Maths

If g(x) = x² + x - 1 and (g∘f)(x) = 4x² – 10x + 5, then f(3/4) is equal to

0 Follower 4 Views

Contributor-Level 10

g (f (x) = f² (x) + f (x) - 1.
g (f (5/4) = f² (5/4) + f (5/4) - 1.
Given g (f (5/4) = 5/4, let f (5/4) = y.
-5/4 = y² + y - 1 (There appears to be a typo in the image's solution)
y² + y - 1 + 5/4 = 0
y² + y + 1/4 = 0
(y + 1/2)² = 0
y = -1/2.
So, f (5/4) = -1/2.

0 0

New answer posted

a year ago

Let α be a root of the equation x² + x + 1 = 0 and the matrix A = (1/√3) [, [1, α, α²], [1, α², α⁴]], then the matrix A³¹ is equal to:

0 Follower 12 Views

Contributor-Level 10

A = 1/3 [ 1; 1 ω ω² 1 ω² ω ]
A² = A * A = 1/9 [ . ]
(The calculation in the image shows A² is the identity matrix, let's verify)
A² leads to I (Identity matrix).
So A² = I.
A³ = A² * A = I * A = A.
A? = (A²)² = I² = I.
A³? = (A²)¹? = I¹? = I.

0 0

New answer posted

a year ago

Let a vector αi + βj be obtained by rotating the vector √3i + j by an angle 45° about the origin in counterclockwise direction in the first quadrant. Then the area of triangle having vertices (α, β), (0, β) and (0, 0) is equal to :

0 Follower 17 Views

Contributor-Level 9

Let z = αi + βj be a complex number & √3i + j = (√3 + i) where I = √-1.
|z| = |z - (√3 + i)| and |z| = |√3 + i| = 2.
This implies z lies on the perpendicular bisector of the segment from 0 to √3+i and on a circle of radius 2 centered at the origin.
z = (√3 + i) ( (1+i)/√2 )
z = (1/√2) [ (√3 - 1) + I (√3 + 1)]
∴ α = (√3 - 1)/√2, β = (√3 + 1)/√2
Area of required triangle = (1/2) * base * height = (1/2) * |α| * |β| * 2 (This seems to be area of triangle with vertices 0, z, and another point)
The provided calculation: Area = (1/2) * | (√3-1)/√2| * | (√3+1)/√2| = (1/2) * (3-1)/2 = 1/2.

0 0

New answer posted

a year ago

If y(α) = √((tan α + cot α)/(1 + tan²α)) + 1/sin²α, α ∈ (3π/4, π) then dy/dα at α = 5π/6 is:

0 Follower 17 Views

Contributor-Level 10

y = √ (2cos²α / (sinα cosα) + 1/sin²α)
y = √ (2cotα + cosec²α)
y = √ (2cotα + 1 + cot²α) = √ (1 + cotα)²) = |1 + cotα|.
Given α is in a range where 1+cotα is negative, y = -1 - cotα.
dy/dα = - (-cosec²α) = cosec²α.
At α = 5π/6, dy/dα = cosec² (5π/6) = (1/sin (5π/6)² = (1/ (1/2)² = 2² = 4.

0 0

New answer posted

a year ago

If Re((z-1)/(2z+i)) = 1, where z = x + iy, then the point (x, y) lies on a:

0 Follower 23 Views

Contributor-Level 10

Given Re (z-1)/ (2z+i) = 1, where z = x + iy.
(z-1)/ (2z+i) = [ (x-1) + iy] / [2x + I (2y+1)]
To rationalize, multiply the numerator and denominator by the conjugate of the denominator [2x - I (2y+1)].
Numerator = [ (x-1) + iy] * [2x - I (2y+1)] = 2x (x-1) - I (x-1) (2y+1) + i2xy + y (2y+1)
Real part of the numerator = 2x (x-1) + y (2y+1).
Denominator = (2x)² + (2y+1)².
Re (z-1)/ (2z+i) = [2x (x-1) + y (2y+1)] / [ (2x)² + (2y+1)²] = 1.
2x² - 2x + 2y² + y = 4x² + 4y² + 4y + 1.
0 = 2x² + 2y² + 2x + 3y + 1.
So, 2x² + 2y² + 2x + 3y + 1 = 0.

