Ncert Solutions Maths class 11th

. xdy - ydx - x² (xdy + ydx) + 3x? dx = 0
⇒ (xdy - ydx)/x² - (xdy + ydx) + 3x²dx = 0 ⇒ d (y/x) - d (xy) + d (x³) = 0
Integrate both side, we get
y/x - xy + x³ = c
Put x = 3, y = 3
⇒ 1 - 9 + 27 = c
c = 19
Put x = 4
y/4 - 4y = 19 - 64
⇒ y = 12

for reflexive (x, x)
x³ - 3x²x + 3x³ = 0
. reflexive
For symmetric
(x, y) ∈ R
x³ - 3x²y - xy² + 3y³ = 0
⇒ (x - 3y) (x² - y²) = 0
For (y, x)
(y - 3x) (y² - x²) = 0
⇒ (3x - y) (x² - y²) = 0
Not symmetric

Let the equation of circle be
x (x-1/2) + y² + λy = 0
=> x² + y² - x/2 + λy = 0
Radius = √ (1/16 + λ²/4) = 2
=> λ² = 63/4 => (x-1/4)² + (y+λ/2)² = 4
∴ This circle and parabola
y-α = (x-1/4)² touch each other, so
α = -λ/2 + 2 => α-2 = -λ/2 => (α-2)² = λ²/4 = 63/16
(4α–8)² = 63

As slope of line joining (1, 2) and (3, 6) is 2 given diameter is parallel to side
∴ a = √ (3-1)²+ (6-2)² = √20 and
b/ (a/2) = 4/a => b = 8/√5
Area
ab = 2√5 * 8/√5 = 16

T? = r/ (2r²)²+1)
= r/ (2r²+1)²- (2r)²)
= (1/4) * (4r)/ (2r²+2r+1) (2r²-2r+1)
S? = 1/4 Σ? ¹? [ 1/ (2r²-2r+1) - 1/ (2r²+2r+1) ]
=> S? = (1/4) * (220/221) = 55/221 = m/n
∴ m+n = 276

First we arrange 5 red cubes in a row and assume x? , x? , x? , x? , x? and x? number of blue cubes between them
Here, x? + x? + x? + x? + x? + x? = 11
and x? , x? , x? , x? ≥ 2
so x? + x? + x? + x? + x? + x? = 3
No. of solutions =? C? = 56

Let e? = t then equation reduces to
t³ – 11t – 45/t + 81/t² = 0
=> 2t? – 22t³ + 81t – 45 = 0 . (i)
If roots of
e³? – 11e²? – 45e? + 81/2 = 0
=> α? + α? + α? = ln45 => p = 45

| p | q | p^q | ¬ (p^q) | (¬ (p^q)vq | pvq | p? q | p? (pvq) |
|-|-|-|-|-|-|-|-|
| T | T | T | F | T | T | T | T |
| T | F | F | T | T | T | F | T |
| F | T | F | T | T | T | T | T |
| F | F | F | T | T | F | T | T |
Tautology, Tautology

x²/a² + y²/b² = 1
(-4/√5)²)/a² + (2/√5)²)/b² = 1
=> 32/a² + 9/b² = 1
=> 32/ (5a²) + 9/b² = 1 . (i)
From (i)
6/b² + 9/b² = 1 => b²=15 & a²=16
a²+b² = 15+16=31