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Class 12th Integrals

If ∫ (e²x + 2eˣ – e⁻ˣ – 1)e^(eˣ+e⁻ˣ)dx = g(x)e^(eˣ+e⁻ˣ) + c where c is a constant of integration, then g(0) is equal to:

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alok kumar singh

Contributor-Level 10

I = ∫ (e²? + 2e? - e? - 1)e^ (e? +e? ) dx
I = ∫ (e²? + e? - 1)e^ (e? +e? ) dx + ∫ (e? - e? )e^ (e? +e? ) dx
I = ∫ (e? + 1 - e? )e^ (e? +e? ) dx + e^ (e? +e? )
(e? - e? + 1)dx = du
I = e^ (e? +e? ) + e^ (e? +e? ) = e^ (e? +e? ) (e? + 1) then g (x) = e? + 1
g (0) = 2

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New answer posted

a year ago

Class 12th Maths

Let λεR. The system of linear equations.
2x₁ – 4x₂ + λx₃ = 1
x₁ - 6x₂ + x₃ = 2
λx₁ – 10x₂ + 4x₃ = 3
Is inconsistent for:

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alok kumar singh

Contributor-Level 10

Here D = | 2 -4 λ |
| 1 -6 1 | = (λ-3) (3λ+2)
| λ -10 4 |
D = 0 ⇒ λ = 3, -2/3

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Class 12th Maths

A plane P meets the coordinate axes at A, B and C respectively. The centroid of ? ABC is given to be (1,1,2). Then the equation of the line through this centroid and perpendicular to the plane P is:

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Raj Pandey

Contributor-Level 9

Let A (α, 0,0), B (0, β, 0), C (0,0, γ), then the centroid is G (α/3, β/3, γ/3) = (1,1,2).
α = 3, β = 3, γ = 6
∴ Equation of plane is x/α + y/β + z/γ = 1
⇒ x/3 + y/3 + z/6 = 1
⇒ 2x + 2y + z = 6
∴ Required line passing through G (1,1,2) and normal to the plane is (x-1)/2 = (y-1)/2 = (z-2)/1.

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a year ago

Class 12th Maths

Let θ = π/5 and A = [cosθ sinθ; -sinθ cosθ]. If B = A + A?, then det(B) :

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Raj Pandey

Contributor-Level 9

A = [cosθ, sinθ], [-sinθ, cosθ]
A² = [cos2θ, sin2θ], [-sin2θ, cos2θ]
⇒ A? = [cos4θ, sin4θ], [-sin4θ, cos4θ]
B = [cos4θ, sin4θ], [-sin4θ, cos4θ] + [cosθ, sinθ], [-sinθ, cosθ]
= [cos4θ + cosθ, sin4θ + sinθ], [- (sin4θ + sinθ), cos4θ + cosθ]
det (B) = (cos4θ + cosθ)² + (sin4θ + sinθ)²
= (cos²4θ + sin²4θ) + (cos²θ + sin²θ) + 2 (cos4θcosθ + sin4θsinθ)
= 1 + 1 + 2cos (4θ - θ)
= 2 + 2cos3θ
Given 3θ = 3π/5
|B| = 2 + 2cos (3π/5)
= 2 + 2 (- (√5-1)/4) = 2 - (√5-1)/2 = (4-√5+1)/2 = (5-√5)/2 ∈ (1,2)

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a year ago

Class 12th Physics Ray Optics and Optical Instruments

A compound microscope consists of an objective lens of focal length 1 cm and an eye piece of focal length 5 cm with a separation of 10 cm. The distance between an object and the objective lens, at which the strain on the eye is minimum is n/40 cm. The value of n is

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alok kumar singh

Contributor-Level 10

Image by objective must be at focus of eye piece.
1/v - 1/u = 1/f?
1/5 - 1/u = 1 ; u = -5/4 cm
Hence, N = 50.

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a year ago

Class 12th Physics Wave Optics

A beam of electrons of energy E scatters from a target having atomic spacing of IÅ °. The first maximum intensity occurs at θ = 60°. Then E (in eV) is (Planck constant h = electron mass m = 9.1 * 10?³¹ kg )

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alok kumar singh

Contributor-Level 10

We know, P = √2Em and λ = h/√2Em
For 1st maxima, 2nd sinθ = λ
Put (1) and get the answer
E = 50.47eV.
B = µ? I (200)/2R? R? = 20 cm; R? = 1 cm
e = -dΦ/dt = π (R? )² (500) dB/dt
= π (R? )² (500) (µ? I/2R? ²) (200) (10t-2)
= π (1/10000) (500) (200) (µ? ) (8) (5)/ (2)
= 16π² x 10 x 5 x 10?
= 800π² x 10?
e = 0.7887mv
4/x = 0.8; x = 5.

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a year ago

Class 12th Electromagnetic Induction

Two concentric circular coils, C₁ and C₂ are placed in the XY plane. C₁ has 500 turns, and a radius of 1 cm. C₂ has 200 turns and radius of 20 cm. C₂ carries a time dependent current I(t) = (5t² - 2t + 3)A where t is in s. The emf induced in C₁ (in mV ), at the instant t = 1 s is x/4. The value of x is

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alok kumar singh

Contributor-Level 10

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New answer posted

a year ago

Class 12th Physics

A particle of mass 200MeV/c² collides with a hydrogen atom at rest. Soon after the collision the particle comes to rest, and the atom recoils and goes to its first excited state. The initial kinetic energy of the particle (in eV) is N/4. The value of N is: (Given the mass of the hydrogen atom to be 1GeV/c²)

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alok kumar singh

Contributor-Level 10

M? = 200MeV/C², m = 1Gev/C²
Initial velocity of particle is 'V'.
Final velocity of hydrogen atom is 'V'
M? V? = mV
V = M? V? /m
Also, 1/2 M? V? ² = 1/2 mV² + 3/4 x 13.6
Put (1) and get answer
1/2 M? V? ² = 51/4 eV
Hence, N = 51.

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New answer posted

a year ago

Class 12th Physics

Two capacitors of capacitances C and 2C are charged to potential differences V and 2 V, respectively. These are then connected in parallel in such a manner that the positive terminal of one is connected to the negative terminal of the other. The final energy of this configuration is:

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