Since, fourth term of

Given equation is ln(x + y) = 2xy …(1)

ln x + y = 2xy …1

For x = 0

ln(0 + y) = 2.0.y = 0

⇒ lny = 0

⇒ y = 1

Let the third vertex be A(α,β)

Using the diagram, OA⊥BC

⇒ Slope of OA × Slope BC = −1

If P(α, β) be the centre of C₂ of radius r₂ = 1

We know that, if two circles with centres A and P and radii r₁ and r₂ touches each other externally, then distance between their centres AP is equal to the sum of their radii i.e. AP = r₁ + r₂

The locus is obtained by replacing (α, β) by (x, y)

Hence, the locus is x² + y² − 4x − 2y − 20 = 0

Any point on the first line is (4r+k, 2r+1,r−1), and any point on the second line is (r′+k+1 ,−r′,2r′+1) for some values of r and r'. The lines are intersecting if these two points coincide i.e

4r + k = r′ + k + 1, 2r + 1 = −r′, r − 1 = 2r′ + 1 for some r and r'

⇒ 4r − r′ = 1, 2r + r′ = −1, r − 2r′ = 2

Now, 4r − r′ = 1, 2r + r′ = −1 ⇒ r = 0, r′ = −1 which satisfy r − 2 r′ = 2.

⇒ The given lines are intersecting for all real values of k.

n(A × A) = 9

Number of relations containing (1, 2), (2, 1), (a, a) = 2³

Number of relations containing (1, 2), (2, 1), (1, 3), (3, 1), (a, a) = 2³

Number of relations containing (1, 2), (2, 1), (2, 3), (3, 2), (a, a) = 2³

Number of relations containing (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2), (a, a) = 2³

⇒ Total number of symmetric relations = 32.

As we know, ~(p → q) = p ∧ ~q

Hence, ~(p→(q ∧ r)) = p ∧(~(q ∧ r))

⇒~(p → (q ∧ r)) = p∧(~q ∨ ~r)