IGNOU MTE-09 Solved Question Paper PDF Download

IGNOU MTE-09 Previous Year Solved Question Paper in Hindi

IGNOU MTE-09 Previous Year Solved Question Paper in English

Q1. (a) Draw the graph of the function f, defined by: f(x) = |x-2| + [8x], x ∈ [1/8, 3]. (b) Let f be a function defined by: f(x) = x sin(1/x), x ∈ (0, 10]. Check, whether or not, f is uniformly continuous on the interval (0, 10]. (c) Let f:[0, 1]→R be defined by f(x) = x² – 2. Let P₁ = {0, 1/4, 1/2, 3/4, 1} and P₂ = {0, 1/3, 1/2, 2/3, 1} be two partitions of the interval [0, 1]. Find L(P₁, f) and U(P₂, f).

Ans. (a) The given function is f(x) = |x-2| + [8x] for x ∈ [1/8, 3]. To draw the graph, we analyze the two component functions: the absolute value function |x-2| and the greatest integer function [8x]. The greatest integer function [8x] changes its value whenever 8x is an integer.

• For 1/8 ≤ x < 2/8, [8x] = 1
• For 2/8 ≤ x < 3/8, [8x] = 2
• …
• For 23/8 ≤ x < 24/8 = 3, [8x] = 23
• For x = 3, [8x] = 24

The absolute value function |x-2| changes its definition at x=2.

For x ∈ [1/8, 2), |x-2| = -(x-2) = 2-x.

For x ∈ [2, 3], |x-2| = x-2.

Now we can define the function on different sub-intervals:

• For x ∈ [1/8, 1/4): f(x) = (2-x) + 1 = 3-x
• For x ∈ [1/4, 2): f(x) = (2-x) + [8x]
• At x = 2: f(2) = |2-2| + [16] = 0 + 16 = 16
• For x ∈ (2, 3]: f(x) = (x-2) + [8x]

The graph will consist of segments of straight lines, with jumps in value wherever the greatest integer function jumps. There will be discontinuities at x = k/8 for integer k.

For example:

• On [1/8, 1/4), the graph is y = 3-x.
• On [1/4, 3/8), the graph is y = (2-x) + 2 = 4-x.

The graph consists of line segments on each interval [k/8, (k+1)/8) with jump discontinuities. The behavior also changes at x=2. The final graph is a combination of a “V-shape” and a “step” function, resulting in a piecewise linear graph with many jumps.

(b) The given function is f(x) = x sin(1/x) for x ∈ (0, 10]. We want to check for uniform continuity on this interval. A key theorem states that a function that is continuous on a closed and bounded interval is uniformly continuous on that interval. Our interval (0, 10] is not closed. Let’s check if the function can be extended continuously to the closed interval [0, 10]. This requires calculating the limit as x → 0⁺. lim (x→0⁺) f(x) = lim (x→0⁺) x sin(1/x) We know that the sine function is bounded: -1 ≤ sin(1/x) ≤ 1. For x > 0, we can multiply by x: -x ≤ x sin(1/x) ≤ x. By the Squeeze Theorem, since lim (x→0⁺) (-x) = 0 and lim (x→0⁺) x = 0, we have: lim (x→0⁺) x sin(1/x) = 0. Since the limit exists and is finite, we can define a new function F on the closed interval [0, 10] as: F(x) = { x sin(1/x) if x ∈ (0, 10], 0 if x=0 } Now, F(x) is continuous on (0, 10] as it is a product/composition of continuous functions. F(x) is also continuous at x=0 because lim (x→0⁺) F(x) = 0 = F(0). Therefore, F(x) is continuous on the closed and bounded interval [0, 10]. By the theorem mentioned earlier, F(x) is uniformly continuous on [0, 10]. A function that is uniformly continuous on a set is also uniformly continuous on any subset. Therefore, the original function f(x) is uniformly continuous on (0, 10].

(c) Given function f(x) = x² – 2 on the interval [0, 1]. First, let’s find the derivative: f'(x) = 2x. For x ∈ [0, 1], f'(x) ≥ 0, which means f is a monotonically increasing function . For an increasing function on any sub-interval [xᵢ₋₁, xᵢ], the infimum mᵢ is f(xᵢ₋₁) and the supremum Mᵢ is f(xᵢ).

