IGNOU MTE-01 Solved Question Paper PDF Download

IGNOU MTE-01 Previous Year Solved Question Paper in Hindi

IGNOU MTE-01 Previous Year Solved Question Paper in English

Q1. Which of the following statements are True or False. Justify your answers with a short proof or a counter-example: (i) The function, f(x) = |x-2| is differentiable at x=2. (ii) sin(x-π) is an odd function of x. (iii) If x=c is a critical point of the function f, then f is derivable at x=c. (iv) y = -2x² is increasing in the interval, ]-4,-2]. (v) d/dx [∫(from 1 to cosx) tan²(2t)dt] = cos x tan²(2 cos x).

Ans. (i) False . Justification: To check the differentiability of the function f(x) = |x-2| at x=2, we compute the left-hand and right-hand derivatives. Left-Hand Derivative (L.H.D.): lim (h→0⁻) [f(2+h) – f(2)] / h = lim (h→0⁻) [|2+h-2| – |2-2|] / h = lim (h→0⁻) |h|/h = -h/h = -1. Right-Hand Derivative (R.H.D.): lim (h→0⁺) [f(2+h) – f(2)] / h = lim (h→0⁺) [|2+h-2| – |2-2|] / h = lim (h→0⁺) |h|/h = h/h = 1. Since L.H.D. ≠ R.H.D. (-1 ≠ 1), the function is not differentiable at x=2. Its graph has a sharp corner at x=2.

(ii) True . Justification: A function g(x) is said to be an odd function if g(-x) = -g(x). Here, g(x) = sin(x-π). We know that sin(θ-π) = -sin(π-θ) = -sin(θ). Therefore, g(x) = -sin(x). Now, let’s check the condition: g(-x) = sin(-x-π) = sin(-(x+π)) = -sin(x+π). Since sin(x+π) = -sin(x), we have g(-x) = -(-sin(x)) = sin(x). Also, -g(x) = -(-sin(x)) = sin(x). Since g(-x) = -g(x), sin(x-π) is an odd function.

(iii) False . Counter-example: A critical point x=c for a function f is a point where either f'(c) = 0 or f'(c) is undefined. Consider the function f(x) = |x|. It has a critical point at x=0, as the function attains its minimum value there. However, as shown in part (i), the modulus function is not differentiable at its ‘corner’. At x=0, the left-hand derivative is -1 and the right-hand derivative is 1. Since they are not equal, f(x) = |x| is not derivable at x=0, even though it is a critical point.

(iv) True . Justification: A function is increasing on an interval if its derivative is positive on that interval. The given function is y = -2x². The derivative is y’ = dy/dx = -4x. In the interval ]-4, -2], x is a negative number. Let’s take x = -3, which is in this interval. Then y’ = -4(-3) = 12 > 0. In fact, for any x in ]-4, -2], we have -4 < x ≤ -2. Multiplying by -4 reverses the inequality signs: (-4)(-4) > -4x ≥ (-4)(-2) => 16 > -4x ≥ 8. Thus, in the interval ]-4, -2], y’ = -4x is positive (8 ≤ y’ < 16). Since y’ > 0, the function is increasing on this interval.

(v) False . Justification: We use the second part of the Fundamental Theorem of Calculus (Leibniz rule): d/dx [∫(from a(x) to b(x)) f(t) dt] = f(b(x)) b'(x) – f(a(x)) a'(x). Here, f(t) = tan²(2t), a(x) = 1 (a constant), and b(x) = cos x. So, a'(x) = 0 and b'(x) = -sin x. Applying the rule: d/dx [∫(from 1 to cosx) tan²(2t) dt] = tan²(2 cos x) (-sin x) – tan²(2 1) * 0 = -sin x * tan²(2 cos x). This is not equal to the given expression cos x tan²(2 cos x). Hence, the statement is false.

