Maths NCERT Exemplar Solutions Class 11th Chapter One

( (1+i)/ (1-i) )^ (m/2) = ( (1+i)/ (i-1) )^ (n/3) = 1
⇒ ( (1+i)²/2 )^ (m/2) = ( (1+i)²/ (-2) )^ (n/3) = 1
⇒ (i)^ (m/2) = (-i)^ (n/3) = 1
⇒ m/2 = 4k? and n/3 = 4k?
⇒ m = 8k? and n = 12k?
Least value of m = 8 and n = 12
∴ GCD = 4
∴ GCD = 4

(0.16)^log? (1/3 + 1/3² + . to ∞)
= (4/25)^log? (1/2)
= ( (5/2)? ² )^log? /? (1/2) = (5/2)^ (-2 log? /? (1/2)
= (5/2)^ (log? /? ( (1/2)? ² ) = 4

lim_ (x→0) (1/x? ) {1 - cos (x²/2) - cos (x²/4) + cos (x²/2)cos (x²/4)} = 2?
⇒ lim_ (x→0) ( (1 - cos (x²/2) (1 - cos (x²/4) / x? ) = 2?
⇒ 2? = 2? ⇒ k = 8

∴ center lies on x + y = 2 and in 1st quadrant center = (a, 2-a) where a > 0 and 2-a > 0 ⇒ 0 < a < 2
∴ circle touches x = 3 and y = 2
⇒ |3-a| = |2 - (2-a)| = radius
⇒ |3-a| = |a| ⇒ a = 3/2
∴ radius = a
⇒ Diameter = 2a = 3.

A = [ x 1 ]
[ 1 0 ]
A² = [ x 1 ] [ x 1 ] = [ x²+1 x ]
[ 1 0 ] [ 1 0 ] [ x 1 ]
A? = [ x²+1 x ] [ x²+1 x ]
[ x 1 ] [ x 1 ]
= [ (x²+1)²+x² x (x²+1)+x ]
[ x (x²+1)+x²+1 ]
a? = (x² + 1)² + x² = 109
⇒ x = ±3
a? = x² + 1 = 10

$\begin{array}{l}Here,A=\left\{0,1,2\right\},Bisasethavingallrealnumbersfrom0to2.SoA\ne B.\\ Hence,thestatementis\text{'}False\text{'}.\end{array}$

$\begin{array}{l}WecanwrittenthegivensetsisRoasterform\\ R=\left\{\dots ,-8,-6,-4,-2,0,2,4,6,8,\dots \right\}\\ andT=\left\{\dots ,-18,-12,-6,0,6,12,18\dots \right\}\\ SinceeveryelementofTispresentinR.So,T\subset R.\\ Hence,thestatementis\text{'}True\text{'}.\end{array}$

$\begin{array}{l}True,Sinceeveryintegerisarationalnumber\\ \therefore Z\subset Q,SoQ\cup Z=Q.\end{array}$

$False,Sincethetwosetsdonotcontainthesameelements.$