RBSE Class 12th 2010 Business Mathematics & Statistics-SS-36-1-2010 Previous Year Papers

हल:

सर्वप्रथम, \(\cos^2 x - \sin^2 x = (\cos x - \sin x)(\cos x + \sin x)\)

अतः \(\frac{\cos^2 x - \sin^2 x}{\cos x + \sin x} = \cos x - \sin x\)

अब \(\frac{d}{dx}(\cos x - \sin x) = -\sin x - \cos x\)

अतः अभीष्ट मान \(-\sin x - \cos x\) है।

हल:

\(\lim_{x \to 0} \frac{e^x - e^{-x}}{x}\)

\(e^x\) का विस्तार: \(e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\)

\(e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \cdots\)

अतः \(e^x - e^{-x} = 2x + \frac{2x^3}{3!} + \cdots\)

\(\frac{e^x - e^{-x}}{x} = 2 + \frac{2x^2}{3!} + \cdots\)

जब \(x \to 0\), तब सीमा = 2

अतः \(\lim_{x \to 0} \frac{e^x - e^{-x}}{x} = 2\)

Solution:

Given, selling price per unit = Rs. 0 (This appears to be an error in the question; typically price is given. Assuming price is Rs. p per unit, but here it is 0, which implies revenue is 0. However, based on standard interpretation, let quantity = x units. Then total revenue (R) = 0 × x = 0. Variable cost = 30% of total revenue = 30% of 0 = 0. Fixed cost = Rs. 35,000. Total cost (C) = Fixed cost + Variable cost = 35,000 + 0 = 35,000. So total cost function is constant: C = 35,000.

Correct Answer: Total cost function C(x) = 35,000 (constant).

Solution:

We know that sin² x + cos² x = 1 (trigonometric identity).

So, dy/dx = 1.

Integrating both sides with respect to x:

∫ dy = ∫ 1 dx

y = x + C, where C is the constant of integration.

Correct Answer: y = x + C

Solution:

Given y = log(sec x – tan x).

First derivative: dy/dx = (1/(sec x – tan x)) × (sec x tan x – sec² x) = (sec x (tan x – sec x))/(sec x – tan x) = – sec x (sec x – tan x)/(sec x – tan x) = – sec x.

So, dy/dx = – sec x.

Second derivative: d²y/dx² = – sec x tan x.

Correct Answer: d²y/dx² = – sec x tan x

Solution:

For continuity at x = 2, left-hand limit (LHL) = right-hand limit (RHL) = f(2).

LHL = lim_x→2⁻ f(x) = lim_x→2 (kx + 1) = 2k + 1.

RHL = lim_x→2⁺ f(x) = 5.

f(2) = k(2) + 1 = 2k + 1.

For continuity: 2k + 1 = 5 ⇒ 2k = 4 ⇒ k = 2.

Correct Answer: k = 2

Solution:

First, check continuity at x = 2:

LHL = lim_x→2⁻ (x + 2) = 4.

RHL = lim_x→2⁺ (x² + x – 2) = 4 + 2 – 2 = 4.

f(2) = 2 + 2 = 4. So function is continuous at x = 2.

Now check differentiability:

Left-hand derivative (LHD) at x = 2: f'(2⁻) = d/dx (x + 2) = 1.

Right-hand derivative (RHD) at x = 2: f'(2⁺) = d/dx (x² + x – 2) = 2x + 1. At x = 2, RHD = 2(2) + 1 = 5.

Since LHD (1) ≠ RHD (5), the function is not differentiable at x = 2.

Correct Answer: The function is not differentiable at x = 2.

Solution:

Let \(y = \tan^{-1}\left(\frac{1 - \tan x}{1 + \tan x}\right)\).

We know \(\tan\left(\frac{\pi}{4} - x\right) = \frac{1 - \tan x}{1 + \tan x}\).

So, \(y = \tan^{-1}\left[\tan\left(\frac{\pi}{4} - x\right)\right] = \frac{\pi}{4} - x\) (for principal values).

Differentiating with respect to \(x\):

\(\frac{dy}{dx} = -1\).

Solution:

Let \(I = \int e^{2x} \sin x \, dx\).

Using integration by parts (taking \(e^{2x}\) as first function and \(\sin x\) as second):

\(I = e^{2x}(-\cos x) - \int 2e^{2x}(-\cos x) \, dx = -e^{2x}\cos x + 2\int e^{2x}\cos x \, dx\).

