Ncert Solutions Maths class 11th

Equation of tangent to y²=4 (x+1) is y=m (x+1)+1/m.
Equation of tangent to y²=8 (x+2) is y=m' (x+2)+2/m'.
m'=-1/m.
Solving for intersection point, x+3=0.

sinx+sin4x + sin2x+sin3x = 0
2sin (5x/2)cos (3x/2) + 2sin (5x/2)cos (x/2) = 0
2sin (5x/2) [cos (3x/2)+cos (x/2)] = 0
4sin (5x/2)cosxcos (x/2)=0.
sin (5x/2)=0 ⇒ 5x/2=kπ ⇒ x=2kπ/5. x=0, 2π/5, 4π/5, 6π/5, 8π/5, 2π.
cosx=0 ⇒ x=π/2, 3π/2.
cos (x/2)=0 ⇒ x=π.
Sum = 9π.

S? = 3n/2 [2a+ (3n-1)d]. S? = 2n/2 [2a+ (2n-1)d].
S? =3S? ⇒ 3n/2 [2a+ (3n-1)d] = 3 (2n/2) [2a+ (2n-1)d].
2a+ (3n-1)d = 2 [2a+ (2n-1)d] ⇒ 2a+ (n-1)d=0.
S? /S? = (4n/2 [2a+ (4n-1)d]) / (2n/2 [2a+ (2n-1)d]) = 2 [- (n-1)d+ (4n-1)d]/ [- (n-1)d+ (2n-1)d] = 2 (3n)/ (n)=6.

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.

Sol. Σ? k(k+1)/2 = (1/2)Σ(k²+k) = (1/2)[ (5051101/6) + (5051/2) ]
The solution uses k=1 to 20.
(1/2)[ (202141/6) + (2021/2) ] = (1/2)[2870+210] = 1540

D≥0
(a-10)² - 4(2)(33/2 - 2a) ≥ 0
a²-4a-32 ≥ 0 ⇒ a∈(-∞, -4] U [8,∞)