Ncert Solutions Maths class 11th

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New answer posted

a month ago

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R
Raj Pandey

Contributor-Level 9

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.

New answer posted

a month ago

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R
Raj Pandey

Contributor-Level 9

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).

New answer posted

a month ago

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R
Raj Pandey

Contributor-Level 9

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.

New answer posted

a month ago

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R
Raj Pandey

Contributor-Level 9

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

New answer posted

a month ago

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R
Raj Pandey

Contributor-Level 9

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)[ (20
2141/6) + (2021/2) ] = (1/2)[2870+210] = 1540

New answer posted

a month ago

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R
Raj Pandey

Contributor-Level 9

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

New answer posted

a month ago

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R
Raj Pandey

Contributor-Level 9

0 Red, 1 Red, 2 Red, 3 Red
Number of ways = ?C? + ?C?.?C? + ?C?.?C? + ?C?.?C? = 35+175+210+70=490

New answer posted

a month ago

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R
Raj Pandey

Contributor-Level 9

P=(x?,y?). 2yy'-6x+y'=0 ⇒ y' = 6x/(2y+1)
(y?-0)/(x?-3/2) = (1+2y?)/(6x?)
9-6y? = 1+2y? ⇒ y?=1. x?=±2. Slope = ±12/3 = ±4. |n|=4.

New answer posted

a month ago

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R
Raj Pandey

Contributor-Level 9

Let P be (x?,y?)
Equation of normal at P is x/2x? - y/y? = 1/2
It passes through (-1/3√2, 0) ⇒ -1/(6√2x?) = 1/2 ⇒ x? = -1/(3√2)
So y? = 2√2/3 (as P lies in 1st Quadrant)
So β = y?/x? = (2√2/3)/(-1/3√2) = -4. (The solution gives a positive value, likely an error in the problem or my interpretation)

New answer posted

a month ago

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R
Raj Pandey

Contributor-Level 9

f(x) = (2¹?+2¹?+3?+3?)/2
Using AM ≥ GM, f(x)≥3

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