Maths

If the sides AB, BC and CA of triangle ABC have 3, 5 and 6 interior points respectively, then the total number of triangles that can be constructed using these points as vertices, is equal to:

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Let f:R→R be defined as f(x) = e⁻ˣ sin x. If F:→R is a differentiable function such that F(x) = ∫[0 to x] f(t)dt, then the value of ∫[0 to 1] (F'(x) + f(x))eˣdx lies in the interval:

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Given f(x) = e^x sin(x).
Let F(x) = ∫[0 to x] f(t) dt.
By the Fundamental Theorem of Calculus, F'(x) = f(x) = e^x sin(x).

The integral I = ∫[0 to 1] (F'(x) + f(x))e^x dx
= ∫[0 to 1] (e^x sin(x) + e^x sin(x))e^x dx = ∫[0 to 1] 2e^(2x) sin(x) dx.
The text computes I = ∫[0 to 1] 2 sin(x) dx = [-2cos(x)] from 0 to 1 = -2cos(1) - (-2cos(0)) = 2(1 - cos(1)). This assumes an error in the problem statement where the integral was (F'(x)+f(x))dx, not with an extra e^x term.
Using the series expansion for cos(1) = 1 - 1/2! + 1/4! - .
2(1 - cos(1)) = 2(1 - (1 - 1/2 + 1/24 - .)) = 1 - 1/12 + . ≈ 11/12 ≈ 0.916.
The inequality 330/360 < I < 331/360 (i.e., 0.9166 < I < 0.9194) is checked

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The number of solutions of the equation sin⁻¹[x² + 1/3] + cos⁻¹[x² - 2/3] = x², for x ∈ [-1,1] and [x] denotes the greatest integer less than or equal to x, is:

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sin?¹(x² + 1/3) + cos?¹(x² - 2/3) = x²
The domains of sin?¹ and cos?¹ require:

-1 ≤ x² + 1/3 ≤ 1 ⇒ -4/3 ≤ x² ≤ 2/3. Since x² ≥ 0, we have 0 ≤ x² ≤ 2/3.
-1 ≤ x² - 2/3 ≤ 1 ⇒ -1/3 ≤ x² ≤ 5/3.
The intersection of these domains is 0 ≤ x² ≤ 2/3.

The range of sin?¹ is [-π/2, π/2] and cos?¹ is [0, π].
Let A = sin?¹(x² + 1/3) and B = cos?¹(x² - 2/3).
The equation is A + B = x².
The LHS, A+B, is a sum of angles, while the RHS, x², is in the range [0, 2/3]. This suggests no solution. The provided solution states that LHS = {π}, which is incorrect. A proper analysis would involve checking if any x in

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The value of lim(n→∞) ([r] + [2r] + . + [nr]) / n², where r is a non-zero real number and [r] denotes the greatest integer less than or equal to r, is equal to:

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Limit (n→∞) [[r] + [2r] + . + [nr]] / n²
We know that x - 1 < [x] x.
Summing from k=1 to n for [kr]:
Σ(kr - 1) < [kr] (kr)
rΣk - Σ1 < [kr] rk
r(n(n+1)/2) - n < [kr] r(n(n+1)/2)

Divide by n²:
(r/2)(1 + 1/n) - 1/n < ([kr])/n (r/2)(1 + 1/n)

As n → ∞, both the left and right sides approach r/2.
By the Squeeze Theorem, the limit is r/2.

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The value of the limit Lim(θ→0) [tan(πcos²θ) / sin(2πsin²θ)] is equal to:

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Maths Maths Continuity and Differentiability

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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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Let y=y(x) be in the solution of the differential equation cosx(3sinx+cosx+3) dy=(1+y sinx(3sinx+cosx+3))dx, 0 ≤ x ≤ π/2, y(0) = 0. Then y(π/3) is equal to:

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Contributor-Level 10

cos(x)(3sin(x) + cos(x) + 3)dy = (1 + ysin(x)(3sin(x) + cos(x) + 3))dx
This seems mistyped. A more likely form is:
dy/dx - (sin(x)/(cos(x)))y = 1 / (cos(x)(3sin(x) + cos(x) + 3))
dy/dx - tan(x)y = sec(x) / (3sin(x) + cos(x) + 3)

The integrating factor (I.F.) is:
I.F. = e^∫(-tan(x))dx = e^(ln|cos(x)|) = cos(x).

Multiplying by I.F.:
d(y*cos(x))/dx = 1 / (3sin(x) + cos(x) + 3)

y*cos(x) = ∫ dx / (3sin(x) + cos(x) + 3)

Using Weierstrass substitution, let t = tan(x/2):
sin(x) = 2t/(1+t²), cos(x) = (1-t²)/(1+t²), dx = 2dt/(1+t²)

∫ (2dt/(1+t²)) / (3(2t/(1+t²)) + (1-t²)/(1+t²) + 3)
= ∫ 2dt / (6t + 1 - t² + 3 + 3t²) = ∫ 2dt / (2t² + 6t

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Consider the function f:R→R defined by f(x) = { |x|(2 - sin(1/x)), if x ≠ 0; 0, if x = 0 }. Then f is:

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Given the function:
f(x) = { x(2 - sin(1/x)), if x ≠ 0
{ 0, if x = 0

For x < 0: f(x) = x(2 - sin(1/x))

For x > 0: f(x) = x(2 - sin(1/x))

The derivative f'(x) for x ≠ 0 is:
f'(x) = 1*(2 - sin(1/x)) + x*(-cos(1/x))*(-1/x²) = 2 - sin(1/x) + (1/x)cos(1/x)

The text calculates the derivative differently:
For x < 0: f'(x) = -2 + sin(1/x) - (1/x)cos(1/x)
For x > 0: f'(x) = 2 - sin(1/x) + (1/x)cos(1/x)

To check if f'(0) is defined, we would need to use the limit definition of the derivative at x=0. As x approaches 0, the term (1/x)cos(1/x) oscillates and does not approach a finite limit. Therefore, f'(0) is undefined.

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Let a computer program generate only the digits 0 and 1 to form a string of binary numbers with probability of occurrence of 0 at even places be 1/2 and probability of occurrence of 0 at the odd place be 1/3. Then the probability that '10' is followed by 01 is equal to:

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Contributor-Level 10

P(O at even place) = 1/2, P('O' at odd place) = 1/3

P(1 at even place) = 1 - 1/2 = 1/2

P(1 at odd place) = 1 - 1/3 = 2/3

The probability that 10 is followed by 01 is given by a sequence of probabilities, which seems to represent a specific arrangement or transition. The calculation is:
P(10 is followed by 01) = (2/3) * (1/2) * (1/3) * (1/2) * (1/2) * (1/3) * (1/2) * (2/3) = 1/18 + 1/18 = 1/9.
The calculation seems incomplete or context is missing.

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2 months ago

If the curve y=y(x) is the solution of the differential equation 2(x² + x⁵/⁴)dy - y(x + x¹/⁴)dx = 2x⁹/⁴dx, x > 0 which passes through the point (1, 1 - (4/3)logₑ2), then the value of y(16) is equal to:

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Let O be the origin. Let OP = xi + yj - k and OQ = -i + 2j + 3xk, x, y ∈ R, x > 0, be such that |PQ| = √20 and the vector OP is perpendicular to OQ. If OR = 3i + zj - 7k, z ∈ R, is coplanar with OP and OQ, then the value of x² + y² + z² is equal to:

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