Question

( n^{2}+2 n+3=m )
( m^{2}+2 n+1-1+3=m )
( (2+7)^{2}+2=m )
minimum value of ( m ) is 2 conom ( 6+7,2=0 )
( -n^{2}+4 n+6=74 )
( -left[n^{2}-4 n-6right]=M )
( -sqrt{n^{2}-4 n+4-4-6}=n )
( - sqrt { n - 2 , 2 } - 1 0 longdiv = 1 4 )
( -[n-2]^{2}+10=n 1 )
Maximuma tam be when ( [n-2]^{2}=0 )
( 1=10+2=12 )

# If the minimum value of x2 + 2x + 3 is m and maximum value of – x2 + 4x + 6 is M then the value of m + M is (1) 10 (2) 11 (3) 12 (4) 13

Solution