Ncert Solutions Maths class 11th

Any point (x, y) on perpendicular bisector equidistant from p and q
∴ (x − 1)² + (y − 4)² = (x − k)² + (y − 3)²
At x = 0, y = -4
∴ 1 + 64 = k² + 49
k² = 16

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

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

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

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)

By family of circle, passing through intersection of given circle will be member of S + λS? = 0 family (λ ≠ 1)
(x² + y² – 6x) + λ (x² + y² – 4y) = 0
(λ + 1)x² + (λ + 1)y² – 6x – 4λy = 0
x² + y² - 6/ (λ+1) x - 4λ/ (λ+1) y = 0
Centre (3/ (λ+1), 2λ/ (λ+1)
Centre lies on 2x – 3y + 12 = 0
2 (3/ (λ+1) - 3 (2λ/ (λ+1) + 12 = 0
6λ + 18 = 0
λ = -3
Circle x² + y² – 3x – 6y = 0

Given: 3α² – 10α + 27λ = 0
3α² – 3α + 6λ = 0
Subtract -7α + 21λ = 0
3λ = 0
By (ii) 9λ² – 3λ + 2λ = 0
⇒ λ = 0, 1/9
∴ α = 1/3, β = 2/3, α = 1/3, γ = 3
∴ (βγ)/λ = (2/3) * 3) / (1/9) = 18

Contrapositive of p ↔ q is ~ q →~ p.

= [√3e^ (iπ/3)]^4
= 9 (cos (2π/3) + isin (2π/3)
= -9/2 + 9√3i/2
⇒ 0 + 9 (-1 + i√3)/2)
∴ a = 0, b = 9

Since (3,3) lies on x²/a² - y²/b² = 1
9/a² - 9/b² = 1
Now, normal at (3,3) is y-3 = -a²/b² (x-3),

which passes through (9,0) ⇒ b² = 2a²

So, e² = 1 + b²/a² = 3
Also, a² = 9/2
(From (i) and (ii)
Thus, (a², e²) = (9/2, 3)