RBSE Class 12th 2011 Business Mathematics & Statistics-SS-36-1-2011 Previous Year Papers

सही उत्तर: (अ) 0

व्याख्या: फलन f(x) = e^3x सभी x के लिए संतत है। x → 0 पर संततता के लिए, f(0) = lim_x→0 e^3x = e⁰ = 1. दिए गए विकल्पों में 1 नहीं है, अतः प्रश्न में त्रुटि हो सकती है। यदि f(x) = e^3x - 1 हो, तो f(0) = 0 सही होगा। दिए गए विकल्पों में (अ) 0 सबसे उपयुक्त है।

हल: यदि फलन f(x) = 2032 (एक अचर फलन) है, तो यह सभी x के लिए सतत है। अतः f(0) = 2032.

सही उत्तर: (C) 2032

हल: गुणनफल नियम से, d/dx [e^x cos x] = e^x cos x + e^x (– sin x) = e^x (cos x – sin x).

सही उत्तर: (अ) e^x (cos x – sin x)

हल: ∫ 2e^x dx = 2e^x, ∫ (3/x) dx = 3 log |x|, ∫ x² dx = x³/3. अतः समाकल = 2e^x + 3 log |x| – (x³/3) + c.

सही उत्तर: (A) 2e^x + 3 log |x| – (x³/3) + c

हल: मानक समाकल सूत्र से, ∫ [1/(1 + x²)] dx = tan^–1 x + c.

सही उत्तर: (अ) tan^–1 x + c

Solution: Let y = (e^x + 1) / (e^x - 1). Using quotient rule:

dy/dx = [ (e^x - 1)(e^x) - (e^x + 1)(e^x) ] / (e^x - 1)²

= [ e^x(e^x - 1 - e^x - 1) ] / (e^x - 1)²

= [ e^x(-2) ] / (e^x - 1)²

= -2e^x / (e^x - 1)²

Solution: This is of the form 0/0. Using L'Hôpital's rule:

lim_x→1 (1/x) / 1 = lim_x→1 1/x = 1

Alternatively, using the standard limit: lim_x→1 log_e x / (x-1) = 1.

Solution: Revenue function R(x) = x * p(x) = x(30 - x) = 30x - x².

Profit function P(x) = R(x) - C(x) = (30x - x²) - (1000 + 3x) = 30x - x² - 1000 - 3x = -x² + 27x - 1000.

Thus, profit function is P(x) = -x² + 27x - 1000.

Solution: The given equation is y = x(dy/dx) + a/(dy/dx). Multiply both sides by (dy/dx):

y(dy/dx) = x(dy/dx)² + a

Rearranging: x(dy/dx)² - y(dy/dx) + a = 0

The highest order derivative is dy/dx, so order = 1. The power of dy/dx is 2, so degree = 2.

Solution: y = log_a (x + √(a² + x²)) - log_a a = log_a [(x + √(a² + x²))/a]

Using change of base: y = ln[(x + √(a² + x²))/a] / ln a

dy/dx = (1/ln a) * [1/(x + √(a² + x²))] * [1 + (x/√(a² + x²))]

= (1/ln a) * [1/(x + √(a² + x²))] * [(√(a² + x²) + x)/√(a² + x²)]

= (1/ln a) * [1/√(a² + x²)]

= 1 / [√(a² + x²) ln a]

Solution: For continuity at x = 0, we need lim_x→0 f(x) = f(0) = k.

lim_x→0 sin3x / x = lim_x→0 (3 sin3x)/(3x) = 3 * lim_t→0 sin t / t = 3 * 1 = 3.

Therefore, k = 3.

Solution:

For continuity at \( x = 0 \), we require \( \lim_{x \to 0} f(x) = f(0) = k \).

\( \lim_{x \to 0} \frac{\sin 3x}{x} = \lim_{x \to 0} 3 \cdot \frac{\sin 3x}{3x} = 3 \cdot 1 = 3 \).

Hence, \( k = 3 \).

Solution:

Left-hand derivative at \( x = 0 \):

\( Lf'(0) = \lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0} \frac{(2 - h) - 2}{h} = \lim_{h \to 0} \frac{-h}{h} = -1 \).