Q2. (a) Evaluate: ∫xlnxdx. (b) Using the trapezoidal rule, evaluate ∫(from 3 to 9) (1/x) dx by dividing the interval, [3,9] into six equal sub-intervals. (c) Determine the values of p and q for which the function f is given by: f(x) = { [p sinx + q sin2x + sin3x] / x³, x≠0; 1, x=0 } is continuous at x = 0.

Ans. (a) Evaluation of ∫xlnxdx: We use integration by parts, whose formula is ∫u dv = uv – ∫v du. According to the ILATE rule, we will choose u = ln x and dv = x dx. Then, du = (1/x) dx and v = ∫x dx = x²/2. Substituting these values into the formula: ∫x ln x dx = (ln x) (x²/2) – ∫(x²/2) (1/x) dx = (x²/2) ln x – (1/2) ∫x dx = (x²/2) ln x – (1/2) * (x²/2) + C = (x²/2) ln x – x²/4 + C. Thus, ∫x ln x dx = (x²/4)(2 ln x – 1) + C .

(b) Using the trapezoidal rule: We need to evaluate ∫(from 3 to 9) (1/x) dx. The interval is [a, b] = [3, 9] and the number of sub-intervals is n = 6. The width of each sub-interval is h = (b-a)/n = (9-3)/6 = 1. The x-values will be: x₀=3, x₁=4, x₂=5, x₃=6, x₄=7, x₅=8, x₆=9. The function is y = f(x) = 1/x. The corresponding y-values are: y₀ = 1/3 y₁ = 1/4 y₂ = 1/5 y₃ = 1/6 y₄ = 1/7 y₅ = 1/8 y₆ = 1/9 The trapezoidal rule formula is: ∫f(x)dx ≈ (h/2) [y₀ + yₙ + 2(y₁ + y₂ + … + yₙ₋₁)] Substituting the values: ∫(from 3 to 9) (1/x) dx ≈ (1/2) [1/3 + 1/9 + 2(1/4 + 1/5 + 1/6 + 1/7 + 1/8)] ≈ 0.5 * [0.3333 + 0.1111 + 2(0.25 + 0.2 + 0.1667 + 0.1429 + 0.125)] ≈ 0.5 * [0.4444 + 2(0.8846)] ≈ 0.5 [0.4444 + 1.7692] = 0.5 2.2136 = 1.1068 . (The actual value is ln(9) – ln(3) = ln(3) ≈ 1.0986, so our approximation is quite close).

(c) Determining the values of p and q: For the function f to be continuous at x=0, we must have lim (x→0) f(x) = f(0). Here f(0) = 1. So we need lim (x→0) [p sinx + q sin2x + sin3x] / x³ = 1. As x→0, this is of the form 0/0, so we can use Maclaurin expansions: sin x = x – x³/3! + O(x⁵) sin 2x = 2x – (2x)³/3! + O(x⁵) = 2x – 8x³/6 + O(x⁵) sin 3x = 3x – (3x)³/3! + O(x⁵) = 3x – 27x³/6 + O(x⁵) Substituting these into the numerator: p(x – x³/6) + q(2x – 8x³/6) + (3x – 27x³/6) + O(x⁵) = (p + 2q + 3)x – (p/6 + 8q/6 + 27/6)x³ + O(x⁵) So the limit is: lim (x→0) [ (p + 2q + 3)x – (p + 8q + 27)/6 * x³ ] / x³ = lim (x→0) [ (p + 2q + 3)/x² – (p + 8q + 27)/6 ] For the limit to be finite, the coefficient of 1/x² must be zero. Therefore, p + 2q + 3 = 0 —(1) With this condition, the limit becomes: lim (x→0) – (p + 8q + 27)/6 = 1

• (p + 8q + 27) = 6

p + 8q + 27 = -6

p + 8q = -33

—(2)

Now we solve the system of equations (1) and (2):

Subtracting (1) from (2): (p + 8q) – (p + 2q) = -33 – (-3)

6q = -30 =>

q = -5

.

