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Concept:

Greatest Integer Function and Definite Integral:

• The greatest integer function, denoted by [x], gives the largest integer less than or equal to x.
• To integrate a greatest integer function, divide the integral into intervals where the function is constant.
• The function inside the integral is f(x) = [1 / e^x−1] = [e^1−x].
• We need to evaluate ∫₀³ [e^1−x] dx = α − logₑ2.

 

Calculation:

f(x) = [e^1−x] is a decreasing function

f(0) = [e¹] = [2.718] = 2

f(1−ln2) = e^{1−(1−ln2)} = e^ln2 = 2

⇒ boundary point

f(x) = 2 for x ∈ [0, 1−ln2)

f(1) = [e⁰] = [1] = 1

f(x) = 1 for x ∈ [1−ln2, 1)

f(x) < 1 for x ≥ 1 ⇒ [f(x)] = 0

Now break the integral accordingly:

∫₀³ [e^1−x] dx = ∫₀^1−ln2 2 dx + ∫_1−ln2¹ 1 dx + ∫₁³ 0 dx

⇒ 2(1 − ln2) + (1 − (1 − ln2)) + 0

⇒ 2 − 2ln2 + ln2 = 2 − ln2

Given: ∫₀³ [e^1−x] dx = α − ln2

Comparing both sides:

α − ln2 = 2 − ln2 ⇒ α = 2

Now, α³ = 2³ = 8

∴ The value of α³ is 8.

Concept:

Second-Order ODE and Initial Conditions:

• When f''(x)= f(x), multiplying both sides by f'(x) and integrating yields a first integral.
• Use the given initial conditions f(0)=0 and f'(0)=3 to determine the constant of integration.

 

Calculation:

Start from the differential equation:

$f''(x)=f(x)$

Multiply both sides by f'(x) and integrate:

$f'(x),f''(x)=f'(x),f(x) ;\implies;\int f'(x)f''(x),dx=\int f'(x)f(x),dx$

$\frac{(f'(x))^2}{2}=\frac{(f(x))^2}{2}+C ;\implies;(f'(x))^2=(f(x))^2+C'$

Apply initial conditions x=0:

$(f'(0))^2=(f(0))^2+C';\implies;3^2=0^2+C';\implies;C'=9$

Thus:

$(f'(x))^2=(f(x))^2+9 ;\implies;f'(x)=\sqrt{(f(x))^2+9}$

Separate variables y=f(x):

$\int\frac{dy}{\sqrt{y^2+9}}=\int dx ;\implies;\ln\bigl|y+\sqrt{y^2+9}\bigr|=x+C$

Use f(0)=0 to find C:

$\ln\bigl|\sqrt{9}\bigr|=0+C;\implies;\ln3=C$

Hence:

$\ln\bigl|y+\sqrt{y^2+9}\bigr|=x+\ln3 ;\implies;y+\sqrt{y^2+9}=3e^x$

At x=ln3), let y=f(ln3). Then:

$f(\ln3)+\sqrt{(f(\ln3))^2+9}=3e^{\ln3}=9$

Set y=f(ln3):

$y+\sqrt{y^2+9}=9 ;\implies;\sqrt{y^2+9}=9-y ;\implies;y^2+9=(9-y)^2=81-18y+y^2$

Solve for y:

$9=81-18y ;\implies;18y=72 ;\implies;y=4$

Therefore:

$f(\ln3)=4 ;\implies;9,f(\ln3)=9\times4=36$

Hence, the correct answer is 36.

Concept:

Area of Region:

• The area between curves can be found by integrating the difference of the functions over the given interval.
• Use definite integrals and apply limits appropriately to find the enclosed area.

Calculation:

Given region: ${(x,y): |4 - x^2| \leq y \leq x^2, y \leq 4, x \geq 0}$

Area calculation involves three integrals:

$A = \int_0^4 \sqrt{4 + y} , dy - \int_0^2 \sqrt{4 - y} , dy - \int_2^4 \sqrt{y} , dy$

Evaluating each integral:

$\int_0^4 \sqrt{4 + y} , dy = \left[ \frac{2}{3} (4 + y)^{3/2} \right]_0^4 = \frac{2}{3} \left( 8^{3/2} - 4^{3/2} \right)$

$\int_0^2 \sqrt{4 - y} , dy = \left[ \frac{2}{3} (4 - y)^{3/2} \right]_0^2 = \frac{2}{3} \left( 2^{3/2} - 4^{3/2} \right)$

$\int_2^4 \sqrt{y} , dy = \left[ \frac{2}{3} y^{3/2} \right]_2^4 = \frac{2}{3} \left( 4^{3/2} - 2^{3/2} \right)$

Substitute the values:

$A = \frac{2}{3} (16 \sqrt{2} - 8) + \frac{2}{3} (2 \sqrt{2} - 8) - \frac{2}{3} (8 - 2 \sqrt{2})$

$= \frac{40 \sqrt{2} - 48}{3} = \frac{80 \sqrt{2}}{6} - 16$

Given: $\frac{80 \sqrt{2}}{\alpha} - \beta = \frac{80 \sqrt{2}}{6} - 16$

Comparing terms,

$\alpha = 6, \quad \beta = 16$

$\alpha + \beta = 6 + 16 = 22$

Hence, the correct answer is 22.

Given:

Three distinct numbers are selected randomly from the set {1,2,3,…,40}.

The selected numbers are in increasing Geometric Progression (G.P.).

The probability of this event is m/n where      \gcd(m,n) = 1

gcd(m,n) 4= 1.

Calculation: 

Total number of ways to select 3 distinct numbers from 40 numbers = (⁴⁰₃) = 9880

Let the three numbers in increasing G.P. be: a, ar, ar^2, where a, r ∈ N 

    

r>1 (since increasing).