Right-hand derivative at \( x = 0 \):

\( Rf'(0) = \lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0} \frac{(2 + h) - 2}{h} = \lim_{h \to 0} \frac{h}{h} = 1 \).

Since \( Lf'(0) \neq Rf'(0) \), the function is not differentiable at \( x = 0 \).

Solution:

Let \( y = \log_e \left( \frac{a + b \sin x}{a - b \sin x} \right) \).

Using properties of logarithms: \( y = \log_e (a + b \sin x) - \log_e (a - b \sin x) \).

Differentiating with respect to \( x \):

\( \frac{dy}{dx} = \frac{b \cos x}{a + b \sin x} - \frac{-b \cos x}{a - b \sin x} = \frac{b \cos x}{a + b \sin x} + \frac{b \cos x}{a - b \sin x} \).

\( \frac{dy}{dx} = b \cos x \left( \frac{1}{a + b \sin x} + \frac{1}{a - b \sin x} \right) = b \cos x \cdot \frac{2a}{a^2 - b^2 \sin^2 x} \).

Hence, \( \frac{d}{dx} \left( \log_e \left( \frac{a + b \sin x}{a - b \sin x} \right) \right) = \frac{2ab \cos x}{a^2 - b^2 \sin^2 x} \).

Solution:

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

Substitute \( t = e^x \), then \( dt = e^x \, dx \).

\( I = \int \frac{dt}{\sqrt{4 - t^2}} = \sin^{-1} \left( \frac{t}{2} \right) + C = \sin^{-1} \left( \frac{e^x}{2} \right) + C \).

Solution:

Let \( I = \int \sin^{-1} x \, dx \).

Using integration by parts, take \( u = \sin^{-1} x \) and \( dv = dx \).

Then \( du = \frac{1}{\sqrt{1 - x^2}} \, dx \) and \( v = x \).

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

For the integral, substitute \( t = 1 - x^2 \), then \( dt = -2x \, dx \), so \( x \, dx = -\frac{dt}{2} \).

\( \int \frac{x}{\sqrt{1 - x^2}} \, dx = \int \frac{1}{\sqrt{t}} \left( -\frac{dt}{2} \right) = -\frac{1}{2} \int t^{-1/2} \, dt = -\frac{1}{2} \cdot 2\sqrt{t} + C = -\sqrt{1 - x^2} + C \).

Thus, \( I = x \sin^{-1} x + \sqrt{1 - x^2} + C \).

Solution:

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

Complete the square: \( 2x^2 + x + 4 = 2 \left( x^2 + \frac{x}{2} + 2 \right) = 2 \left[ \left( x + \frac{1}{4} \right)^2 + 2 - \frac{1}{16} \right] = 2 \left[ \left( x + \frac{1}{4} \right)^2 + \frac{31}{16} \right] \).

So, \( I = \frac{1}{2} \int \frac{dx}{\left( x + \frac{1}{4} \right)^2 + \left( \frac{\sqrt{31}}{4} \right)^2} \).

Using the formula \( \int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \), we get:

\( I = \frac{1}{2} \cdot \frac{1}{\frac{\sqrt{31}}{4}} \tan^{-1} \left( \frac{x + \frac{1}{4}}{\frac{\sqrt{31}}{4}} \right) + C = \frac{2}{\sqrt{31}} \tan^{-1} \left( \frac{4x + 1}{\sqrt{31}} \right) + C \).

Solution:

Revenue \( R = p \cdot x = \frac{30 - x}{2} \cdot x = \frac{30x - x^2}{2} \).

To maximize revenue, find \( \frac{dR}{dx} = \frac{30 - 2x}{2} = 15 - x \).

Set \( \frac{dR}{dx} = 0 \): \( 15 - x = 0 \Rightarrow x = 15 \).

Check second derivative: \( \frac{d^2R}{dx^2} = -1 < 0 \), so \( x = 15 \) gives maximum revenue.

Maximum revenue: \( R = \frac{30(15) - (15)^2}{2} = \frac{450 - 225}{2} = \frac{225}{2} = 112.5 \) units.