Substitute q = -5 into (1):

p + 2(-5) + 3 = 0 => p – 10 + 3 = 0 => p – 7 = 0 =>

p = 7

.

Thus, the required values are

p = 7

and

q = -5

.

Q3. (a) Evaluate: lim(x→0) [√(1+x) – √(1-x)]/x (b) Find the length of the arc of the cycloid, x = a(t+sint), y = a(1-cost) between t=-π and t=π. (c) If Iₙ = ∫(from 0 to π/2) cosⁿx sin(nx) dx (n>1), then find a reduction formula for Iₙ, and hence evaluate I₂.

Ans. (a) Evaluation of the limit: Given lim(x→0) [√(1+x) – √(1-x)]/x. If we substitute x=0, we get the indeterminate form 0/0. We can solve this by multiplying the numerator and denominator by the conjugate of the numerator. = lim(x→0) { [√(1+x) – √(1-x)]/x * [√(1+x) + √(1-x)] / [√(1+x) + √(1-x)] } = lim(x→0) [ (1+x) – (1-x) ] / [ x(√(1+x) + √(1-x)) ] = lim(x→0) [ 2x ] / [ x(√(1+x) + √(1-x)) ] For x ≠ 0, we can cancel x: = lim(x→0) 2 / (√(1+x) + √(1-x)) Now substituting x=0: = 2 / (√(1+0) + √(1-0)) = 2 / (1+1) = 2/2 = 1 .

(b) Length of the arc: The equations of the cycloid are: x = a(t+sint), y = a(1-cost). The formula for arc length is L = ∫(from t₁ to t₂) √[(dx/dt)² + (dy/dt)²] dt. First, we find the derivatives with respect to t: dx/dt = a(1 + cost) dy/dt = a(sin t) Now we calculate (dx/dt)² + (dy/dt)²: = a²(1 + cost)² + a²(sin t)² = a²(1 + 2cost + cos²t + sin²t) = a²(1 + 2cost + 1) = a²(2 + 2cost) = 2a²(1 + cost) Using the trigonometric identity 1 + cost = 2cos²(t/2): = 2a²(2cos²(t/2)) = 4a²cos²(t/2) So, √[(dx/dt)² + (dy/dt)²] = √(4a²cos²(t/2)) = 2a|cos(t/2)|. For the interval t from -π to π, t/2 is between -π/2 and π/2. In this interval, cos(t/2) ≥ 0, so |cos(t/2)| = cos(t/2). Now we integrate for the length: L = ∫(from -π to π) 2a cos(t/2) dt = 2a [ 2sin(t/2) ](from -π to π) = 4a [ sin(π/2) – sin(-π/2) ] = 4a [ 1 – (-1) ] = 4a(2) = 8a .

(c) Reduction formula and evaluation of I₂: Given Iₙ = ∫(from 0 to π/2) cosⁿx sin(nx) dx. We first try to establish a relationship between Iₙ and Iₙ₋₁. Consider Iₙ – Iₙ₋₁ = ∫ [cosⁿx sin(nx) – cosⁿ⁻¹x sin((n-1)x)] dx = ∫ cosⁿ⁻¹x [cos x sin(nx) – sin((n-1)x)] dx Using the identity sin((n-1)x) = sin(nx-x) = sin(nx)cos(x) – cos(nx)sin(x), we get cos(x)sin(nx) – sin((n-1)x) = cos(nx)sin(x). = ∫ cosⁿ⁻¹x [cos(nx)sin(x)] dx. So, Iₙ – Iₙ₋₁ = ∫(from 0 to π/2) cosⁿ⁻¹x sinx cos(nx) dx. Let’s call this K. Now, apply integration by parts to Iₙ: Let u = cosⁿx, dv = sin(nx)dx. Then du = -n cosⁿ⁻¹x sinx dx, v = -cos(nx)/n. Iₙ = [-cosⁿx cos(nx)/n](from 0 to π/2) – ∫(from 0 to π/2) [-cos(nx)/n] * [-n cosⁿ⁻¹x sinx] dx Iₙ = [0 – (-1*1/n)] – ∫(from 0 to π/2) cosⁿ⁻¹x sinx cos(nx) dx Iₙ = 1/n – K Substituting K = Iₙ – Iₙ₋₁: Iₙ = 1/n – (Iₙ – Iₙ₋₁) 2Iₙ = 1/n + Iₙ₋₁ Iₙ = (1/2)Iₙ₋₁ + 1/(2n) . This is the required reduction formula.