For partition P₁ = {0, 1/4, 1/2, 3/4, 1}: The sub-intervals are [0, 1/4], [1/4, 1/2], [1/2, 3/4], [3/4, 1]. All have length Δx = 1/4. L(P₁, f) = Σ mᵢ Δxᵢ = f(0)(1/4) + f(1/4)(1/4) + f(1/2)(1/4) + f(3/4)(1/4) = (1/4) * [f(0) + f(1/4) + f(1/2) + f(3/4)] = (1/4) * [(0² – 2) + ((1/4)² – 2) + ((1/2)² – 2) + ((3/4)² – 2)] = (1/4) * [-2 + (1/16 – 2) + (1/4 – 2) + (9/16 – 2)] = (1/4) * [-8 + 1/16 + 4/16 + 9/16] = (1/4) [-8 + 14/16] = (1/4) [-8 + 7/8] = (1/4) * [(-64+7)/8] = -57/32. So, L(P₁, f) = -57/32 .

For partition P₂ = {0, 1/3, 1/2, 2/3, 1}: We need to find U(P₂, f). The sub-intervals are [0, 1/3], [1/3, 1/2], [1/2, 2/3], [2/3, 1]. The lengths are Δx₁ = 1/3, Δx₂ = 1/6, Δx₃ = 1/6, Δx₄ = 1/3. U(P₂, f) = Σ Mᵢ Δxᵢ = f(1/3)Δx₁ + f(1/2)Δx₂ + f(2/3)Δx₃ + f(1)Δx₄ = ((1/3)² – 2)(1/3) + ((1/2)² – 2)(1/6) + ((2/3)² – 2)(1/6) + (1² – 2)(1/3) = (1/9 – 2)(1/3) + (1/4 – 2)(1/6) + (4/9 – 2)(1/6) + (-1)(1/3) = (-17/9)(1/3) + (-7/4)(1/6) + (-14/9)(1/6) – 1/3 = -17/27 – 7/24 – 14/54 – 1/3 = -17/27 – 7/24 – 7/27 – 9/27 = (-17 – 7 – 9)/27 – 7/24 = -33/27 – 7/24 = -11/9 – 7/24 = [(-11 8) – (7 3)] / 72 = (-88 – 21) / 72 = -109/72. So, U(P₂, f) = -109/72 .

Q2. (a) Find whether or not the following sequence is convergent: aₙ = (3n + 4n²) / (5n² – n). (b) Represent the number 2 + i√5 on the real plane. (c) Evaluate: lim (n→∞) Σ (r=1 to 5n) [n² / (8n+r)³]

Ans. (a) To check the convergence of the sequence aₙ = (3n + 4n²) / (5n² – n), we find its limit as n → ∞. We can divide both the numerator and the denominator by the highest power of n, which is n²: aₙ = (3n/n² + 4n²/n²) / (5n²/n² – n/n²) aₙ = (3/n + 4) / (5 – 1/n) Now, we take the limit as n → ∞: lim (n→∞) aₙ = lim (n→∞) [ (3/n + 4) / (5 – 1/n) ] Since lim (n→∞) 1/n = 0, we get: lim (n→∞) aₙ = (0 + 4) / (5 – 0) = 4/5 Since the limit is a finite and unique real number (4/5), the given sequence is convergent and converges to 4/5.

(b) The question asks to represent a number on the “real plane”. Assuming the number is the complex number z = 2 + √(-5) = 2 + i√5, it is represented on the Argand plane (complex plane), which is isomorphic to the real plane R². A complex number z = x + iy is represented by the point (x, y) in the Cartesian plane. Here, we have z = 2 + i√5. So, the real part is x = 2 and the imaginary part is y = √5. The value of √5 is approximately 2.236. Thus, we need to plot the point (2, √5) ≈ (2, 2.236) on the plane. To represent it: 1. Draw a Cartesian coordinate system (x-axis and y-axis). Label the x-axis as the ‘Real Axis’ and the y-axis as the ‘Imaginary Axis’. 2. Move 2 units along the positive Real Axis. 3. Move √5 (approx. 2.24) units up, parallel to the Imaginary Axis. 4. The point (2, √5) in the first quadrant represents the complex number 2 + i√5.

