Mathematics Test 254

Let I = ∫ (2x¹² + 5x⁹) / (x⁵ + x³ + 1)³ dx

= ∫ (2/x³ + 5/x⁶) / (1 + 1/x² + 1/x⁵)³ dx

Put 1 + 1/x² + 1/x⁵ = t

⇒ (-2/x³ - 5/x⁶) dx = dt

Then, I = -∫ dt / t³ = 1 / 2t² + c

= 1 / 2(1 + 1/x² + 1/x⁵)² + c

= x¹⁰ / 2(x⁵ + x³ + 1)² + c

Given, points P and Q are on the ellipse defined by 9x² + 4y² = 36, which simplifies to (x² / 4) + (y² / 9) = 1. This is the standard form of the equation of an ellipse centered at the origin, with semi-major axis a = 3 along the y-axis and semi-minor axis b = 2 along the x-axis.

OP is the distance from the origin O to point P, which is given by:

OP = r₁ = √((2√3 / √7)² + (6 / √7)²) = √(12/7 + 36/7) = √(48/7) = 2√(12/7).

1. Representing the given point P (2√3 / √7, 6 / √7) in polar coordinates:
   P = (r₁ cosθ, r₁ sinθ)

   Substituting into the ellipse equation:
   (r₁² cos²θ) / 4 + (r₁² sin²θ) / 9 = 1

   Simplifying:
   (cos²θ) / 4 + (sin²θ) / 9 = 7/48 ...(equation 1)

2. Representing R as (-r₂ sinθ, r₂ cosθ) (since RS is perpendicular to PQ):

   Substituting into the ellipse equation:
   (r₂² sin²θ) / 4 + (r₂² cos²θ) / 9 = 1

   Simplifying:
   (sin²θ) / 4 + (cos²θ) / 9 = 1 / r₂² ...(equation 2)

3. From equations (1) and (2),

   1 / r₁² + 1 / r₂² = 1/4 + 1/9 = 13/144

4. Since PQ and RS are perpendicular and pass through the origin,
   1 / PQ² + 1 / RS² = (1 / r₁² + 1 / r₂²)

5. Substituting values of r₁ and r₂,
   1 / PQ² + 1 / RS² = 13 / 144 = p / q.

Let the equation of the parabola be y² = 4ax.

Point P be (at², 2at) and the point T be (h, k).

Equation of tangent at P is

ty = x + at².

It passes through T(h, k):

tk = h + at² .....(i)

Slope of SP = (2at - 0) / (at² - a) = 2t / (t² - 1).

TL is perpendicular to SP.

Then equation of TL is

2ty + (t² - 1)x - 2kt - (t² - 1)h = 0 .....(ii)

SL = perpendicular distance of S(a, 0) from (ii).

SL = (a + h) ........(iii)

Equation of directrix is x = -a ........(iv)

TN = perpendicular distance of T(h, k) from (iv)

TN = (h(1) + k(0) + a) / √(1² + 0²)

TN = h + a ........(v)

From (iii) and (v)

SL = TN

The roots of the equation ax² + bx + c = 0 are of opposite signs if they are real and their product is negative.

i.e. if b² − 4ac ≥ 0 and ca < 0

i.e. if b² − 4ac ≥ 0 and ac < 0

i.e. if ac < 0

(Q when ac < 0,b² − 4ac is automatically ≥ 0)

Hence, the roots of the given equation are of opposite signs if 2(a² − 2a) < 0

i.e. if a² − 2a < 0

i.e. 0 < a < 2.

f(x) = e^{{4{x}+3}=e{4(x}}=e{4x}}

because {4{x} + 3} = {4{x}} and 4{x} = 4x − 4[x]⟹{4{x}} = {4x}

p → (q → r) ≡ ∼p∨(q → r)

(q → r) ≡ ∼p∨(q → r)

≡ ∼p∨(∼q∨r)

≡ [(∼p)∨(∼q)]∨r

≡ ∼(p∧q)∨r

≡ p∧q → r

∴ given statement is a tautology.

y² = 16x

Tangent at P (16, 16) is 2y = x + 16 ...(1)

Normal at P (16, 16) is y = -2x + 48 ...(2)

i.e. A is (-16, 0); B is (24, 0)

Now, centre of circle is (4, 0).

Now, m_PC = 4/3

m_PB = -2