Question

( f(x)=(1+2 m) x^{2}-2(1+3 m) x+4(1+m) )
For ( F(x) ) to ( b e+v e )
( cdot cdot frac{1<0,4(1+2 m)>0}{sqrt{4}} )
( quad 4(1+3 m)^{2}-4(1+2 m) cdot 9(1+m)<0 )
( Rightarrow 4left[1+9 m^{2}+6 m-4left(1+3 m+2 m^{2}right)<0right. )
( Rightarrow quad m^{2}-6 m-3<0 )
( m=frac{6 pm sqrt{36+12}}{2} )
( m=frac{6 pm 4 sqrt{3}}{2}=3 pm 2 sqrt{2} )
( m in(3-2 sqrt{3}, 3+2 sqrt{3}) & m>- )
( m=0,1,2,3,4,5,6 )
"No, is integralis:

# ard w nicati Mana Knov (1) 2-13 (2) - 2+V2 (4) 4-312 (3) 4-213 moniosos The number of integral values of m for which the quadratic expression, [JEE(Main) -2019] (1 + 2m) x² – 2(1 + 3m)x + 4(1 + m), xe R, is always positive, is: (4) 8 (3) 3 (2) 7 (1) 6 22. ips 23. If a and ß be the roots of the equation x' - 2x + 2 = 0, the least value of n for which

Solution