(c) We need to evaluate the limit: L = lim (n→∞) Σ (r=1 to 5n) [n² / (8n+r)³] This is a limit of a sum in the form of a definite integral. We aim to express it as a Riemann sum. L = lim (n→∞) Σ (r=1 to 5n) n² / [n³(8 + r/n)³] L = lim (n→∞) Σ (r=1 to 5n) (1/n) * [1 / (8 + r/n)³] Let us substitute x = r/n and dx = 1/n. As n → ∞, the sum becomes an integral. We need to determine the limits of integration: Lower limit: When r = 1, x = 1/n. As n → ∞, x → 0. Upper limit: When r = 5n, x = 5n/n = 5. So, the limit can be written as the definite integral: L = ∫ (from 0 to 5) [1 / (8+x)³] dx L = ∫ (from 0 to 5) (8+x)⁻³ dx Now we integrate: L = [ (8+x)⁻² / -2 ] (from 0 to 5) L = -1/2 * [ 1/(8+x)² ] (from 0 to 5) L = -1/2 * [ (1/(8+5)²) – (1/(8+0)²) ] L = -1/2 * [ 1/13² – 1/8² ] L = -1/2 * [ 1/169 – 1/64 ] L = -1/2 [ (64 – 169) / (169 64) ] L = -1/2 * [ -105 / 10816 ] L = 105 / 21632 .

Q3. (a) Test the convergence of the series: 1/3 + 4/9 + 9/27 + … (b) Evaluate the following limit, if it exists: lim (x→0) [x³ / (tan³x + tanx – x)] (c) Prove that every infinite set contains a denumerable set.

Ans. (a) The given series is 1/3 + 4/9 + 9/27 + … Let’s find the n-th term (aₙ) of the series. The numerators are 1, 4, 9, …, which are n². The denominators are 3, 9, 27, …, which are 3ⁿ. So, the n-th term is aₙ = n² / 3ⁿ. To test the convergence of this series, we will use the Ratio Test . aₙ₊₁ = (n+1)² / 3ⁿ⁺¹ Now we compute the limit L = lim (n→∞) |aₙ₊₁ / aₙ| L = lim (n→∞) | ( (n+1)² / 3ⁿ⁺¹ ) / ( n² / 3ⁿ ) | L = lim (n→∞) | ( (n+1)² / (3 3ⁿ) ) ( 3ⁿ / n² ) | L = lim (n→∞) | (1/3) * ( (n+1)² / n² ) | L = (1/3) * lim (n→∞) | ((n+1)/n)² | L = (1/3) * lim (n→∞) | (1 + 1/n)² | L = (1/3) (1 + 0)² = (1/3) 1 = 1/3 According to the Ratio Test, if L < 1, the series converges. Here, L = 1/3 < 1. Therefore, the given series is convergent .

(b) We need to evaluate the limit L = lim (x→0) [x³ / (tan³x + tanx – x)]. As x → 0, the numerator x³ → 0. The denominator tan³x + tanx – x → tan³(0) + tan(0) – 0 = 0. This is an indeterminate form 0/0. We can use the Maclaurin series expansion for tan(x) near x=0: tan(x) = x + x³/3 + 2x⁵/15 + O(x⁷) Substitute this into the denominator: Denominator = (x + x³/3 + …)³ + (x + x³/3 + …) – x For x→0, the lowest order terms dominate. The leading term of (x + x³/3 + …)³ is x³. So, the denominator is approximately x³ + (x + x³/3 – x) = x³ + x³/3 = 4x³/3 for small x. Now compute the limit: L = lim (x→0) [ x³ / (4x³/3) ] L = lim (x→0) [ 3/4 ] = 3/4 . Alternatively, applying L’Hôpital’s Rule three times would yield the same result but would be more tedious.