Now we evaluate I₂. Using the formula, I₂ = (1/2)I₁ + 1/(4). We first need to calculate I₁. I₁ = ∫(from 0 to π/2) cos x sin x dx = (1/2) ∫(from 0 to π/2) sin(2x) dx = (1/2) [-cos(2x)/2](from 0 to π/2) = (-1/4) [cos(π) – cos(0)] = (-1/4) [-1 – 1] = (-1/4)(-2) = 1/2. So, I₁ = 1/2. Now, I₂ = (1/2)I₁ + 1/4 = (1/2)(1/2) + 1/4 = 1/4 + 1/4 = 1/2 . (Alternatively, evaluating directly: I₂ = ∫cos²x sin(2x) dx = ∫cos²x(2sinxcosx)dx = 2∫cos³x sinx dx = 2[-cos⁴x/4](0 to π/2) = -1/2(0-1) = 1/2.)

Q4. (a) Find an approximate value of ln(0.9), upto 3 decimal places, using Maclaurin’s expansion. (b) Evaluate: ∫dx / [(1+x)√(1-x²)]. (c) Find the area of the region bounded by the curve, 16y² = x²(4-x).

Ans. (a) Approximate value of ln(0.9): We cannot expand ln(x) using Maclaurin’s expansion as it’s undefined at x=0. Instead, we expand ln(1-x) and then substitute x=0.1. Let f(x) = ln(1-x). The Maclaurin series is f(x) = f(0) + f'(0)x + f”(0)x²/2! + f”'(0)x³/3! + … We compute the derivatives: f(x) = ln(1-x) => f(0) = ln(1) = 0 f'(x) = -1/(1-x) => f'(0) = -1 f”(x) = -1/(1-x)² => f”(0) = -1 f”'(x) = -2/(1-x)³ => f”'(0) = -2 Substituting these values into the series: ln(1-x) = 0 + (-1)x + (-1)x²/2 + (-2)x³/6 + … ln(1-x) = -x – x²/2 – x³/3 – … Now, to find ln(0.9) = ln(1 – 0.1), we set x = 0.1: ln(0.9) ≈ -(0.1) – (0.1)²/2 – (0.1)³/3 ≈ -0.1 – (0.01)/2 – (0.001)/3 ≈ -0.1 – 0.005 – 0.000333… ≈ -0.105333… Rounding to 3 decimal places, we get -0.105 .

(b) Evaluation of the integral: The given integral is I = ∫dx / [(1+x)√(1-x²)]. We know that √(1-x²) = √[(1-x)(1+x)]. I = ∫dx / [(1+x)√(1-x)√(1+x)] = ∫dx / [(1+x)^(3/2) (1-x)^(1/2)] This form is complex. Let’s try a trigonometric substitution. Let x = cosθ, then dx = -sinθ dθ. 1+x = 1+cosθ and 1-x² = 1-cos²θ = sin²θ. I = ∫(-sinθ dθ) / [(1+cosθ)√(sin²θ)] For -1 < x < 1, sinθ > 0, so √sin²θ = sinθ. I = ∫(-sinθ dθ) / [(1+cosθ)sinθ] = -∫dθ / (1+cosθ) Using the half-angle formula, 1+cosθ = 2cos²(θ/2): I = -∫dθ / [2cos²(θ/2)] = -(1/2)∫sec²(θ/2) dθ = -(1/2) * [tan(θ/2) / (1/2)] + C = -tan(θ/2) + C Now we substitute back from θ to x. We have tan²(θ/2) = (1-cosθ)/(1+cosθ) = (1-x)/(1+x). So, tan(θ/2) = √[(1-x)/(1+x)]. Thus, the value of the integral is -√[(1-x)/(1+x)] + C .

