Class 11th

log x = t
(t² - 1/t)¹²
T? = ¹²C? (t²)¹²? (-1/t)? = ¹²C? t²? ³? (-1)?
For constant term r = 8 & coefficient ¹²C?
= (12*11*10*9) / 24 = 45*11 = 495

tan? ¹ (x) = t
t² - 4t + 3 > 0
t ∈ (-∞, 1) U (3, ∞)
tan? ¹ (x) ∈ (-π/2, 1)
x ∈ (-∞, tan1)
the largest integral x = 1

z - 2² = z + 2²
⇒ x=0
Hence minimum '0'

(Var)new = k² (Var)old
= 4 * 5 = 20

  $x^2+y^2-2x-6y+6=0$ centre (1, 3)

$\begin{array}{l}r=\sqrt[]{1+9-6}=2\\ CM=\sqrt[]{1+4}=\sqrt[]{5}\end{array}$           

$r=\sqrt[]{5+4}=3$

~ (p v (~ p v q)
= p ^ ( (~p v q) = ~p ^ (p ^ ~q)

Sum of elements $A\cap \left(B\cup C\right)=274*400$  

In set B numbers of the form 9k + 2 are {101, 109, .992}

  $\therefore sum=\frac{100}{2}\left(101+992\right)=\frac{100*1093}{2}......\left(i\right)$         

Another possible number is 9k + 5 forms are {104, .995}

  $\therefore sum=\frac{100}{2}\left(104+995\right)=\frac{100}{2}*1099.........\left(ii\right)$

$\therefore Total=\frac{100}{2}*\left[1093+1099\right]=100*1096=274*4*100=274*400$           

$\therefore$ possible value of $\mathcal{l}$ = 5

  $\frac{4}{sinx}+\frac{1}{1-sinx}=a\forall x\in \left(0,\frac{\pi }{2}\right)$

Let sin x = t, t  $\in$ (0, 1)

$g\left(t\right)=\frac{4}{t}+\frac{1}{1-t}$         

$g\text{'}\left(t\right)=0\Rightarrow t=\frac{2}{3}$           

$g\text{'}\text{'}\left(\frac{2}{3}\right)>0$           

$\therefore g{\left(t\right)}_{minimum}=\frac{4}{2/3}+\frac{1}{1-2/3}=9$          

Minimum value of a for which solution exist = 9

T? = (1/3)T?
⇒ T? , T? , T? G.P.
T = 1/3
T? = 1/243
ar? = 1/243
a (1/3)? = 1/243
a/729 = 1/243 ⇒ a = 3

Sum of infinite series = T? + T? + T?
= (T? ) / (1-r²) = (3 * 1) / (1 - 1/9) = 3 / (8/9) = 27/8

y² = 4x
(x - 3)² + y² = 9
y = mx + 1/m
(3, 0), r = 3

|3m + 1/m| / √ (1+m²) = 3

9m² + 1/m² + 6 = 9 (1+m²)
9m² + 1/m² + 6 = 9 + 9m²
1/m² = 3 ⇒ m = ±1/√3
m = 1/√3 in first quadrant
y = x/√3 + √3 ⇒ √3y = x + 3