Maths

Let x = 4 be a directrix to an ellipse whose centre is at the origin and its eccentricity is 1/2. If P(1, β), β > 0 is a point on this ellipse, then the equation of the normal to it at P is:

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Contributor-Level 10

Given : a/e = 4 and 1/4 = 1 - b²/a²
Solving : a = 2, b = √3
Parametric co - ordinates are
(2cosθ, √3sinθ) = (1, β)
∴ θ = 60°
∴ Equation of normal is axsecθ − bycosecθ = a² − b²
⇒ 4x - 2y = 1

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New answer posted

a year ago

The angle of elevation of a cloud C from a point P, 200 m above a still lake is 30°. If the angle of depression of the image of C in the lake from the point P is 60°, then PC (in m ) is equal to

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Contributor-Level 10

tan 30° = x/y ⇒ y = √3x
and tan 60° = (x+400)/y ⇒ √3y = x+400
= x + 400
Solving (i) and (ii), we get
2x = 400, x = 200
sin 30° = x/PC = 200/PC ⇒ PC = 400

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New answer posted

a year ago

In a game two players A and B take turns in throwing a pair of fair dice starting with player A and total of scores on the two dice, in each throw is noted. A wins the game if he throws a total of 6 before B throws a total of 7 and B wins the game if he throws a total of 7 before A throws a total of six. The game stops as soon as either of the players wins. The probability of A winning the game is:

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Contributor-Level 10

sum 6 → (1,5), (5,1), (3,3), (2,4), (4,2)
sum 7 → (1,6), (6,1), (5,2), (2,5), (3,4), (4,3)
= P (A) + P (? ) · P (B) · P (A) + P (? )P (B)P (? ) · P (B) · P (A) + …
This is infinite G.P. with common ratio P (? ) * P (B)
Probability of A wins = P (A) / (1 - P (? )P (B? )
= (5/36) / (1 - (31/36)* (30/36) = 30/61

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New answer posted

a year ago

The distance of the point (1, −2,3) from the plane x − y + z = 5 measured parallel to the line x/2 = y/3 = z/-6 is:

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Contributor-Level 10

Equation PQ
(x-1)/2 = (y+2)/3 = (z-3)/ (-6) = λ
Let Q = (2λ + 1, 3λ − 2, −6λ + 3)
Q lies on x - y + z = 5
⇒ (2λ + 1) − (3λ − 2) + (−6λ + 3) = 5
⇒ λ = -1/7
Q = (5/7, -17/7, 15/7)

∴ PQ = √ (2/7)² + (3/7)² + (6/7)²)

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New answer posted

a year ago

Let f: (0,∞) → (0,∞) be a differentiable function such that f(1) = e and lim(t→x) (t²f²(x)-x²f²(t))/(t-x) = 0. If f(x) = 1, then x is equal to:

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Contributor-Level 10

Applying L'Hôpital's Rule
Lim (t→x) [2tf² (x) – x² (2f (t)f' (t)] / 1
∴ 2xf² (x) – x² (2f (x)f' (x) = 0
⇒ f (x) – xf' (x) = 0
⇒ f' (x)/f (x) = 1/x ⇒ lnf (x) = lnx + C
At x=1, c=1
∴ lnf (x) = lnx + 1
when f (x) = 1
then lnx = -1
x = 1/e

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New answer posted

a year ago

The minimum value of 2ˢⁱⁿˣ + 2ᶜᵒˢˣ is:

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Maths Differential Equations

Contributor-Level 10

Since AM of two positive quantities ≥ their G.M.
(2^sinx + 2^cosx)/2 ≥ √ (2^sinx * 2^cosx)
= √ (2^ (sinx+cosx)
= √2^ (√2cos (x-π/4)
≥ √2^ (-√2) ⇒ 2^sinx + 2^cosx ≥ 2 · 2^ (-1/√2) = 2^ (1-1/√2)

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New answer posted

a year ago

The solution of differential equation dy/dx + (y+3x)/loge(y+3x) + 3 = 0 is:
(where C is a constant of integration.)

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Contributor-Level 10

dy/dx = (y-3x)/ln (y-3x) - 3
dy/dx + 3 = (y+3x)/ln (y+3x)
d/dx (y+3x) = (y+3x)/ln (y+3x)
∫ (ln (y+3x)/ (y+3x) d (y+3x) = ∫ dx
Let ln (y+3x) = t
1/ (y+3x) d (y+3x) = dt
∫ tdt = ∫ dx
t²/2 = x+c
(ln (y+3x)²/2 = x+c

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New answer posted

a year ago

Maths Determinants

Suppose the vectors x₁, x₂, and x₃ are the solutions of the system of linear equations, Ax = b when the vector b on the right side is equal to b₁, b₂ and b₃ respectively. If x₁ = [1; 1; 1], x₂ = [0; 2; 1], x₃ = [0; 0; 1], b₁ = [1; 0; 0], b₂ = [0; 2; 0] and b₃ = [0; 0; 2], then the determinant of A is equal to :

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Contributor-Level 10

Let A =
| a? |
| b? |
| c? |

Ax? = B?
a? + a? + a? = 1
b? + b? + b? = 0
c? + c? + c? = 0
Similar 2a? + a? = 0 and a? = 0
2b? + b? = 2, b? = 0
2c? + c? = 0, c? = 2
∴ a? = 0, b? = 1, c? = -1,
a? = 1, b? = -1, c? = -1
A =
| 1 0 |
| -1 0 |
| -1 -1 2 |
∴ |A| = 2

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New answer posted

a year ago

The area (in sq. units) of the largest rectangle ABCD whose vertices A and B lie on the x-axis and vertices C and D lie on the parabola, y = x² – 1 below the x-axis, is:

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Contributor-Level 10

A (α, 0), B (-α, 0)
⇒ D (α, α² − 1)
Area (ABCD) = (AB) (AD)
⇒ S = (2α) (1 − α²) = 2α – 2α³
dS/dα = 2 - 6α²
= 0 ⇒ α² = 1/3

⇒ α = 1/√3
Area = 2α – 2α³ = 2/√3 - 2/ (3√3)
= 4/ (3√3)

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New answer posted

a year ago

The integral ∫(from π/6 to π/3) tan³x · sin²3x(2sec²x · sin²3x + 3tanx · sin6x)dx is

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