(c) Finding the area: The equation of the curve is 16y² = x²(4-x), or y² = x²(4-x)/16. Since y² ≥ 0, we must have x²(4-x) ≥ 0. As x² is always ≥ 0, this requires 4-x ≥ 0, which means x ≤ 4. The curve intersects the x-axis at x=0 and x=4. The curve is symmetric about the x-axis due to the y² term. The bounded region forms a loop between x=0 and x=4. The area A = ∫(from 0 to 4) 2|y| dx A = 2 ∫(from 0 to 4) √[x²(4-x)/16] dx = 2 ∫(from 0 to 4) (|x|/4)√(4-x) dx Since x is in [0, 4], |x| = x. A = (1/2) ∫(from 0 to 4) x√(4-x) dx. Use the substitution: Let u = 4-x, so x = 4-u and dx = -du. When x=0, u=4. When x=4, u=0. A = (1/2) ∫(from 4 to 0) (4-u)√u (-du) = (1/2) ∫(from 0 to 4) (4-u)u^(1/2) du = (1/2) ∫(from 0 to 4) [4u^(1/2) – u^(3/2)] du = (1/2) [4 * (u^(3/2))/(3/2) – (u^(5/2))/(5/2)](from 0 to 4) = (1/2) [ (8/3)u^(3/2) – (2/5)u^(5/2) ](from 0 to 4) = (1/2) [ (8/3)(4)^(3/2) – (2/5)(4)^(5/2) – 0 ] = (1/2) [ (8/3)(8) – (2/5)(32) ] = (1/2) [ 64/3 – 64/5 ] = (1/2) 64 [1/3 – 1/5] = 32 [(5-3)/15] = 32 (2/15) = 64/15 square units.

Q5. Trace the curve, y²x² = (x+1)³, stating all the properties, used for tracing it.

Ans. We will analyze the following properties to trace the curve y²x² = (x+1)³:

1. Symmetry:

• The equation contains an even power of y (y²), so if (x, y) is on the curve, then (x, -y) is also on the curve. Thus, the curve is symmetric about the x-axis .
• Replacing x with -x changes the equation, so it is not symmetric about the y-axis.

2. Origin:

• Substituting x=0 and y=0, we get 0 = (0+1)³ = 1, which is false. Therefore, the curve does not pass through the origin .

3. Intercepts:

• x-axis intercept: Set y=0, which gives (x+1)³ = 0, meaning x=-1. So, the curve cuts the x-axis at (-1, 0) .
• y-axis intercept: Set x=0, which gives 0 = 1, an impossibility. This means the curve never cuts the y-axis.

4. Asymptotes:

• Vertical Asymptote: Vertical asymptotes can occur where the denominator is zero. The equation can be written as y² = (x+1)³/x². As x → 0, the denominator x² → 0 and the numerator (x+1)³ → 1. In this case, y² → ∞. Thus, x=0 (the y-axis) is a vertical asymptote .
• Horizontal/Oblique Asymptote: As x → ∞, y² = (x+1)³/x² ≈ x³/x² = x. So, y ≈ ±√x. Since y does not tend to a finite value as x → ∞, there is no horizontal asymptote. For oblique asymptote, m = lim(x→∞) y/x = lim(x→∞) ±√((x+1)³/x⁴) = 0. c = lim(x→∞) (y-mx) = lim(x→∞) y = ∞. So, there is no oblique asymptote either.