Now, \(\int e^{2x}\cos x \, dx = e^{2x}\sin x - \int 2e^{2x}\sin x \, dx = e^{2x}\sin x - 2I\).

Substituting: \(I = -e^{2x}\cos x + 2[e^{2x}\sin x - 2I] = -e^{2x}\cos x + 2e^{2x}\sin x - 4I\).

\(5I = e^{2x}(2\sin x - \cos x)\).

\(I = \frac{e^{2x}}{5}(2\sin x - \cos x) + C\).

Solution:

Let \(I = \int \frac{x^2 + 1}{x^4 + 1} \, dx\).

Divide numerator and denominator by \(x^2\):

\(I = \int \frac{1 + \frac{1}{x^2}}{x^2 + \frac{1}{x^2}} \, dx = \int \frac{1 + \frac{1}{x^2}}{(x - \frac{1}{x})^2 + 2} \, dx\).

Let \(t = x - \frac{1}{x}\), then \(dt = (1 + \frac{1}{x^2})dx\).

\(I = \int \frac{dt}{t^2 + 2} = \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{t}{\sqrt{2}}\right) + C\).

\(I = \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{x - \frac{1}{x}}{\sqrt{2}}\right) + C = \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{x^2 - 1}{\sqrt{2}x}\right) + C\).

Solution:

Let \(I = \int \frac{dx}{\sqrt{4 + 3x - 2x^2}}\).

Rewrite denominator: \(4 + 3x - 2x^2 = -2\left(x^2 - \frac{3}{2}x - 2\right) = -2\left[(x - \frac{3}{4})^2 - \frac{9}{16} - 2\right] = -2\left[(x - \frac{3}{4})^2 - \frac{41}{16}\right] = 2\left[\frac{41}{16} - (x - \frac{3}{4})^2\right]\).

So, \(I = \int \frac{dx}{\sqrt{2\left[\frac{41}{16} - (x - \frac{3}{4})^2\right]}} = \frac{1}{\sqrt{2}} \int \frac{dx}{\sqrt{\left(\frac{\sqrt{41}}{4}\right)^2 - (x - \frac{3}{4})^2}}\).

Using formula \(\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\frac{x}{a}\right) + C\):

\(I = \frac{1}{\sqrt{2}} \sin^{-1}\left(\frac{x - \frac{3}{4}}{\frac{\sqrt{41}}{4}}\right) + C = \frac{1}{\sqrt{2}} \sin^{-1}\left(\frac{4x - 3}{\sqrt{41}}\right) + C\).

Solution:

Profit function: \(P(x) = R - C = 2x - (50 + 0.10x^2) = 2x - 50 - 0.10x^2\).

For maximum profit, \(\frac{dP}{dx} = 0\) and \(\frac{d^2P}{dx^2} < 0\).

\(\frac{dP}{dx} = 2 - 0.20x = 0 \Rightarrow x = 10\).

\(\frac{d^2P}{dx^2} = -0.20 < 0\) (maximum).

Thus, the company should produce 10 units for maximum profit.

Solution:

Marginal cost \(MC = \frac{dC}{dx} = 5 - 3x^2\) (assuming \(0x\) is a typo and should be 0).

Total cost function: \(C(x) = \int MC \, dx = \int (5 - 3x^2) \, dx = 5x - x^3 + k\), where \(k\) is constant.

Given \(C(2) = 22\): \(5(2) - (2)^3 + k = 22 \Rightarrow 10 - 8 + k = 22 \Rightarrow k = 20\).

So, total cost function: \(C(x) = 5x - x^3 + 20\).

Average cost function: \(AC(x) = \frac{C(x)}{x} = \frac{5x - x^3 + 20}{x} = 5 - x^2 + \frac{20}{x}\).

