Question

please solve this for me

Answer 1

here you figure is not clear !!!
by the way I solve if any problems then ask

1) domain of 1/√( |x| -x )
here |x| > x
this is possible only when ,
x < 0 e.g domain € ( -∞, 0)


2) 1/√( 16 -x²)

for defining of function ,
16 -x² >0
(4 -x )(4 + x) >0
-4 < x < 4

3) f(x) = log{( x -3)(6 -x)}
for log to be defined
{(x -3)(6-x )} >0
3 < x < 6

4) f(x ) = 1/( 1 -2cosx )

range means [ minf(x) , maxf(x) ]

maxf(x ) = 1/( 1-2(-1) ) = 1/3
min f(x) = 1/(1-2) = -1

Range€[ -1, 1/3]

5) 1 +3cos2x

-1≤ cos2x ≤ 1
-3 ≤ 3cos2x≤ 3
-2 ≤ 1 + 3cos2x ≤ 4

so , range € [ -2, 4]

6) domain of f(x ) = tan[]¹(5x)
5x €R
so, x € R and domain€( -∞ , ∞)


7) domain of f(x) =√(x -2) + 1/log(4-x)
x≥ 2 and 4 -x >0 & 4-x ≠1
x≥ 2 , and x < 4 x ≠ 3
2≤ x < 4 & x ≠ 3

e.g x €[ 2, 3) U(3, 4)

8)f(x ) = sin[]¹x + cos[]¹x + tan[]¹x
but we know,
sin[]¹x + cos[]¹x =π/2
and tan[]¹x € (-π/2, π/2)

now ,
-π/2< tan[]¹x < π/2
π/2 -π/2 < π/2 +tan[]¹x < π/2 +π/2
range € (0, π)


9) √(x² +x ) and tan²a /√(x² +x ) both are positive so, use AM≥ GM
√(x² +x) +tan²a/(√(x²+x) ≥ 2