5. Region of Existence:

• From the equation y² = (x+1)³/x², for y to be real, y² must be non-negative. Since x² is always positive (at x≠0), we require (x+1)³ ≥ 0.
• This implies x+1 ≥ 0, or x ≥ -1 .
• Thus, the curve exists only for x ≥ -1 and x ≠ 0. The region is [-1, 0) U (0, ∞).

6. Special Points and Tangents:

• At the point (-1, 0), let’s check the nature of the tangent. Near (-1, 0), x is close to -1. y² = (x+1)³/x² ≈ (x+1)³ / (-1)² = (x+1)³. This is the equation of a cusp , with the tangent at the cusp being x+1 = 0 or x = -1 (parallel to the y-axis).

Summary for Sketching:

• The curve starts at x=-1.
• There is a cusp at (-1, 0) with a vertical tangent.
• The curve is symmetric about the x-axis.
• x=0 (the y-axis) is a vertical asymptote, which the curve approaches from both sides.
• No part of the curve exists for x < -1.
• As x increases towards infinity, y also increases, behaving like ±√x.

Sketch: The curve starts at the cusp at (-1, 0), splitting into two branches, one going up and one down. These branches rise towards the y-axis (the asymptote) as x approaches 0 from -1. For x > 0, the curve comes down from ±∞ along the y-axis and then extends to the right, again with two branches symmetric about the x-axis, resembling the shape of a parabola lying on its side. [A mental or actual sketch of the curve would be drawn here, illustrating all these points: a cusp at (-1,0), an asymptote at x=0, and the parabolic-like behavior for x>0.]

Q6. (a) Find the derivative of (cosec x)^(cot x) + (cot x)^(cosec x) with respect to x. (b) Evaluate: ∫(from 0 to π) dx / (1 + 2sin²x). (c) Prove that: tan⁻¹x > x/(1+x²) ∀ x>0.

Ans. (a) Differentiation: Let y = (cosec x)^(cot x) + (cot x)^(cosec x). We can treat this as y = u + v, where u = (cosec x)^(cot x) and v = (cot x)^(cosec x). Then dy/dx = du/dx + dv/dx.

Derivative of u: u = (cosec x)^(cot x) Take logarithm: log u = cot x * log(cosec x). Differentiating with respect to x: (1/u)(du/dx) = [-cosec²x log(cosec x)] + [cot x (1/cosec x) * (-cosec x cot x)] (1/u)(du/dx) = -cosec²x * log(cosec x) – cot²x du/dx = u [-cosec²x log(cosec x) – cot²x] du/dx = (cosec x)^(cot x) * [-cosec²x log(cosec x) – cot²x]

Derivative of v: v = (cot x)^(cosec x) Take logarithm: log v = cosec x * log(cot x). Differentiating with respect to x: (1/v)(dv/dx) = [-cosec x cot x log(cot x)] + [cosec x (1/cot x) * (-cosec²x)] (1/v)(dv/dx) = -cosec x cot x * log(cot x) – cosec³x / cot x (1/v)(dv/dx) = -cosec x cot x * log(cot x) – cosec²x sec x dv/dx = v * [-cosec x cot x log(cot x) – cosec²x sec x] dv/dx = (cot x)^(cosec x) * [-cosec x cot x log(cot x) – cosec²x sec x]

The final answer, dy/dx, is the sum of the two expressions found for du/dx and dv/dx.