Solution:

कुल आय में वृद्धि = ∫₀²⁰ MR dx = ∫₀²⁰ (100 – 2x) dx

= [100x – x²]₀²⁰ = (100×20 – 20²) – (0 – 0) = 2000 – 400 = 1600

अतः कुल आय में 1600 इकाई की वृद्धि होगी।

Solution:

फलन y = ½ cos 2x का आवर्तकाल π है। x ∈ [–π, π] में दो पूर्ण आवर्त होंगे।

मुख्य बिंदु:

• x = –π पर, y = ½ cos(–2π) = ½
• x = –3π/4 पर, y = ½ cos(–3π/2) = 0
• x = –π/2 पर, y = ½ cos(–π) = –½
• x = –π/4 पर, y = ½ cos(–π/2) = 0
• x = 0 पर, y = ½ cos 0 = ½
• x = π/4 पर, y = ½ cos(π/2) = 0
• x = π/2 पर, y = ½ cos π = –½
• x = 3π/4 पर, y = ½ cos(3π/2) = 0
• x = π पर, y = ½ cos 2π = ½

इन बिंदुओं को जोड़कर एक कोसाइन वक्र प्राप्त होता है जिसका आयाम ½ है।

Solution:

माना sin x = t, तब cos x dx = dt

जब x = 0, t = 0; जब x = π/2, t = 1

समाकल = ∫₀¹ dt / [(1 + t)(2 + t)]

आंशिक भिन्नों में वियोजन:

1/[(1 + t)(2 + t)] = 1/(1 + t) – 1/(2 + t)

समाकल = ∫₀¹ [1/(1 + t) – 1/(2 + t)] dt

= [log|1 + t| – log|2 + t|]₀¹

= [log(1 + t)/(2 + t)]₀¹

= log(2/3) – log(1/2) = log(2/3 × 2/1) = log(4/3)

अतः अभीष्ट मान = log(4/3)

Solution:

प्रतिच्छेद बिंदु ज्ञात करने के लिए:

y² = 4x और x² = 4y ⇒ y = x²/4

(x²/4)² = 4x ⇒ x⁴/16 = 4x ⇒ x⁴ = 64x ⇒ x(x³ – 64) = 0

⇒ x = 0 या x = 4

जब x = 0, y = 0; जब x = 4, y = 4

प्रतिच्छेद बिंदु (0,0) और (4,4) हैं।

अभीष्ट क्षेत्रफल = ∫₀⁴ (√(4x) – x²/4) dx

= ∫₀⁴ (2√x – x²/4) dx

= [2 × (2/3)x^(3/2) – x³/12]₀⁴

= [(4/3)×8 – 64/12] – 0

= 32/3 – 16/3 = 16/3 वर्ग इकाई

Solution:

कुल आय R = p × x = (10 – x/2)x = 10x – x²/2

कुल लागत C = x²/2 + 5x + 5

लाभ P = R – C = (10x – x²/2) – (x²/2 + 5x + 5)

= 10x – x²/2 – x²/2 – 5x – 5

= 5x – x² – 5

अधिकतम लाभ के लिए, dP/dx = 0

dP/dx = 5 – 2x = 0 ⇒ x = 2.5

द्वितीय अवकलज जाँच: d²P/dx² = –2 < 0, अतः x = 2.5 पर लाभ अधिकतम है।

अधिकतम लाभ = 5(2.5) – (2.5)² – 5 = 12.5 – 6.25 – 5 = 1.25

अतः उत्पादन की 2.5 इकाई पर अधिकतम लाभ 1.25 रुपये होगा।

Solution:

कुल आय R = 6500x

कुल लागत C = 200x² + 3500x + 10000

लाभ-हानि रहित बिन्दु पर, R = C

6500x = 200x² + 3500x + 10000

200x² + 3500x – 6500x + 10000 = 0

200x² – 3000x + 10000 = 0

200 से भाग देने पर: x² – 15x + 50 = 0

(x – 5)(x – 10) = 0

⇒ x = 5 या x = 10

अतः लाभ-हानि रहित बिन्दु 5 इकाई और 10 इकाई पर हैं।

Solution (First equation):

Given: dy/dx + xy = x

This is a linear differential equation of the form dy/dx + P(x)y = Q(x), where P(x) = x and Q(x) = x.

Integrating factor (I.F.) = e^{∫P dx} = e^{∫x dx} = e^x²/2

Solution: y × I.F. = ∫(Q × I.F.) dx

⇒ y e^x²/2 = ∫ x e^x²/2 dx

Let t = x²/2, then dt = x dx

⇒ y e^x²/2 = ∫ e^t dt = e^t + C = e^x²/2 + C

⇒ y = 1 + C e^-x²/2

Solution (Second equation - OR):

Given: x² dy/dx + xy = 1

Divide by x²: dy/dx + (1/x)y = 1/x²

This is linear with P(x) = 1/x and Q(x) = 1/x².