Thus, production level is 15 units and maximum revenue is 112.5.

Solution:

Total cost \( C(x) = \int MC \, dx = \int \left( \frac{3}{x} + 2 \right) dx = 3 \ln x + 2x + C_0 \).

Fixed cost is 6, so when \( x = 0 \), \( C(0) = 6 \). However, \( \ln 0 \) is undefined, so we consider fixed cost as the constant of integration: \( C(x) = 3 \ln x + 2x + 6 \).

Total cost for 9 units: \( C(9) = 3 \ln 9 + 2(9) + 6 = 3 \ln 9 + 18 + 6 = 3 \ln 9 + 24 \).

Average cost \( AC = \frac{C(9)}{9} = \frac{3 \ln 9 + 24}{9} = \frac{\ln 9}{3} + \frac{8}{3} \).

Since \( \ln 9 = \ln 3^2 = 2 \ln 3 \), we have \( AC = \frac{2 \ln 3}{3} + \frac{8}{3} = \frac{2}{3} \ln 3 + \frac{8}{3} \).

Solution:

Given MC = 80 – 43x + 9 = 89 – 43x. Total cost (TC) = ∫ MC dx = ∫ (89 – 43x) dx = 89x – (43/2)x² + k, where k is fixed cost = 6. So TC = 89x – 21.5x² + 6. For 9 units, TC = 89(9) – 21.5(81) + 6 = 801 – 1741.5 + 6 = –934.5. Average cost = TC/x = –934.5/9 = –103.83 (approx).

Solution:

Increase in cost = ∫₀³⁰⁰ MC dx = ∫₀³⁰⁰ 2000/√(2x+25) dx. Let u = 2x+25, du = 2 dx, dx = du/2. When x=0, u=25; x=300, u=625. So integral = ∫₂₅⁶²⁵ 2000/√u * (du/2) = 1000 ∫₂₅⁶²⁵ u⁻¹/² du = 1000 [2√u]₂₅⁶²⁵ = 2000 (√625 – √25) = 2000 (25 – 5) = 2000 × 20 = 40000. Increase in cost = Rs. 40,000.

Solution:

For x = -3, y = 2⁻³ = 1/8 = 0.125; x = -2, y = 1/4 = 0.25; x = -1, y = 1/2 = 0.5; x = 0, y = 1; x = 1, y = 2; x = 2, y = 4; x = 3, y = 8. Plot these points and draw a smooth curve passing through them. The graph is an increasing exponential curve passing through (0,1) and approaching the x-axis as x decreases.

Solution:

Let sin x = t, then cos x dx = dt. When x=0, t=0; x=π/2, t=1. Integral = ∫₀¹ 2 / [(1+t)(2+t)] dt. Use partial fractions: 2/[(1+t)(2+t)] = A/(1+t) + B/(2+t). Solving gives A=2, B=-2. So integral = ∫₀¹ [2/(1+t) – 2/(2+t)] dt = 2[ln|1+t| – ln|2+t|]₀¹ = 2[(ln2 – ln3) – (ln1 – ln2)] = 2[ln2 – ln3 + ln2] = 2[2ln2 – ln3] = 2 ln(4/3) = ln(16/9).

Solution:

The curves are y² = ax (a right-opening parabola) and x² + y² = 4ax (a circle with center (2a,0) and radius 2a). Intersection: substitute y² = ax into circle: x² + ax = 4ax → x² – 3ax = 0 → x(x – 3a) = 0 → x=0 or x=3a. For x=0, y=0; for x=3a, y²=3a² → y=±√3 a. Area between x-axis and curves from x=0 to x=3a: For x from 0 to 3a, y from parabola = √(ax) and from circle = √(4ax – x²). The required area = ∫₀³ᵃ [√(4ax – x²) – √(ax)] dx. Compute separately: ∫√(4ax – x²) dx = ∫√(4a² – (x-2a)²) dx = (1/2)[(x-2a)√(4ax – x²) + 4a² sin⁻¹((x-2a)/(2a))] + C. Evaluate from 0 to 3a: at x=3a, √(12a² – 9a²)=√3 a, (x-2a)=a, so term = (1/2)[a·√3 a + 4a² sin⁻¹(1/2)] = (1/2)[√3 a² + 4a²(π/6)] = (√3/2)a² + (π/3)a². At x=0, √(0)=0, (x-2a)=-2a, sin⁻¹(-1)=-π/2, term = (1/2)[0 + 4a²(-π/2)] = -πa². So definite = [(√3/2)a² + (π/3)a²] – [-πa²] = (√3/2)a² + (π/3)a² + πa² = (√3/2)a² + (4π/3)a². Next, ∫√(ax) dx = (2/3)√a x^(3/2). From 0 to 3a: (2/3)√a (3a)^(3/2) = (2/3)√a · 3√3 a^(3/2) = 2√3 a². So area = [(√3/2)a² + (4π/3)a²] – 2√3 a² = (√3/2 – 2√3)a² + (4π/3)a² = (-3√3/2)a² + (4π/3)a² = a²(4π/3 – 3√3/2).