(b) Evaluation of the integral: Let I = ∫(from 0 to π) dx / (1 + 2sin²x). We can use the property ∫(from 0 to 2a) f(x)dx = 2∫(from 0 to a) f(x)dx, if f(2a-x) = f(x). Here 2a = π, so a = π/2. The function is f(x) = 1/(1+2sin²x). f(π-x) = 1/(1 + 2sin²(π-x)) = 1/(1 + 2sin²x) = f(x). So, the property applies: I = 2 ∫(from 0 to π/2) dx / (1 + 2sin²x). Now, divide the numerator and denominator by cos²x: I = 2 ∫(from 0 to π/2) sec²x dx / (sec²x + 2tan²x) I = 2 ∫(from 0 to π/2) sec²x dx / (1 + tan²x + 2tan²x) = 2 ∫(from 0 to π/2) sec²x dx / (1 + 3tan²x). Let t = tan x, so dt = sec²x dx. The limits change: when x=0, t=0; when x=π/2, t→∞. I = 2 ∫(from 0 to ∞) dt / (1 + 3t²) = 2 ∫(from 0 to ∞) dt / [1 + (√3 t)²]. This is of the form ∫dx/(a²+x²) = (1/a)tan⁻¹(x/a). I = 2 * [ (1/√3) tan⁻¹(√3 t) ](from 0 to ∞) = (2/√3) [ lim(t→∞) tan⁻¹(√3 t) – tan⁻¹(0) ] = (2/√3) [ π/2 – 0 ] = π/√3 .

(c) Proof of the inequality: We need to prove that tan⁻¹x > x/(1+x²) for all x > 0. Consider a function f(x) = tan⁻¹x – x/(1+x²). We want to show that f(x) > 0 for x > 0. First, let’s find the value of f(0): f(0) = tan⁻¹(0) – 0/(1+0) = 0 – 0 = 0. Now, let’s calculate f'(x) to check if the function is increasing or decreasing. f'(x) = d/dx(tan⁻¹x) – d/dx(x/(1+x²)) = 1/(1+x²) – [ (1+x²)(1) – x(2x) ] / (1+x²)² (using the quotient rule) = 1/(1+x²) – (1 – x²) / (1+x²)² = [ (1+x²) – (1 – x²) ] / (1+x²)² = (1 + x² – 1 + x²) / (1+x²)² = 2x² / (1+x²)². For any x > 0, x² > 0 and (1+x²)² > 0. Therefore, f'(x) = (positive)/(positive) = positive. Since f'(x) > 0 for all x>0, the function f(x) is strictly increasing for x>0. Since f(0)=0 and the function is strictly increasing for x>0, it must be that for any x>0, the value of f(x) is greater than f(0). Thus, f(x) > 0, or tan⁻¹x – x/(1+x²) > 0 for all x>0. Hence, tan⁻¹x > x/(1+x²) for all x>0.

Q7. (a) A manufacturer can sell x items per day at a price of p rupees each, where p = 125 – (1/5)x. The cost of production for x items is given by 500+13x + (1/10)x². Find the number of items she should produce to have maximum profit, assuming that all the items produced are sold. (b) Differentiate sin⁻¹(2x/(1+x²)) with respect to tan⁻¹(2x/(1-x²)). (c) Evaluate: ∫(x³+4x²+3x+8)/(x³-x²+4x-4) dx.

Ans. (a) Maximum Profit: Revenue, R(x) = Number of items × price per item R(x) = x * p = x(125 – x/5) = 125x – x²/5. Cost, C(x) = 500 + 13x + x²/10. Profit, P(x) = Revenue – Cost = R(x) – C(x) P(x) = (125x – x²/5) – (500 + 13x + x²/10) P(x) = 125x – x²/5 – 500 – 13x – x²/10 P(x) = (125-13)x – (x²/5 + x²/10) – 500 P(x) = 112x – (2x²/10 + x²/10) – 500 P(x) = 112x – (3/10)x² – 500. For maximum profit, we set the first derivative of the profit function to zero, dP/dx = 0. dP/dx = d/dx (112x – (3/10)x² – 500) = 112 – (3/10)(2x) = 112 – (3/5)x. Setting dP/dx = 0 => 112 – (3/5)x = 0 => 112 = (3/5)x. x = 112 * 5 / 3 = 560/3 ≈ 186.67. To confirm this is a maximum, we check the second derivative: d²P/dx² = -3/5 < 0. Since the second derivative is negative, x = 560/3 maximizes the profit. Since the number of items must be an integer, we check the profit for the integers closest to 186.67, which are x=186 and x=187. Because the profit function is a downward-opening parabola, the maximum integer value will be one of these. P(186) = 112(186) – 0.3(186)² – 500 = 20832 – 10378.8 – 500 = 9953.2 P(187) = 112(187) – 0.3(187)² – 500 = 20944 – 10490.7 – 500 = 9953.3 The profit is slightly higher for x = 187. Therefore, she should produce 187 items for maximum profit.

