Class 11th
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New answer posted
2 months agoContributor-Level 10
To find the sum Σ[r=0 to 6] (?C?)².
This is the coefficient of x? in the expansion of (1+x)?(x+1)? = (1+x)¹².
By the binomial theorem, (1+x)¹² = Σ[k=0 to 12] ¹²C? x?.
The coefficient of x? is ¹²C?.
¹²C? = (12 * 11 * 10 * 9 * 8 * 7) / (6 * 5 * 4 * 3 * 2 * 1) = 11 * 2 * 3 * 2 * 7 = 924.
New answer posted
2 months agoContributor-Level 10
g? = g - ω²R ⇒ 0 = g - ω²R ⇒ ω = √ (g/R)
⇒ T = 2π/ω = 2π√ (R/g) = 2 * 3.14 * √ (6400 * 10³)/10)
New answer posted
2 months agoContributor-Level 10
I = ∫[0 to 10] [sin(2πx)] / e^(x-[x]) dx.
The period of the integrand involves [sin(2πx)] which depends on the sign of sin(2πx) and {x} = x - [x], which has a period of 1.
Let f(x) = [sin(2πx)] / e^{x}.
The integral is ∫[0 to 10] f(x) dx = 10 * ∫[0 to 1] f(x) dx due to the periodicity of {x} and the integer period of sin(2πx).
In the interval (0, 1/2), sin(2πx) is between 0 and 1, so [sin(2πx)] = 0.
In the interval (1/2, 1), sin(2πx) is between -1 and 0, so [sin(2πx)] = -1.
At x=0, 1/2, 1, the value is 0.
So, ∫[0 to 1] f(x) dx = ∫[0 to 1/2] 0 dx + ∫[1/2 to 1] -1/e^x dx
= 0 + [-e^(-x) * (-1)] from 1/2 to 1 = [e^(-x)] from 1
New answer posted
2 months agoContributor-Level 10
(A) sin (ωt) + cos (ωt) = √2 sin (ωt + π/4) ⇒ T = 2π/ω
(B) sin² (ωt) = 1/2 - (1/2)cos (2ωt) ⇒ T = 2π/ (2ω) = π/ω
(C) 3cos (π/4 - 2ωt) ⇒ T = 2π/ (2ω) = π/ω
(D) cos (ωt) + cos (2ωt) + cos (3ωt)
Time period of cos (ωt) = 2π/ω
Time period of cos (2ωt) = 2π/ (2ω)
Time period of cos (3ωt) = 2π/ (3ω)
Time period of combined function = 2π/ω
New answer posted
2 months agoContributor-Level 10
A = Area swept ⇒ dA/dt = (1/2)r² (dθ/dt) = (1/2) (Mr²ω)/M = L / (2M)
New answer posted
2 months agoContributor-Level 10
Parabola: y² = 4x - 20 = 4(x - 5). Vertex at (5,0).
Line: The text seems to derive the tangent equation y = x - 4. This is not a tangent to the given parabola. The standard tangent to y²=4aX is Y=mX+a/m. Here X=x-5, a=1. So y = m(x-5)+1/m.
The other curve is an ellipse: x²/a² + y²/b² = 1.
The text says x²/2 + (x-4)²/b² = 1. This assumes a² = 2.
x²/2 + (x²-8x+16)/b² = 1
x²(1/2 + 1/b²) - (8/b²)x + (16/b² - 1) = 0.
For tangency, the discriminant (D) of this quadratic equation must be zero.
D = (8/b²)² - 4(1/2 + 1/b²)(16/b² - 1) = 0.
64/b? - 4(8/b² - 1/2 + 16/b? - 1/b²) = 0.
16/b? - (7/b² - 1/2 + 16/b?) = 0.
-7/b² + 1/2 = 0
New answer posted
2 months agoContributor-Level 10
PV? = C ⇒ V? (dP/dV) + P (γV^ (γ-1) = 0 ⇒ dP/dV = -γ (P/V) ⇒ dP/P = -γ (dV/V)
New answer posted
2 months ago
Contributor-Level 10
U = mV (r) = -Cm/r
F = -dU/dr = -Cm/r² ⇒ The force which provides required centripetal force
⇒ mv²/r = Cm/r² ⇒ r = C/v²
New answer posted
2 months ago
Contributor-Level 10
(v/v? ) + (x/x? ) = 1 ⇒ v = - (v? /x? )x + v?
⇒ a = dv/dt = - (v? /x? ) (v) = - (v? /x? ) [- (v? /x? )x + v? ] ⇒ a = (v? ²/x? ²)x - v? ²/x?
- Concept involved: Graph of kinematics
- Topic: Kinematics
- Difficulty level: Moderate
- Note: IIT-Jee-2005
- Point of Error: Writing Equation of straight line and differentiation
New answer posted
2 months agoContributor-Level 10
Limit (θ→0) [tan(πcos²θ) / sin(2πsin²θ)]
Let θ → 0. Then cos²θ → 1 and sin²θ → 0.
Let u = πsin²θ. As θ → 0, u → 0.
cos²θ = 1 - sin²θ = 1 - u/π.
The expression becomes:
Limit (u→0) [tan(π(1 - u/π)) / sin(2u)]
= Limit (u→0) [tan(π - u) / sin(2u)]
= Limit (u→0) [-tan(u) / sin(2u)]
= Limit (u→0) [-tan(u) / (2sin(u)cos(u))]
= Limit (u→0) [-(sin(u)/cos(u)) / (2sin(u)cos(u))]
= Limit (u→0) [-1 / (2cos²(u))] = -1 / (2 * 1²) = -1/2.
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