(c) To prove that every infinite set contains a denumerable set. Theorem: Every infinite set S has a denumerable subset. Proof: Let S be an infinite set. A set is infinite if it is not empty and not finite. Since S is infinite, it is non-empty. Therefore, an element exists in it. 1. Since S is non-empty, we can choose an element a₁ ∈ S. 2. Consider the set S₁ = S – {a₁}. Since S is infinite, removing one element leaves it infinite. Thus S₁ is infinite, and in particular, non-empty. So we can choose an element a₂ ∈ S₁. Note that a₂ ≠ a₁. 3. Similarly, consider the set S₂ = S₁ – {a₂} = S – {a₁, a₂}. This set is also infinite and non-empty. We can therefore choose an element a₃ ∈ S₂, where a₃ ≠ a₁ and a₃ ≠ a₂. 4. We can continue this process indefinitely. At the n-th step, we have constructed a set of n distinct elements {a₁, a₂, …, aₙ}. The set Sₙ = S – {a₁, a₂, …, aₙ} is still infinite, so we can choose an element aₙ₊₁ ∈ Sₙ. This process never terminates because S is infinite. By this process, we construct a sequence of elements a₁, a₂, a₃, …, aₙ, … where all the aᵢ are distinct. Consider the set A = {a₁, a₂, a₃, …}. This set is a subset of S by construction. There exists a bijection f: N → A between the set of natural numbers N and the set A, defined by f(n) = aₙ. This function is one-to-one because all aₙ are distinct, and it is onto because every element in A corresponds to some n. A set that is in bijection with N is denumerable . Therefore, A is a denumerable subset of S. Thus, we have proved that every infinite set contains a denumerable subset. (This proof relies on the Axiom of Choice).

Q4. (a) Examine the continuity of the function f:[2, 5]→R defined by f(x) = [x] / (4x-3) where [x] denotes the greatest integer ≤x. (b) Show that Rₙ(x), the Lagrange’s form of remainder in the Maclaurin’s series expansion of cos(3x), tends to zero as n→∞. Hence obtain the Maclaurin’s infinite expansion of cos(3x).

Ans. (a) The function is f(x) = [x] / (4x – 3) defined on the interval [2, 5]. Potential points of discontinuity are where the denominator is zero or where the greatest integer function [x] jumps (i.e., at integers). 1. Denominator Check: The denominator 4x – 3 = 0 when x = 3/4. Since 3/4 is not in the interval [2, 5], there is no discontinuity from this cause. 2. Greatest Integer Function Check: In the interval [2, 5], the value of [x] jumps at x = 3, x = 4, and x = 5. We must check continuity at these integer points.

Continuity at x = 3:

• Left-Hand Limit (LHL): lim (x→3⁻) f(x) = lim (x→3⁻) [x] / (4x – 3). As x approaches 3 from the left (e.g., 2.99), [x] = 2. LHL = 2 / (4(3) – 3) = 2 / (12 – 3) = 2/9 .
• Right-Hand Limit (RHL): lim (x→3⁺) f(x) = lim (x→3⁺) [x] / (4x – 3). As x approaches 3 from the right (e.g., 3.01), [x] = 3. RHL = 3 / (4(3) – 3) = 3 / 9 = 1/3 .
• Function Value: f(3) = [3] / (4(3) – 3) = 3 / 9 = 1/3 .

Since LHL ≠ RHL, the function is

discontinuous at x = 3

.

Continuity at x = 4:

• LHL: lim (x→4⁻) f(x) = lim (x→4⁻) [x] / (4x – 3). As x → 4⁻, [x] = 3. LHL = 3 / (4(4) – 3) = 3 / (16 – 3) = 3/13 .
• RHL: lim (x→4⁺) f(x) = lim (x→4⁺) [x] / (4x – 3). As x → 4⁺, [x] = 4. RHL = 4 / (4(4) – 3) = 4 / 13 = 4/13 .

Since LHL ≠ RHL, the function is

discontinuous at x = 4

.

Continuity at x = 5: Since 5 is the endpoint of the interval, we only check the left-hand limit and compare it to f(5).

