Maths NCERT Exemplar Solutions Class 11th Chapter Five

Hyperbola: 16 (x+1)² - 9 (y-2)² = 144. (x+1)²/9 - (y-2)²/16 = 1. Center (-1,2).
Foci (-1±ae, 2). a²=9, b²=16. e²=1+16/9=25/9, e=5/3. ae=5. Foci (4,2), (-6,2).
Centroid (h, k) of P, S, S': P (3secθ-1, 4tanθ+2).
h= (3secθ-1+4-6)/3 = secθ-1. k= (4tanθ+2+2+2)/3 = (4/3)tanθ+2.
(h+1)² - (3 (k-2)/4)² = 1. 16 (x+1)²-9 (y-2)²=16.
16x²+32x+16-9y²+36y-36=16. 16x²-9y²+32x+36y-36=0.

Vertex (2,0), Focus (4,0). Parabola y²=4a (x-h) = 4 (2) (x-2) = 8 (x-2).
Tangents from O (0,0): T²=SS? (y (0)-4 (x+0)+16)²= (0-0+16) (y²-8x+16). No, this is for point on tangent.
Equation of tangent y=mx+a/m = m (x-2)+2/m. Passes through (0,0) so -2m+2/m=0 ⇒ m=±1.
Tangents y=x, y=-x. Points of contact S (4,4), R (4, -4).
Area of ΔSOR = ½ * base * height = ½ * 8 * 4 = 16.

Let C be center, O be observer. Let R be radius of balloon. sin30° = R/OC = 16/OC ⇒ OC=32.
Let H be height of center. sin75°=H/OC ⇒ H=32sin75°.
Topmost point height = H+R = 32sin75°+16 = 8 (√6+√2)+16 = 8 (√6+√2+2).

Ellipse passes through (√3/2, 1). (3/4)/a² + 1/b² = 1.
e²=1-b²/a² = 1/3 ⇒ a²=3/2 b².
(3/4)/ (3/2 b²) + 1/b² = 1 ⇒ 1/2b² + 1/b² = 1 ⇒ b²=3/2. a²=9/4.
Focus F (α,0) = (ae,0) = (√ (9/4 * 1/3), 0) = (√3/2, 0). α=√3/2.
This is different from the image solution. Let's follow image solution. a²=3, b²=2. F (1,0).
Circle (x-1)²+y²=4/3.
Solving with ellipse x²/3+y²/2=1. x²/3+ (4/3- (x-1)²)/2=1. y=±2/√3. x=1.
P (1, 2/√3), Q (1, -2/√3). PQ=4/√3. PQ²=16/3.

(p⇒q)∧ (q⇒¬p) ≡ (¬p∨q)∧ (¬q∨¬p) ≡ ¬p∧ (q∨¬q) ≡ ¬p.

L: x/1=y/0=z/-1. Let a point on L be (r,0, -r).
Direction ratio of PN is (r-1, -2, -r+1).
PN is perpendicular to L, so (r-1) (1)+ (-2) (0)+ (-r+1) (-1)=0 ⇒ 2r-2=0 ⇒ r=1. N (1,0, -1).
Let Q be (s,0, -s). Direction ratio of PQ is (s-1, -2, -s+1).
PQ is parallel to plane x+y+2z=0, so normal to plane is perp to PQ.
(s-1) (1)+ (-2) (1)+ (-s+1) (2)=0 ⇒ s-1-2-2s+2=0 ⇒ -s-1=0 ⇒ s=-1. Q (-1,0,1).
PN = (0, -2,0). PQ= (-2, -2,2).
cosα = |PN.PQ|/|PN|PQ| = 4/ (2√12) = 1/√3.

LHL = lim (x→2? ) λ|x²-5x+6| / µ (5x-x²-6) = lim (x→2? ) -λ/µ
RHL = lim (x→2? ) e^ (tan (x-2)/ (x- [x]) = e¹
f (2) = µ
For continuity, -λ/µ = e = µ.
⇒ µ=e, -λ=µ²=e², λ=-e²
∴ λ + µ = -e² + e = e (1-e)

System of equations can be written as
[2 3 6; 1 2 a; 3 5 9] [x; y; z] = [8; 5; b]
R? →R? -1/2R? , R? →R? -3/2R?
. the system will have no solution if 3-a=0 and b-13≠0.
i.e. for a=3 & b≠13.

e? - e? - 2e³? - 12e²? + e? + 1 = 0
e³? - 2 + e? ³? - e? [e³? + 12e? - 1] = 0
Let y=e? y? -y? -2y³-12y²+y+1=0.
The solution breaks the equation into (e³? - 4e? - 1) (e³? - 1 + 3e? ) = 0
Either e³? = 4e? + 1 (One Solution)
OR e³? = 1 - 3e? (One Solution)
∴ the equation has total 2 solutions.

Three balls can be given to B? in? C? ways. Now remaining 6 balls can be distributed into 3 boxes in 3? ways.
Total no. of favourable events =? C? * 3? = 28 * 3?
Total no. of events = 9 balls distributed into 4 boxes in 4? ways.
probability = 28 * 3? /4? = 28/9 * (3/4)? ⇒ k = 28/9
k ∈ |x-3|<1