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\(\int\frac{4t^2+10}{(t^2+3)(t^2+4)}dt\) = \(\int\frac{4(t^2+3)+2((t^2+4)-(t^2+3))}{(t^2+3)(t^2+4)}dt\)

\(=\int(\frac4{t^2+4}-\frac2{t^2+3}+\frac3{t^2+4})dt\)

\(=\int\frac{6}{t^2+4}dt-\int\frac{2}{t^2+3}dt\) 

\(=\frac62tan^{-1}\frac{t}2-\frac2{\sqrt3}tan^{-1}\frac{t}{\sqrt3}+c\) 

 = 3 tan^-1\(\frac{t}2\) \(-\frac{2}{\sqrt3}tan^{-1}\frac{t}{\sqrt3}+c\)

I =\(\int\limits_{\pi/2}^{\pi/2}\)|sin x + cos x|dx

 = \(\int\limits_{-\pi/2}^{-\pi/4}\)-(sin x + cos x)dx + \(\int\limits_{-\pi/4}^{\pi/2}\)(sin x + cos x)dx

 = \(-[-cos x + sin x]_{-\pi/2}^{-\pi/4}dx\) + \([-cos x+sin x]_{-\pi/4}^{\pi/2}\)

 = -(-cos π/4 - sin π/4) + (-cos π/2  - sin π/2) + (- cos π/2 + sin π/2) - (-cos π/4 - sin π/4)

(\(\because\) cos(-θ) + cos θ and sin(-θ) = -sin θ)

 = cos π/4 + sin π/4 - cos π/2 - sin π/2 - cos π/2 + sin π/2 + cos π/4 + sin π/4

 = 2 cos π/4 + 2 sin π/4 - 2 cos π/2

 = 2 x \(\frac1{\sqrt2}\) + 2 x \(\frac1{\sqrt2}\) - 2 x 0

 = \(\sqrt2+\sqrt2\)

 = 2\(\sqrt2\) 

Correct option (a) 1/2

Explanation:

Differentiate both sides w.r.t x

⇒ f(x) = 1 + (0 - xf(x)) ⇒f(x) = 1/1 + x

f(1) = 1/2

(a) log 2 sq.units

Area = \(\int\limits _1^2\)1/x dx

= \((logx)_1^2\)

= log 2 - log 1

= log 2(Since log 1 = 0)

Given, marginal Revenue function MR = 9 – 4x²

Revenue function, R = ∫(MR) dx + k

R = \(\int\)(9 - 4x²)dx + k

R = 9x - 4/3x³ + k

Since R = 0 when x = 0, k = 0

R = 9 x - 4/3 x³

Demand funcatiuon P = R/x

P = 9 - 4/3x²

(a) e³ – 1 

Area = \(\int\limits_0^3\)e^xdx = \((e^x)_0^3\) = e³ - 1

(a) log 5

Area = \(\int\limits_1^5 1/xdx = (logx)_1^5 = log 5\)

C = ∫(MC) dx + k 

= ∫(3x² – 2x + 8) dx + k 

= x³ – x² + 8x + k

No fixed cost ⇒ k = 0 

So total cost, C = x³ – x² + 8x

(d) log|x + 1| + k

Revenue function R = ∫1/(x + 1)dx = log|x + 1| + k

(a) – (iii) 

(b) – (iv) 

(c) – (i) 

(d) – (ii)

p₀ = 35 – 2(3) – 3³ 

= 35 – 6 – 9 

= 20

MC = C'(x) = 5 + 0.13x 

C(x) = ∫C'(x) dx + k₁ 

= ∫(5 + 0.13x) dx + k₁ 

= 5x + 0.13/2 x² + k₁ 

When quantity produced is zero, fixed cost is 120 

(i.e) When x = 0, C = 120 ⇒ k₁ = 120 

Cost function is 5x + 0.065x² + 120 

Now given MR = R'(x) = 18 

R(x) = ∫18 dx + k₁ = 18x + k₂ 

When x = 0, R = 0 ⇒ k₂ = 0 

Revenue = 18x 

Profit P = Total Revenue – Total cost 

= 18x – (5x + 0.065x² + 120) 

Profit function = 13x – 0.065x² – 120

Given C'(x) = 10.6x 

C'(x) = ∫(10.6x) dx + k 

= 10.6 x²/2 + k 

= 5.3x² + k 

The fixed cost is Rs. 50 

(i.e.) when x = 0, c = 50 ⇒ k = 50 

Hence cost function C = 5.3x² + 50 

Now, total revenue = number of units sold × price/unit 

Since x be the number of units sold. 

The selling price is Rs.5 per unit. 

Revenue R(x) = 5x 

Profit P = Total revenue – Total cost 

= 5x – (5.3x² + 50) 

= 5x – 5.3x² – 50

(a) MR 

MR = k (constant) 

Revenue function R = ∫(MR) dx + c₁ 

= ∫kdx + c₁ 

= kx + c₁ 

When R = 0, x = 0, ⇒ c₁ = 0 

R = kx

Demand function p = R/x = kx/x = k constant 

⇒ p = MR

(a) 2

D(x) = 16 – x² , S(x) = 2x² + 4 

Under equilibriuim, D(x) = S(x) 

⇒ 16 – x² = 2x² + 4 

⇒ 3x² = 12 

⇒ x = ±2. 

Since x cannot be negative x = 2.

(c) 40/3

D(x) = S(x) in equilibrium 

20 – 5x = 4x + 8 

9x = 12 

x = 4/3 = x₀ 

p₀ = 20 – 5(4/3)

= 20 - 20/3

= 40/3

(a) P = ∫(MR – MC) dx + k

(d) 54x – 9x²/2 + k

Profit = ∫(MR – MC) dx + k 

= ∫(30 – 60) – (-24 + 3x) dx + k

= ∫(54 – 9x) dx + k 

= 54x – 9x²/2 + k

(c) 9 

The maximum profit occurs when MR – MC = 0 

⇒ 36x – 3x² – 81 = 0 

⇒ x² – 12x + 27 = 0 

⇒ (x – 9)(x – 3) = 0 

⇒ x = 9, 3

Now = dp/dx 36x² – 3x – 81 ⇒ d²p/dx² = 36 – 9x 

At x = 9, d²p/dx² = 36 – 81 < 0 

At x = 3, d²p/dx² = 36 – 27 > 0 

Therefore profit is maximum when x = 9.

f'(x) = e^x 

Integrating both sides of the equation, 

∫ f'(x) dx = ∫ e^x dx 

⇒ f(x) = e^x + c 

given f(0) = 2 

2 = e⁰ + c 

⇒ c = 1 

Thus f(x) = e + 1

∫2 cos x – 3 sin x + 4 sec²x – 5 cosec²x 

= 2 ∫ cos x dx – 3 ∫sin x dx + 4 ∫ sec²x – 5 ∫ cosec²x dx 

= 2 sin x + 3 cos x + 4 tan x + 5 cot x + c

The correct answer is : (d) cos x

(c) 9 log|x – 3| – log|x + 1| + c 

(i) ∫1/(p + qx) dx = (1/q) log(p + qx) + c

(ii) ∫cos(4x + 5) dx = sin(4x + 5)/4 

= (1/4) sin(4x + 5) + c 

(i) ∫cosec²(7 - 11x) dx = -cot(7 - 11x)/-11 

= (1/11) cot(7 - 11x) + c

(ii) ∫sec(3 + x) tan(3 + x) dx = sec(3 + x) + c 

(i) ∫cosec(3 – 2x) cot(3 – 2x) dx = -cosec(3 - 2x)/-2 

= (1/2) cosec(3 - 2x) + c

(ii) ∫e^{3x + 2} dx = e^{3x + 2}/3 + c 

Dividing numerator and denominator by ex, we get

∫(e^x - e^-x)/(e^x + e^-x)dx

Now d/dx(e^x + e^-x) 

= e^x - e^-x 

= log(e^x + e^-x) +c

(c) \(\int\limits_{a}^{b}\)f(x)dx

∫1/√((x + 3)² - 2²)dx 

log|(x + 3) + √(x² + 6x + 5| + c

(d) \(\int\limits_{0}^{∞}\)e^-xx^{n - 1}dx

(c) 2^x/log2 + c

(a) (n – 1)! 

Γ(n) = Γ(n – 1) + 1 = (n – 1)!

