathematicssaakaar-2-0-batch-for-iit-jam-2024-mathematics-327842mathematics-781061real-analysis-564326.pdf

1 SAAKAAR 2.0 BATCH Real Analysis DPP-3...

1 SAAKAAR 2.0 BATCH Real Analysis DPP-32  x 2 ; x   6. Let f : [a, b] → R be a continuous function and let 1. Consider the function f ( x )   , then f(a) < f(b). Then by intermediate value theorem 2 x ; x   c (1) f([a, b]) = [f(a), g(b)] (1) 𝑓 is continuous at exactly 2 points (2) f([a, b])  [f(a), g(b)] (2) 𝑓 is continuous at exactly three points (3) f([a, b]) ⊆ [f(a), g(b)] (3) 𝑓 is continuous at infinitely many points (4) f([a, b]) ≠ [f(a), g(b)] (4) 𝑓 is no where continuous 7. Let f: R →R be defined by f(t) = t2 and let U be any 2. Consider the function 𝑓 : [3,4] → , non-empty open subset of R. Then  x 2 ; x   3, 4  (1) f(U) is open f ( x)   , then (2) f1(U) is open  2 x ; x  c  3, 4 (3) f(U) is closed (1) 𝑓 is continuous at exactly 2 points (4) f1(U) is closed (2) 𝑓 is continuous at exactly three points (3) 𝑓 is continuous at infinitely many points 8. Let f:[0,1] → R be the continuous function defined (4) 𝑓 is no where continuous by f ( x ) = ( x − 1)( x − 2 ). Then the maximal subset  x ; x   ( x − 3)( x − 4 ) 3. Consider the function f ( x ) =  c , then of R on which f has a continuous extension is 0 ; x   (1) (–∞, 3) (1) 𝑓 has unique point of continuity (2) (–∞, 3) ∪ (4, ∞) (2) 𝑓 has uncountable number of points of (3) R \ {3,4} discontinuity (4) R (3) 𝑓 is discontinuous at all rational points (4) 𝑓 is discontinuous at all irrational points 9. The continuous function f : R → R defined by f(x) = (x2 + 1)2003 is 4. Let 𝑓:  → , be a continuous function such that (1) Onto but not one-one 𝑓(𝑥) − 2 = 0 is true for all 𝑥 ∈ {𝑚 + 𝑛 2 : 𝑚, 𝑛 ∈ (2) One-one but not onto }, then f ( 5)= (3) Both one-one and onto (4) Neither one-one nor onto (1) Must be an even prime number (2) Can be any prime number 10. Let f : R → R be a continuous function. If f(Q) ⊆ (3) Can not be an even integer N, then (4) Can not be an odd integer (1) f(R) = N 5. Let 𝑓:  → , be a continuous function. Define a (2) f(R) ⊆ N but f need not be constant (3) f is unbounded A set 𝐴 = {𝛼 ∈ : 𝑝 𝛼 = 0, for some non-constant (4) f is a constant function polynomial 𝑝(𝑥) with rational coefficient}. Suppose 𝑓(𝑥) = 0 for all 𝑥 ∈ 𝐴, Then (1) 𝑓 must be a periodic function (2) 𝑓 must be an even function (3) 𝑓 must be an odd function (4) 𝑓 must be a bounded function  2 Answer Key 1. (1) 2. (4) 3. (1,2 4) 4. (1, 4) 5. (1, 2, 3, 4) 6. (2) 7. (2) 8. (3) 9. (4) 10. (4)  3 Hint & Solutions 1. (1) 6. (2) The only points of continuities of 𝑓 is 𝑥 such that Let y  [f (a), f (b)]  f(a)  y  f (b) 𝑥2 − 2𝑥 = 0, and hence 𝑥 (𝑥 – 2) = 0 or that 𝑥 = 0,2 and hence there are exactly two points where the   x  [a, b] such that y = f(x) (by IVP) given function is continuous.  y  f ([a, b])  [f (a), f(b)]  f ([a, b]) 2. (4) For (1) and (3) define The only points of continuities of 𝑓 is 𝑥 such that 𝑥2 − 2𝑥 = 0, and hence 𝑥 (𝑥 – 2) = 0 or that 𝑥 = 0,2 and  4x, 0  x  1 / 2 f (x) =   since 0,2 ∉ [3,4], therefore 𝑓 is not continuous at 3 − 2x, 1 / 2  x  1  any point.  f(0) = 0 < f(1) = 1 3. (1,2, 4) But f(1/2) = 2[f(0), f(1)] = [0, 1] The only point where the given function is (4) f:[0, 1] → , f(x) = x  f([0, 1]) = [f(0), f(1)] continuous is given by 𝑥 = 0, and hence the set of points of discontinuities is  − {0}, which is an 7. (2) uncountable set. Also 𝑓 is discontinuous at all irrational points. (1) f(t) = t2 is a continuous function and under a continuous map f–1 (U) is open in  for every 4. (1, 4) open subset U of . Since we know that the set 𝑆 = {𝑚 + 𝑛 2 : 𝑚, 𝑛 ∈ } is s dense subset of , and the given function is (2) Take U = , the U is open in  continuous on  therefore 𝑓 (𝑥) − 2 = 0 for all 𝑥 ∈ but f(U) = [0, ) which is not open. , and hence 𝑓 (𝑥) = 2 for all 𝑥 ∈ . (3) Take U = (0, 1)  f(U) = (0, 1) Therefore 𝑓 ( 5) = 2 8. (3) Polynomials are continuous, given function is not 5. (1, 2, 3, 4) define only on 3 and 4. Given function is continuous on , and it vanishes 9. (4) at all algebraic numbers hence it must vanish on . Range of function does not attain negative values [i.e., not onto] Therefore 𝑓(𝑥) = 0 for all 𝑥 ∈  f(–1) = f(1) [ i.e., not 1–1] Which periodic, bounded, even, odd 10. (4) For (a) and (b): under continuous map image of interval is interval, for (c) take constant function. PW Web/App - https://smart.link/7wwosivoicgd4 Library - https://smart.link/sdfez8ejd80if