(b) Differentiation with respect to another function: Let u = sin⁻¹(2x/(1+x²)) and v = tan⁻¹(2x/(1-x²)). We need to find du/dv, which is equal to (du/dx) / (dv/dx). Use the substitution x = tanθ. This is valid for |x|<1. u = sin⁻¹(2tanθ / (1+tan²θ)) = sin⁻¹(sin(2θ)). For -π/2 ≤ 2θ ≤ π/2, which means -π/4 ≤ θ ≤ π/4, or -1 ≤ x ≤ 1, we have u = 2θ = 2tan⁻¹x. v = tan⁻¹(2tanθ / (1-tan²θ)) = tan⁻¹(tan(2θ)). For -π/2 < 2θ < π/2, which means -1 < x < 1, we have v = 2θ = 2tan⁻¹x. So, for -1 < x < 1, we have u = 2tan⁻¹x and v = 2tan⁻¹x. Clearly, u = v. Now we differentiate: du/dx = d/dx(2tan⁻¹x) = 2/(1+x²). dv/dx = d/dx(2tan⁻¹x) = 2/(1+x²). du/dv = (du/dx) / (dv/dx) = [2/(1+x²)] / [2/(1+x²)] = 1 . The derivative of sin⁻¹(2x/(1+x²)) with respect to tan⁻¹(2x/(1-x²)) is 1.

(c) Evaluation of the integral: Let I = ∫(x³+4x²+3x+8)/(x³-x²+4x-4) dx. The degree of the numerator and denominator is the same, so we perform long division first: (x³+4x²+3x+8) = 1 * (x³-x²+4x-4) + (5x²-x+12). So, the integrand is 1 + (5x²-x+12)/(x³-x²+4x-4). I = ∫ [1 + (5x²-x+12)/(x³-x²+4x-4)] dx. Next, we factor the denominator: x³-x²+4x-4 = x²(x-1) + 4(x-1) = (x-1)(x²+4). Now use partial fractions for the remainder term: (5x²-x+12)/[(x-1)(x²+4)] = A/(x-1) + (Bx+C)/(x²+4) 5x²-x+12 = A(x²+4) + (Bx+C)(x-1) Set x=1: 5-1+12 = A(1+4) => 16 = 5A => A = 16/5. Comparing coefficients of x²: 5 = A+B => B = 5 – A = 5 – 16/5 = 9/5. Comparing constant terms: 12 = 4A – C => C = 4A – 12 = 4(16/5) – 12 = 64/5 – 60/5 = 4/5. So the integral becomes: I = ∫ [1 + (16/5)/(x-1) + ((9/5)x + 4/5)/(x²+4)] dx I = ∫1 dx + (16/5)∫1/(x-1) dx + (1/5)∫(9x+4)/(x²+4) dx I = x + (16/5)ln|x-1| + (9/5)∫x/(x²+4) dx + (4/5)∫1/(x²+4) dx I = x + (16/5)ln|x-1| + (9/5)(1/2)∫2x/(x²+4) dx + (4/5)(1/2)tan⁻¹(x/2) I = x + (16/5)ln|x-1| + (9/10)ln(x²+4) + (2/5)tan⁻¹(x/2) + C .