(c) 2/3x^3/2 + 2x^1/2 + c

MR = 14 – 6x + 9x² 

R = ∫(14 – 6x + 9x²) dx + k 

= 14x – 3x² + 3x³ + k 

Since R = 0, when x = 0, k = 0 

So revenue function R = 14x – 3x² + 3x³ 

Demand function P = R/x = 14 – 3x + 3x²

(i) ∫(1 + x²)^-1 dx = ∫1/(1 + x²) dx = tan^-1 x + c

(ii) ∫(1 - x²)^-1/2 dx = ∫1/√(1 - x²) dx = sin^-1 x + c 

∫f'(x) dx = ∫(4x – 5) dx 

f(x) = 4x²/2 - 5x + c

f(x) = 2x² – 5x + c 

But f(2) = 1 

2(2)² – 5(2) + c = 1 

8 – 10 + c = 1 

c = 3 

Thus, f(x) = 2x² – 5x + 3

f'(x) = – 6x + c 

f'(x) = 6x² – 6x + c 

But f'(1) = 5 

6(1)² – 6(1) + c = 5 

c = 5 

f” (x) = 6x² – 6x + 5

∫f” (x) dx = ∫(6x² - 6x + 5) dx 

f(x) = 6x³/3 - 6x²/2 + 5x + c 

f(x) = 2x³ – 3x² + 5x + c 

But f(1) = 30 

2(1)³ – 3(1)² + 5(1) + c = 30 

2 – 3 + 5 + c = 30

c = 26 

f(x) = 2x³ – 3x² + 5x + 26

(i) ∫12³ dx = 12³∫dx = 12³ x + c

(ii) ∫x²⁴/x²⁵ dx = ∫1/x dx = log|x| + c 

(iii) ∫e^x dx = e^x + c

Let I = ∫e^x (sec x + sec x tan x) dx

Take f(x) = sec x 

f‘ (x) = sec x tan x 

This is of the form of ∫e^x[f(x) + f'(x)] dx = e^x 

f(x) + c 

= e^x sec x + c

(i) ∫1/sin² x dx = ∫cosec² x dx = -cot x + c

(ii) ∫tan x/cos x dx = ∫sec x tan x dx = sec x + c

(iii) ∫cos x/sin² x dx = ∫cosec x cot x dx = - cosec x + c

(iv) ∫1/cos² x dx = ∫sec² x dx = tan x + c

f'(x) = ∫(9x² – 6x) dx

f(x) = 9x³/3 - 6x²/2 + c

f(x) = 3x³ – 3x² + c 

But f(0) = -3 

3(0)³ – 3(0)² + c = -3

c = -3 

Thus, f(x) = 3x³ – 3x² – 3 

f(x) = 3(x³ – x² – 1)

Given f’ (x) = 2x – 7 

⇒ f (x) = ∫(2x - 7) dx

= x² – 7x + c 

Given f(1) = 0 ⇒ 1 – 7 + c = 0 

⇒ c = 6 

So f(x) = x² – 7x + 6

(i) ∫√x⁷ dx = ∫(x⁷)^1/2 dx = ∫x^7/2 dx = x^{7/2 + 1}/(7/2 + 1) = x^9/2/(9/2) = (2/9)x^9/2 + c

(ii) ∫(x¹⁰)^1/7 dx = ∫x^10/7 dx = x^{10/7 + 1}/(10/7 + 1) = x^17/7/(17/7) = (7/17)x^17/7 + c

f” (x) = 6x + 6 

⇒ f ‘ (x) = (6x + 6) dx 

= 6x²/2 + 6x + c 

= 3x² + 6x + c

Given f ‘(0) = -5 ⇒ 0 + c = -5 

⇒ c = -5 

∴ f ‘(x) = 3x² + 6x – 5 

So f(x) = ∫(3x² + 6x - 5) dx 

= 3x³/3 + 6x²/2 – 5x + c 

(i.e.,) f(x) = x³ + 3x² – 5x + c 

Given f(1) = 6 

⇒ 1 + 3 + 5 + c = 6 

⇒ c – 1 = 6 ⇒ c = 7 

So f(x) = x³ + 3x² – 5x + 7

(i) ∫1/x⁵ dx = ∫x^-5 dx = x^{-5 + 1}/(-5 + 1) = x^-4/-4 = -1/4x⁴ + c

(ii) ∫x^-1 dx = ∫1/x dx = log x + c

(iii) ∫1/x^-5/2 dx = x^{-5/2 + 1}/(-5/2 + 1) = x^-3/2/(-3/2) = -2/3x^3/2 + c 

(i) ∫e^{3x - 6} dx = (1/3)e^{3x - 6} + c

(ii) ∫e^{8 - 7x} dx = (-1/7)e^{8 - 7x} + c

(iii) ∫1/(6 - 4x) dx = (-1/4) log (6 - 4x) + c

(i) ∫1/(√(1 - (4x)²) dx = (1/4) sin^-1 4x + c

(ii) ∫1/√(1 - 81x²) dx = ∫1/√(1 - (9x)²) dx = (1/9) sin^-1 9x + c

(iii) ∫1/(1 + 36x²) dx = ∫1/(1 + (6x)²) dx = (1/6) tan^-1 6x + c 

(i) ∫sin 3x dx = (-1/3) cos 3x + c

(ii) ∫cos (5 – 11x) dx = (-1/11) sin (5 - 11x) + c

(iii) ∫cosec² (5x – 7) dx = (-1/5) cot (5x - 7) + c

(i) ∫sec² x/5 dx = 5 tan x/5 + c

(ii) ∫cosec (5x + 3) cot (5x + 3) dx = -1/5 cosec(5x + 3) + c

(iii) ∫sec (2 – 15x) tan (2 – 15x) dx = sec(2 - 15x)/-15 + c = (-1/15)sec(2 - 15x) + c