• LHL: lim (x→5⁻) f(x) = lim (x→5⁻) [x] / (4x – 3). As x → 5⁻, [x] = 4. LHL = 4 / (4(5) – 3) = 4 / (20 – 3) = 4/17 .
• Function Value: f(5) = [5] / (4(5) – 3) = 5 / 17 = 5/17 .

Since LHL ≠ f(5), the function is

discontinuous at x = 5

.

For any non-integer point c ∈ (2, 5), [x] is constant in a small neighborhood of c, and 4x-3 is a continuous and non-zero function. Thus, f(x) is continuous at all non-integer points. Conclusion: The function f(x) is continuous on [2, 5] except at x = 3, 4, and 5.

(b) Let f(x) = cos(3x). The Lagrange form of the remainder for the Maclaurin series is given by: Rₙ(x) = [ fⁿ⁺¹(c) / (n+1)! ] * xⁿ⁺¹, for some c between 0 and x. First, we find the derivatives of f(x) = cos(3x): f'(x) = -3sin(3x) f”(x) = -3²cos(3x) f”'(x) = 3³sin(3x) In general, the k-th derivative fᵏ(x) is of the form ±3ᵏcos(3x) or ±3ᵏsin(3x). In any case, we know that |sin(u)| ≤ 1 and |cos(u)| ≤ 1. Therefore, the magnitude of the k-th derivative is bounded: |fᵏ(x)| ≤ 3ᵏ This is also true for fⁿ⁺¹(c): |fⁿ⁺¹(c)| ≤ 3ⁿ⁺¹. Now consider the magnitude of Rₙ(x): |Rₙ(x)| = | [ fⁿ⁺¹(c) / (n+1)! ] * xⁿ⁺¹ | |Rₙ(x)| ≤ [ |fⁿ⁺¹(c)| / (n+1)! ] * |x|ⁿ⁺¹ |Rₙ(x)| ≤ [ 3ⁿ⁺¹ / (n+1)! ] * |x|ⁿ⁺¹ = (3|x|)ⁿ⁺¹ / (n+1)! We need to show that lim (n→∞) |Rₙ(x)| = 0. This is equivalent to showing lim (n→∞) (3|x|)ⁿ⁺¹ / (n+1)! = 0. Let a = 3|x|. We need to show that for any fixed real number a, lim (n→∞) aⁿ⁺¹ / (n+1)! = 0. This is a standard limit. The factorial (n+1)! grows much faster than the exponential aⁿ⁺¹. For any fixed x, the term aⁿ/n! → 0 as n → ∞. This can be shown using the ratio test on the series Σaⁿ/n!, which shows the series converges, implying the term must go to zero. Thus, lim (n→∞) Rₙ(x) = 0.

Now, to obtain the Maclaurin infinite expansion of cos(3x), we evaluate the derivatives at x=0: f(0) = cos(0) = 1 f'(0) = -3sin(0) = 0 f”(0) = -9cos(0) = -9 f”'(0) = 27sin(0) = 0 f⁴(0) = 81cos(0) = 81 … The Maclaurin series is: f(x) = Σ (from n=0 to ∞) [fⁿ(0)/n!] * xⁿ cos(3x) = f(0) + f'(0)x + [f”(0)/2!]x² + [f”'(0)/3!]x³ + [f⁴(0)/4!]x⁴ + … cos(3x) = 1 + 0 x + (-9/2!)x² + 0 x³ + (81/4!)x⁴ + … cos(3x) = 1 – 9/2! x² + 81/4! x⁴ – … cos(3x) = 1 – (3x)²/2! + (3x)⁴/4! – (3x)⁶/6! + … So, the Maclaurin infinite expansion for cos(3x) is: cos(3x) = Σ (from n=0 to ∞) (-1)ⁿ * (3x)²ⁿ / (2n)!

Q5. (a) Examine the following series for convergence: Σ (n=1 to ∞) [((n+1)/(3n-2))ⁿ] (b) Prove that every strictly increasing onto function is invertible. (c) Examine the function: f(x) = (x+2)³(x-5)² for extreme values.

