Question

Hey please provide me trick to find range of jee problem ..

Please I stuck in every range

Answer 1

2)

in 2 nd question

we know that value inside root is positive

16 - 4^ ( x^2 -x) >0

16 > 4^ ( x^2 - x)

4^2 >4^ ( x^2 -x)

that means

2> x^2 - x

x^2 - x - 2<0

x^2 +x -2x -2<0

x( x+1) -2( x+1)<0

( x-2)( x+1)<0

xbelongs To( -1,2)

now take minimum value of 4^ (x^2 -x)

in between (-1,2) x

its 1 for x= 0

So 1/√16 -1 = 1/√15

So 1/√15, infinity

as by putting -1 and 2 its 1/0 that is infinity



3)1/2 - cos3x

take minimum value of cos3x = -1

1/2 +1 = 1/3

max value = 1

1/2 -1 = 1

so (1/3,1)