Solution:

Profit P(x) = R – C = 3x – (100 + 0.015x²) = 3x – 100 – 0.015x². For maximum, dP/dx = 3 – 0.03x = 0 → x = 100 units. Second derivative d²P/dx² = –0.03 < 0, so maximum. Maximum profit = 3(100) – 100 – 0.015(10000) = 300 – 100 – 150 = Rs. 50.

Solution:

Revenue R(x) = 9000x. At break-even, R(x) = C(x) → 9000x = 500x² + 4500x + 10000 → 500x² – 4500x + 10000 = 0 → Divide by 500: x² – 9x + 20 = 0 → (x – 4)(x – 5) = 0 → x = 4 or x = 5. So break-even points are at 4 units and 5 units.

हल:

दिया गया अवकल समीकरण है:

dy/dx = x log₂ x

⇒ dy = x log₂ x dx

दोनों पक्षों का समाकलन करने पर:

∫ dy = ∫ x log₂ x dx

⇒ y = ∫ x log₂ x dx

माना log₂ x = t, तो x = 2^t और dx = 2^t log 2 dt

y = ∫ 2^t · t · 2^t log 2 dt = log 2 ∫ t · 2^{2t} dt

खण्डशः समाकलन से:

y = log 2 [ t · (2^{2t}/2 log 2) - ∫ (2^{2t}/2 log 2) dt ]

y = (t · 2^{2t})/2 - (2^{2t})/(4 log 2) + C

t = log₂ x पुनः रखने पर:

y = (x² log₂ x)/2 - x²/(4 log 2) + C

या y = (x²/2) log₂ x - x²/(4 log 2) + C

हल:

दिया गया अवकल समीकरण है:

dy/dx = e^x + x e^y

⇒ dy/dx - x e^y = e^x

यह रैखिक अवकल समीकरण नहीं है। इसे चर पृथक्करण विधि से हल करते हैं:

dy/dx = e^x + x e^y

⇒ dy/dx = e^x (1 + x e^{y-x})

माना y - x = v, तो dy/dx = dv/dx + 1

dv/dx + 1 = e^x (1 + x e^v)

⇒ dv/dx = e^x + x e^{x+v} - 1

यह सरल नहीं है। वैकल्पिक विधि:

dy/dx - e^x = x e^y

e^{-y} dy/dx - e^{x-y} = x

माना e^{-y} = u, तो du/dx = -e^{-y} dy/dx

-du/dx - e^x u = x

⇒ du/dx + e^x u = -x

समाकलन गुणांक = e^{∫ e^x dx} = e^{e^x}

u · e^{e^x} = ∫ -x e^{e^x} dx + C

यह समाकल सरल नहीं है। अतः दिए गए समीकरण का हल जटिल है।

हल:

f(x) = (x-1)(x-2)(x-3) = x³ - 6x² + 11x - 6

f(x) एक बहुपदीय फलन है, अतः यह [0,4] में सतत तथा (0,4) में अवकलनीय है।

f(0) = (0-1)(0-2)(0-3) = (-1)(-2)(-3) = -6

f(4) = (4-1)(4-2)(4-3) = (3)(2)(1) = 6

लैग्रान्ज प्रमेय के अनुसार, कम से कम एक c ∈ (0,4) इस प्रकार होगा कि:

f'(c) = [f(4) - f(0)] / (4 - 0) = [6 - (-6)] / 4 = 12/4 = 3

अब f'(x) = 3x² - 12x + 11

f'(c) = 3c² - 12c + 11 = 3

⇒ 3c² - 12c + 8 = 0

⇒ c = [12 ± √(144 - 96)] / 6 = [12 ± √48] / 6 = [12 ± 4√3] / 6 = 2 ± (2√3)/3

c₁ = 2 + 2√3/3 ≈ 3.155, c₂ = 2 - 2√3/3 ≈ 0.845

दोनों मान (0,4) में स्थित हैं। अतः लैग्रान्ज मध्य मान प्रमेय सत्यापित होता है।

हल:

z = x³ - 18x² + 96x

dz/dx = 3x² - 36x + 96

क्रान्तिक बिन्दुओं के लिए dz/dx = 0

3x² - 36x + 96 = 0

⇒ x² - 12x + 32 = 0

⇒ (x - 4)(x - 8) = 0

⇒ x = 4, x = 8

दोनों मान [0,9] में स्थित हैं।

अब अन्तराल के अन्त बिन्दुओं और क्रान्तिक बिन्दुओं पर मान ज्ञात करते हैं:

z(0) = 0 - 0 + 0 = 0

z(4) = 64 - 288 + 384 = 160

z(8) = 512 - 1152 + 768 = 128

z(9) = 729 - 1458 + 864 = 135

अतः अन्तराल [0,9] में:

अधिकतम मान: 160 (x = 4 पर)

न्यूनतम मान: 0 (x = 0 पर)

हल:

∫₁² x² dx = lim_{n→∞} (b-a)/n ∑_{r=1}^{n} f(a + r·(b-a)/n)

यहाँ a = 1, b = 2, h = (b-a)/n = 1/n

f(x) = x²

∫₁² x² dx = lim_{n→∞} (1/n) ∑_{r=1}^{n} (1 + r/n)²

= lim_{n→∞} (1/n) ∑_{r=1}^{n} (1 + 2r/n + r²/n²)

= lim_{n→∞} (1/n) [∑_{r=1}^{n} 1 + (2/n)∑_{r=1}^{n} r + (1/n²)∑_{r=1}^{n} r²]

= lim_{n→∞} (1/n) [n + (2/n)·n(n+1)/2 + (1/n²)·n(n+1)(2n+1)/6]

= lim_{n→∞} [1 + (n+1)/n + (n+1)(2n+1)/(6n²)]

= lim_{n→∞} [1 + 1 + 1/n + (2n² + 3n + 1)/(6n²)]

= lim_{n→∞} [2 + 1/n + 1/3 + 1/(2n) + 1/(6n²)]

= 2 + 0 + 1/3 + 0 + 0

= 7/3

अतः ∫₁² x² dx = 7/3

हल:

यहाँ वर्षों का अन्तराल समान (5 वर्ष) है। अतः हम अन्तर्वेशन सूत्र का प्रयोग करेंगे।

माना x = वर्ष, y = लाभ (₹ 1000 में)

x₀ = 1980, x₁ = 1985, x₂ = 1990, x₃ = 1995

y₀ = 36.06, y₁ = 39.2, y₂ = 42.8, y₃ = 47.38

h = 5

अग्र अन्तर सारणी:

x = 1986 के लिए, p = (x - x₀)/h = (1986 - 1980)/5 = 6/5 = 1.2

न्यूटन के अग्र अन्तर्वेशन सूत्र से:

y(1986) = y₀ + p·Δy₀ + p(p-1)/2! · Δ²y₀ + p(p-1)(p-2)/3! · Δ³y₀

= 36.06 + 1.2(3.14) + (1.2 × 0.2)/2 × 0.46 + (1.2 × 0.2 × (-0.8))/6 × 0.52

= 36.06 + 3.768 + 0.0552 - 0.01664

= 39.86656

अतः वर्ष 1986 के लिए आनुमानित लाभ ≈ ₹ 39,867 (लगभग) है।