Mathematics Test - 6

Key Idea: Use

(a × b)·c = (b × c)·a = (c × a)·b

∴ w·(u × r) = -v·(u × w)

We have, foot of the perpendicular drawn from the point (0, 0, 0) to the plane is (4 , - 2, -5).

∴ Direction ratios of normal to the required plane is (0-4, 0 + 2, 0 + 5 i.e -4, 2, 5)

∴ Required aquation of plane passing through (4, - 2, 5) is a(x - x₁ ) + b(y - y₁ ) + c(z - z₁ ) = 0

= (-4) (x - 4) + (-2) (y + 2) + (5)(z + 5) = 0

= -4x + 2y+ 5z + 16 + 4 + 25 = 0

⇒ 4x - 2y - 5z = 45

Given have. x^y = e^x-y

taking log on both sides, we get

y log x = (x - y) log e = (x - y)

when x = 1, then y (log 1) = (1 - y)

⇒ y = 1

On differentiating both sides,

when x = 1 the

Key Idea: A statement that is neither a tautology nor a contradiction is a contingency.

Option (1) (p v q) v ~ q

= p v(q v ~ q) = p v T

= T, which is a tautology.

Option (2) (p v q) v ~

= (p v ~ p) v q = T v q

= T which is tautology.

Option (3)(p v q) ∧ ~ q

= (p ∧ ~ q) v (q ∧ ~ q) = (p ∧ ~ g) v F

= (p ∧ ~ q)

if p → T, q → T

Then p∧ ~ q → F

if p → T, q → F

Then p∧ ~ q → T

So, (p∧ ~ q) is a contingency.

Therefore statement (p v q)∧ ~ q is contingency.

We have equation of line y = x and equation of circle x² + y² - 2x = 0

Now. intersecting points of given line and circle, x² + x² - 2x = 0

⇒ 2x² - 2x = 0

⇒ 2x(x - 1) = 0

⇒ x = 0, 1

when x = 0 then y = 0 and when x = It then y = 1

∴ Coordinates of endpoints of diameter AB are (0, 0) and (1, 1)

Required equation of the circle with diameter AB

(x - 0)(x - 1) + (y - 0) (y - 1) = 0

x² - 2 + y² - y = 0

⇒ x² + y² - x - y = 0

put 1 - x = t ⇒ x = 1 - t

⇒ -dx = dt

⇒ dx = -dt

when x = 0, then t = 1 and

when x =1 , then t = 0

We have

Let

put x = a sin² θ

⇒ dx = a(2 sinθ cos θ)dθ

when, x = 0, θ = 0 and x = a , θ = π/2

∴ k = πa

We have a differential equation

dθ/dt = -k(θ - θ₀ ), where k is constant

⇒ (dθ/dt) + kθ = kθ₀

Which is linear differential equation in the form of

(dy/dx) + Py = Q

IF = e^∫kt = e^kt

before the required solution,

(θ)(e^kt ) = ∫(e^kt x kθ₀ )dt

⇒ θe^kt = e^kt θ₀ + a

⇒ θ = θ₀ + ae^-kt

We have, x = 2 + 27t - t³

⇒ dx/dt = 27 - 3t²

Since, the direction will be reverse,

∴ dx/dt = 0

⇒ 27 - 3t² = 0

⇒ t = 3

∴ Distance x = 2 + 27t - t³

= 2 + 27 x 3 - (3)³

= 2 + 81 - 27 = 56 units

Let, θ = 18º then, 2θ = 36º = 90º - 54° = 90º - 3θ

Now. sin2θ = sin(90º - 3θ)

⇒ 2sinθ cosθ = cos 3θ

⇒ 2 sinθ cosθ = 4cos³ θ- 3cosθ

⇒ 2sinθ cosθ = cosθ(4cos² θ - 3)

⇒ 2sinθ = 4 cos² θ - 3(as cos θ ≠ 0)

⇒ 2sinθ = 4 - 4 sin² θ - 3

⇒ 4sin² + θ + 2sinθ - 1 = 0

⇒

But as sinθ > 0 we have sinθ > 0 we have

A =

Therefor, |A| =

= (1 + 2i)(1 - 2i) + i²

= 1 - (2i)² + (-1) = 1 + 4 -1 = 4

∴ A(adj A) = |A| = 4i

On adding Eqs. (i) and (ii), we get

Given equation of circle x² + y² - 6x + 4y -12 = 0 ...(i)

Centre of circle (i) te (3, 2)

Equation of circle concentric with the circle (i) and touching the Y- axis is

(x - 3)² + (y - 2)² = (3)²

⇒ x² + 9 - 6x + y² + 4 - 4y = 9

⇒ x² + y² - 6x - 4y + 4 = 0

We have , 4 sin^-1 x 6 cos^-1 x = 3π

⇒ 4 sin^-1 x + 3π - 6 sin^-1 x = 3π

⇒ -2 sin^-1 x = 0 ⇒ sin^-1 x = 0

⇒ x = sin(0) = 0

We have

= k sin(B + C)

= k sin(180º - A)

= k sin A = a

We have equation of hyperbola

25x² - 9y² = 225

On comparing with we get

A² = 9, b² = 25

Therefore: