Ncert Solutions Maths class 11th

MOTHER
1? E
2? H
3? M
4? O
5? R
6? T
So position of word MOTHER in dictionary
2*5!+2*4!+3*3!+2!+1
=240+48+18+2+1
=309

Slope of tangent is 2, Tangent of hyperbola
x²/4-y²/2=1 at the point (x? , y? ) is xx? /4-yy? /2=1 (T=0)
Slope: x? /2y? =2 ⇒ x? =4y?
(x? , y? ) lies on hyperbola
⇒ x? ²/4-y? ²/2=1
From (1) and (2)
(4y? )²/4-y? ²/2=1 ⇒ 4y? ²-y? ²/2=1
⇒ 7y? ²=2 ⇒ y? ²=2/7
Now x? ²+5y? ² = (4y? )²+5y? ² = 21y? ² = 21*2/7=6

Let three terms of G.P. are a/r, a, ar product=27
⇒ a³=27 ⇒ a=3
S=3/r+3r+3
For r>0
(3/r+3r)/2 ≥ √9 = 3
⇒ 3/r+3r≥6
For r<0, 3/r+3r-6
From (1) and (2)
S∈ (-∞, -3]∪ [9, ∞)

α and β are roots of 5x²+6x-2=0
⇒ 5α²+6α-2=0
⇒ 5α? ²+6α? ¹-2α? =0
(By multiplying α? )
Similarly 5β? ²+6β? ¹-2β? =0
By adding (1) and (2)
5S? +6S? -2S? =0
For n=4
5S? +6S? =2S?

Let p denotes statement
p: I reach the station in time.
q: I will catch the train.
Contrapositive of p→q is q→p
q→p: I will not catch the train, then I do not reach the station in time.

|x|/2 + |y|/3 = 1
x²/4 + y²/9 = 1
Area of Ellipse = πab = 6π
Required area = π*2*3 - (Area of quadrilateral)
= 6π - 1/2*6*4
= 6π-12
= 6 (π-2)

The value of (1+sin (2π/9)+icos (2π/9)/ (1+sin (2π/9)-icos (2π/9)³
= (1+cos (5π/18)+isin (5π/18)/ (1+cos (5π/18)-isin (5π/18)³
= (2cos² (5π/36)+2isin (5π/36)cos (5π/36)/ (2cos² (5π/36)-2isin (5π/36)cos (5π/36)³
= (cos (5π/36)+isin (5π/36)/ (cos (5π/36)-isin (5π/36)³
= (e^ (i5π/36)/e^ (-i5π/36)³ = (e^ (i5π/18)³ = e^ (i5π/6) = cos (5π/6)+isin (5π/6)
= -√3/2 + i/2

Let L be the common normal to parabola
y = x²+7x+2 and line y = 3x-3
⇒ slope of tangent of y=x²+7x+2 at P=3
⇒ dy/dx|for p = 3
⇒ 2x+7=3 ⇒ x=-2 ⇒ y=-8
So P (-2, -8)
Normal at P: x+3y+C=0
⇒ C=26 (satisfies the line)
Normal: x+3y+26=0

Determinant = -2-sin2x. Max -1, min -3. (m, M)= (-3, -1).