Ans. (a) The given series is Σ aₙ where aₙ = ((n+1)/(3n-2))ⁿ. Since the term is raised to the n-th power, the Root Test is the most suitable method. According to the Root Test, we compute the limit: L = lim (n→∞) |aₙ|¹/ⁿ L = lim (n→∞) | ((n+1)/(3n-2))ⁿ |¹/ⁿ L = lim (n→∞) | (n+1)/(3n-2) | To evaluate the limit, we divide the numerator and denominator by n: L = lim (n→∞) | (1 + 1/n) / (3 – 2/n) | As n → ∞, 1/n → 0 and 2/n → 0. L = | (1 + 0) / (3 – 0) | = 1/3 The Root Test states that if L < 1, the series converges. Since L = 1/3 < 1, the given series is convergent .

(b) Let f: A → B be a strictly increasing and onto function. For a function to be invertible, it must be both one-to-one (injective) and onto (surjective), i.e., it must be a bijection. We are given that the function f is onto . We need to prove that f is one-to-one . A function f is one-to-one if for any two distinct elements x₁ and x₂ in the domain, it follows that f(x₁) ≠ f(x₂). Let x₁, x₂ ∈ A with x₁ ≠ x₂. Since A is an ordered set (like a subset of R), we can assume either x₁ < x₂ or x₂ < x₁. Let’s assume x₁ < x₂. Since f is a strictly increasing function, by definition, if x₁ < x₂, then f(x₁) < f(x₂). Since f(x₁) < f(x₂), it is clear that f(x₁) ≠ f(x₂). This shows that distinct elements in the domain always map to distinct elements in the codomain. Therefore, the function f is one-to-one. Since f is both one-to-one and onto, it is a bijection. Every bijective function is invertible. Hence, every strictly increasing onto function is invertible.

(c) To examine the function f(x) = (x+2)³(x-5)² for extreme values, we need to find its critical points. Critical points occur where f'(x) = 0 or f'(x) is undefined. First, calculate f'(x) using the product rule: f'(x) = d/dx[(x+2)³] (x-5)² + (x+2)³ d/dx[(x-5)²] f'(x) = 3(x+2)²(1) (x-5)² + (x+2)³ 2(x-5)(1) Now, factor out the common terms, which are (x+2)² and (x-5): f'(x) = (x+2)²(x-5) [ 3(x-5) + 2(x+2) ] f'(x) = (x+2)²(x-5) [ 3x – 15 + 2x + 4 ] f'(x) = (x+2)²(x-5) (5x – 11) Set f'(x) = 0 to find the critical points: (x+2)²(x-5)(5x – 11) = 0 This gives us the critical points: x = -2, x = 5, x = 11/5 = 2.2. We now use the First Derivative Test to determine the nature of these points by checking the sign of f'(x):

• For x < 11/5 (but not x=-2): e.g., x=0. f'(0) = (2)²(-5)(-11) > 0. (Increasing)
• For 11/5 < x < 5: e.g., x=3. f'(3) = (5)²(-2)(4) < 0. (Decreasing)
• For x > 5: e.g., x=6. f'(6) = (8)²(1)(19) > 0. (Increasing)

Let’s analyze each critical point:

• At x = 11/5: The sign of f'(x) changes from + to -. Therefore, f has a local maximum at x = 11/5.
• At x = 5: The sign of f'(x) changes from – to +. Therefore, f has a local minimum at x = 5.
• At x = -2: Let’s check the sign of f’ around -2. For x < -2 (e.g., x=-3), f'(-3) = (-1)²(-8)(-26) > 0. For -2 < x < 11/5 (e.g., x=0), f'(0) > 0. The sign of f'(x) does not change at x = -2. Therefore, x = -2 is neither a local maximum nor minimum. It is a stationary point of inflection .

Q6. (a) Prove that a non-empty subset of R is closed if its complement is open. (b) Show that the equation: 2x³ – 3x² + x – 4 = 0 has a real root in the interval [1, 2]. (c) Prove that the function f defined by: f(x) = {2, if x is rational; -2, if x is irrational} is discontinuous at each real number, using sequential definition of the continuity.

