Class 11th

A= {1,2,3,4,5.100}
B= {4,7,10,13,16,19.}
C= {2,4,6,8,10.}
B−C= {7,13,19, .97}
A∩ (B−C)= {7,13,19, .97}
sum of elements=832

Direction ratio of line (1, -1, -6)
Equation of line (x−3)/1 = (y+4)/-1 = (z+5)/-6 =k
x=k+3, y=−k−4, z=−6k−5
Solving with plane k=−2
⇒x=1, y=−2, z=7
⇒Distance=√ (3−1)²+6²+3²=√49=7

N=2¹? 5¹? 11¹¹ 13¹¹
5→4n+1 type→number of choice=11
11→4n+3 type→number of choice=6
13→4n+1 type→number of choice=12
. Number of divisor of 4n+1 type=11*6*12=792

Z = (3+2icosθ)/ (1-3icosθ) = (3−6cos²θ)+i (11cosθ) / (1+9cos²θ)
Real part = 0
⇒3−6cos²θ=0
⇒θ=45?
⇒sin²3θ+cos²θ=1/2+1/2=1

$\overrightarrow{A}*\overrightarrow{A}=AAsin\theta \widehat{n}$

$=AAsin0°\widehat{n}$

= 0 [since Angle between the vectors are zero degree]

$\overrightarrow{A}*\overrightarrow{A}=0$

e? +e³? −4e²? −e? +1=0
Divide by e²?
⇒ (e²? +e? ²? )− (e? +e? )−4=0
Put e? +e? =t>0
e²? +e? ²? +2=t²
t²−2−t−4=0
⇒t²−t−6=0
⇒t=3, t=−2 but t≠−2
⇒t=3
⇒e? +e? =3
Number of solution=2

p: there exist M>0
Such that x≥M for all x∈S
Obviously ~p: M>0 such that x. Negation of 'there exists' is 'for all'.

Any point on line (1)
x=α+k
y=1+2k
z=1+3k
Any point on line (2)
x=4+Kβ
y=6+3K
Z=7+3K?
⇒1+2k=6+3K, as the intersect
∴1+3k=7+3K?
⇒K=1, K? =−1
x=α+1; x=4−β
⇒y=3; y=3
z=4; z=4
Equation of plane
x+2y−z=8
⇒α+1+6−4=8 . (i)
and 4−β+6−4=8 . (ii)
Adding (i) and (ii)
α+5−β+12−8=16
α−β+17=24
⇒α−β=7

Lt? →? x/ (1−sinx)¹/? − (1+sinx)¹/? )
= 2x/ (1−sinx)¹/? − (1+sinx)¹/? ) Multiply by conjugate
= 4x/ (1−sinx)¹/²− (1+sinx)¹/²) Multiply by conjugate
= 8x/ (1−sinx−1−sinx) Multiply by conjugate
= 4x/sinx = −4