Hints/Solutions for Tutorial Problem Set - 3 1. Let x ∈ R \ Q. Then there exists a sequence (r n ) in Q such that r n → x. So f(r n ) = r n → x = [x] = f(x). Hence fis not continuous at x. Again, let y ∈ Q. Then there exists a sequence (t n ) in R \ Q such that t n < y for all n ∈ N and t n → y. For each n ∈ N, f(t n ) = [t n ] ≤ y − 1 ify ∈ Z, [t n ] ≤ [y] < y ify ∈ Z. In either case f(t n ) → f(y) = y. Hence fis not continuous at y. Therefore fis not continuous at any point ofR. 2. Let x > 0. By hypothesis f(x) = f(x 1/2 ) = f(x 1/4 ) = ··· = f(x 1/2 n ) for all n ∈ N. Since x 1/2 n → 1 (as (x 1/2 n ) is a subsequence of (x 1/n ) and x 1/n → 1) and since fis continuous at 1, f(x 1/2 n ) → f(1). It fol lo ws that f(x) = f(1). Al so f(−x) = f((−x) 2 ) = f(x 2 ) = f(x). Hence f(x) = f(1) for all x(= 0) ∈ R. Si nce fis continuous at 0, f(0) = lim x→0 f(x) = f(1). Thus f(x) = f(1) for all x ∈ R. Conse- quently f: R → R is a constant function. 3. (i) Let g(x) = f(x + 1 2 ) − f(x) for all x ∈ [0, 1 2 ]. Since fis continuous, g : [0, 1 2 ] → R is co nt inuous. Als o g(0) = f( 1 2 ) − f(0) and g( 1 2 ) = f(1) − f( 1 2 ) = −g(0), since f(0) = f(1). If g(0) = 0, then we can take x 1 = 1 2 and x 2 = 0. Otherwise, g( 1 2 ) and g(0) are of opposite signs and hence by the intermediate value theorem, there exists c ∈ (0, 1 2 ) such that g(c) = 0, i.e. f(c + 1 2 ) = f(c). We take x 1 = c + 1 2 and x 2 = c. (ii) Let g(x) = f(x + 1 3 ) − f(x) for all x ∈ [0, 2 3 ]. Since fis continuous, g : [0, 2 3 ] → R is continuous. Also g(0)+ g( 1 3 ) + g( 2 3 ) = f(1) − f(0) = 0. If at lea st one ofg(0), g( 1 3 ) and g( 2 3 ) is 0, then the resu lt follows immediatel y . Otherwise, at least two ofg(0), g( 1 3 ) and g( 2 3 ) are of opposite signs and hence by the intermediate value theorem, there exists c ∈ (0, 2 3 ) such that g(c) = 0, i.e. f(c+ 1 3 ) = f(c). We take x 1 = c+ 1 3 and x 2 = c. 4. Let f(x) = p(x) − g(x) for all x ∈ R. Since both p and g are continuous, f: R → R is continuous. Since g is bounded, there exists M > 0 such that |g(x)| ≤ Mfor all x ∈ R. Let p(x) = a 0 x n +a 1 x n−1 +··· +a n−1 x+a n for all x ∈ R, where a i ∈ R for i = 0, 1,...,n, n ∈ N is odd and a 0 = 0. So p(x) = a 0 x n (1 + a 1 a 0 · 1 x + ··· + a n−1 a 0 · 1 x n−1 + a n a 0 · 1 x n ) for all x(= 0) ∈ R. We assume that a 0 > 0. (Th e cas e a 0 < 0 is almost simi lar .) Then lim x→∞ p(x) = ∞ and lim x→−∞ p(x) = −∞ (since n is odd). So there exist x 1 > 0 and x 2 < 0 such that p(x 1 ) > Mand p(x 2 ) < −M. Hence f(x 1 ) > 0 and f(x 2 ) < 0. By the intermediate value theorem, there exists x 0 ∈ (x 2 , x 1 ) such that f(x 0 ) = 0, i.e. p(x 0 ) = g(x 0 ). (For (i), we take g(x) = 0 for all x ∈ R. F or (ii), we tak e p(x) = x 9 − 4x 6 + x 5 − 17 and g(x) = si n3x − 1 1+x 2 for all x ∈ R and we note that |g(x)| ≤ 2 for all x ∈ R. For (iii), given y ∈ R, we take g(x) = y for all x ∈ R.) 5. (i) is false. If possible, let there exist a con tinuous function f: [0, 1] → (0, 1) which is onto . Then there exi sts x 0 ∈ [0, 1] such that f(x 0 ) ≤ f(x) for all x ∈ [0, 1]. Sinc e 0 < 1 2 f(x 0 ) < 1 and since fis onto, there exists c ∈ [0, 1] such that f(c) = 1 2 f(x 0 ). From above, we get f(x 0 ) ≤ f(c), i.e. f(x 0 ) ≤ 1 2 f(x 0 ), which is not possible, since f(x 0 ) > 0. (i i) is true . The function f: (0, 1) → [0, 1], defined by f(x) = | sin(2πx)| for all x ∈ (0, 1), is continuous. It is also ont o, because sin : [ π 2 , π] → [0, 1] is onto and if x ∈ [ π 2 , π], then x 2π ∈ (0, 1) and f( x 2π ) = sin x.