0 0

New answer posted

a year ago

If y = y(x) is the solution of the differential equation, dy/dx + 2ytanx = sinx, y(π/3) = 0, then the maximum value of the function y(x) over R is equal to :

0 Follower 5 Views

Contributor-Level 9

dy/dx + 2y tan (x) = sin (x)
I.F. = e^ (∫2tan (x)dx) = e^ (2ln (sec (x) = sec² (x)
Solution is y sec² (x) = ∫sin (x)sec² (x)dx = ∫sec (x)tan (x)dx
⇒ y sec² (x) = sec (x) + c
y (π/3) = 0 ⇒ 0 * sec² (π/3) = sec (π/3) + c ⇒ 0 = 2 + c ⇒ c = -2.
∴ y = (sec (x) - 2) / sec² (x)
Now let g (t) = (t - 2)/t² = 1/t - 2/t² for |t| ≥ 1.
g' (t) = -1/t² + 4/t³
g' (t) = 0 ⇒ t = 4.
g' (t) = 2/t³ - 12/t? g' (4) < 0, hence maximum.
∴ g (t)max = g (4) = (4 - 2)/4² = 2/16 = 1/8.

0 0

New answer posted

a year ago

Let P be a plane passing through the points (2,1,0), (4,1,1) and (5,0,1) and R be any point (2,1,6). Then the image of R in the plane P is:

0 Follower 8 Views

Contributor-Level 10

The equation of the line is (x-2)/1 = (y-1)/1 = (z-6)/-2.
Let this be equal to k. So, a point on the line is (k+2, k+1, -2k+6).
This point lies on the plane x + y - 2z = 3.
(k+2) + (k+1) - 2 (-2k+6) = 3
2k + 3 + 4k - 12 = 3
6k - 9 = 3
6k = 12 ⇒ k = 2.
The point of intersection is (2+2, 2+1, -2 (2)+6) = (4, 3, 2).

0 0

New answer posted

a year ago

Let the position vectors of two points P and Q be 3i - j + 2k and i + 2j - 4k respectively. Let R and S be two points such that the direction rations of lines PR and QS are (4, -1, 2) and (-2, 1, -2), respectively. Let lines PR and QS intersect at T. If the vector TA is perpendicular to both PR and QS and the length of vector TA is √5 units, then the modulus of a position vector of A is :

0 Follower 5 Views

Contributor-Level 9

PR (line): r = (3i - j + 2k) + λ (4i - j + 2k) - (I)
QS (line): r = (i + 2j - 4k) + μ (-2i + j - 2k) - (II)
If they intersect at T then:
3 + 4λ = 1 - 2μ
-1 - λ = 2 + μ
2 + 2λ = -4 - 2μ
Solving the first two equations gives λ = 2 & μ = -5. These values satisfy the third equation.
∴ T (11, -3, 6)
Also, OT is coplanar with lines PR and QS.
⇒ TA ⊥ OT
|OT| = √166
|TA| = √5
|OA| = √ (|OT|² + |TA|²) = √171

0 0

New answer posted

a year ago

Maths Determinants

If the system of linear equations 2x + 2ay + az = 0, 2x + 3by + bz = 0, 2x + 4cy + cz = 0 where a, b, c ∈ R are non-zero and distinct; has a non-zero solution, then:

0 Follower 4 Views

Contributor-Level 10

0 0

New answer posted

a year ago

Let the functions f: R → R and g: R → R be defined as :
f(x) = {x+2, x<0; x, x0} and g(x) = {x, x<1; 3x-2, x1}
Then, the number of points in R where (fog)(x) is NOT differentiable is equal to :

0 Follower 9 Views