I.F. = e^{∫(1/x) dx} = e^{ln x} = x

Solution: y × x = ∫(1/x² × x) dx = ∫(1/x) dx = ln|x| + C

⇒ y = (ln|x| + C)/x

Solution:

Given f(x) = x² - 5x + 6, interval [2, 3].

1. f(x) is a polynomial, so it is continuous on [2, 3] and differentiable on (2, 3).

2. Check f(2) = 4 - 10 + 6 = 0

f(3) = 9 - 15 + 6 = 0

Thus f(2) = f(3) = 0.

3. By Rolle's theorem, there exists at least one c ∈ (2, 3) such that f'(c) = 0.

f'(x) = 2x - 5

Set f'(c) = 0 ⇒ 2c - 5 = 0 ⇒ c = 2.5

Since 2.5 ∈ (2, 3), Rolle's theorem is verified.

Solution:

The general term of the series is r²/(r³ + n²) for r = 1, 2, ..., n.

We need: S = lim_n→∞ Σ_r=1ⁿ r²/(r³ + n²)

Divide numerator and denominator by n²:

S = lim_n→∞ Σ_r=1ⁿ (r²/n²) / (r³/n² + 1)

Let x = r/n, then as n→∞, Σ becomes an integral from 0 to 1.

r²/n² = x², and r³/n² = n x³, which tends to ∞ unless x=0. This approach needs correction.

Alternatively, rewrite as:

S = lim_n→∞ (1/n) Σ_r=1ⁿ (r²/n) / (r³/n² + 1)

Let x = r/n, then r = nx, r² = n²x², r³ = n³x³.

Term = (n²x²)/(n³x³ + n²) = x²/(nx³ + 1)

S = lim_n→∞ Σ (1/n) × [x²/(nx³ + 1)]

As n→∞, nx³ → ∞ for x>0, so term → 0 for x>0. At x=0, term = 0.

Thus S = 0.

Answer: 0

Solution:

Given f(x) = sin x (1 + cos x) = sin x + sin x cos x = sin x + (1/2) sin 2x

f'(x) = cos x + cos 2x

For maxima/minima, f'(x) = 0:

cos x + cos 2x = 0

Using cos 2x = 2cos²x - 1:

cos x + 2cos²x - 1 = 0

2cos²x + cos x - 1 = 0

(2cos x - 1)(cos x + 1) = 0

⇒ cos x = 1/2 or cos x = -1

For x ∈ (0, π), cos x = 1/2 gives x = π/3, and cos x = -1 gives x = π.

Now f''(x) = -sin x - 2 sin 2x

At x = π/3: f''(π/3) = -sin(π/3) - 2 sin(2π/3) = -√3/2 - 2(√3/2) = -√3/2 - √3 = -3√3/2 < 0

Since f''(π/3) < 0, the function has a maximum at x = π/3.

Hence proved.

Solution:

We need to find the premium at age 25 using interpolation. Given data points:

Age (x): 20, 24, 28, 32

Premium (y): 427, 58, 77, 996

Note: The data seems inconsistent (premium drops from 427 to 58 then rises). Assuming the values are correct, we use Lagrange interpolation for x = 25.

Using Lagrange's formula:

y(25) = Σ y_i × L_i(25), where L_i(x) = Π_j≠i (x - x_j)/(x_i - x_j)

For x=25:

L₀(25) = (25-24)(25-28)(25-32) / [(20-24)(20-28)(20-32)] = (1)(-3)(-7) / [(-4)(-8)(-12)] = 21 / (-384) = -21/384

L₁(25) = (25-20)(25-28)(25-32) / [(24-20)(24-28)(24-32)] = (5)(-3)(-7) / [(4)(-4)(-8)] = 105 / 128

L₂(25) = (25-20)(25-24)(25-32) / [(28-20)(28-24)(28-32)] = (5)(1)(-7) / [(8)(4)(-4)] = -35 / (-128) = 35/128

L₃(25) = (25-20)(25-24)(25-28) / [(32-20)(32-24)(32-28)] = (5)(1)(-3) / [(12)(8)(4)] = -15 / 384

y(25) = 427 × (-21/384) + 58 × (105/128) + 77 × (35/128) + 996 × (-15/384)

= (-8967/384) + (6090/128) + (2695/128) + (-14940/384)

= (-8967 - 14940)/384 + (6090 + 2695)/128

= (-23907)/384 + (8785)/128

= (-23907)/384 + (26355)/384

= 2448/384 = 6.375

Answer: The annual net premium at age 25 is approximately 6.38 (units as per table).