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11) given,
the domain of log (x³-x) :
f(x) is defined for x³-x>0 because if x³-x = 0
then the value of log0 is not defined.
x³-x>0 => x(x²-1)>0 => x(x-1)(x+1)>0
So Domain = (-1,0) U (1, infinity)
12) given,
f(x) is defined when (x-2)>0 and (3-x)>0
=> (x-2)>0 and (x-3)<0
=> x>2 and x<3
Domain = (2,3)
1) given,
f(x) is defined [x]²-[x]-6>0
=>[x]²-3[x]+2[x]-6>0
=>[x] ([x]-3) +2 ([x]-3)>0
=>([x]-3) ([x]+2) >0
=>[x]<-2 and [x]>3
=>x<-2 and x≥4
so Domain = (-infinity,2) U [4, infinity)
HOPE U UNDERSTAND
PLS MARK IT AS BRAINLIEST
the domain of log (x³-x) :
f(x) is defined for x³-x>0 because if x³-x = 0
then the value of log0 is not defined.
x³-x>0 => x(x²-1)>0 => x(x-1)(x+1)>0
So Domain = (-1,0) U (1, infinity)
12) given,
f(x) is defined when (x-2)>0 and (3-x)>0
=> (x-2)>0 and (x-3)<0
=> x>2 and x<3
Domain = (2,3)
1) given,
f(x) is defined [x]²-[x]-6>0
=>[x]²-3[x]+2[x]-6>0
=>[x] ([x]-3) +2 ([x]-3)>0
=>([x]-3) ([x]+2) >0
=>[x]<-2 and [x]>3
=>x<-2 and x≥4
so Domain = (-infinity,2) U [4, infinity)
HOPE U UNDERSTAND
PLS MARK IT AS BRAINLIEST
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priyanka05:
tq
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