Ans. (a) This statement is the standard definition of a closed set in topology. We will prove it starting from the limit point definition of a closed set. Proof: Let S be a non-empty subset of R and assume its complement, Sᶜ = R \ S, is open. We want to prove that S is closed. A set is closed if it contains all of its limit points. Let x be any limit point of S. We must show that x ∈ S. We will prove this by contradiction. Assume that x ∉ S. If x ∉ S, then by definition, x ∈ Sᶜ. We are given that Sᶜ is an open set. By the definition of an open set, for every point in it, there exists an ε > 0 such that the ε-neighborhood of the point is entirely contained within the set. Therefore, there exists an ε > 0 such that the open interval (x – ε, x + ε) ⊂ Sᶜ. This means that the neighborhood (x – ε, x + ε) contains no points of S. But this contradicts the fact that x is a limit point of S, because the definition of a limit point states that every neighborhood of it must contain at least one point of S other than x itself. This contradiction arose from our initial assumption that x ∉ S. Therefore, our assumption must be false. Hence, x ∈ S. Since x was an arbitrary limit point of S, this means S contains all its limit points. Therefore, S is a closed set.

(b) We want to show that the equation 2x³ – 3x² + x – 4 = 0 has a real root in the interval [1, 2]. We will use the Intermediate Value Theorem (IVT) . Let f(x) = 2x³ – 3x² + x – 4. 1. Since f(x) is a polynomial function, it is continuous for all real numbers. Therefore, it is continuous on the closed interval [1, 2]. 2. Now, we evaluate the function at the endpoints of the interval: f(1) = 2(1)³ – 3(1)² + 1 – 4 = 2 – 3 + 1 – 4 = -4 f(2) = 2(2)³ – 3(2)² + 2 – 4 = 2(8) – 3(4) + 2 – 4 = 16 – 12 – 2 = 2 3. We observe that f(1) is negative and f(2) is positive. So, f(1) < 0 < f(2). The IVT states that if a function f is continuous on a closed interval [a, b], and k is any number between f(a) and f(b), then there must exist at least one number c in the open interval (a, b) such that f(c) = k. In our case, a=1, b=2, and we choose k=0, which is between f(1)=-4 and f(2)=2. The theorem guarantees that there exists a number c ∈ (1, 2) such that f(c) = 0. This number c is a real root of the equation. Thus, the equation 2x³ – 3x² + x – 4 = 0 has a real root in the interval [1, 2].

(c) The given function is the Dirichlet function (modified): f(x) = { 2, if x is rational; -2, if x is irrational } We want to prove f is discontinuous at every real number c, using the sequential definition of continuity. A function f is continuous at c if for every sequence (xₙ) that converges to c, the sequence (f(xₙ)) converges to f(c). Let c be any real number. We consider two cases.

Case 1: c is a rational number. In this case, by definition, f(c) = 2. To show f is discontinuous at c, we need to find a sequence (xₙ) converging to c such that lim (n→∞) f(xₙ) ≠ 2. We know that the irrational numbers are dense in the real line. This means between any two distinct real numbers, there is an irrational number. Thus, we can construct a sequence of irrational numbers (xₙ) that converges to c. For example, for each n ∈ N, choose an irrational xₙ such that c < xₙ < c + 1/n. By the Squeeze Theorem, lim (n→∞) xₙ = c. Since every term in the sequence (xₙ) is irrational, f(xₙ) = -2 for all n. Therefore, lim (n→∞) f(xₙ) = lim (n→∞) -2 = -2. But f(c) = 2. Since lim (n→∞) f(xₙ) ≠ f(c), the function f is discontinuous at c.

Case 2: c is an irrational number. In this case, by definition, f(c) = -2. Now we will construct a sequence (yₙ) converging to c such that lim (n→∞) f(yₙ) ≠ -2. We know that the rational numbers are also dense in the real line. Thus, we can construct a sequence of rational numbers (yₙ) that converges to c. (For example, by taking the decimal expansion of c and truncating it at the n-th place). Since every term in the sequence (yₙ) is rational, f(yₙ) = 2 for all n. Therefore, lim (n→∞) f(yₙ) = lim (n→∞) 2 = 2. But f(c) = -2. Since lim (n→∞) f(yₙ) ≠ f(c), the function f is discontinuous at c. Since c was an arbitrary real number, and we have shown discontinuity in both cases (rational and irrational), we conclude that the function f is discontinuous at every real number .