Since all numbers are distinct integers between 1 and 40,

⇒ 1 ≤ a< ar < ar² ≤ 40

Since r is the common ratio, for the sequence to be an increasing G.P. with integers,  r must be a positive integer greater than 1

For  r=2:

⇒ a ≤ $\frac{40}{2^2} = 10$ 

So possible a=1,2,…,10 → 10 sequences

For r = 3

⇒ a ≤ 40/9 ≈ 4

So Possible a= 1, 2, 3, 4 → 4 sequences

For r = 4

⇒ a ≤ 40/16 = 2.5

Possible  a = 1, 2

a= 1,2 → 2 sequences

For r = 5

⇒ a ≤ 40/25 = 1.6 

Possible  a = 1

a=1 → 1 sequence

For r = 6 

⇒ a ≤ 40/36 = 1.11

Possible a= 1 → 1 sequence

For r ≥7 a ≤ $\frac{40}{r^2}$​ < 1, no sequences.

Total number of favorable sequences = 10 + 4 + 2 + 1 + 1 = 18

⇒ Now P = $\frac{18}{(^{40}_3)} = \frac{18}{9880} = \frac{9}{4940}$

Here, gcd(94940) = 1 so

⇒ m/n = 9/4940

and m + n  = 9 + 4940 = 4949

Hence, the Correct answer is 4949.

Concept:

Circle Equation and Radius:

• The radius of a circle touching a line can be found using the perpendicular distance from the center to the line.
• The circle equation is $(x - a)^2 + (y - b)^2 = r^2$, where $(a,b)$ is the center and $r$ is the radius.

Calculation:

Center of circle is $(a, 0)$.

Radius $r$ is distance from center to line $x + y = 3$:

$r = \left| \frac{a - 0 - 3}{\sqrt{2}} \right| = \left| \frac{a - 3}{\sqrt{2}} \right|$

Equation of circle:

$(x - a)^2 + y^2 = \left( \frac{a - 3}{\sqrt{2}} \right)^2$

Circle passes through point $(-9,4)$:

$(-9 - a)^2 + 4^2 = \frac{(a - 3)^2}{2}$

Expanding and simplifying:

$(a + 9)^2 + 16 = \frac{(a - 3)^2}{2}$

$2(a^2 + 18a + 81 + 8) = a^2 - 6a + 9$

$2a^2 + 36a + 194 = a^2 - 6a + 9$

$a^2 + 42a + 185 = 0$

Factorizing:

$(a + 37)(a + 5) = 0 \implies a = -37, -5$

Calculating radii:

$r_1 = \left| \frac{-37 - 3}{\sqrt{2}} \right| = \frac{40}{\sqrt{2}} = 20\sqrt{2}$

$r_2 = \left| \frac{-5 - 3}{\sqrt{2}} \right| = \frac{8}{\sqrt{2}} = 4\sqrt{2}$

Absolute difference between squares of radii:

$|r_1^2 - r_2^2| = |(20\sqrt{2})^2 - (4\sqrt{2})^2| = |800 - 32| = 768$

Hence, the correct answer is 768.

Calculation:

The direction of propagation of light is perpendicular to the wave front and is symmetric about x, y, and z axis.

Angle made by the light with x, y & z axis is same.

Thus, cosα = cosβ = cosγ (α, β, and γ are the angles made by light with x, y, and z axes respectively)

Also, cos²α + cos²β + cos²γ = 1 [Sum of direction cosines]

⇒ α = cos⁻¹(1/√3)

Calculation:

(P + a/V²)(V - b) = RT

Therefore, [a] = [P][V²] = M L⁻¹ T⁻² L⁶ = M L⁵ T⁻²

[b] = [V] = L³

[ab⁻²] = M L⁵ T⁻² L⁻⁶ = M L⁻¹ T⁻²

⇒ Dimension of energy density

Calculation:

W_F = 20 × 1 = 20 J

Therefore, ΔKE = 20 J = (1/2) I ω²

I = M R² = 10 × 0.1² = 0.1 kg m²

Therefore, 20 = (1/2) × 0.1 × ω²

Implying, ω = 20 rad/sec

Calculation:

Location of image of A :-

(1/v) - (1/u) = (1/f)

⇒ (1/v) - (1/(v - 30)) = (1/20)

⇒ (1/v) = (1/60)

⇒ v = 60 cm

The magnification m = 2.

Since the size of the object is small with respect to the location, hence

dv = m² du ⇒ dv = 4 × 1 = 4 cm

h_i = m h₀ ⇒ h_i(dy) = 2 × 2 = 4 cm

Therefore, Angle made with principle a×is = -45°

Calculation:

Given:

Surface charge density on Plate 1, σ₁ = +σ

Surface charge density on Plate 2, σ₂ = -2σ

Point charge, q = +q at midpoint between plates

Distance between plates = d (charge at midpoint, so at d/2 from each plate)

Electric field due to Plate 1 (positive charge):

⇒ E₁ = σ / (2ε₀) directed away from Plate 1

Electric field due to Plate 2 (negative charge):

⇒ E₂ = 2σ / (2ε₀) = σ / ε₀, directed toward Plate 2 (since negative charge)

Since charge q is at midpoint, the net electric field E_net is vector sum:

⇒ E_net = E₁ (rightward) + E₂ (rightward, since Plate 2's field points toward it)

⇒ E_net = (σ / 2ε₀) + (σ / ε₀) = (3σ) / (2ε₀)

Force on point charge q:

⇒ F = q × E_net

⇒ F = q × (3σ) / (2ε₀)

∴ Force on charge q = (3σ q) / (2ε₀)