Ncert Solutions Maths class 12th

MD²+MC² is minimum when M is the midpoint of projection of CD on AB.

p∨ (¬p∧q) = (p∨¬p)∧ (p∨q)=T∧ (p∨q)=p∨q. Negation is ¬p∧¬q.

(80+1)? /8 = (¹C?80? +.+¹C? )/8. Remainder is 1/8.

The expression inside log is a GP sum: (1/3)/ (1-1/3) = 1/2.
log? (1/2) = 1/2.
The series S = 1+2/3+6/3²+.
S - S/3 = 1 + 4/3² + 4/3³ + . = 1 + (4/9)/ (1-1/3) = 1+2/3=5/3.
(2/3)S = 5/3 ⇒ S=5/2.
l = (5/2)/ (1/2)=5. l²=25.

S = {n ∈ N | [n i; 0 1] [a b; c d] = [a b; c d] ∀a, b, c, d ∈ R}
[na+ic, nb+id; c, d] = [a, b;c, d]
This must be an identity matrix. n=1. The question seems to have typos. Let's follow the solution logic.
The solution implies the matrix must be [1 0; 0 1] or [-1 0; 0 -1]. And n must be a multiple of 8.
2-digit multiples of 8 are 16, 24, ., 96. Total 11 numbers.

Required probability = (? C? +? C? )/¹¹C? = 25/165=5/33.

Required area = 4 [∫? ¹/² 2y²dy + ½*½*1] = 5/6.

α+β=64; αβ=256=2?
(α³/β? )¹/? + (β³/α? )¹/? = (α+β)/ (αβ)? /? = 64/32=2.

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.

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.