Q7. Which of the following statements are true? Give reasons for your answers: (a) Every bounded sequence is divergent. (b) The set Q of rational numbers is an open subset of R. (c) The function f(x) = 8x² + |x+3| is differentiable at x=-2. (d) The function f given by: f(x) = {(2ˣ – 2⁻ˣ)/x if x≠0, 3 if x=0} is continuous at x=0. (e) The necessary condition for a function f to be integrable is that it is continuous.

Ans. (a) Every bounded sequence is divergent. False. Reason: This statement is incorrect. A bounded sequence can be either convergent or divergent. For example, the sequence aₙ = 1/n is bounded (since 0 < aₙ ≤ 1 for all n) and it converges to 0. Another example is the constant sequence aₙ = 5, which is bounded and converges to 5. The statement claims every bounded sequence is divergent, which is disproven by these counterexamples.

(b) The set Q of rational numbers is an open subset of R. False. Reason: A set U is open in R if for every point x ∈ U, there exists an ε > 0 such that the open interval (x – ε, x + ε) is entirely contained in U. Let q be any rational number in Q. Any open interval (q – ε, q + ε) around q, no matter how small ε is, will always contain irrational numbers (due to the density of irrationals in R). Therefore, no neighborhood of q can be entirely contained in Q. Thus, Q is not an open set.

(c) The function f(x) = 8x² + |x+3| is differentiable at x=-2. True. Reason: The differentiability of a function involving an absolute value |g(x)| is typically questioned at points where g(x) = 0. For |x+3|, this critical point is at x = -3. We are checking for differentiability at x = -2. In a neighborhood of x = -2 (for instance, the interval (-2.5, -1.5)), the term x+3 is always positive. (e.g., at x=-2, x+3 = 1 > 0). When x+3 > 0, we have |x+3| = x+3. So, in a neighborhood of x = -2, the function can be written as: f(x) = 8x² + x + 3 This is a polynomial function, which is differentiable everywhere. Its derivative is: f'(x) = 16x + 1 At x = -2, the derivative is f'(-2) = 16(-2) + 1 = -31. Since the derivative exists at x = -2, the function is differentiable at this point.

(d) The function f given by: f(x) = {(2ˣ – 2⁻ˣ)/x if x≠0, 3 if x=0} is continuous at x=0. False. Reason: For a function f to be continuous at x=0, we must have lim (x→0) f(x) = f(0). We are given f(0) = 3. Now let’s calculate the limit: lim (x→0) (2ˣ – 2⁻ˣ)/x This is an indeterminate form of type 0/0 (since 2⁰ – 2⁰ = 1-1=0). We can use L’Hôpital’s Rule: lim (x→0) [ d/dx(2ˣ – 2⁻ˣ) / d/dx(x) ] = lim (x→0) [ 2ˣln(2) – 2⁻ˣ(-ln(2)) ] / 1 = [ 2⁰ln(2) + 2⁰ln(2) ] / 1 = ln(2) + ln(2) = 2ln(2) = ln(4) So, the limit of the function as x approaches 0 is ln(4). Since lim (x→0) f(x) = ln(4) ≠ 3 = f(0), the function f is not continuous at x=0.

(e) The necessary condition for a function f to be integrable is that it is continuous. False. Reason: This statement confuses necessary and sufficient conditions. Continuity on a closed interval is a sufficient condition for a function to be Riemann-integrable, but it is not a necessary one. A necessary condition must be true for all integrable functions. However, there are many functions that are integrable but not continuous. Counterexample: Consider the step function f(x) defined on [-1, 1] by f(x) = { 0 if -1 ≤ x < 0, 1 if 0 ≤ x ≤ 1 }. This function is discontinuous at x=0, but it is integrable on [-1, 1]. Its integral is 1. The necessary and sufficient condition for Riemann integrability is that the set of discontinuities of the function has measure zero.